## KSEEB SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Ex 1.3

KSEEB SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Ex 1.3 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Exercise 1.3.

## Karnataka SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Exercise 1.3

Question 1.
FInd the sum of the following APs:
i) 2, 7, 12, ……… to 10 terms
ii) -37, -33, -29,………. to 12 terms.
iii) 0.8, 1.7, 2.8 to 100 terms.
iv) $$\frac{1}{15}, \frac{1}{12}, \frac{1}{10} \ldots, \text { to } 11$$ terms
Solution:
i) Sum of 2, 7, 12, ………. 10?
a = 2, d = 7 – 2 = 5, n = 10
$$S_{n}=\frac{n}{2}[2 a+(n-1) d]$$
$$s_{10}=\frac{10}{2}[2 \times 2+(10-1) 5]$$
= 5[4 + 9 × 5]
= 5[4 + 45]
= 5 × 49
∴ S10 = 245

ii) -37, -33, -29, to 12……. terms.
Solution:
a = -37, d = -33 – (-37) = -33 + 37
n = 12, d = 4
$$S_{n}=\frac{n}{2}[2 a+(n-1) d]$$
$$S_{12}=\frac{12}{2}[2 \times -37+(12-1) 4]$$
= 6[-74 + 11 × 4]
= 6[-74 + 44]
= 6 × -30
∴ S12 = -180.

iii) 0.6, 1.7, 2.8,……… , to 100 terms.
a = 0.6, d = 1.7 – 0.6 = 1.1
n = 100, S100 =?
$$S_{n}=\frac{n}{2}[2 a+(n-1) d]$$
$$S_{100}=\frac{100}{2}[2 \times 0.6+(100-1)(1.1)]$$
= 50[1.2 + 99(1.1)]
= 50[1.2 + 108.9]
= 50 × 110.1
∴ S100 = 5505

iv) $$\frac{1}{15}, \frac{1}{12}, \frac{1}{10} \ldots, \text { to } 11$$ terms
Solution:

The LCM of 12 and 18 is 36.

Question 2.
Find the sums given below:
i) $$7+10 \frac{1}{2}+14+\ldots \ldots+84$$
ii) 34 + 32 + 30 + ……….+ 10
iii) -5 + (-8) + (-11) + ……….. +(-230)
Solution:

ii) 34 + 32 + 30 + ………. + 10
Solution:
a = 34, d = 32 – 34 =-2, an = 10,
Sn =?
a + (n – 1)d = an
34 + (n – 1)(-2) = 10
34 – 2n + 2 = 10
-2n + 36 = 10
-2n = 10 – 36
-2n = -26
2n = 26
$$n=\frac{26}{2}$$
∴ n = 13
$$S_{n}=\frac{n}{2}[a+l]$$
$$\mathrm{S}_{23}=\frac{13}{2}[34+10]$$
$$=\frac{13}{2} \times 44$$
= 13 × 22
∴ S23 = 286.

iii) -5 + (-8) + (-11) +……… + (-230)
Solution:
a = -5, d = -8 – (-5) = -8 + 5
a, = -230.
Sn =? d = -3
a+(n – 1)d = an
-5 + (n – 1)(-3) = -230
-5 – 3n + 3 = -230
-3n – 2 = -230
-3n = -230 +2
-3n = -228
3n = 228
∴ n = 228/3
∴ n = 76.
$$\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left[\mathrm{a}+\mathrm{a}_{\mathrm{n}}\right]$$
$$S_{ 76 }=\frac { 76 }{ 2 } \left[ -5-230 \right]$$
$$=\frac{76}{2} \times -235$$
= 38 × -235
∴ S76= -8930.

Question 3.
In an AP:
i) given a = 5. d = 3. an = 50, find ‘n’ and Sn.
ii) given a = 7, a13 = 35, find d’ and S13.
iii) given a12= 37. d = 3. find ‘a’ and S12.
iv) given a3 = 15, S10 = 125. find ‘d’ and a10.
v) given d = 5. S9 = 72, fInd ‘a’ and a9.
vi) given a = 2. d = 8, Sn = 90, fInd ‘n and a11.
vii) given a = 8, an = 62. Sn = 210, find ‘n’ and ‘d’.
viii) given an = 4, d = 2, Sn = -14. find ‘n’ and a.
ix) given a = 3, n = 8, S = 192, find ‘d’.
x) given L = 28, S = 144 and there are total 9 terms. Find ‘a’.
Solution:
i) a = 5, d = 3, an = 50. n=?, Sn =?
a + (n – 1)d = an
5 + (n – 1) 3 = 50
5 + 3n – 3 = 50
3n + 2 = 50
3n = +50 – 2
3n = 48
$$n=\frac{48}{3}$$
∴ n = 16
$$S_{n}=\frac{n}{2}\left[a+a_{n}\right]$$
$$S_{16}=\frac{16}{2}[5+50]$$
$$=\frac{16}{2}[55]$$
= 8 × 55
∴ S16= 440.

ii) a = 7, a13 = 35. d =?. S13 =?
Solution:
a+(n – 1) d = an
7 + (13 – 1) d = 35
7 + 12d = 35
12d = 35 – 7
12d = 28
$$d=\frac{28}{12}$$
$$d=\frac{7}{3}$$
$$S_{n}=\frac{n}{2}\left[a+a_{n}\right]$$
$$S_{n}=\frac{7}{3} \times \frac{1}{2}[7+35]$$
$$\mathrm{s}_{\mathrm{n}}=\frac{7}{6} \mathrm{I} 42 \mathrm{l}$$
= 7 × 7
∴ Sn = 49

iii) a12= 37, d = 3, a =? S12 =?
a + (n – 1)d = an
a + (12 -1) 3 = 37
a + 11 × 3 = 37
a + 33 = 37
∴ a = 37 – 33
∴ a = 4
$$S_{n}=\frac{n}{2}\left[a+a_{n}\right]$$
$$\mathrm{S}_{12}=\frac{12}{2}[4+37]$$
= 6[4 + 37]
∴ S2 = 246

iv) a3 = 15, S10 = 125. d =?, a10=?
Solution:
a3 = a + 2d = 15
∴ a= 15 – 2d
$$S_{n}=\frac{n}{2}[2 a+(n-1) d]$$
$$s_{10}=\frac{10}{2}[2(15-2 d) \div(10-1) d]=125$$
= 5[30 – 4d + 9d] = 125
= 5[30 + 5d] = 125
150 + 25d = 125
25d = 125 + 150
25d = -25
$$\mathrm{d}=\frac{-25}{25}$$
∴ d = -1.
Substittuting the value of ‘d’
a = 15 – 2d
= 15 – 2(-1)
= 15 + 2
∴ a = 17.
∴ a10 = a + 9d
= 17 + 9(-1)
= 17 – 9
∴ a10 = 8

v) d = 5, S9 = 72. a =? a9 =?
Solution:
$$\mathrm{s}_{\mathrm{n}}=\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]$$
$$s_{9}=\frac{9}{2}[2 \times a+(9-1) 5]=72$$
$$\frac{9}{2}[2 a+8 \times 5]=72$$
$$\frac{9}{2}[2 a+40]=72$$
$$2 a+40=72 \times \frac{2}{9}$$
2a + 40 = 16
2a = 16 – 40
2a = -24
$$a=\frac{-24}{2}$$
∴ a = -12
a9 = a + (n – 1) d
= -12 + (9 – 1) (5)
= -12 + 8 × 5
= -12 + 40
∴ a9 = 28

vi) a = 2, d = 8 Sn = 90, n =?, a3 =?
Solution:
$$S_{n}=\frac{n}{2}[2 a+(n-1) d]$$
$$=\frac{n}{2}[2 \times 2+(n-1) 8]=90$$
$$\frac{n}{2}[4+8 n-8]=90$$
n (8n – 4) = 90 × 2/1
8n2 – 4n = 180
8n2 – 4n = 1800
2n2 – n – 45 = 0

2n2– 10n + 9n – 45 = 0
2n(n – 5) + 9(n – 5) = 0
(n – 5)(2n + 9) = 0
If n – 5 = 0 then n = 5
∴ an = a + (n – 1) d
a5= 2 + (5 – 1) 8
= 2 + 4 × 8
= 2 + 32
∴ a5 = 34

vii) a = 8, an = 62, Sn = 210, n =? d =?
Solution:
$$S_{n}=\frac{n}{2}\left[a+a_{n}\right]=210$$
$$\frac{n}{2}[8+62]=210$$
$$\frac{n}{2} \times 70=21$$
$$n=210 \times \frac{2}{70}$$
∴ n = 6
an = 62
a6= a + 5d = 62
8 + 5d = 62
5d = 62 – 8
5d = 54
$$d=\frac{54}{5}$$

viii) a = 4, d = 2, Sn = -14, n =?, a =?
Solution:
an = a + (n – 1) d = 4
a + (n – 1) 2 = 4
a + 2n – 2 = 4
a + 2n = 4 + 2
a = 2n = 6 …………. (i)
$$S_{n}=\frac{n}{2}\left[a+a_{n}\right]=210$$
$$\frac{n}{2}[-2 n+6+4]=-14$$
$$\frac{n}{2}[-2 n+10]=-14$$
-2n2 + 10n = -28
-2n2 + 10n + 28 = 0
2n2 – 10n – 28 = 0 – 14
2n2 – 5n – 14 = 0

n2 – 7n + 2n – 14 = 0
n(n – 7) + 2 (n – 7) = 0
(n – 7) (n + 2) = 0
If n – 7 = 0 then. n = 7
a = -2n + 6
= -2 × 7 + 6
= -14 + 6
∴ a = -8
∴ n = 7, a = -8.

ix) a = 3, n = 8, S = 192, d =?
$$S_{n}=\frac{n}{2}[2 a+(n-1) d]$$
$$S_{8}=\frac{8}{2}[2 \times 3+(8-1) d]=192$$
$$\frac { 8 }{ 2 } [6+7\times d]=192$$
4 (6 + 7d) = 192
$$6+7 d=\frac{192}{4}$$
6 + 7d = 48
7d = 48 – 6
7d = 42
$$d=\frac{42}{7}$$
∴ d = 6

x) l = an = 28, S = 144, n = 9, a =?
Solution:
$$s_{n}=\frac{n}{2}\left[a+a_{n}\right]$$
$$s_{9}=\frac{9}{2}[a+28]=144$$
$$\frac{9}{2}[a+28]=144$$
$$a+28=144 \times \frac{2}{9}$$
a + 28 = 32
a = 32 – 28
∴ a = 4.

Question 4.
How many terms of the AP: 9, 17, 25, … must be taken to give a sum of 636?
Solution:
9 + 17 + 25 + ……. + an = 636
a = 9, d = 17 – 9 = 8, Sn= 636, n =?
$$S_{n}=\frac{n}{2}[2 a+(n-1) d]$$
$$S_{n}=\frac{n}{2}[2 \times 9+(n-1) 8]=636$$
$$\frac{n}{2}[18+8 n-8]=636$$
$$\frac{n}{2}[8 n+10]=636$$
n(8n + 10) = 636 × 2
8n2 + 10n = 1272
8n2+ 10n – 1272 = 0
4n2 + 5n – 636 = o
4n2+ 5n – 48n – 636 = 0

n(4n + 5) – 12(4n + 5) = 0
(4n + 5) (n – 12) = 0
1f n – 12 = 0 then, n = 12
∴ n = 12.

Question 5.
The first term of an AP is 5. the last term is 45 and the sum is 400. Find the number of terms and the common difference.
Solution:
a = 5, an = 45, Sn= 400, n =?, d =?
$$S_{n}=\frac{n}{2}\left[a+a_{n}\right]$$
$$=\frac{n}{2}[5+45]=400$$
$$\frac{n}{2} \times 50=400$$
n × 50 = 800
$$n=800 \times \frac{1}{50}$$
∴ n = 16
an = a + (n – 1) d
a16= 5 + (16 – 1)d = 45
5 + 15d = 45
15d = 45 – 5
15d = 40
$$d=\frac{40}{15}=\frac{8}{3}$$
$$\mathrm{n}=16, \mathrm{d}=\frac{8}{3}$$

Question 6.
The first and the last terms of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?
Solution:
a = 17, an= 350, d = 9, n =?, Sn =?
an = a + (n – 1) d
17 + (n – 1)9 = 350
17 + 9n – 9 = 350
9n + 8 = 350
9n = 350 – 8
9n = 342
$$\quad n=\frac{342}{9}$$
∴ n = 38
$$S_{n}=\frac{n}{2}\left[a+a_{n}\right]$$
$$S_{38}=\frac{38}{2}[17+350]$$
= 19 × 367
∴ S38 = 6973
∴ n = 38, S38 = 6973

Question 7.
Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.
Solution:
d = 7, a22 = 149, S22 =?
a = a + (n – 1)d
a + (22 – 1)7 = 149
a + 21 × 7 = 149
a + 147 = 149
=149 – 147
∴ a = 2
$$S_{n}=\frac{n}{2}\left[a+a_{n}\right]$$
$$S_{22}=\frac{22}{2}[2+149]$$
= 11[151]
∴ S22 = 1661

Question 8.
Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
Solution:
If a2 = 14, a3 = 18, then S11 =?
a2 = 14
a + d = 14 …………. (i)
a3 = 18
a + 2d = 18 …………… (ii)
from equation (1) – equatIon (11),

d =4.
Substituting the value of ‘d’ In eqn. (i)
a + d = 14
a + 4 = 14
a = 14 – 4
∴ a = 10
a = 10, d = 4
$$S_{n}=\frac{n}{2}[2 a+(n-1) d]$$
$$S_{51}=\frac{51}{2}[2 \times 10+(51-1) 4]$$
$$=\frac{51}{2}[20+50 \times 4]$$
$$=\frac{51}{2}[20+200]$$
$$=\frac{51}{2} \times 220$$
= 51 × 110
∴ S51 = 5610

Question 9.
If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first ‘n terms.
Solution:
S7 = 49, S17 = 289, S11 =?
$$S_{n}=\frac{n}{2}[2 a+(n-1) d]$$
$$S_{7}=\frac{7}{2}[2 a+(7-1) d]$$
$$=\frac{7}{2}[2 a+6 d]=49$$
[latex2+6 d=49 \times \frac{2}{7}[/latex]
2a + 6d = 14 …………… (i)
$$S_{n}=\frac{n}{2}[2 a+(n-1) d]$$
$$S_{17}=\frac{17}{2}[2 a+(n-1) d]$$
$$=\frac{17}{2}[2 \mathrm{a}+16 \mathrm{d}]$$
$$\frac{17}{2}[2 a+16 d]=289$$
$$2 a+16 d=289 \times \frac{2}{17}$$
∴ 2a + 16d = 34 …………. (ii)

10d = 20
∴ d=2
Substituting the value of d in equation (i), we have
2a + 6d = 14
2a + 6 × 2 = 14
2a + 12 = 14
2a = 14 – 12
2a = 2
$$a=\frac{2}{2}=1$$
a = 1, d = 2, Sn =?
$$S_{n}=\frac{n}{2}[2 a+(n-1) d]$$
$$S_{n}=\frac{n}{2}[2 \times 1+(n-1) 2]$$
$$=\frac{\mathrm{n}}{2}[2+2 \mathrm{n}-2]$$
$$=\quad \frac{\mathrm{n}}{2}[2 \mathrm{n}]$$
∴ Sn = n2

Question 10.
Show that a1, a2, a3, … a11. … form an AP where a11 is defIned as below:
i) a = 3 + 4n
ii) a = 9 – 5n
Solution:
i) an= 3 + 4n
a1 = 3 + 4(1)
= 3 + 4
∴ an = 7
∴ a = 7
a1 = 3 + 4n
a2 = 3 + 4 × 2
= 3 + 8
∴ a2 = 11
an = 3 + 4n
a3 = 3 + 4 × 3
= 3 + 12
a3 = 15
∴ a1, a2, a3, ………….
7, 11, 15, ……….
d = a2 – a1 = 11 – 7 = 4
d = a3 – a2= 15 – 11 = 4
Here, the value of ‘d’ is constant.
∴ an = 3 + 4n forms an Arithmetic Progression.

ii) an = 9 – 5n
a1= 9 – 5 × 1
= 9 – 5
∴ a1 = 4
a1 = 9 – 5n
a1 = 9 – 5 × 2
= 9 – 10
∴ a2 = -1
an= 9 – 5n
a3 = 9 – 5 × 3
= 9 – 15
∴ a3 = -6
a1, a2, a3, …………… an
4, -1, -6, …………
d = a2 – a1 = -1 – 4 = -5
d=a3 – a2 = -6 – (-1) = -5
Here, the value of ‘d’ is constant.
∴ an = 9 – 5n form an Arithmetic Progression.

Question 11.
lf the sum of the first n terms of an AP is 4n – n2, what is the first term (that is SI)? What is the sum of first two termš? What is the second term? Similarly. find the 3rdrd the 10th and the nth terms.
Solution:
If Sn = 4n – n2, then
i) S1 = a =?
ii) S2 =?
iii) a2 =?
iv) a3 =?
v) a10 =?
vi) an =?
(i) Sn = 4n – n2
S1 = 4(1) – 12
= 4 – 1
S1 = 3
∴ S1 = a = 3.

(ii) S2 = 4n – n2
S2 = 4(2) – 22
= 8 – 4
S2 = 4
∴ S2 = 4

(iii) We have S2 = a1 + a2 = 4
= 3 + a2 = 4
∴ a2 = 4 – 3
∴a2 = 1

(iv) Sn = 4n – n2
S3= 4(3) – 32
= 12 – 9
S3 = 3
a1 + a2 + a3 = 3
3 + 1 + a3 = 3
4 + a3= 3
∴ a3 = 3 – 4
∴ a 3 = -1

(v) d = a3 – a3 = -1 -1 = -2
a10= a + 9d
= 3 + 9(-2)
= 3 – 18
a10= -15
∴ a10 = -15

(vi) an = a + (n – 1)d
= 3 + (n – 1)(-2)
= 3 + 2n + 2
∴ an = 5 – 2n
∴ an = 5 – 2n

Question 12.
Find the sum of the first 40 positive integers divisible by 6.
Solution:
6 + 12 + 18 + 24 + 40 term
Here a = 6, d = 2 – a1 = 12 – 6 = 6
n = 40, S40 =?
$$S_{n}=\frac{n}{2}[2 a+(n-1) d]$$
$$S_{40}=\frac{40}{2}[2 \times 6+(40-1) 6]$$
= 20[12 + 39 × 6]
= 20[12 + 234]
= 20 × 246
∴ S40 = 4920.

Question 13.
Find the sum of first 15 multiples of 8,
Solution:
Sum of the first 15 multiples of 8 ?
8 + 16 + 24+ ……… 15 terms.
Here, a = 8, d = a2 – a1= 16 – 8 = 8
n = 15, S15 =?
$$S_{n}=\frac{n}{2}[2 a+(n-1) d]$$
$$S_{15}=\frac{15}{2}[2 \times 8+(15-1) 8]$$
$$=\frac{15}{2}[16+14 \times 8]$$
$$=\frac{15}{2}[16+112]$$
$$=\frac{15}{2} \times 128$$
= 15 × 64
∴ S15 = 960

Question 14.
Fnd the sum of the odd numbers between 0 and 50.
Solution:
Odd numbers between 0 and 50 are
1 + 3 + 5 + 7 + ……… + 49 =?
Here. a = 1, d = a3 – a1= 3 – 1 = 2
an = 49, n =?, S =?
an = a + (n – 1) d = 49
= 1 + (n – 1)2 = 49
1 + 2n – 2 = 49
2n – 1 = 49
2n = 49 + 1
2n = 50
$$n=\frac{50}{2}$$
∴ n = 25
$$S_{n}=\frac{n}{2}\left[a+a_{n}\right]$$
$$\mathrm{s}_{25}=\frac{25}{2}[1+49]$$
$$=\frac{25}{2} \times 50$$
= 25 × 25
∴ S25 = 625

Question 15.
A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: Rs. 200 for the first day. Rs. 250 for the second day. Rs. 300 for the third day. etc.. the penalty for each succeeding day being Rs. 50 more than for the preceeding day. How much money the contractor has to pay as penalty. If he has delayed the work by 30 days?
Solution:
Penalty for the 1st Day 2nd day 3rd day … 30th day
Rs. 200 Rs. 250 Rs. 300 Rs.?
200 + 250 + 300 + 30 days
a = 200, d = 250 – 200 = 50, n = 20,
S30 =?
$$S_{n}=\frac{n}{2}[2 a+(n-1) d]$$
$$\dot{\mathrm{S}}_{30}=\frac{30}{2}[2 \times 200+(30-1) 50]$$
= 15[400 + 29 × 50]
= 15 [400 + 1450]
= 15 × 1850
∴ S30= Rs. 27750.

Question 16.
A sum of Rs. 700 is to be used to give seven cash prizes to students of a school for thler overall academic performance. If each prize is Rs. 20 less than its preceding prize, find the value of each of the prizes.
Solution:
Let the first prize be ‘a’.
and Second prize is a – 20 and
the third prize Is a 40.
a, (a – 20), (a – 40) ……….. n = 7
a = a, d = a2 – a1 = a – 20 – a d = -20
n = 7, S7 =?
$$\frac{\mathrm{n}}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]=\mathrm{Sn}$$
$$\frac{7}{2}[2 \times a+(7-1)(-20)]=700$$
$$\frac{7}{2}[2 a+6(-20)]=700$$
$$\frac{7}{2}[2 a-120]=700$$
$$2 a-120=700 \times \frac{2}{7}$$
2a – 120 = 200
∴ 2a = 200 + 120
2a = 320
$$\quad a=\frac{320}{2}=R s .160$$
∴ Each prizes are
Rs. 160, 140, 120, 100, 80, 60, 40

Question 17.
In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant. will be the same as the class. In which they are studying. e.g., a section of Class I will plant 1 tree. a section of Class Il will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?
Solution:

a = 3. d = 6 – 3 = 3, n = 12, S12 =?
$$S_{n}=\frac{n}{2}[2 a+(n-1) d]$$
$$\mathrm{S}_{12}=\frac{12}{2}[2 \times 3+(12-1) 3]$$
= 6[6 + 11 × 3]
= 6[6 + 33]
= 6 × 39
∴ S12 = 234
∴ Total number of trees from 3 sections of each class upto 12 class is 234.

Question 18.
A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm. 1.0 cm. 1.5 cm, 2.0 cm as shows In fig. What Is the total length of such a spiral made up of thirteen consecutive semicircles? (Take $$\pi=\frac{22}{7}$$)

[Hint: Length of successive semicircles is l1, l2, l3, l4 with centres at A, B, A. B respectively.]
Solution :
$$: l_{1}=\pi \times \frac{1}{2}, 12=\pi \times 1.13=\pi \times \frac{3}{2}$$
$$l_{1}=\frac{\pi}{2}, \quad l_{2}=\mathrm{p}, \quad l_{3}=\frac{3}{2} \pi$$
∴ Arithmetic Progression,
l1, l2, l3, l4, ………..

Question 19.
200 logs are stacked In the following manner. 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on (see the fig. given below). In how many rows are the 200 logs placed and how many logs are In the top row?

Solution:
20, 19, 18, …..
a = 20, d = 19 – 20 = -1
Sn = 200, n =?, an =?
$$S_{n}=\frac{n}{2}[2 a+(n-1) d]$$
$$200=\frac{\mathrm{n}}{2}[2 \times 20+(\mathrm{n}-1)(-1)]$$
$$200=\frac{\mathrm{n}}{2}[40-\mathrm{n}+1]$$
$$200=\frac{n}{2}[41-n]$$
∴ 400 = n(41 – n)
400 = 4n – n2
∴ n2 – 41n + 400 = 0
n2 – 25n – 16n + 400 = 0
n(n – 25) – 16(n – 25) = 0
(n – 25) (n – 16) =
1f n – 16 = 0 then, n = 16
∴ an = a + (n – 1) d
a16= 20 + (16 – 1) (-1)
= 20 + 15(-1)
= 20 – 15
∴ a16 = 5
∴ 200 logs are placed In 16 rows and there are 5 logs in the top row.

Question 20.
In a potato race, a bucket is placed at the starting point, which Is 5 m from the first potato. and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line (see fig, given below)

A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops ¡tin the bucket, runs back to pick up the next potato, runs to the bucket to drop It in, and she continues in the same way until al) the potaotes are in the bucket. What is the total distance the competitor has to run?
(Hint: To pick up the first potato and the second potato, the total distance (in metres) run by a competitor is 2 × 5 +2 × (5 + 3)]
Solution:
Total distance competitor taken to pick up the first potato = 5 + 5m
= 2 × 5m.
= 10m.
Total distance taken by compeUtor to pick up the second potato
= 5 + 3 + 3 + 5m.
= 2 × 5 + 2 × 3
= 2(5 + 3)
= 2 × 8
= 16m.
∴ 10m. 16m, 22m 10th potato
a = 10, d = 16 – 10 = 6m.
n = 10, S10 =?
$$S_{n}=\frac{n}{2}[2 a+(n-1) d]$$
$$S_{10}=\frac{10}{2}[2 \times 10+(10-1) 6]$$
= 5[20 + 9 × 6]
= 5 [20 + 54]
= 5 × 74
∴ S10 = 370m.
∴ Total distance the competitor has to run to pick up 10 potatoers is 370 m.

Nearest tenth is the first digit after the decimal point.

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Ex 1.3 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Exercise 1.3, drop a comment below and we will get back to you at the earliest.

## KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2

KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.2.

## Karnataka SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.2

Find the Value of x is used to consider unknown value.

Question 1.
Form the pair of linear equations in the following problems, and find their solutions graphically.
(i) 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.
(ii) 5 pencils and 7 pens together cost Rs. 50, whereas 7 pencils and 5 pens together cost Rs. 46. Find the cost of one pencil and that of one pen.
Solution:
(i) Strength of the X Std. is 10
Number of boys be ‘y’, then
number of girls be ‘x’ .
x + y = 10 …….. (i)
x = y + 4
∴ x – y = 4 ………. (ii)
From equations (i) + (ii)

$$\quad x=\frac{14}{2}=7$$
Substituting the value of ‘x’ in eqn. (i),
x + y = 10
7 + y = 10
y = 10 – 7
y = 4.
∴ Number of girls, x = 7
Number of boys, y = 4.

(ii) Cost of each pencil be Rs. ‘x’
Cost of pen be Rs. ‘y’
5x + 7y = 50 ………. (i)
7x + 5y = 46 ……….. (ii)
From equation (i) + equation (ii)

∴ x + y = 8 …………… (iii)
from Eqn. (ii) – Eqn. (i),

∴ -x + y = 2 …………. (iv)
Eqn. (iii) + Eqn. (iv)

∴ y = 5
Substituting the value of ‘y’ in eqn. (i)
x + y = 8
x + 5 = 8
∴ x = 8 – 5 x = 3
∴Cost of each pencil is Rs. 3.
Cost of each pen is Rs. 5.

Question 2.
On comparing the ratios $$\frac{a_{1}}{a_{2}}, \frac{b_{1}}{b_{2}} \text { and } \frac{c_{1}}{c_{2}}$$ find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:
i) 5x – 4y + 8 = 0
7x + 6y – 9 = 0
ii) 9x + 3y + 12 = 0
18x + 6y + 24 = 0
iii) 6x – 3y + 10 = 0
2x – y + 9 = 0
Solution:
i) 5x – 4y + 8 = 0 a1= 5, b1 = -4, C1= 8
7x + 6y – 9 = 0 a2 = 7, b2= 6, c2= -9
$$\frac{a_{1}}{a_{2}}=\frac{5}{7} \quad \frac{b_{1}}{b_{2}}=\frac{-4}{6} \quad \frac{c_{1}}{c_{2}}=\frac{8}{-9}$$
Here, $$\frac{\mathbf{a}_{1}}{\mathbf{a}_{2}} \neq \frac{\mathbf{b}_{1}}{\mathbf{b}_{2}}$$
∴ Lines representing the pairs of linear equations intersect at a point.

ii) 9x + 3y + 12 = 0
18x + 6y + 24 = 0
a1 =9, b1 = 3, c1 = 12
a2 = 18, b2 = 6, c2 = 24
$$\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{9}{18}=\frac{1}{2} \quad \frac{\mathrm{b}_{1}}{\mathrm{b}_{2}}=\frac{3}{6}=\frac{1}{2}$$
$$\frac{c_{1}}{c_{2}}=\frac{12}{24}=\frac{1}{2}$$
Here, $$\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{\mathrm{b}_{1}}{\mathrm{b}_{2}}=\frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}$$
∴ Representation of lines graphically are coincident.

iii) 6x – 3y + 10 = 0
2x – y + 9 = 0
Here, a1 = 6, b1 = -3, c1 = 10
a2 = 2, b2 = -1, c2 = 9
$$\frac{a_{1}}{a_{2}}=\frac{6}{2}=\frac{3}{1} \quad \frac{b_{1}}{b_{2}}=\frac{-3}{-1}=\frac{3}{1}$$
$$\frac{c_{1}}{c_{2}}=\frac{10}{9}$$
Here, $$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$$
∴ Representation of lines graphically is parallel.

Question 3.
On compairing the ratios $$\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}, \frac{\mathrm{b}_{1}}{\mathrm{b}_{2}}$$ and $$\frac{c_{1}}{c_{2}}$$ find out whether the following pair of linear equations are consistent, or inconsistent.
i) 3x + 2y = 5; 2x – 3y = 7
ii) 2x – 3y = 8; 4x-6y = 9
iii) $$\frac{3}{2} x+\frac{5}{3} y=7 : 9 x-10 y=14$$
iv) 5x – 3y = 11: -10x – 6y = -22
v) $$\frac{4}{3} x+2 y=8 ; 2 x+3 y=12$$
Solution:
i) 3x + 2y = 5 ⇒ 3x + 2y – 5 = 0
2x – 3y = 7 ⇒ 2x – 3y – 7 = 0
Here, a1 = 3, b1 = 2, c1 = -5
a2 = 2, b2 = -3, c2 = -7
$$\frac{a_{1}}{a_{2}}=\frac{3}{2} \quad \frac{b_{1}}{b_{2}}=-\frac{2}{3} \quad \frac{c_{1}}{c_{2}}=\frac{-5}{-7}=\frac{5}{7}$$
Here, $$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$$
∴ Graphical representation is intersecting lines.
∴ Pair of linear equations are consistent.

ii) 2x – 3y = 8 ⇒ 2x – 3y – 8 = 0
4x – 6y = 9 ⇒ 4x – 6y – 9 = 0
a1 = 2, b1 = -3, c1 = -8
a2 = 4, b2 = -6, c2 = -9
$$\frac{a_{1}}{a_{2}}=\frac{2}{4}=\frac{1}{2} \quad \frac{b_{1}}{b_{2}}=\frac{-3}{-6}=\frac{1}{2}$$
$$\frac{c_{1}}{c_{2}}=\frac{-8}{-9}=\frac{8}{9}$$
Here, $$\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{\mathrm{b}_{1}}{\mathrm{b}_{2}} \neq \frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}$$
∴ Graphical representation is parallel lines.
∴ Equations are inconsistent.

iii) $$\frac{3}{2} x+\frac{5}{3} y=7 \quad \frac{3}{2} x+\frac{5}{3} y-7=0$$
9x – 10y = 14 ⇒ 9x – 10y – 14 = 0
$$a_{1}=\frac{3}{2}, \quad b_{1}=\frac{5}{3}, \quad c_{1}=-7$$
a2 = 9, b2 = -10, c2 = -17
$$\frac{a_{1}}{a_{2}}=\frac{3}{2} \times \frac{1}{9} \quad \frac{b_{1}}{b_{2}}=\frac{5}{3} \times \frac{-1}{6}$$
$$\frac{c_{1}}{c_{2}}=\frac{-7}{-14}=\frac{7}{14}=\frac{1}{2}$$
Here, $$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$$
∴ Pair of equations are consistent

iv) 5x – 3y = 11 ⇒ 5x – 3y – 11 = 0
-10x + 6y = – 22 ⇒ -10x + 6y + 22 = 0
a1 = 5, b1 = -3, c1 = -11
a2 = -10, b2 = 6, c2 = -22
$$\frac{a_{1}}{a_{2}}=\frac{5}{-10}=-\frac{1}{2} \quad \frac{b_{1}}{b_{2}}=\frac{-3}{6}=-\frac{1}{2}$$
$$\frac{c_{1}}{c_{2}}=\frac{-11}{22}=-\frac{1}{2}$$
Here, $$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$$
∴ Pair of equations are consistent
∴ Graphical representation is coninciding.

v) $$\frac{4}{3} x+2 y=8 \quad \frac{4}{3} x+26-8=0$$
2x + 3y = 12 ⇒ 2x + 3y – 12 = 0
$$a_{1}=\frac{4}{3}, \quad b_{1}=2, \quad c_{1}=-8$$
a2 = 2, b2 = 3, c2 = -12
$$\frac{a_{1}}{a_{2}}=\frac{4}{3} \times \frac{1}{2}=\frac{1}{6} \quad \frac{b_{1}}{b_{2}}=\frac{2}{3}$$
$$\frac{c_{1}}{c_{2}}=\frac{-8}{-12}=\frac{2}{3}$$
Here, $$\frac{\mathbf{a}_{1}}{\mathbf{a}_{2}} \neq \frac{\mathbf{b}_{1}}{\mathbf{b}_{2}}$$
∴ Pair of equations are consistent

Question 4.
Which of the following pairs of linear equations are consistent/inconsistent? If consistent obtain the solution graphically
(i) x + y = 5, 2x + 2y = 10
(ii) x-y = 8 3x-3y= 16
(iii) 2x + y – 6 = 0 4x – 2y – 4 = 0
(iv) 2x – 2y – 2 = 0 4x – 3y – 5 = 0
Solution:
(i) x + y = 5 ⇒ x + y – 5 = 0
2x + 2y = 10 ⇒ 2x + 2y – 10 = 0
a1 = 1, b1 = 1, c1 = -5
a2 = 2, b2 = 2, c2 = -10
$$\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{1}{2} \quad \frac{\mathrm{b}_{1}}{\mathrm{b}_{2}}=\frac{1}{2} \quad \frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}=\frac{-5}{-10}=\frac{1}{2}$$
Here, $$\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{\mathrm{b}_{1}}{\mathrm{b}_{2}}=\frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}$$
∴ Pair of equations are consistent
(i) x + y = 5
y = 5 – x

 x 0 2 4 y = 5 – x 5 3 1

(ii) 2x + 2y = 10
x + y = 5
y = 5 – x

 x 0 2 5 y = 5 – x 5 3 0

∴ We can give any value for ‘x’, i.e., solutions are infinite.
∴ Solution, P (5, 0) x = 5, y = 0

(ii) x – y = 8 ⇒ x – y – 8 = 0
3x – 3y = 16 ⇒ 3x – 3y – 16 = 0
Here, $$\frac{a_{1}}{a_{2}}=\frac{1}{3} \quad \frac{b_{1}}{b_{2}}=\frac{-1}{-3}=\frac{1}{3}$$
$$\frac{c_{1}}{c_{2}}=\frac{-8}{-16}=\frac{1}{2}$$
$$\quad \frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{\mathrm{b}_{1}}{\mathrm{b}_{2}} \neq \frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}$$
∴ Linear equations are inconsistenttent.
∴ Algebraically it has no solution.
Graphical representation → Parallel Lines.
(i) x – y = 8
-y = 8 – x
y = -8 + x

 x 8 10 9 y = -8+x 0 2 1

(ii) 3x – 3y = 16
-3y = 16 – 3x
3y = -16 + 3x
$$\quad y=\frac{-16+3 x}{3}$$

 x 6 8 $$y=\frac{-16+3 x}{3}$$ 0.8 2.6

No solution because it is inconsistent

(iii) 2x + y – 6 = 0
4x – 2y – 4 = 0
Here a1 = 2, b1 = 1, c1 = -6
a2 = 4, b2 = -2, c2 = -4
$$\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{2}{4}=\frac{1}{2} \quad \frac{\mathrm{b}_{1}}{\mathrm{b}_{2}}=\frac{1}{-2}$$
$$\frac{c_{1}}{c_{2}}=\frac{-6}{-4}=\frac{3}{2}$$
Here, $$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$$
Pair of equations are consistent. Algebraically both lines intersect.
Graphical Representation :
(i) 2x + y = 6
y = 6 – 2x

 x 0 2 y = 6 – 2x 6 2

(ii) 4x – 2y – 4 = 0
4x – 2y = 4
-2y = 4 – 4x
2y = -4 + 4x
$$\quad y=\frac{-4+4 x}{2}$$

 x 1 3 $$y=\frac{-4+4 x}{2}$$ 0 4

Solution: intersecting point, P (2, 2) i.e., x = 2, y = 2

(iv) 2x – 2y – 2 = 0
4x – 3y – 5 = 0
a1 = 2, b1 = -2, c1 = -2
a2 = 4, b2 = -3, c2 = -5
$$\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{2}{4}=\frac{1}{2} \quad \frac{\mathrm{b}_{1}}{\mathrm{b}_{2}}=\frac{-2}{-3}=\frac{2}{3}$$
$$\frac{c_{1}}{c_{2}}=\frac{-2}{-5}=\frac{2}{5}$$
Here, $$\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}} \neq \frac{\mathrm{b}_{1}}{\mathrm{b}_{2}}$$
Pair of equations are consistent.
∴ Algebraically both lines intersect.
Graphical Representation :
(i) 2x – 2y – 2 =0
2x – 2y = 2
-2y = 2 – 2x
2y = -2 + 2x
$$\quad y=\frac{-2+2 x}{2}$$
∴ y = – 1 + x

 x 2 4 y= -1+ x 1 3

(ii) 4x – 3y – 5 = 0
4x – 3y = 5
-3y = 5 – 4x
3y = -5 + 4x
$$\quad y=\frac{-5+4 x}{3}$$

 x 2 5 $$y=\frac{-5+4 x}{3}$$ 1 5

Solution: P(2, 1) i.e., x = 2, y = 1

Question 5.
Half the perimeter of a rectangular garden, whose length is 4m more than its width, is 36m. Find the dimensions of the garden.

Solution:
Length of rectangular garden be ‘x’ m.
Breadth of rectangular garden be ‘y’ m, then
Length of the garden is 4m more than its width.
x = y + 4 ……….. (i)
x – y = 4
Half of the circumference is 36 m.
$$\frac{2 x+2 y}{2}=36$$
2x + 2y = 72
x + y = 36 …………… (ii)
∴ x – y = 4 (i)
x + y = 36 (ii)
From eqn. (i) + eqn. (ii)

$$\quad x=\frac{40}{2}$$
∴ x = 20 m.
Substituting the value of ’x’ in eqn. (i)
x – y = 4
20 – y = 4
-y = 4 – 20
-y = -16
y = 16 m.
∴ Length of the garden = 20 m.
Breadth of the garden = 16 m.

Question 6.
Given the linear equation 2x + 3y – 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is :
(i) intersecting lines
(ii) parallel lines
(iii) coincident lines.
Solution:
Linear equation is 2x + 3y – 8 = 0.
(i) Linear equations for intersecting lines:
2x + 3y – 8 = 0
3x + 2y – 7 = 0
a1 = 2, b1 = 3, c1 = -8
a2 = 3, b2 = 2, c2 = -7
$$\frac{a_{1}}{a_{2}}=\frac{2}{3} \quad \frac{b_{1}}{b_{2}}=\frac{3}{2}$$
Here, when $$\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}} \neq \frac{\mathrm{b}_{1}}{\mathrm{b}_{2}}$$ Geometrical representation is intersecting lines.

(ii) For Parallel lines :
2x + 3y – 8 = 0
2x + 3y – 12 = 0
a1 = 2, b1 = 3, c1= -8
a1 = 2, b1 = 3, c1 = -12
$$\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{2}{2}=\frac{1}{1} \quad \frac{\mathrm{b}_{1}}{\mathrm{b}_{2}}=\frac{3}{3}=\frac{1}{1}$$
$$\frac{c_{1}}{c_{2}}=\frac{-8}{-12}=\frac{8}{12}=\frac{2}{3}$$
Here, $$\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{\mathrm{b}_{1}}{\mathrm{b}_{2}} \neq \frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}$$ hence
Graphical representation is parallel lines.

(iii) For Intersecting lines :
2x + 3y – 8 = 0
4x + 6y – 16 = 0
a1 = 2, b1 = 3, c1 = -8
a1 = 4, b1 = 6, c1 = -16
$$\frac{a_{1}}{a_{2}}=\frac{2}{4}=\frac{1}{2} \quad \frac{b_{1}}{b_{2}}=\frac{3}{6}=\frac{1}{2}$$
$$\frac{c_{1}}{c_{2}}=\frac{-8}{-16}=\frac{1}{2}$$
Here,
∴ Lines are intersecting.

Question 7.
Draw the graphs of the equations x – y + 1 =0 and 3x + 2y – 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.
Solution:
x – y + 1 = 0 ………. (i)
3x – 2y – 12 = 0 ……….. (ii)
x – y + 1 = 0
-y = -x -1
y = x + 1

 x 2 4 y = x + 1 3 5

3x + 2y – 12 = 0
2y = -3x + 12
$$\quad y=\frac{-3 x+12}{2}$$

 x 0 2 $$y=\frac{-3 x+12}{2}$$ 6 3

Graphs of these two equations intersect at A. Vertices formed for ∆ABC are,
A (2, 3), B (-1, 0), C (4, 0).

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.2, drop a comment below and we will get back to you at the earliest.

## Karnataka SSLC Maths Model Question Paper 4 Kannada Medium

Karnataka SSLC Maths Model Question Paper 4 Kannada Medium

## Karnataka SSLC Maths Model Question Paper 4 Kannada Medium

ವಿಷಯ : ಗಣಿತ
ಸಮಯ: 3 ಗಂಟೆಗಳು
ಗರಿಷ್ಠ ಅಂಕಗಳು: 80

I. ಕೆಳಗಿನ ಪ್ರಶ್ನೆಗಳಿಗೆ ಅಥವಾ ಅಪೂರ್ಣ ಹೇಳಿಕೆಗೆ ನಾಲ್ಕು ಪರ್ಯಾಯ ಉತ್ತರಗಳನ್ನು ನೀಡಲಾಗಿದೆ. ಇವುಗಳಲ್ಲಿ ಸೂಕ್ತವಾದ ಉತ್ತರವನ್ನು ಆರಿಸಿ, ಕ್ರಮಾಕ್ಷರದೊಡನೆ ಪೂರ್ಣ ಉತ್ತರವನ್ನು ಬರೆಯಿರಿ. (8 × 1 = 8)

Question 1.
ಕೆಳಗಿನ ಸಂಖ್ಯೆಗಳ ಗಣಗಳಲ್ಲಿ ಸಮರೂಪ ತ್ರಿಭುಜಗಳಾಗುವಂತಹ ಜೋಡಿಯು
(A) (3, 4, 6) (9, 12, 24)
(B) (3, 4, 6) (9, 12, 18)
(C) (2,4, 6) (2,3, 14)
(D) (5, 10, 15) (10, 30, 45)

Question 2.
ಚತುರ್ಥಕದ ವಿಸ್ತೀರ್ಣವನ್ನು ಕಂಡುಹಿಡಿಯುವ ಸೂತ್ರ
(A) πr2
(B) $$\frac { 1 }{ 2 }$$ πr2
(C) $$\frac { 1 }{ 3 }$$ πr2
(D) $$\frac { 1 }{ 4 }$$ πr2

Question 3.
(2, 3) ಮತ್ತು (4, 5) ಬಿಂದುಗಳನ್ನು ಸೇರಿಸುವ ರೇಖಾಖಂಡದ ಮಧ್ಯಬಿಂದುವಿನ ನಿರ್ದೇಶಾಂಕಗಳು
(A) (3, 4)
(B) (4, 5)
(C) (5, 6)
(D) (6, 7)

Question 4.
ಧನ ಪೂಣಾಂಕ ‘a’ ಅನ್ನು ಪೂಣಾಂಕ ‘b’ ನಿಂದ ಭಾಗಿಸಿದಾಗ ಅನನ್ಯ ಪೂರ್ಣಾಂಕಗಳಾದ ‘q’ ಮತ್ತು ‘r’ ಗಳು ಇರುತ್ತವೆ. ಇದಕ್ಕೆ ಸರಿಹೊಂದುವ ಹೇಳಿಕೆ.
(A) b = a × q + r
(B) q = a × b + r
(C) a = b × q – r
(D) a = b × q + r

Question 5.
tan260° ಯ ಬೆಲೆ
(A) $$\frac { 1 }{ 3 }$$
(B) $$\frac { 1 }{ \surd 3 }$$
(C) 3
(D) √3

Question 6.
m ನ ಯಾವ ಧನಾತ್ಮಕ ಬೆಲೆಗೆ 3x2 + ka + 3 = 0 ಸಮೀಕರಣದ ಮೂಲಗಳು ಸಮವಾಗಿರುತ್ತವೆ.
(A) 2
(B) 3
(C) 5
(D) 6

Question 7.
ಒಂದು ಪ್ರಯೋಗದ ಎಲ್ಲಾ ಪ್ರಾಥಮಿಕ ಘಟನೆಗಳ ಸಂಭವನೀಯತೆಗಳ ಮೊತ್ತವು
(A) 0
(B) 1
(C) 2
(D) 3

Question 8.
ಒಂದು ಸಿಲಿಂಡರ್‌ನ ಪಾದದ ವಿಸ್ತೀರ್ಣ 24 cm2 ಮತ್ತು ಎತ್ತರ 10 cm ಆದರೆ ಸಿಲಿಂಡರ್‌ನ ಘನಫಲ
(A) 24 cm3
(B) 48 cm3
(C) 240 cm3
(D) 480 cm3

II. ಈ ಕೆಳಗಿನ ಪ್ರಶ್ನೆಗಳಿಗೆ ಉತ್ತರಿಸಿ: (6 × 1 = 6)

Question 9.
ಥೇಲ್ಸ್‌ನ ಪ್ರಮೇಯವನ್ನು ನಿರೂಪಿಸಿ,

Question 10.
135 ಮತ್ತು 225 ರ ಮ.ಸಾ.ಅ. ವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 11.
ಬಹುಪದೋಕ್ತಿ x2 – 3x +5 ರ ಶೂನ್ಯತೆಗಳ ಮೊತ್ತವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 12.
ಬಾಹ್ಯ ಬಿಂದುವಿನಿಂದ ವೃತ್ತಕ್ಕೆ ಎಳೆದ ಸ್ಪರ್ಶಕದ ಉದ್ದವು 8 cm ಮತ್ತು ವೃತ್ತಕೇಂದ್ರ ಹಾಗು ಬಾಹ್ಯಬಿಂದುವಿನ ದೂರ 10 cm ಆದರೆ ವೃತ್ತದ ತ್ರಿಜ್ಯವೆಷ್ಟು?

Question 13.
sin A = $$\frac { 3 }{ 4 }$$ ಆದರೆ cosec A ನ ಬೆಲೆಯೇನು?

Question 14.
ಹಂತ ವಿಚಲನಾ ವಿಧಾನದಿಂದ ಸರಾಸರಿಯನ್ನು ಕಂಡುಹಿಡಿಯುವ ಸೂತ್ರವನ್ನು ಬರೆಯಿರಿ.

III. ಈ ಕೆಳಗಿನ ಪ್ರಶ್ನೆಗಳಿಗೆ ಉತ್ತರಿಸಿ: (16 × 2 = 32)

Question 15.
ವಾರ್ಷಿಕ ಸಂಬಳ ₹ 5000 ಮತ್ತು ಪ್ರತಿವರ್ಷಕ್ಕೆ ಹೆಚ್ಚುವರಿ ಬತ್ಯ ₹ 200 ಇರುವ ಕೆಲಸಕ್ಕೆ ಸುಬ್ಬರಾವ್ 1995 ರಲ್ಲಿ ಸೇರಿದರು. ಯಾವ ವರ್ಷದಲ್ಲಿ ಅವರ ಸಂಬಳ ₹ 7000 ಆಗುತ್ತದೆ?

Question 16.
ಚಿತ್ರದಲ್ಲಿ LM || CB ಮತ್ತು LN || CD ಆದರೆ $$\frac { AM }{ AB }$$ = $$\frac { AN }{ AD }$$ ಎಂದು ಸಾಧಿಸಿ.

Question 17.
ಈ ಜೋಡಿ ಸಮೀಕರಣಗಳನ್ನು ವರ್ಜಿಸುವ ವಿಧಾನದಿಂದ ಬಿಡಿಸಿ, 3x + 4y = 10 & 2x – 2y = 2.

Question 18.
ಕೆಳಗಿನ ರೇಖಾತ್ಮಕ ಸಮೀಕರಣಗಳ ಜೋಡಿಗಳು ಪ್ರತಿನಿಧಿಸುವ ಸರಳರೇಖೆಗಳು ಒಂದು ಬಿಂದುವಿನಲ್ಲಿ ಛೇದಿಸುತ್ತವೆಯೇ? ಸಮಾಂತರವಾಗಿವೆಯೆ? ಅಥವಾ ಐಕ್ಯಗೊಂಡಿವೆಯೆ? ಕಂಡುಹಿಡಿಯಿರಿ
9x + 3y + 12 = 0; 18x + 6y + 24 = 0.

Question 19.
ಪರಧಿಯು 22 cm ಇರುವ ಒಂದು ಅರ್ಧ ವೃತ್ತದ ವಿಸ್ತೀರ್ಣವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 20.
4 cm ಮತ್ತು 6 cm ತ್ರಿಜ್ಯಗಳಿರುವ ಏಕಕೇಂದ್ರೀಯ ವೃತ್ತಗಳಿವೆ. 6 cm ತ್ರಿಜ್ಯದ ವೃತ್ತದ ಮೇಲಿನ ಒಂದು ಬಿಂದುವಿನಿಂದ 4 cm ತ್ರಿಜ್ಯದ ವೃತ್ತಕ್ಕೆ ಸ್ಪರ್ಶಕವನ್ನು ರಚಿಸಿ.

Question 21.
(2, -5) ಮತ್ತು (-2, 9) ರಿಂದ ಸಮಾನ ದೂರದಲ್ಲಿರುವ x-ಅಕ್ಷದ ಮೇಲಿನ ಬಿಂದುವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 22.
ಶೃಂಗಬಿಂದುಗಳು (1, -3), (4, 1) ಮತ್ತು (2, 3) ಆಗಿರುವ ತ್ರಿಭುಜದ ವಿಸ್ತೀರ್ಣವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 23.
ಶೂನ್ಯತೆಗಳ ಮೊತ್ತ $$\frac { -1 }{ 4 }$$ ಮತ್ತು ಗುಣಲಬ್ಧ $$\frac { 1 }{ 4 }$$ ಆಗಿರುವ ಒಂದು ವರ್ಗ ಬಹುಪದೋಕ್ತಿಯನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 24.
x2 + 6x – 7 = 0 ಸಮೀಕರಣವನ್ನು ವರ್ಗಪೂರ್ಣಗೊಳಿಸುವ ವಿಧಾನದಿಂದ ಬಿಡಿಸಿ.

Question 25.
ಗೋಪುರದ ಪಾದದಿಂದ 30m ದೂರದ ನೆಲದ ಮೇಲಿನ ಒಂದು ಬಿಂದುವಿನಿಂದ, ಗೋಪುರದ ತುದಿಯನ್ನು ನೋಡಿದಾಗ ಉಂಟಾಗುವ ಉನ್ನತ ಕೋನವು 30° ಆದರೆ, ಗೋಪುರದ ಎತ್ತರವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 26.
ಒಂದು ಕಟ್ಟಡದ ಮೇಲಿನಿಂದ ಹಾಗೂ ಕೆಳಗಿನಿಂದ ಬೆಟ್ಟದ ತುದಿಯನ್ನು ಗಮನಿಸಿದಾಗ ಉಂಟಾದ ಉನ್ನತ ಕೋನವು 45° ಮತ್ತು 60° ಆಗಿದೆ. ಕಟ್ಟಡದ ಎತ್ತರ 24 m ಆದರೆ ಬೆಟ್ಟದ ಎತ್ತರವೇನು?

Question 27.
ಈ ಕೆಳಗಿನ ದತ್ತಾಂಶಗಳಿಗೆ ಸರಾಸರಿಯನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

ಅಥವಾ
ಈ ಕೆಳಗಿನ ದತ್ತಾಂಶಗಳಿಗೆ ಬಹುಲಕವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 28.
ಒಂದು ಚೀಲದಲ್ಲಿ 3 ಕೆಂಪು ಚೆಂಡುಗಳು ಮತ್ತು 5 ಕಪ್ಪು ಚೆಂಡುಗಳಿವೆ. ಚೀಲದಿಂದ ಯಾದೃಚ್ಛಿಕವಾಗಿ ಒಂದು ಚೆಂಡನ್ನು ತೆಗೆಯಲಾಗಿದೆ. ತೆಗೆದ ಚಂಡು ಕೆಂಪು ಆಗಿರುವ ಸಂಭವನೀಯತೆ ಎಷ್ಟು?

Question 29.
64 cm3 ಘನಫಲವನ್ನು ಹೊಂದಿರುವ 2 ವರ್ಗ ಘನಗಳ ಮುಖಗಳನ್ನು ಸೇರಿಸಿ ಒಂದು ಆಯತ ಘನಾಕೃತಿ ಮಾಡಿದೆ. ಈ ಘನಾಕೃತಿಯ ಮೇಲೈ ವಿಸ್ತೀರ್ಣವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
ಅಥವಾ
14 cm ಎತ್ತರವಿರುವ ಒಂದು ಕುಡಿಯುವ ನೀರಿನ ಗಾಜಿನ ಲೋಟವು ಶಂಕುವಿನ ಭಿನ್ನಕದ ರೂಪದಲ್ಲಿದೆ. ಅದರ ಎರಡು ವೃತ್ತಾಕಾರದ ಪಾದಗಳ ವ್ಯಾಸಗಳು 4 cm ಮತ್ತು 2 cm ಗಳಾಗಿವೆ. ಗಾಜಿನ ಲೋಟ ಸಾಮರ್ಥ್ಯವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 30.
12 ಮತ್ತು 15 ರ ಲ.ಸಾ.ಅ. ಮತ್ತು ಮ.ಸಾ.ಅ. ವನ್ನು ಅವಿಭಾಜ್ಯ ಅಪವರ್ತನ ವಿಧಾನದಿಂದ ಕಂಡುಹಿಡಿಯಿರಿ.

IV. ಈ ಕೆಳಗಿನ ಪ್ರಶ್ನೆಗಳಿಗೆ ಉತ್ತರಿಸಿ: (6 × 3 = 18)

Question 31.
ವೃತ್ತದ ಮೇಲಿನ ಯಾವುದೇ ಬಿಂದುವಿನಲ್ಲಿ ಎಳೆದ ಸ್ಪರ್ಶಕವು, ಸ್ಪರ್ಶ ಬಿಂದುವಿನಲ್ಲಿ ಎಳೆದ ತ್ರಿಜ್ಯಕ್ಕೆ ಲಂಬವಾಗಿರುತ್ತದೆ ಎಂದು ಸಾಧಿಸಿ.
ಅಥವಾ
ಒಂದು ಸಮಾಂತರ ಚತುರ್ಭುಜದಲ್ಲಿ ವೃತ್ತವು ಅಂತಸ್ಥವಾದಾಗ ಸಮಾಂತರ ಚತುರ್ಭುಜವು ವಜ್ರಾಕೃತಿಯಾಗುತ್ತದೆ ಎಂದು ಸಾಧಿಸಿ.

Question 32.
ಪಾದ 8 cm ಮತ್ತು ಎತ್ತರ 4 cm ಇರುವ ಒಂದು ಸಮದ್ವಿಬಾಹು ತ್ರಿಭುಜವನ್ನು ರಚಿಸಿ, ನಂತರ ಮತ್ತೊಂದು ತ್ರಿಭುಜವನ್ನು ಅದರ ಬಾಹುಗಳು ಮೊದಲು ರಚಿಸಿದ ಸಮದ್ವಿಬಾಹು ತ್ರಿಭುಜದ ಅನುರೂಪ ಬಾಹುಗಳ 1$$\frac { 1 }{ 2 }$$ ರಷ್ಟಿರುವಂತೆ ರಚಿಸಿ,

Question 33.
ಎರಡು ಕ್ರಮಾಗತ ಬೆಸ ಧನ ಪೂರ್ಣಾಂಕಗಳ ವರ್ಗಗಳ ಮೊತ್ತವು 290 ಆದರೆ ಆ ಪೂಣಾರ್ಂಕಗಳನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
ಅಥವಾ
ಒಂದು ಆಯತಾಕಾರದ ಹೊಲದ ಕರ್ಣವು ಅದರ ಚಿಕ್ಕ ಬಾಹುವಿಗಿಂತ 60 m ಹೆಚ್ಚಾಗಿದೆ. ಅದರ ದೊಡ್ಡ ಬಾಹುವು ಚಿಕ್ಕ ಬಾಹುವಿಗಿಂತ 30 m ಹೆಚ್ಚಾಗಿದ್ದರೆ, ಆ ಹೊಲದ ಬಾಹುಗಳ ಉದ್ದಗಳನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 34.

ಅಥವಾ
sec A (1 – sin A) (sec A + tan A) = 1 ಎಂದು ಸಾಧಿಸಿ.

Question 35.
ಕೆಳಗಿನ ದತ್ತಾಂಶಗಳಿಗೆ ಓಜೀವ್ ನಕ್ಷೆಯನ್ನು ರಚಿಸಿ, (ಕಡಿಮೆ ವಿಧಾನ)

Question 36.
3x2 – x – 4 ಎಂಬ ಬಹುಪದೋಕ್ತಿಯ ಶೂನ್ಯತೆಗಳನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ ಹಾಗು ಶೂನ್ಯತೆಗಳು ಮತ್ತು ಸಹಗುಣಕಗಳ ನಡುವಿನ ಸಂಬಂಧವನ್ನು ತಾಳೆ ನೋಡಿ.
ಅಥವಾ
x4 – 3x2 + 4x + 5 ನ್ನು x2 + 1 – x ದಿಂದ ಭಾಗಿಸಿ ಭಾಗಲಬ್ದ ಮತ್ತು ಶೇಷವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

V. ಈ ಕೆಳಗಿನ ಪ್ರಶ್ನೆಗಳಿಗೆ ಉತ್ತರಿಸಿ: (4 × 4 = 16)

Question 37.
ಸಮಾಂತರ ಶ್ರೇಢಿಯ ಮೂರು ಪದಗಳ ಮೊತ್ತವು 15 ಮತ್ತು ಅವುಗಳ ಅಂತ್ಯಪದಗಳ ವರ್ಗಗಳ ಮೊತ್ತವು 58 ಆಗಿದೆ. ಶ್ರೇಢಿಯ ಆ ಮೂರು ಪದಗಳನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
ಅಥವಾ
ಸಮಾಂತರ ಶ್ರೇಢಿಯ ಮೊದಲ 5 ಪದಗಳ ಮೊತ್ತವು ಮುಂದಿನ 5 ಪದಗಳ ಮೊತ್ತದ ನಾಲ್ಕನೇ ಒಂದು ಭಾಗದಷ್ಟಿದೆ. ಮೊದಲ ಪದ 2 ಆದರೆ, a20 = -112 ಎಂದು ಸಾಧಿಸಿ ಮತ್ತು S20 ನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 38.
ಎರಡು ಸಮರೂಪ ತ್ರಿಭುಜಗಳ ವಿಸ್ತೀರ್ಣಗಳ ಅನುಪಾತವು ಅವುಗಳ ಅನುರೂಪ ಬಾಹುಗಳ ವರ್ಗಗಳ ಅನುಪಾತಕ್ಕೆ ಸಮನಾಗಿರುತ್ತದೆ ಎಂದು ಸಾಧಿಸಿ.

Question 39.
ಒಂದು ಬಾವಿಯ ವ್ಯಾಸ 3 m ಮತ್ತು ಆಳ 14 m ಇರುವಂತೆ ತೋಡಿದೆ. ಭೂಮಿಯಿಂದ ತೆಗೆದ ಮಣ್ಣನ್ನು ಬಾವಿಯ ಸುತ್ತಲು ಸಮವಾಗಿ ಹರಡಿ 4 m ಅಗಲವಿರುವ ವೃತ್ತಾಕಾರದ ಕಟ್ಟೆಯನ್ನು ಕಟ್ಟಿದೆ. ಕಟ್ಟೆಯ ಎತ್ತರವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 40.
ನಕ್ಷೆಯ ಮೂಲಕ ಸಮೀಕರಣಗಳನ್ನು ಬಿಡಿಸಿ.
2x + y = 8
x + 2y = 7

Solutions

I.
Solution 1.
(B) (3, 4, 6) (19, 12, 18)

Solution 2.
(D) $$\frac { 1 }{ 4 }$$ πr2

Solution 3.
(A) (3, 4)

Solution 4.
(D) a = b × q + r

Solution 5.
(C) 3

Solution 6.
(D) 6

Solution 7.
(B) 1

Solution 8.
(C) 240 cm3

II.
Solution 9.
ತ್ರಿಭುಜದ ಎರಡು ಬಾಹುಗಳನ್ನು ಎರಡು ವಿಭಿನ್ನ ಬಿಂದುಗಳಲ್ಲಿ ಛೇದಿಸುವಂತೆ ಒಂದು ಬಾಹುವಿಗೆ ಸಮಾಂತರವಾಗಿ ಎಳೆದ ಸರಳರೇಖೆಯು ಉಳಿದೆರಡು ಬಾಹುಗಳನ್ನು ಸಮಾನುಪಾತದಲ್ಲಿ ವಿಭಾಗಿಸುತ್ತದೆ.

Solution 10.

Solution 11.

Solution 12.
OA2 = OP2 – AP2 =102 – 82
⇒ OA2 = 100 – 64
⇒ OA2 = 100 – 64
⇒ OA = √36
⇒ OA = 6 cm
ವೃತ್ತದ ತ್ರಿಜ್ಯ = 6 cm

Solution 13.
cosec A = $$\frac { 4 }{ 3 }$$

Solution 14.

III.
Solution 15.
5000, 5200, 5400, ……… 7000.
a = 5000, d = 200 an = 7000, n = ?
an = a + (n – 1) d
⇒ 7000 = 5000 + (n – 1) 200
⇒ 7000 – 5000 = (n – 1) 200
⇒ 2000 = 200 (n – 1)
⇒ (n – 1) = 10
⇒ n = 10 + 1
⇒ n = 11
11 ನೇ ವರ್ಷದಲ್ಲಿ ಅವರ ಸಂಬಳ ₹ 7000 ಆಗುತ್ತದೆ.

Solution 16.

Solution 17.
3x + 4y = 10 …. (1)
2x – 2y = 2 …… (2)
ಸಮೀಕರಣ 2ನ್ನು 2ರಿಂದ ಗುಣಿಸಿದಾಗ
3x + 4y = 10
4x – 4y = 4
…………………
7x = 14
x = 2
3x + 4y = 10
3(2) + 4y = 10
4y = 10 – 6
y = 1
∴ x = 2, y = 1

Solution 18.

Solution 19.

ಅರ್ಧ ವೃತ್ತದ ವಿಸ್ತೀರ್ಣ = $$\frac { 77 }{ 4 }$$ cm2

Solution 20.
R = 6 cm, r = 4 cm

Solution 21.

Solution 22.

Solution 23.
ಬಹುಪದೋಕ್ತಿ = ax2+ bx + c ಆಗಿರಲಿ
ಶೂನ್ಯತೆಗಳು α ಮತ್ತು β ಆಗಿರಲಿ

Solution 24.
x2 + 6x – 7 = 0
⇒ x2 + 6x = 7
⇒ x2 + 6x + 32 – 32 = 7
⇒ (x + 3)2 – 9 = 7
⇒ (x + 3)2 = 16
⇒ x + 3 = ±√16
⇒ x + 3 = ±4
⇒ x = ±4 – 3
⇒ x = +4 – 3 ಅಥವಾ x = -4 – 3
⇒ x = 1 ಅಥವಾ x = -7

Solution 25.

Solution 26.

Solution 27.

Solution 28.
ಚೀಲದಲ್ಲಿರುವ ಒಟ್ಟು ಚೆಂಡುಗಳು = n(S) = 3 + 5 = 8
ಕೆಂಪು ಚೆಂಡುಗಳು ಸಂಖ್ಯೆ = 3 = n(A)
ಕೆಂಪು ಚೆಂಡು ತೆಗೆಯುವ ಸಂಭವನೀಯತೆ = $$\frac { n(A) }{ n(S) }$$
ಕೆಂಪು ಚೆಂಡು ತೆಗೆಯುವ ಸಂಭವನೀಯತೆ = $$\frac { 3 }{ 8 }$$

Solution 29.
ವರ್ಗಗಳ ಘನಫಲ = 64cm3
ಬಾಹುವಿನ ಅಳತೆ = $$\sqrt [ 3 ]{ 64 }$$ = 4 cm
ಆಯತ ಘನದ ಅಗಲ b = 4 cm
ಆಯತ ಘನದ ಉದ್ದ l = 8 cm (4 + 4)
ಆಯತ ಘನದ ಎತ್ತರ h = 4 cm
ಆಯತ ಘನದ ಮೇಲೈ ವಿಸ್ತೀರ್ಣ = 2 (lb + bh + hl)
= 2(8 × 4 + 4 × 4 + 4 × 8)
= 2(32 + 16 + 32)
= 160 cm2
ಆಯತ ಘನದ ಮೇಲೈ ವಿಸ್ತೀರ್ಣ = 160 cm2
ಅಥವಾ
14 cm ಎತ್ತರವಿರುವ ಒಂದು ಕುಡಿಯುವ ನೀರಿನ ಗಾಜಿನ ಲೋಟವು ಶಂಕುವಿನ ಭಿನ್ನಕದ ರೂಪದಲ್ಲಿದೆ. ಅದರ ಎರಡು ವೃತ್ತಾಕಾರದ ಪಾದಗಳ ವ್ಯಾಸಗಳು 4 cm ಮತ್ತು 2 cm ಗಳಾಗಿವೆ. ಗಾಜಿನ ಲೋಟ ಸಾಮರ್ಥ್ಯವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Solution 30.
12 = 2 × 2 × 5
15 = 3 × 5
ಮ.ಸಾ.ಅ. = 5
ಲ.ಸಾ.ಅ. = 2 × 2 × 3 × 5
ಲ.ಸಾ.ಅ. = 60

IV.
Solution 31.

ದತ್ತ: ‘O’ ವೃತ್ತಕೇಂದ್ರ, XY ಸ್ಪರ್ಶಕ, Pಸ್ಪರ್ಶಬಿಂದು
ಸಾಧನೀಯ: OP ⊥ XY
ರಚನೆ: P ಯನ್ನು ಹೊರತುಪಡಿಸಿ XY ಮೇಲೆ ಮತ್ತೊಂದು ಬಿಂದು Q ಆಗಿರಲಿ, OQ ಸೇರಿಸಿ.
ಸಾಧನೆ: OQ ವೃತ್ತವನ್ನು ಬಿಂದುವಿನಲ್ಲಿ ಛೇದಿಸಿದೆ.
OP = OR (ಒಂದೇ ವೃತ್ತದ ತ್ರಿಜ್ಯಗಳು)
OQ = OR + RQ
OQ > OR
OQ > OP (OP = OR)
OP ಯು O ನಿಂದ ಸ್ಪರ್ಶಕಕ್ಕೆ ಕನಿಷ್ಟ ದೂರವಾಗಿದೆ
OP ⊥ XY
ಅಥವಾ

ಸಾಧನೀಯ: ABCD ಒಂದು ವಜ್ರಾಕೃತಿ. (AB = BC = CD = AD)
ಸಾಧನೆ: AB = CD ಮತ್ತು AD = BC …… (1)
AP = AS, BP = BQ, CQ = CR, DS = DR
(ಬಾಹ್ಯಬಿಂದುವಿನಿಂದ ವೃತ್ತಕ್ಕೆ ಎಳೆದ ಸ್ಪರ್ಶಕಗಳು)
AB + CD = AP + PB + DR + CR
⇒ AB + CD = AS + BQ + DS + CQ
⇒ AB + CD = AS + DS + BQ + CQ
⇒ AB + CD = AD + BC
⇒ AB = AD ….. (2)
∴ AB = BC = CD = AD (ಸಮೀಕರಣ (1) & (2) ರಿಂದ)

Solution 32.

Solution 33.
ಕ್ರಮಾಗತ ಬೆಸ ಧನ ಪೂರ್ಣಾಂಕಗಳು x & x + 2
x2 + (x + 2)2 = 290
⇒ x2 + x2 + 4x + 4 – 290 = 0
⇒ 2x2 + 4x – 286 = 0
⇒ x2 + 2x – 143 = 0
⇒ x2 + 13x – 11x – 143 = 0
⇒ x (x + 13) – 11(x + 13) = 0
⇒ (x + 13) (x – 11) = 0
⇒ x + 13 = 0 ಅಥವಾ x – 11 = 0
⇒ x = -13 ಅಥವಾ x = 11
ಕ್ರಮಾಗತ ಬೆಸ ಧನ ಪೂರ್ಣಾಂಕಗಳು 11 & 13
[∴ x = 11 & x + 2 = 11 + 2 = 13]
ಅಥವಾ

ಚಿಕ್ಕ ಬಾಹು x ಆಗಿರಲಿ
AC2 = AB2 + BC2
⇒ (x + 60)2 = x2 + (x + 30)2
⇒ x2 + 3600 + 120x = x2 + x2 + 900 + 60x
⇒ x2 + 3600 + 120x – 2x2 – 900 – 60x = 0
⇒ -x2 + 2700 + 60x = 0
⇒ x2 – 60x – 2700 = 0
⇒ x2 – 90x + 30x – 2700 = 0
⇒ x (x – 90) + 30 (x – 90) = 0
⇒ x – 90 = 0 ಅಥವಾ x + 30 = 0
⇒ x = 90 ಅಥವಾ x = -30
ಚಿಕ್ಕ ಬಾಹುವಿನ ಉದ್ದ = x = 90 m
ದೊಡ್ಡ ಬಾಹುವಿನ ಉದ್ದ = x + 30 = 90 + 30 = 120 m

Solution 34.

Solution 35.

Solution 36.

V.
Solution 37.
ಸಮಾಂತರ ಶ್ರೇಢಿಯ ಮೂರು ಪದಗಳು
a – d, a, a + d
a – d + a + a + d = 15
⇒ 3a = 15
⇒ a = 5
(a – d)2 + (a + d)2 = 58
⇒ a2 + d2 – 2ad + a2 + d2 + 2ad = 58
⇒ 2a2 + 2d2 = 58
⇒ 2(a2 + d2) = 58
⇒ a2 + d2 = 29
⇒ 52 + d2 = 29
⇒ d2 = 29 – 25
⇒ d2 = 4
⇒ d = ± 2
⇒ d = 2 ಅಥವಾ d = -2
ಮೂರು ಪದಗಳು
∴ a – d = 5 – 2 = 3
∴ a = 5
∴ a + d = 5 + 2 = 7
ಅಥವಾ

Solution 38.

Solution 39.

Solution 40.

## Karnataka SSLC Maths Model Question Paper 3 Kannada Medium

Karnataka SSLC Maths Model Question Paper 3 Kannada Medium

## Karnataka SSLC Maths Model Question Paper 3 Kannada Medium

ವಿಷಯ : ಗಣಿತ
ಸಮಯ: 3 ಗಂಟೆಗಳು
ಗರಿಷ್ಠ ಅಂಕಗಳು: 80

I. ಕೆಳಗಿನ ಪ್ರಶ್ನೆಗಳಿಗೆ ಅಥವಾ ಅಪೂರ್ಣ ಹೇಳಿಕೆಗೆ ನಾಲ್ಕು ಪರ್ಯಾಯ ಉತ್ತರಗಳನ್ನು ನೀಡಲಾಗಿದೆ. ಇವುಗಳಲ್ಲಿ ಸೂಕ್ತವಾದ ಉತ್ತರವನ್ನು ಆರಿಸಿ, ಕ್ರಮಾಕ್ಷರದೊಡನೆ ಪೂರ್ಣ ಉತ್ತರವನ್ನು ಬರೆಯಿರಿ. (8 × 1 = 8)

Question 1.
ಎರಡು ಸಮರೂಪ ತ್ರಿಭುಜಗಳ ವಿಸ್ತೀರ್ಣಗಳ ಅನುಪಾತ 16 : 25 ಆದರೆ ಅವುಗಳ ಬಾಹುಗಳ ಅನುಪಾತ
(A) 3 : 4
(B) 4 : 5
(C) 5 : 6
(D) 6 : 7

Question 2.
7 cm ತ್ರಿಜ್ಯವಿರುವ ವೃತ್ತದಲ್ಲಿ 60° ಕೋನವನ್ನು ಹೊಂದಿರುವ ತ್ರಿಜ್ಯಾಂತರ ಖಂಡದ ವಿಸ್ತೀರ್ಣ
(A) 25.67 cm2
(B) 35.32 cm2
(C) 15.25 cm2
(D) 77 cm2

Question 3.
ಮೂಲಬಿಂದು ಮತ್ತು (6, 8) ಬಿಂದುವಿನ ನಡುವಿನ ದೂರ
(A) 6 ಮೂ.ಮಾ.
(B) 8 ಮೂ.ಮಾ.
(C) 10 ಮೂ.ಮಾ.
(D) 14 ಮೂ.ಮಾ.

Question 4.
(15, 20) ರ ಮ.ಸಾ.ಅ. 5 ಆದರೆ ಅವುಗಳ ಲ.ಸಾ.ಅ.
(A) 15
(B) 20
(C) 40
(D) 60

Question 5.
$$\frac { { sin18 }^{ 0 } }{ { cos72 }^{ 0 } }$$ ಯ ಬೆಲೆ
(A) 1
(B) 0
(C) -1
(D) 2

Question 6.
x2 – 25 = 0 ಈ ಸಮೀಕರಣದ ಮೂಲಗಳು
(A) (+5, -5)
(B) (+5, +5)
(C) (-5, -5)
(D) (25, -25)

Question 7.
ಒಂದು ನಿರ್ದಿಷ್ಟ ದಿನದಲ್ಲಿ ಮಳೆ ಬೀಳುವ ಸಂಭವನೀಯತೆಯು 0.64 ಆಗಿದೆ. ಅದೇ ದಿನ ಮಳೆ ಬೀಳದಿರುವ ಸಂಭವನೀಯತೆ.
(A) -0.64
(B) 64
(C) 0.36
(D) -0.36

Question 8.
ಶಂಕುವಿನ ಭಿನ್ನಕದ ಪಾರ್ಶ್ವ ಮೇಲೈ ವಿಸ್ತೀರ್ಣ ಕಂಡುಹಿಡಿಯುವ ಸೂತ್ರ

II. ಈ ಕೆಳಗಿನ ಪ್ರಶ್ನೆಗಳಿಗೆ ಉತ್ತರಿಸಿ: (6 × 1 = 6)

Question 9.
ಪೈಥಾಗೊರಸ್ ಪ್ರಮೇಯವನ್ನು ನಿರೂಪಿಸಿ,

Question 10.
ಒಂದು ಸಂಖ್ಯೆಯನ್ನು 14 ರಿಂದ ಭಾಗಿಸಿದಾಗ ಶೇಷ 5 ಆದರೆ 7ರಿಂದ ಅದೇ ಸಂಖ್ಯೆಯನ್ನು ಭಾಗಿಸಿದಾಗ ದೊರೆಯುವ ಶೇಷವೆಷ್ಟು?

Question 11.
α ಮತ್ತು β ಗಳು ax2 + bx + c ಎಂಬ ಒಂದು ವರ್ಗ ಬಹುಪದೋಕ್ತಿಯ ಶೂನ್ಯತೆಗಳಾದರೆ αβ ದ ಬೆಲೆಯೇನು?

Question 12.
‘O’ ಕೇಂದ್ರವುಳ್ಳ ವೃತ್ತಕ್ಕೆ PA ಮತ್ತು PB ಗಳು ಬಾಹ್ಯ ಬಿಂದು P ನಿಂದ ಎಳೆದ ಸ್ಪರ್ಶಕಗಳು ಮತ್ತು ∠APB = 60° ಹಾಗು AP = 8 cm ಆದಾಗ AB ಜ್ಯಾದ ಉದ್ದವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 13.
(1 + sin290°)2 ನ ಬೆಲೆಯನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 14.
12, 16, 20, 24, 28 ರ ಸರಾಸರಿಯನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

III. ಈ ಕೆಳಗಿನ ಪ್ರಶ್ನೆಗಳಿಗೆ ಉತ್ತರಿಸಿ: (16 × 2 = 32)

Question 15.
a = 5, d = 3, an = 50 ಆದರೆ Sn ಕಂಡುಹಿಡಿಯಿರಿ.

Question 16.
ಚಿತ್ರದಲ್ಲಿ ∆ABCಯ ಎತ್ತರಗಳಾದ AD ಮತ್ತು CE ಗಳು ಪರಸ್ಪರ P ಬಿಂದುವಿನಲ್ಲಿ ಛೇದಿಸುತ್ತವೆ. ಆದರೆ ∆AEP ~ ∆CDP ಎಂದು ಸಾಧಿಸಿ.

ಅಥವಾ
ABCD ಚತುರ್ಭುಜದಲ್ಲಿ $$\frac { AO }{ BO }$$ = $$\frac { CO }{ DO }$$ ಆಗುವಂತೆ ಕರ್ಣಗಳು ಪರಸ್ಪರ 0ಬಿಂದುವಿನಲ್ಲಿ ಛೇದಿಸುತ್ತವೆ. ಆದರೆ ABCD ಯು ಒಂದು ತ್ರಾಪಿಜ್ಯ ಎಂದು ಸಾಧಿಸಿ.

Question 17.

Question 18.
5x – 3y = 11, -10x + 6y = -22 ಈ ಸಮೀಕರಣಗಳು ಸ್ಥಿರವಾಗಿವೆಯೆ? ಅಥವಾ ಅಸ್ಥಿರವಾಗಿವೆಯೆ? ಎಂಬುದನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 19.
ಚಿತ್ರದಲ್ಲಿ ಕೇಂದ್ರ O ಇರುವ ಎರಡು ಏಕಕೇಂದ್ರೀಯ ವೃತ್ತಗಳ ತ್ರಿಜ್ಯಗಳು ಕ್ರಮವಾಗಿ 7 cm ಮತ್ತು 14 cm ಇವೆ. ∠AOC = 40° ಆದರೆ ಛಾಯಕೃತ ಭಾಗದ ವಿಸ್ತೀರ್ಣವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 20.
4 cm ತ್ರಿಜ್ಯದ ವೃತ್ತಕ್ಕೆ ಸ್ಪರ್ಶಕಗಳ ನಡುವಿನ ಕೋನ 70° ಇರುವಂತೆ ಒಂದು ಜೊತೆ ಸ್ಪರ್ಶಕಗಳನ್ನು ಎಳೆಯಿರಿ.

Question 21.
AB ವ್ಯಾಸವಾಗಿರುವ ವೃತ್ತದ ಕೇಂದ್ರ (2, -3) ಮತ್ತು B ಯು (1, 4) ಆದರೆ A ಬಿಂದುವಿನ ನಿರ್ದೇಶಾಂಕಗಳನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 22.
(8, 1), (k, -4), (2, -5) ಎಂಬ ಬಿಂದುಗಳು ಸರಳರೇಖಾಗತವಾಗಿದ್ದರೆ k ಯ ಬೆಲೆಯನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 23.
ಶೂನ್ಯತೆಗಳ ಮೊತ್ತ ಹಾಗು ಗುಣಲಬ್ದಗಳು ಕ್ರಮವಾಗಿ 4 ಮತ್ತು 1 ಆಗಿರುವ ಒಂದು ವರ್ಗ ಬಹುಪದೋಕ್ತಿಯನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 24.
2x2 + 3x – 5 = 0 ಈ ವರ್ಗ ಸಮೀಕರಣವನ್ನು ಸೂತ್ರದಿಂದ ಬಿಡಿಸಿ,

Question 25.
20 m ಎತ್ತರದ ಕಟ್ಟಡವೊಂದರ ಮೇಲೆ ಸ್ಥಾಪಿಸಲಾದ ಪ್ರಸರಣೆಯ ಗೋಪುರವೊಂದರ ಮೇಲುದಿ ಮತ್ತು ಪಾದಗಳನ್ನು ನೆಲದ ಮೇಲಿನ ಒಂದು ಬಿಂದುವಿನಿಂದ ನೋಡಿದಾಗ ಉನ್ನತ ಕೋನಗಳು ಕ್ರಮವಾಗಿ 60° ಮತ್ತು 45° ಇದೆ. ಪ್ರಸರಣೆಯ ಗೋಪುರದ ಎತ್ತರ ಕಂಡುಹಿಡಿಯಿರಿ.

Question 26.
50√3 m ಎತ್ತರದಲ್ಲಿರುವ ಒಂದು ಕಟ್ಟಡದ ಮೇಲಿನಿಂದ ನೆಲದ ಮೇಲಿರುವ ಒಂದು ಕಾರನ್ನು ನೋಡಿದಾಗ ಉಂಟಾದ ಅವನತ ಕೋನವು 60° ಆಗಿರುತ್ತದೆ. ಹಾಗಾದರೆ ಕಟ್ಟಡದಿಂದ ಕಾರಿಗಿರುವ ದೂರವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 27.
ಕೆಳಗಿನ ದತ್ತಾಂಶಗಳಿಗೆ ಮಧ್ಯಾಂಕವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 28.
ಒಂದು ದಾಳವನ್ನು ಒಂದು ಸಲ ಎಸೆಯಲಾಗಿದೆ. (i) ಒಂದು ಅವಿಭಾಜ್ಯ ಸಂಖ್ಯೆ (ii) 1 ಮತ್ತು 5ರ ನಡುವಿನ ಒಂದು ಸಂಖ್ಯೆಯನ್ನು ಪಡೆಯುವ ಸಂಭವನೀಯತೆಯನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 29.
ಸಮನಾದ ತ್ರಿಜ್ಯವುಳ್ಳ ಒಂದು ಶಂಕುವನ್ನು ಒಂದು ಅರ್ಧಗೋಳಾಕೃತಿಯ ಮೇಲೆ ಜೋಡಿಸಿ ಒಂದು ಆಟಿಕೆಯನ್ನು ಮಾಡಲಾಗಿದೆ. ಶಂಕುವಿನ ಭಾಗದ ವ್ಯಾಸವು 6 cm ಮತ್ತು 4 cm ಎತ್ತರ ಇದ್ದರೆ ಈ ಘನವಸ್ತುವಿನ ಮೇಲೈ ವಿಸ್ತೀರ್ಣವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
ಅಥವಾ
ಒಂದು ಶಂಕುವಿನ ಭಿನ್ನಕದ ಓರೆ ಎತ್ತರ 10 cm, ಅದರ ವೃತ್ತಾಕಾರದ ಪಾದಗಳ ತ್ರಿಜ್ಯಗಳು ಕ್ರಮವಾಗಿ 8 cm ಮತ್ತು 6 cm ಆಗಿದೆ. ಆ ಭಿನ್ನಕದ ವಕ್ರಮೇಲೈ ವಿಸ್ತೀರ್ಣವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 30.
3 + 2√5 ಒಂದು ಅಭಾಗಲಬ್ಧ ಸಂಖ್ಯೆ ಎಂದು ಸಾಧಿಸಿ.

IV. ಈ ಕೆಳಗಿನ ಪ್ರಶ್ನೆಗಳಿಗೆ ಉತ್ತರಿಸಿ: (6 × 3 = 18)

Question 31.
ಬಾಹ್ಯಬಿಂದುವಿನಿಂದ ವೃತ್ತಕ್ಕೆ ಎಳೆದ ಸ್ಪರ್ಶಕಗಳ ಉದ್ದವು ಸಮನಾಗಿರುತ್ತವೆ ಎಂದು ಸಾಧಿಸಿ.
ಅಥವಾ
ಎರಡು ಏಕಕೇಂದ್ರೀಯ ವೃತ್ತಗಳ ತ್ರಿಜ್ಯಗಳು 5 cm ಮತ್ತು 3 cm ಆಗಿವೆ. ಚಿಕ್ಕ ವೃತ್ತಕ್ಕೆ ಸ್ಪರ್ಶಿಸುವಂತೆ ದೊಡ್ಡ ವೃತ್ತದ ಹ್ಯಾದ ಉದ್ದವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 32.
BC = 7 cm, ∠A = 45°, ∠B = 105° ಇರುವಂತೆ ABC ತ್ರಿಭುಜವನ್ನು ರಚಿಸಿ ನಂತರ ಮತ್ತೊಂದು ತ್ರಿಭುಜವನ್ನು ಅದರ ಬಾಹುಗಳು, ∆ABCಯ ಅನುರೂಪ ಬಾಹುಗಳ $$\frac { 3 }{ 4 }$$ ರಷ್ಟಿರುವಂತೆ ರಚಿಸಿ,

Question 33.
3 ವರ್ಷಗಳ ಹಿಂದಿನ ರೆಹಮಾನನ ವಯಸ್ಸು ಮತ್ತು 5 ವರ್ಷಗಳ ನಂತರದ ಅವನ ವಯಸ್ಸು ಇವುಗಳ ವ್ಯತ್ಯಮಗಳ ಮೊತ್ತ $$\frac { 1 }{ 3 }$$ ಆದರೆ ಅವನ ಈಗಿನ ವಯಸ್ಸನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
ಅಥವಾ
ಒಂದು ರೈಲು 360km ದೂರವನ್ನು ಏಕರೂಪ ಜವದೊಂದಿಗೆ ಕ್ರಮಿಸುತ್ತದೆ. ಅದರ ಜವವು 5km/hr ಹೆಚ್ಚಾಗಿದ್ದರೆ, ಅಷ್ಟೇ ದೂರವನ್ನು ಕ್ರಮಿಸಲು ಅದು 1 ಗಂಟೆ ಕಡಿಮೆ ತೆಗೆದುಕೊಳ್ಳುತ್ತಿತ್ತು. ರೈಲಿನ ಜವವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 34.
(cos A + sec A)2 + (sin A + cosec A)2 = 7 + tan2A + cot2A ಎಂದು ಸಾಧಿಸಿ.
ಅಥವಾ
A, B ಮತ್ತು C ಗಳು ∆ABCಯ ಒಳಕೋನಗಳಾದರೆ sin($$\frac { B+C }{ 2 }$$) = cos($$\frac { A }{ 2 }$$) ಎಂದು ಸಾಧಿಸಿ.

Question 35.
ಕೆಳಗಿನ ದತ್ತಾಂಶಗಳಿಗೆ ಓಜೀವ್ ನಕ್ಷೆಯನ್ನು ರಚಿಸಿ,

(ಅಧಿಕ ಇರುವ ವಿಧಾನ)

Question 36.
4u2 – 8u ವರ್ಗ ಬಹುಪದೋಕ್ತಿಯ ಶೂನ್ಯತೆಗಳನ್ನು ಕಂಡುಹಿಡಿದು ಹಾಗು ಶೂನ್ಯತೆಗಳು ಮತ್ತು ಸಹಗುಣಕಗಳ ನಡುವಿನ ಸಂಬಂಧವನ್ನು ತಾಳೆನೋಡಿ.
ಅಥವಾ
p(x) = x5 – 4x3 + x2 + 3x + 1 ನ್ನು g(x) = x3 – 3x + 1 ರಿಂದ ಭಾಗಿಸಿ g(x), p(x) ನ ಅಪವರ್ತನವನ್ನು ಪರೀಕ್ಷಿಸಿ.

V. ಈ ಕೆಳಗಿನ ಪ್ರಶ್ನೆಗಳಿಗೆ ಉತ್ತರಿಸಿ: (4 × 4 = 16)

Question 37.
ಒಂದು ಸಮಾಂತರ ಶ್ರೇಢಿಯ 4ನೇ ಮತ್ತು 8ನೇ ಪದಗಳ ಮೊತ್ತವು 24ಹಾಗು ಅದೇ ಶ್ರೇಢಿಯ 6ನೇ ಮತ್ತು 10ನೇ ಪದಗಳ ಮೊತ್ತವು 44 ಆದರೆ ಶ್ರೇಢಿಯ ಮೊದಲ ಮೂರು ಪದಗಳನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
ಅಥವಾ
ಒಂದು ಸುರುಳಿಯನ್ನು ಕ್ರಮಾಗತ ಅರೆ ವೃತ್ತಗಳಿಂದ ಮಾಡಲಾಗಿದೆ ಅವುಗಳ ಕೇಂದ್ರಗಳು ಪರ್ಯಾಯವಾಗಿ A & B ನಲ್ಲಿದ್ದು A ಕೇಂದ್ರದಿಂದ ಆರಂಭವಾಗಿ ತ್ರಿಜ್ಯಗಳು 0.5 cm, 1 cm, 1.5 cm, 2 cm …….. ಹೀಗೆ ಇದೆ. ಈ ರೀತಿಯ 13 ಕ್ರಮಾಗತ ಅರೆ ವೃತ್ತಗಳಿಂದ ಮಾಡಲ್ಪಟ್ಟ ಒಟ್ಟು ಉದ್ದ ಏನು?

Question 38.
ಒಂದು ಲಂಬಕೋನ ತ್ರಿಭುಜದಲ್ಲಿ ವಿಕರ್ಣದ ಮೇಲಿನ ವರ್ಗವು ಉಳಿದೆರಡು ಬಾಹುಗಳ ವರ್ಗಗಳ ಮೊತ್ತಕ್ಕೆ ಸಮನಾಗಿರುತ್ತದೆ ಎಂದು ಸಾಧಿಸಿ.

Question 39.
ನಕ್ಷೆಯ ಮೂಲಕ ಸಮೀಕರಣಗಳನ್ನು ಬಿಡಿಸಿ, 2x – y = 5 ಮತ್ತು x + 3y = 6.

Question 40.
ಒಂದು ಅರ್ಧಗೋಳಾಕಾರದ ಬಟ್ಟಲಿನ ಒಳಜ್ಯವು 18 cm ಇದ್ದು ಅದರ ತುಂಬ ಹಣ್ಣಿನ ರಸವಿದೆ. ಈ ರಸವನ್ನು 3 cm ತ್ರಿಜ್ಯವಿರುವ ಮತ್ತು 9 cm ಎತ್ತರವಿರುವ ಸಿಲಿಂಡರಿನಾಕೃತಿಯ ಬಾಟಲಿಗಳಿಗೆ ತುಂಬಬೇಕು. ಬಟ್ಟಲನ್ನು ಖಾಲಿ ಮಾಡಲು ಎಷ್ಟು ಬಾಟಲಿಗಳ ಅವಶ್ಯಕತೆ ಇದೆ?

Solutions

I.
Solution 1.
(B) 4 : 5

Solution 2.
(A) 25.67 cm2

Solution 3.
(C) 10 ಮೂ.ಮಾ.

Solution 4.
(D) 60

Solution 5.
(A) 1

Solution 6.
(A) (+5, -5)

Solution 7.
(C) 0.36

Solution 8.
(B) π (r1 + r2) l

II.
Solution 9.
ಒಂದು ಲಂಬಕೋನ ತ್ರಿಭುಜದಲ್ಲಿ ವಿಕರ್ಣದ ಮೇಲಿನ ವರ್ಗವು ಉಳಿದೆರಡು ಬಾಹುಗಳ ಮೇಲಿನ ವರ್ಗಗಳ ಮೊತ್ತಕ್ಕೆ ಸಮ.

Solution 10.
ಶೇಷ – 5

Solution 11.
αβ = $$\frac { c }{ a }$$

Solution 12.

∆APB ನಲ್ಲಿ PA = PB = 8cm
∠PAB = ∠PBA = 60°
∠APB = 60°
∆APB ಒಂದು ಸಮಬಾಹು ತ್ರಿಭುಜ
AB = 8 cm

Solution 13.
(1 + sin290°)2 = (1 + 1)2 = 4

Solution 14.

III.
Solution 15.

Solution 16.

ಸಾಧನೀಯ: ∆AEP ~ ∆CDP
∆AEP & ∆CDP ಗಳಲ್ಲಿ
∠AEP = ∠CDP = 90°
∠APE = ∠CPD (ಶೃಂಗಾಭಿಮುಖ ಕೋನಗಳು)
∆AEP ~ ∆CDP
ಅಥವಾ
ABCD ಚತುರ್ಭುಜದಲ್ಲಿ $$\frac { AO }{ BO }$$ = $$\frac { CO }{ DO }$$ ಆಗುವಂತೆ ಕರ್ಣಗಳು ಪರಸ್ಪರ O ಬಿಂದುವಿನಲ್ಲಿ ಛೇದಿಸುತ್ತವೆ. ಆದರೆ ABCD ಯು ಒಂದು ತ್ರಾಪಿಜ್ಯ ಎಂದು ಸಾಧಿಸಿ.

Solution 17.

Solution 18.

Solution 19.
R = 14 cm, r = 7 cm, θ = 40°
ಛಾಯಕೃತ ಭಾಗದ ವಿಸ್ತೀರ್ಣ

Solution 20.
ತ್ರಿಜ್ಯ = 4cm
ಕೇಂದ್ರ ಕೋನ = 180° – 70° = 110°

Solution 21.

Solution 22.

Solution 23.

Solution 24.

Solution 25.

ಗೋಪುರದ ಎತ್ತರ = AB = x ಆಗಿರಲಿ
ಕಟ್ಟಡದ ಎತ್ತರ = BC = 20 m
tan 45° = $$\frac { BC }{ CD }$$
1 = $$\frac { 20 }{ CD }$$
CD = 20 m
tan 60° = $$\frac { AC }{ CD }$$
⇒ √3 = $$\frac { x+20 }{ 20 }$$
⇒ 20√3 = x + 20
⇒ x + 20 = 20√3
⇒ x = 20(√3 – 1)
ಗೋಪುರದ ಎತ್ತರ = 20(√3 – 1) m

Solution 26.

ಕಟ್ಟಡ ಎತ್ತರ = AB = 50√3 m
ಕಟ್ಟಡದಿಂದ ಕಾರಿಗಿರುವ ದೂರ BC ಆಗಿರಲಿ
tan 60° = $$\frac { AB }{ BC }$$
⇒ √3 = $$\frac { 50\surd 3 }{ BC }$$
⇒ BC = 50m
ಕಟ್ಟಡದಿಂದ ಕಾರಿಗಿರುವ ದೂರ = 50m

Solution 27.

Solution 28.
ಫಲಿತಗಳ ಸಂಖ್ಯೆ = n(S) = 6 {1, 2, 3, 4, 5, 6}
(i) ಒಂದು ಅವಿಭಾಜ್ಯ ಸಂಖ್ಯೆ = A = {2, 3, 5}
n(A) = 3
ಸಂಭವನೀಯ = P(A) = $$\frac { n(A) }{ n(S) }$$ = $$\frac { 3 }{ 6 }$$
(ii) 1 ಮತ್ತು 5ರ ನಡುವಿನ ಒಂದು ಸಂಖ್ಯೆ = B = {2, 3, 4}
n(B) = 3
ಸಂಭವನೀಯತೆ = P(B) = $$\frac { n(A) }{ n(S) }$$ = $$\frac { 3 }{ 6 }$$

Solution 29.

Solution 30.
3 + 2√5 ಒಂದು ಭಾಗಲಬ್ಧ ಸಂಖ್ಯೆಯಾಗಿರಲಿ ಎಂದು ಊಹಿಸೋಣ.

ಭಾಗಲಬ್ಧ ಸಂಖ್ಯೆ = ಭಾಗಲಬ್ಧ ಸಂಖ್ಯೆ
√5 ಒಂದು ಅಭಾಗಲಬ್ಧ ಸಂಖ್ಯೆ
ನಮ್ಮ ಊಹೆ 3 + 2√5 ಒಂದು ಭಾಗಲಬ್ಧ ಸಂಖ್ಯೆ ಎಂಬುದು ತಪ್ಪು
3 + 2√5 ಒಂದು ಅಭಾಗಲಬ್ಧ ಸಂಖ್ಯೆ

IV.
Solution 31.

ದತ್ತ: O ವೃತ್ತಕೇಂದ್ರ PA ಮತ್ತು PB ಗಳ ಬಾಹ್ಯಬಿಂದು P ನಿಂದ ವೃತ್ತಕ್ಕೆ ಎಳೆದ ಸ್ಪರ್ಶಕಗಳು.
ಸಾಧನೀಯ: PA = PB
ಸಾಧನೆ: ಹೇಳಿಕೆಗಳು ಕಾರಣಗಳು
ΔPOA & ΔPOB ಗಳಲ್ಲಿ
OA = OB
∠PAO = ∠PBO = 90°
OP = OP
ΔPOA = ΔPOB
PA = PB
ಒಂದೇ ವೃತ್ತದ ತ್ರಿಜ್ಯಗಳು ತ್ರಿಜ್ಯ ಮತ್ತು ಸ್ಪರ್ಶಕದ ನಡುವಿನ ಕೋನ ಉಭಯಸಾಮಾನ್ಯ ಲಂ.ವಿ.ಬಾ. ಜ್ವಸಿದ್ಧಾಂತ ಸರ್ವಸಮ ತ್ರಿಭುಜದ ಅನುರೂಪ ಭಾಗಗಳು.
ಅಥವಾ

ΔPOB ನಲ್ಲಿ
PB2 = OB2 – OP2
⇒ PB2 = 52 – 32 = 25 – 9 = 16
⇒ PB2 = 16
⇒ PB = 4
AB = AP + PB
⇒ AB = 4 + 4 (∴ AP = PB)
⇒ AB = 8 cm ಜ್ಯಾದ ಉದ್ದ 8 cm

Solution 32.
∠A = 45°, ∠B = 105°, BC = 7 cm
∠C = 180° – 150° = 30°

ರಚಿಸಬೇಕಾದ ತ್ರಿಭುಜ A’BC’

Solution 33.
ರೆಹಮಾನನ ಈಗಿನ ವಯಸ್ಸು x ಆಗಿರಲಿ
ಮೂರು ವರ್ಷಗಳ ಹಿಂದೆ ಅವನ ವಯಸ್ಸು = x – 3 ವರ್ಷ
5 ವರ್ಷಗಳ ನಂತರ ಅವನ ವಯಸ್ಸು = x + 5 ವರ್ಷ

⇒ x2 – 4x – 21 = 0
⇒ x2 – 7x + 3x – 21 = 0
⇒ x (x – 7) + 3 (x – 7) = 0
⇒ (x – 7) (x + 3) = 0
⇒ x – 7 = 0 ಅಥವಾ x + 3 = 0
⇒ x = 7 ಅಥವಾ x = -3
ರೆಹಮಾನನ ಈಗಿನ ವಯಸು 7 ವರ್ಷ
ಅಥವಾ
ರೈಲಿನ ಜನ x km/hr, ಆಗಿರಲಿ
ದೂರ = d = 360 km/hr

Solution 34.

Solution 35.

Solution 36.

V.
Solution 37.

Solution 38.

ದತ್ತ: ΔABC ನಲ್ಲಿ ∠BAC = 90°
ಸಾಧನೀಯ: BC2 = AB2 + AC2
ರಚನೆ: AD ⊥ BC ಗೆ ಎಳೆಯಿರಿ,
ಸಾಧನೆ: ಹೇಳಿಕೆಗಳು ಕಾರಣಗಳು
ΔBAC & ΔBDA ಗಳಲ್ಲಿ
∠ABC = ∠ABD ಉಭಯಸಾಮಾನ್ಯ
∠BAC = ∠BDA = 90° ದತ್ತ ಮತ್ತು ರಚನೆ
ΔBAC ~ ΔBDA ಕೊ .ಕೋ .ನಿ.ಗು.
$$\frac { BA }{ BD }$$ = $$\frac { BC }{ BA }$$
AB2 = BD.BC …….. (1)
∠ACB = ∠ACD ಉಭಯಸಾಮಾನ್ಯ
∠BAC = ∠ADC = 90° ದತ್ತ ಮತ್ತು ರಚನೆ
ΔBAC ~ ΔADC ಕೋ .ಕೋ .ನಿ.ಗು.
$$\frac { AC }{ DC }$$ = $$\frac { BC }{ AC }$$
AC2 = DC × BC ……. (2)
AB2 + AC2 = BD.BC + DC.BC
(1) + (2) ರಿಂದ
BC(BD + DC) = BC × BC
AB2 + AC2 = BC2

Solution 39.

Solution 40.

## Karnataka SSLC Maths Model Question Paper 2 Kannada Medium

Karnataka SSLC Maths Model Question Paper 2 Kannada Medium

## Karnataka SSLC Maths Model Question Paper 2 Kannada Medium

ವಿಷಯ : ಗಣಿತ
ಸಮಯ: 3 ಗಂಟೆಗಳು
ಗರಿಷ್ಠ ಅಂಕಗಳು: 80

I. ಕೆಳಗಿನ ಪ್ರಶ್ನೆಗಳಿಗೆ ಅಥವಾ ಅಪೂರ್ಣ ಹೇಳಿಕೆಗೆ ನಾಲ್ಕು ಪರ್ಯಾಯ ಉತ್ತರಗಳನ್ನು ನೀಡಲಾಗಿದೆ. ಇವುಗಳಲ್ಲಿ ಸೂಕ್ತವಾದ ಉತ್ತರವನ್ನು ಆರಿಸಿ, ಕ್ರಮಾಕ್ಷರದೊಡನೆ ಪೂರ್ಣ ಉತ್ತರವನ್ನು ಬರೆಯಿರಿ. (8 × 1 = 8)

Question 1.
ಚಿತ್ರದಲ್ಲಿ QE = 7.2 cm, PF = 1.8 cm, FR = 5.4cm ಆದರೆ PE ಯು

(A) 2 cm
(B) 2.4 cm
(C) 2.8 cm
(D) 3.2 cm

Question 2.
‘r’ ವೃತ್ತ ತ್ರಿಜ್ಯ ಹಾಗೂ 60° ಕೋನವನ್ನು ಹೊಂದಿರುವ ತ್ರಿಜ್ಯಾಂತರ ಖಂಡದ ವಿಸ್ತೀರ್ಣ

Question 3.
P (-2, -1), ರಲ್ಲಿ ಲಂಬದೂರವು
(A) -2
(B) -1
(C) 1
(D) 2

Question 4.
‘0’ ಮೂಲಬಿಂದು ಮತ್ತು P (4, 3) ರ ನಡುವಿನ ದೂರ (ಏಕಮಾನಗಳಲ್ಲಿ) OP = …….
(A) 5
(B) 4
(C) 3
(D) 2

Question 5.
P (x) = 5x – 10 ರ ಶೂನ್ಯತೆ.
(A) 2
(B) -2
(C) 5
(D) -5

Question 6.
x + $$\frac { 2 }{ x }$$ = 3 ಸಮೀಕರಣದ ಆದರ್ಶರೂಪ
(A) x2 + 2x – 3 = 0
(B) x2 + 3x + 2 = 0
(C) x2 – 3x + 2 = 0
(D) x2 – 2x + 3 = 0

Question 7.
ಎರಡು ನಾಣ್ಯಗಳನ್ನು ಏಕಕಾಲಕ್ಕೆ ಚಿಮ್ಮಿದಾದ, ಶಿರ H ಮತ್ತು ಪ್ರಚ್ಛ T ಇರುವಂತೆ ಸಾಧ್ಯ ಫಲಿತಗಳು
(A) {T, H, H, T}
(B) {TT, HH, HT, TH}
(C) {T, H}
(D) {TT, HH}

Question 8.
ಸಿಲಿಂಡರ್‌ನ ಎತ್ತರ ‘h’ ಮತ್ತು ಪಾದದ ತ್ರಿಜ್ಯ ‘r’ ಆದಾಗ ಸಿಲಿಂಡರ್‌ನ ಪಾರ್ಶ್ವ ಮೇಲೈ ವಿಸ್ತೀರ್ಣ
(A) 2π (r + h)
(B) 2πr (r + h)
(C) 2πrh
(D) $$\frac { 2\pi r }{ h }$$

II. ಈ ಕೆಳಗಿನ ಪ್ರಶ್ನೆಗಳಿಗೆ ಉತ್ತರಿಸಿ: (6 × 1 = 6)

Question 9.
ಒಂದು 5m ಎತ್ತರದ ಏಣಿಯನ್ನು ನೆಲದ ಮೇಲಿನಿಂದ 4 m ಎತ್ತರದ ಕಿಟಕಿಯನ್ನು ತಲುಪುವಂತೆ ಗೋಡೆಗೆ ಒರಗಿಸಿದೆ. ಗೋಡೆಯ ಪಾದದಿಂದ ಏಣಿಯ ಪಾದಕ್ಕಿರುವ ದೂರ ಲೆಕ್ಕಿಸಿ.

Question 10.
ಒಂದು ಬಿಂದುವಿನಿಂದ ವೃತ್ತದ ಮೇಲಿನ ಬಿಂದುವಿಗೆ ಎಳೆದ ಸ್ಪರ್ಶಕಗಳ ಸಂಖ್ಯೆ ಎಷ್ಟು?

Question 11.
‘ಭಾಗ ಪ್ರಮಾಣ ಸೂತ್ರ’ ವನ್ನು ಬರೆಯಿರಿ.

Question 12.
ಯೂಕ್ಲಿಡ್‌ನ ಭಾಗಾಕಾರ ಅನುಪ್ರವೇಯವನ್ನು ತಿಳಿಸಿ.

Question 13.
ಘನಪದೋಕ್ತಿಯು ಹೊಂದಿರಬಹುದಾದ ಗರಿಷ್ಟ ಶೂನ್ಯತೆಗಳ ಸಂಖ್ಯೆ ಎಷ್ಟು?

Question 14.
ಎರಡು ಸಂಖ್ಯೆಗಳ ಮೊತ್ತ 27 ಮತ್ತು ಗುಣಲಬ್ಧ 182 ಆದರೆ ಆ ಸಂಖ್ಯೆಗಳು ಯಾವುವು?

III. ಈ ಕೆಳಗಿನ ಪ್ರಶ್ನೆಗಳಿಗೆ ಉತ್ತರಿಸಿ: (16 × 2 = 32)

Question 15.
ಮೊತ್ತ ಕಂಡುಹಿಡಿಯಿರಿ: 34 + 32 + 30 + …………… + 10,

Question 16.
ಒಂದು ಏಣಿಯ ಪಾದವು ನೆಲದ ಮೇಲೆ ಗೋಡೆಯಿಂದ 2.5m ದೂರದಲ್ಲಿ ಹಾಗೂ ಅದರ ತುದಿಯು ನೆಲದ ಮೇಲಿಂದ 6 m ಎತ್ತರದಲ್ಲಿರುವ ಕಿಟಕಿಯನ್ನು ಮುಟ್ಟುವಂತೆ ಏಣಿಯನ್ನು ಗೋಡೆಗೆ ಒರಗಿಸಿ ಇಡಲಾಗಿದೆ. ಏಣಿಯ ಉದ್ದವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 17.
ಒಂದು ಎರವಲು ಗ್ರಂಥಾಲಯದಲ್ಲಿ ಮೊದಲ 3 ದಿನಕ್ಕೆ ಒಂದು ನಿಗದಿತ ಶುಲ್ಕವಿರುತ್ತದೆ. ಆ ದಿನದ ಪ್ರತಿಯೊಂದೂ ದಿನಕ್ಕೂ ಒಂದು ಹೆಚ್ಚುವರಿ ಶುಲ್ಕವಿರುತ್ತದೆ. ಪುಸ್ತಕವನ್ನು 7 ದಿನ ತನ್ನಲ್ಲಿ ಇರಿಸಿಕೊಂಡಿದ್ದಕ್ಕಾಗಿ ಸರಿತಾ ₹ 27 ನ್ನು ಪಾವತಿಸಿದರೆ, ಪುಸ್ತಕವನ್ನು 5 ದಿನ ಇರಿಸಿಕೊಂಡಿದ್ದಕ್ಕಾಗಿ ಸೂಸಿ ₹ 21 ನ್ನು ಪಾವತಿಸಿದಳು. ನಿಗದಿತ ಶುಲ್ಕ ಮತ್ತು ಪ್ರತಿಯೊಂದು ಹೆಚ್ಚುವರಿ ದಿನದ ಶುಲ್ಕವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 18.
ನೀರಿನ ಒಳಭಾಗದಲ್ಲಿರುವ ಬಂಡೆಗಳ ಬಗ್ಗೆ ಎಚ್ಚರಿಸಲು ಒಂದು ದೀಪಸ್ಥಂಭವು 80° ಕೋನವಿರುವ ತ್ರಿಜ್ಯಾಂತರ ಖಂಡದಲ್ಲಿ 16.5 km ದೂರಕ್ಕೆ ಕೆಂಪು ಬೆಳಕನ್ನು ಹರಡುತ್ತದೆ. ಹಡಗುಗಳನ್ನು ಎಚ್ಚರಿಸುವ ಈ ಭಾಗದ ವಿಸ್ತೀರ್ಣವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
ಅಥವಾ
ಚಿತ್ರದಲ್ಲಿ ತೋರಿಸಿರುವಂತೆ 4 cm ಬಾಹುವುಳ್ಳ ಒಂದು ಚೌಕದ ಪ್ರತೀ ಮೂಲೆಯಲ್ಲಿ 1 cm ತ್ರಿಜ್ಯವಿರುವ ವೃತ್ತ ಚತುರ್ಥಕವನ್ನು ಮತ್ತು 2 cm ವ್ಯಾಸವಿರುವ ಒಂದು ವೃತ್ತವನ್ನು ಕತ್ತರಿಸಿದೆ. ಚೌಕದ ಉಳಿದ ಭಾಗದ ವಿಸ್ತೀರ್ಣವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 19.
ಪಾದ 8 cm ಮತ್ತು ಎತ್ತರ 4 cm ಇರುವ ಒಂದು ಸಮದ್ವಿಬಾಹು ತ್ರಿಭುಜವನ್ನು ರಚಿಸಿ, ನಂತರ ಮತ್ತೊಂದು ತ್ರಿಭುಜವನ್ನು, ಅದರ ಬಾಹುಗಳು ಮೊದಲು ರಚಿಸಿದ ಸಮದ್ವಿಬಾಹು ತ್ರಿಭುಜದ ಅನುರೂಪ ಬಾಹುಗಳ 1$$\frac { 1 }{ 2 }$$ ರಷ್ಟಿರುವಂತೆ ರಚಿಸಿ,

Question 20.
(4, -3) ಮತ್ತು (8, 5) ಬಿಂದುಗಳನ್ನು ಸೇರಿಸುವ ರೇಖಾಖಂಡವನ್ನು ಆಂತರಿಕವಾಗಿ 3 : 1 ಅನುಪಾತದಲ್ಲಿ ವಿಭಾಗಿಸುವ ಬಿಂದುವಿನ ನಿರ್ದೇಶಾಂಕಗಳನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 21.
ಶೃಂಗಗಳು ಈ ಕೆಳಗಿನಂತಿರುವ ತ್ರಿಭುಜಗಳ ವಿಸ್ತೀರ್ಣಗಳನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ. (5, -1), (3, -5), (5, 2)

Question 22.
√5 ಒಂದು ಅಭಾಗಲಬ್ಧ ಸಂಖ್ಯೆ ಎಂದು ಸಾಧಿಸಿ.

Question 23.
(1, 1) ನ್ನು ಕ್ರಮವಾಗಿ ಶೂನ್ಯತೆಗಳ ಮೊತ್ತ ಹಾಗೂ ಗುಣಲಬ್ದವಾಗಿ ಹೊಂದಿರುವ ವರ್ಗಬಹುಪದೋಕ್ತಿಯನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 24.
ಮೂಲಗಳ ಸ್ವಭಾವವನ್ನು ವಿವೇಚಿಸಿ, ವಾಸ್ತವ ಮೂಲಗಳಿದ್ದಲ್ಲಿ, ಅವುಗಳನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ: 2x2 – 3x + 5 = 0

Question 25.
1.6m ಎತ್ತರದ ಪ್ರತಿಮೆಯೊಂದನ್ನು ಒಂದು ಪೀಠದ ಮೇಲ್ಬಾಗದಲ್ಲಿ ಇರಿಸಲಾಗಿದೆ. ನೆಲದ ಮೇಲಿನ ಒಂದು ಬಿಂದುವಿನಿಂದ ಪ್ರತಿಮೆಯ ಮೇಲಿನ ಉನ್ನತ ಕೋನವು 60° ಮತ್ತು ಅದೇ ಬಿಂದುವಿನಿಂದ ಪೀಠದ ಮೇಲ್ತುದಿಯ ಉನ್ನತ ಕೋನವು 45° ಇದ್ದರೆ, ಪೀಠದ ಎತ್ತರವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 26.
10 ಪಂದ್ಯಗಳಲ್ಲಿ ಒಬ್ಬ ಬೌಲರ್‌ನು ಪಡೆದ ವಿಕೆಟ್‌ಗಳ ಸಂಖ್ಯೆಯು ಈ ಕೆಳಗಿನಂತಿದೆ.

Question 27.
ಒಂದು ದಾಳವನ್ನು ಒಂದು ಸಲ ಎಸೆಯಲಾಗಿದೆ. 2 ಮತ್ತು 6ರ ನಡುವೆ ಒಂದು ಸಂಖ್ಯೆಯನ್ನು ಪಡೆಯುವ ಸಂಭವನೀಯತೆಯನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 28.
ಒಂದು ಪಾತ್ರೆಯ ಆಕಾರವು ಟೊಳ್ಳಾದ ಸಿಲಿಂಡರಿನ ಒಂದು ಪಾದದ ಮೇಲೆ ಟೊಳ್ಳಾದ ಅರ್ಧಗೋಳಾಕೃತಿಯನ್ನು ಕೂಡಿಸಿ ಮಾಡಿದೆ. ಅರ್ಧಗೋಳದ ವ್ಯಾಸವು 14 cm ಮತ್ತು ಪಾತ್ರೆಯ ಒಟ್ಟು ಎತ್ತರ 13 cm ಇದೆ. ಈ ಪಾತ್ರೆಯ ಒಳಮೇ ವಿಸ್ತೀರ್ಣವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
ಅಥವಾ
ಒಂದು ಘನ ಸಿಲಿಂಡರಿನ ಎತ್ತರ 2.4 m ಮತ್ತು ವ್ಯಾಸ 1.4 m ಇದೆ. ಇದರಿಂದ ಒಂದೇ ಎತ್ತರ ಮತ್ತು ಒಂದೇ ವ್ಯಾಸವನ್ನು ಹೊಂದಿರುವ ಶಂಕುವಿನಾಕಾರದ ಹಳ್ಳವನ್ನು ಕೊರೆದು ಟೊಳ್ಳಾಗಿಸಿದೆ. ನೂತನ ಘನದ ಒಟ್ಟು ಮೇಲೈ ವಿಸ್ತೀರ್ಣವನ್ನು ಅತ್ಯಂತ ಸಮೀಪದ ಬೆಲೆಗೆ cm2 ನಲ್ಲಿ ಕಂಡುಹಿಡಿಯಿರಿ.

Question 29.
$$\frac { 3 }{ 2 }$$ x + $$\frac { 5 }{ 3 }$$ y = 7: 9x – 10y = 14 ಈ ರೇಖಾತ್ಮಕ ಸಮೀಕರಣಗಳಲ್ಲಿ $$\frac { { a }_{ 1 } }{ { a }_{ 2 } }$$, $$\frac { { b }_{ 1 } }{ { b }_{ 2 } }$$ ಮತ್ತು $$\frac { { c }_{ 1 } }{ { c }_{ 2 } }$$ ಅನುಪಾತಗಳನ್ನು ಹೋಲಿಸುವ ಮೂಲಕ ಜೋಡಿಗಳು ಸ್ಥಿರವಾಗಿವೆಯೇ ಅಥವಾ ಅಸ್ಥಿರವಾಗಿವೆಯೇ ಎಂಬುದನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 30.
150 cm ಎತ್ತರವಿರುವ ಒಬ್ಬ ವ್ಯಕ್ತಿಯು ತನ್ನ ನೆರಳಿನ ತುದಿಯನ್ನು ಗಮನಿಸಿದಾಗ ಅದು ಅವನ ಪಾದದಿಂದ 150√3 cm ದೂರದಲ್ಲಿರುವುದು ಕಂಡುಬರುತ್ತದೆ. ಹಾಗಾದರೆ ಅವನ ನೋಟದಲ್ಲಿ ಉಂಟಾದ ಅವನತ ಕೋನವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

IV. ಈ ಕೆಳಗಿನ ಪ್ರಶ್ನೆಗಳಿಗೆ ಉತ್ತರಿಸಿ: (6 × 3 = 18)

Question 31.
‘ವೃತ್ತದ ಮೇಲಿನ ಯಾವುದೇ ಬಿಂದುವಿನಲ್ಲಿ ಎಳೆದ ಸ್ಪರ್ಶಕವು, ಸ್ಪರ್ಶ ಬಿಂದುವಿನಲ್ಲಿ ಎಳೆದ ತ್ರಿಜ್ಯಕ್ಕೆ ಲಂಬವಾಗಿರುತ್ತದೆ’ ಎಂದು ಸಾಧಿಸಿ.
ಅಥವಾ
ಬಾಹ್ಯ ಬಿಂದುವಿನಿಂದ ವೃತ್ತಕ್ಕೆ ಎಳೆದ ಸ್ಪರ್ಶಕಗಳು ಸಮ ಎಂದು ಸಾಧಿಸಿ

Question 32.
BC = 6 cm, AB = 5cm ಮತ್ತು ∠ABC = 60° ಇರುವಂತೆ ABC ತ್ರಿಭುಜವನ್ನು ರಚಿಸಿ, ನಂತರ ಮತ್ತೊಂದು ತ್ರಿಭುಜವನ್ನು, ಅದರ ಬಾಹುಗಳು ತ್ರಿಭುಜ ABC ಯ ಅನುರೂಪ ಬಾಹುಗಳ $$\frac { 3 }{ 4 }$$ ರಷ್ಟಿರುವಂತೆ ರಚಿಸಿ.

Question 33.
ಮೂರು ವರ್ಷಗಳ ಹಿಂದೆ ರೆಹಮಾನನ ವಯಸ್ಸು (ವರ್ಷಗಳಲ್ಲಿ) ಮತ್ತು 5 ವರ್ಷಗಳ ನಂತರದ ಅವನ ವಯಸ್ಸು ಇವುಗಳ ವೃಶ್ಯಮಗಳ ಮೊತ್ತ $$\frac { 1 }{ 3 }$$ ಆದರೆ ಅವನ ಈಗಿನ ವಯಸ್ಸನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
ಅಥವಾ
ಮೂರು ಕ್ರಮಾನುಗತ ಸಂಖ್ಯೆಗಳಲ್ಲಿ, ಮೊದಲನೆಯ ವರ್ಗ ಮತ್ತು ಉಳಿದೆರಡು ಸಂಖ್ಯೆಗಳ ಗುಣಲಬ್ದಗಳ ಮೊತ್ತ 154 ಆಗಿದೆ. ಹಾಗಾದರೆ ಆ ಮೂರು ಸಂಖ್ಯೆಗಳನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 34.
ಈ ಕೆಳಗೆ ವ್ಯಾಖ್ಯಾನಿಸಲ್ಪಟ್ಟ ಹೇಳಿಕೆಗಳ ಕೋನಗಳು ಲಘಕೋನಗಳು. ಈ ಕೆಳಗಿನ ಸಮೀಕರಣಗಳನ್ನು ಸಾಧಿಸಿ.

ಅಥವಾ

Question 35.
ಒಂದು ಕಾರ್ಖಾನೆಯ 50 ಕೆಲಸಗಾರರ ದೈನಂದಿನ ಆದಾಯವನ್ನು ಕೆಳಗಿನ ವಿತರಣೆಯು ನೀಡುತ್ತಿದೆ.

ಮೇಲಿನ ವಿತರಣೆಯನ್ನು ಕಡಿಮೆ ಇರುವ ವಿಧಾನದ’ ಸಂಚಿತ ಆವೃತ್ತಿ ವಿತರಣೆಯಾಗಿ ಬದಲಾಯಿಸಿ ಮತ್ತು ಅದರ ಓಜೀವ್ ಎಳೆಯಿರಿ.

Question 36.
ಎರಡನೇ ಬಹುಪದೋಕ್ತಿಯನ್ನು ಮೊದಲನೇ ಬಹುಪದೋಕ್ತಿಯಿಂದ ಭಾಗಿಸಿ ಹಾಗೂ ಮೊದಲನೇ ಬಹುಪದೋಕ್ತಿಯು ಎರಡನೇ ಬಹುಪದೋಕ್ತಿಯ ಅಪವರ್ತನವಾಗಿದೆಯೇ ಎಂದು ಪರೀಕ್ಷಿಸಿ.
t2 – 3 ; 2t4 + 3t3 – 2t2 – 9t – 12
ಅಥವಾ
ಬಹುಪದೋಕ್ತಿ p(x) ನ್ನು ಬಹುಪದೋಕ್ತಿ g(x) ನಿಂದ ಭಾಗಿಸಿ, ಭಾಗಲಬ್ಧ ಮತ್ತು ಶೇಷವನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.
p(x) = x4 – 3x + 4x + 5
g(x) = x2 + 1 – x

V. ಈ ಕೆಳಗಿನ ಪ್ರಶ್ನೆಗಳಿಗೆ ಉತ್ತರಿಸಿ: (4 × 4 = 16)

Question 37.
ಎರಡು ತ್ರಿಭುಜಗಳಲ್ಲಿ ಒಂದು ತ್ರಿಭುಜದ ಮೂರು ಬಾಹುಗಳು ಮತ್ತೊಂದು ತ್ರಿಭುಜದ ಮೂರು ಬಾಹುಗಳೊಡನೆ ಸಮಾನುಪಾತ ಹೊಂದಿದ್ದರೆ, ಅವುಗಳ ಅನುರೂಪ ಕೋನಗಳು ಸಮವಾಗಿರುತ್ತವೆ ಮತ್ತು ಅದರಿಂದಾಗಿ ಆ ಎರಡು ತ್ರಿಭುಜಗಳು ಸಮರೂಪಿಗಳಾಗಿರುತ್ತವೆ ಎಂದು ಸಾಧಿಸಿ.

Question 38.
ನಕ್ಷೆಯ ಮೂಲಕ ಬಿಡಿಸಿ: y = 2x + 1; x = 2y – 5

Question 39.
(4, -1) ಮತ್ತು (-2, -3) ಬಿಂದುಗಳನ್ನು ಸೇರಿಸುವ ರೇಖಾಖಂಡದ ಭಾಜಕ ಬಿಂದುಗಳ ನಿರ್ದೇಶಾಂಕಗಳನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Question 40.
ಒಂದು ಸಮಾಂತರ ಶ್ರೇಢಿಯ ಮೊದಲ ಪದದ ವರ್ಗವು ಅದರ 8ನೇ ಪದಕ್ಕೆ ಸಮವಾಗಿದೆ ಹಾಗೂ 8ನೇ ಪದವು ನಾಲ್ಕನೇ ಪದಕ್ಕಿಂತ 24 ಹೆಚ್ಚಾಗಿದೆ. ಹಾಗಾದರೆ ಶ್ರೇಢಿಯ ಪದಗಳನ್ನು ಬರೆಯಿರಿ.

Solutions

I.
Solution 1.
(B) 2.4 cm

Solution 2.
(C) $$\frac { { \pi r }^{ 2 } }{ 6 }$$

Solution 3.
(B) -1

Solution 4.
(A) 5

Solution 5.
(A) 2

Solution 6.
(C) x2 – 3x + 2 = 0

Solution 7.
(B) {TT, HH, HT, TH}

Solution 8.
(C) 2πrh

II.
Solution 9.
ತ್ರಿವಳಿ: 5, 4, 3
ಏಣಿಯ ಪಾದದಿಂದ ಗೋಡೆಯ
ಪಾದಕ್ಕಿರುವ ದೂರ = 3 ಮೀ.

Solution 10.
ಒಂದು ಬಿಂದುವಿನಿಂದ ವೃತ್ತದ ಮೇಲಿನ ಬಿಂದುವಿಗೆ ಎಳೆದ ಸ್ಪರ್ಶಕಗಳ ಸಂಖ್ಯೆ = 1

Solution 11.
(x1, y1) ಮತ್ತು (x2, y2) ಬಿಂದುಗಳಿಗೆ ಸಂಬಂಧಿಸಿದಂತೆ : ಭಾಗ ಪ್ರಮಾಣ ಸೂತ್ರ (m1 : m2 ಅನುಪಾತದಲ್ಲಿ)

Solution 12.
ಭಾಜ್ಯ = ಭಾಜಕ × ಭಾಗಲಬ್ದ + ಶೇಷ
a = b × q + r

Solution 13.
ಗರಿಷ್ಟ ಶೂನ್ಯತೆಗಳು = 3

Solution 14.
x + y = 27
13 + 14 = 27
xy = 182
13 × 14 = 182
ಮೊದಲ ಸಂಖ್ಯೆ = 13, ಎರಡನೆಯ ಸಂಖ್ಯೆ = 14

Solution 15.
a1 = 38, a16 = 73, a31 = ?
a16 = a + 15d
73 = 38 + 15d
73 = 38 + 15d
d = 3
a31 = a1 + 30d = 38 + 30(3) = 38 + 90 = 128
a31 =128

Solution 16.

AB ಯು ಏಣಿ, CA ಯು ಗೋಡೆ ಮತ್ತು ಕಿಟಕಿಯಾಗಿರಲಿ.
BC = 2.5m ಮತ್ತು CA = 6m
AB2 = BC2 + CA2 = (2.5)2 + (6)2 = 42.25
AB = 6.5
ಏಣಿಯ ಉದ್ದ ಆಗಿದೆ.

Solution 17.
ಮೊದಲ 3 ದಿನದ ಒಂದು ಶುಲ್ಕ = x
ಪ್ರತೀ ದಿನದ ಹೆಚ್ಚುವರಿ ಶುಲ್ಕ = y
x + 4y = 27
x + 2y = 21
2y = 6
y = 3
x + 2y = 21
x + 2(3) = 21
x = 21 – 6
x = 15
ಮೂರು ದಿನಗಳ ಶುಲ್ಕ = x = ₹ 15
ಹೆಚ್ಚುವರಿ ಪ್ರತೀ ದಿನದ ಶುಲ್ಕ = y = ₹ 13

Solution 18.

Solution 19.

Solution 20.

Solution 21.

Solution 22.
√5 ಒಂದು ಭಾಗಲಬ್ಧ ಸಂಖ್ಯೆಯಾಗಿರಲಿ.
√5 = $$\frac { p }{ q }$$ p, q ∈ I
p ಮತ್ತು q ಗಳು 1 ನ್ನು ಹೊರತುಪಡಿಸಿ ಬೇರೆ ಸಾಮಾನ್ಯ ಅಪವರ್ತನ ಹೊಂದಿದ್ದರೆ, ಸಾಮಾನ್ಯ ಅಪವರ್ತನದಿಂದ ಭಾಗಿಸಬಹುದು, ಆದ್ದರಿಂದ p ಮತ್ತು Q ಗಳು ಸಹ ಅವಿಭಾಜ್ಯಗಳೆಂದು ಭಾವಿಸೋಣ
p = q√5
p2 = 5q2 (ಎರಡು ಕಡೆ ವರ್ಗ ಮಾಡಿದಾಗ)
5, p2 ನ್ನು ಭಾಗಿಸುತ್ತದೆ…… (1)
5, p ಮತ್ತು q ಗಳ ಸಾಮಾನ್ಯ ಅಪವರ್ತನವಾಗಿದೆ.
ಏಕೆಂದರೆ, 5, q ನ್ನೂ ಸಹ ಭಾಗಿಸುತ್ತದೆ.
ಅಂದರೆ p = 5r
5q2 = 52r2
q = 5r2 ………(2)
p ಮತ್ತು q ಗಳು ಸಹ ಅವಿಭಾಜ್ಯಗಳು ಎಂಬ ಸತ್ಯಸಂಗತಿಗೆ ವಿರುದ್ಧವಾಗಿದೆ.
√5 ಒಂದು ಅವಿಭಾಜ್ಯ ಸಂಖ್ಯೆಯಾಗಿದೆ.

Solution 23.
ಶೂನ್ಯಗಳ ಮೊತ್ತ = α + β = 1
ಶೂನ್ಯಗಳ ಗುಣಲಬ್ದ = αβ = 1
ವರ್ಗಬಹುಪದೋಕ್ತಿ: x2 – (α + β) x + (αβ)
ಯಲ್ಲಿ ಆದೇಶಿಸಿದರೆ x2 – 1x + 1
(1, 1) ನ್ನು ಕ್ರಮವಾಗಿ ಶೂನ್ಯತೆಗಳ ಮೊತ್ತ ಹಾಗೂ ಗುಣಲಬ್ದವಾಗಿ ಹೊಂದಿರುವ ಯನ್ನು ಕಂಡುಹಿಡಿಯಿರಿ.

Solution 24.
2x2 – 3x + 5 = 0
ax2 + bx + c = 0,
a = 2, b = -3, c = 5
ಶೋಧಕ: b2 – 4ac = 9 – 4(2)(5) = 9 – 40 = -31 < 0
ಮೂಲಗಳು ಸಂಮಿಶ್ರ ಸಂಖ್ಯೆಗಳಾಗಿವೆ.

Solution 25.

Solution 26.
ಆವೃತ್ತಿ ವಿತರಣಾ ಪಟ್ಟಿ

ಬೌಲರ್ ಗರಿಷ್ಠ ಪಂದ್ಯಗಳಾದ 3 ರಲ್ಲಿ ಪಡೆದ ವಿಕೆಟ್‌ಗಳ ಸಂಖ್ಯೆ 2 ಆಗಿದೆ. ಆದ್ದರಿಂದ ದತ್ತಾಂಶಗಳ ಬಹುಲಕ 2,

Solution 27.
ಫಲಿತಗಣ S = {1, 2, 3, 4, 5, 6}
n(S) = 6
2 & 6 ರ ನಡುವಿನ ಸಂಖ್ಯೆಗಳು A = {3, 4, 5}
n(A) = 3
2 & 6 ರ ನಡುವಿನ ಸಂಖ್ಯೆ ಪಡೆಯುವ ಸಂಭವನೀಯತೆ
P(A) = $$\frac { n(A) }{ n(S) }$$ = $$\frac { 3 }{ 6 }$$ = $$\frac { 1 }{ 2 }$$

Solution 28.

Solution 29.

Solution 30.

IV.
Solution 31.

ವೃತ್ತದ ಯಾವುದೇ ಬಿಂದುವಿನಲ್ಲಿ ಎಳೆದ ಸ್ಪರ್ಶಕವು, ಸ್ಪರ್ಶ ಬಿಂದುವಿನಲ್ಲಿ ಎಳೆದ ತ್ರಿಜ್ಯಕ್ಕೆ ಲಂಬವಾಗಿರುತ್ತದೆ’
ದತ್ತ: O ಕೇಂದ್ರವುಳ್ಳ ವೃತ್ತಕ್ಕೆ OP ತ್ರಿಜ್ಯ ಮತ್ತು XY ಸ್ಪರ್ಶಕ ಎಳೆದಿದೆ.
ಸಾಧನೀಯ: OP ಯು XY ಗೆ ಲಂಬವಾಗಿದೆ.
ರಚನೆ: P ಯನ್ನು ಹೊರತುಪಡಿಸಿ XY ಮೇಲೆ ಮತೋಂದು ಬಿಂದು Q ಗುರುತಿಸಿ, OQ ಸೇರಿಸಿ.
ಸಾಧನೆ: Q ಬಿಂದುವು ವೃತ್ತದ ಹೊರಭಾಗದಲ್ಲಿರಬೇಕು.
Q ಬಿಂದುವು ವೃತ್ತದ ಒಳಭಾಗದಲ್ಲಿದ್ದರೆ, XY ವೃತ್ತಕ್ಕೆ ಛೇದಕವಾಗುತ್ತದೆಯೇ ಹೊರತು ವೃತ್ತಕ್ಕೆ ಸ್ಪರ್ಶಕವಾಗುವುದಿಲ್ಲ. ಆದ್ದರಿಂದ OQ ಇದು ವೃತ್ತದ ತ್ರಿಜ್ಯ OP ಗಿಂತ ಉದ್ದವಾಗಿದೆ. ಅಂದರೆ OQ > OP.
P ಬಿಂದುವನ್ನು ಹೊರತುಪಡಿಸಿ, XY ಮೇಲಿನ ಎಲ್ಲಾ ಬಿಂದುಗಳಿಗೂ ಇದು ಅನ್ವಯಿಸುವುದರಿಂದ, O ಬಿಂದುವಿನಿಂದ XYನ ಮೇಲಿನ ಇತರೆ ಬಿಂದುಗಳಿಗಿರುವ ದೂರಕ್ಕಿಂತ OP ಯು ಕನಿಷ್ಟ ಉದ್ದ ಹೊಂದಿದೆ
OP ಯು XY ಗೆ ಲಂಬವಾಗಿದೆ.
ಅಥವಾ

ಬಾಹ್ಯಬಿಂದುವಿನಿಂದ ವೃತ್ತಕ್ಕೆ ಎಳೆದ ಸ್ಪರ್ಶಕಗಳ ಉದ್ದವು ಸಮ.
ದತ್ತ: 0ವೃತ್ತಕೇಂದ್ರ, P ಬಾಹ್ಯಬಿಂದು, PQ & PR ಗಳು ಬಾಹ್ಯಬಿಂದು Pನಿಂದ ಎಳೆದ ಸ್ಪರ್ಶಕಗಳು
ಸಾಧನೀಯ: PQ = PR
ರಚನೆ: OP, OR, OP ಸೇರಿಸಿ
ಸಾಧನೆ: ΔOQP ಮತ್ತು ΔORP ಗಳಲ್ಲಿ
OQ = OR (ತ್ರಿಜ್ಯಗಳು)
∠OQP = ∠ORP = 90°
OP = OP (ಉಭಯಸಾಮಾನ್ಯ)
ΔOQP = ΔORP
PQ = PR

Solution 32.

Solution 33.
ರೆಹಮಾನನ ಈಗಿನ ವಯಸ್ಸು = x ಆಗಿರಲಿ
3 ವರ್ಷಗಳ ಹಿಂದೆ, ಅವನ ವಯಸ್ಸು = x – 3
5 ವರ್ಷಗಳ ನಂತರ ಅವನ ವಯಸು = x + 5

Solution 34.

Solution 35.

Solution 36.

Solution 37.

Solution 38.

Solution 39.

Solution 40.

## Karnataka SSLC Maths Model Question Paper 1 with Answer in Kannada

Students can Download Karnataka SSLC Maths Model Question Paper 1 with Answers in Kannada, Karnataka SSLC Maths Model Question Papers with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

## Karnataka SSLC Maths Model Question Paper 5 with Answers

Students can Download Karnataka SSLC Maths Model Question Paper 5 with Answers, Karnataka SSLC Maths Model Question Papers with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

## Karnataka State Syllabus SSLC Maths Model Question Paper 5 with Answers

Time: 3 Hours
Max Marks: 80

I. In the following questions, four choices are given for each question, choose and write the correct answer along with its alphabet: ( 1 × 8 = 8 )

Question 1.
The pair of linear equations 3a + 4b = k, 9a + 12b = 6 have infinitely many solutions when,.
a) K = -2
b) K = 3
c) K = 2
d) K = -3
c) K = 2
Solution:

Question 2.
n2 – 1 is divisible by 8, if n is
a) Prime numbers
b) Odd integer
c) Even integer
d) Natural number
b) Odd integer
Solution:
n2 – 1
If n is an odd. integer, 1,3,5,
Ex: 12-1 = 1- 1= 0 divisible by 8.
32 – 1=9-1=8 divisible by 8.
52 – 1 = 25 – 1 = 24 divisible by 8.

Question 3.
$$\sqrt{1+\tan ^{2} \theta}$$ = , where 0<θ< 90°
a) secθ
b) cosec θ
c) cos θ
d) sin θ
a) secθ
Solution:
$$\sqrt{1+\tan ^{2} \theta}=\sqrt{\sec ^{2} \theta}=\sec \theta$$

Question 4.
If Q divides the line A(3, 5) and B (7,9) internally in the ratio 2:3, then the co ordinates of Q are.

c) $$\left(\frac{23}{5}, \frac{33}{5}\right)$$
Solution:

Question 5.

If Area of the sector  OPRQ = $$\frac{5}{18}[ /latex] of Area of circle. Then the value of x a) 25° b) 50° c) 75° d) 100° Answer: d) 100° Solution: Question 6. If 1 + 2 + 3 + n terms = 28, then n is equal to a) 28 b) 7 c) 8 d) 56 Answer: b) 7 Solution: [latex]\frac{\mathrm{n}(\mathrm{n}+1)}{2}$$
n(n + 1) = 28 x 2 = 56
n(n + 1) = 7 x 8
∴ n = 7

Question 7.
If E1E2 E3…….E10 are the possible elementary events of a random experiment, then P(E1) + P(E2) + P(E3) + ………P(E10) is equal to
a) 0
b) 1
c) 2
d) 3
b) 1

Question 8.
If we express sec A in terms of sin A, then sec A is equal to

$$\frac{1}{\sqrt{1-\sin ^{2} A}}$$
Solution:

II. Answer the following Questions : ( 1 x 8 = 8 )

Question 9.
What is the $$\frac{p}{q}$$ form of 43.123456789 ?
$$\frac{43123456789}{999999999}$$

Question 10.
Write the quadratic equation formed by the roots √3 + √5 and √3 – √5
x2 -(α + β)x + αβ = 0
X2 – [3 + √5 + 3 – √5]x + (3 + √5)(3 – √5) = 0
x2 – [6] x + (9 – 5) = 0
x2 – 6 + 4 = 0

Question 11.
Find the value of $$\frac{\sin 26}{\sec 64}+\frac{\cos 26}{\csc 64}$$

Question 12.
What is the distance between the points p(cos0, sin0) and Q(sin0, -COS0)?

Question 13.
If a number “x” choosen at random from the numbers -2, -1, 0, 1, 2. What is the probability that x2 < 3 ?
Clearly “x” can take any of the five given values.
∴ n(x) = 5.
If x2 < 3, then x can take the values -1, 0, 1.
∴ This even A = ,{-1, 0, 1}
n(A) = 3.
P(A) = $$\frac{n(A)}{n(S)}=\frac{3}{5}$$

Question 14.
What is the Area of a circle whose perimeter is 44 cms.
2πr = 44
2 x $$\frac{22}{7}$$ x r = 44
r = $$\frac{44 \times 7}{22 \times 2}$$
r = 7
A = πr² = $$\frac{22}{7}$$ x 7 x 7 = 154cm2

Question 15.
A toy was made by scooping out a hemisphere from each end of a solid cylinder. If the height of the cylinder is h cm and base radius is r cms, find the total surface area of the toy.
Total surface Area of the toy = C.S.A of cylinder + 2 [Surface Area of Hemisphere]
= 2πrh + 2πr²
= 2 πr (h + r)

Question 16.
The circumference of a circle exceeds the diameter by 15cm. Find the raidus of the circle.
Circumference = 2r + 15
2πr = 2r + 15
2πr – 2r = 15
2 x $$\frac{22}{7}$$r – 2r = 15
Multiply by 7
2 x 22r – 14r = 105 .
30r = 105
r = 3.5

III. Answer the following : (2 x 8 = 16 )

Question 17.
Prove that √5 + √3 is an irrational number.

∴ √3 is rational, is a contradiction
∴ √5 + √3 is irrational.

Question 18.
AX and BY are perpendiculars to segment XY. If AO = 5cm, BO = 7cm and Area of ∆ AOX = 150 cm2, find the area of ∆ BOY.

In ∆ OYB and ∆ OXA,
We have ∠X = ∠Y
So, by AA – criterion of similarity, we have
∆ YOB ≅ ∆ XOA

OR

In the given figure, BD ⊥ AC. Prove that AB2 + CD2 = AD2 + BC2
Soln: In ∆ BDC,
∠ BDC = 90°

BC2 = BD2 + DC2 (By Pythagoras)
In ∆ BDA, ∠ APB = 90°
AB2 = AD2 + BD2 (By Pythagoras)
AB2 – BC2 = AD2 + BD2 – BD2 – DC2
AB2 + CD2 = AD2 + BC2
Hence proved.

Question 19.
If zeroes of the polynomial p(y) = y3 – 3y2+ y + 1 are a – b, a, a+ b. Find a and b.
Sum of the zeroes of p(y) = y3 – 3y2+ y + 1

(a-b)(a)(a + b)= $$\frac{-1}{1}$$ = -1
(a2 – b2) a = -1
(1 – b2) (1) = -1
– b2 = -1
-b2 = – 1 – 1
-b2 = -2
b2 = 2
b = ±√2

Question 20.
Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length.

Question 21.
Find the zeroes of the polynomial p(y) = y3 – 5y2 – 16y + 80. If zeroes are α, -α and β.
Sum of the zeroes

α + (-α) + β = 5
α – α + β = 5
β = 5

(α )(-α )(β) =-80
2β = -80
α2β = 80
α2(5) = 80
α = 80/5 = 16
α = √16
= ±4

Question 22.
Two pillars of equal height and on either side of a road, which is 100m wide. The angles of elevation of the top of the pillars are 60 and 30 at a point on the road between the pillars. Find the position of the point between the pillars and height of each pillars.

In ∆ PAB,
tab 60 = $$\frac{\mathrm{AP}}{\mathrm{AB}}$$
√3 = h/x
h = √3x
In ∆ BCQ,
tan 30 = $$\frac{\mathrm{CQ}}{\mathrm{BC}}$$
$$\frac{1}{\sqrt{3}}=\frac{h}{100-x}$$
h√3 -= 100 – x
(√3x)(√3x) = 100-x
4x = 100
x = 25
∴ h= √3 X 25 = 25√3
= 25 x 1.73 = 43.3

OR

The three metallic spheres of radii are in the ratio of 3 : 4 : 5 are melted to form a single solid sphere of radius 12cm. Find the radius of the three small metallic spheres.
Volume of the solid sphere = Sum of the volumes of each sphere
$$\frac{4}{3} \pi r^{3}=\frac{4}{3} \pi\left(r_{1}^{3}+r_{2}^{3}+r_{3}^{3}\right)$$
4 x 123 = 4[(3x)3 + (4x)3 + (5x)3]
1728 = 27x3 + 64x3 + 125x3 1728 = 216 x3
x = $$\frac{1728}{216}$$
x3 = 8 x = 3√8 = 2
∴ R1 = 6, R2 8, R3 = 10

Question 23.
In the given fig. OACB is a quadrant of a circle with centre O and radius 3.5cm. If OD = 2cm, find the area of the shaded region

Question 24.
Find the volume of the largest right circular cone that can be cut of a cube of edge 7cm.
Given radius of the base of cone = $$\frac{7}{2}$$ cm
h = height of cone = 7cm
Volume of the cone = $$\frac{1}{3}$$πr²h
Volume = $$\frac{1}{3} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 7$$
= $$\frac{22 \times 7 \times 7}{12}$$
= 89.83cm3

IV. Answer the following : ( 3 x 9 = 27 )

Question 25.
The sum of a two digit number and the number obtained by reversing the order of its digits is 165. If the digits differ by 3, find the number.
Let the digits at units and tens place of the given number be y and x respectively, then
Given number = 10x + y
Number obtained by reversing the order of the digits = 10y + x
(10x + y) + (10y + x) = 165
10x + y+ 10y + x = 165
(11x+11y=165) = 11 ’
x + y = 15 ……(1)
Digits differ by 3, if x > y
x – y = 3 ……..(2)
Solve (1) and (2)

Consider x + y =15
9 + y = 15
y = 15 – 9
y = 6
∴ The number is 96, if y >x, number is 69.

OR

Ten years ago Sudhir was twelve times as old as his son Raghav and ten years, hence, he will be twice as old as his son will be. Find their present ages.
Let the present ages of Sudhir and Raghav be x years and y years respectively. Ten years ago, Father ’s age = (x -10) years Son’s age = (y – 10) years,
x- 10= 12(y – 10)
=> x – 10 = 12y – 120
x – 12y = 120+ 10
x – i2y = – 110
x – 12y + 110 = 0. …….(1)
Ten years later, Father’s age = (x + 10) years
Son’s age = (y + 10)
x + 10 = 2(y + 10)
⇒ x+ 10 = 2y + 20
x – 2y = 20 – 10
x – 2y = 10
x – 2y – 10 = 0 ……(2)
Solve (1) and (2)

y = 12
x – 12y = 110
x – 12(12) = -110
x – 144 = -110
x = -110 + 144 = 34
∴ Present age of Sudhir = 34 years and present age of Raghav = 12 years.

Question 26.
The altitude of a right triangle is 7cm less than its base. If the hypotenuse is 13cm, find the other two sides.
Let the base BC be = x cm
Then its height AB = (x – 7) cm
Hypotenuse AC = 13 cm
In ∆ABC, AC2 = AB2 + BC2
132 = (x – If + (x)2
169 = x2 + 49 – 14x + x2
169 = 2x2 – 14x + 49
2x2 – 14x + 49 – 169 = 0
2x2 – 14x-120 = 0
x2 – 7x – 60 = 0
x2– 12x + 5x-60 = 0
x(x- 12) + 5(x – 12) = 0
(x-12) (x + 5) = 0
(x – 12) = 0 or x + 5 = 0
x = 12 or x = -5
The base of the triangle = 12 cm Length of the altitude = (12 – 7)cm = 5 cm.

Question 27.
X(2, 5), Y (-1, 2) and Z(5, 8) are the co – ordinates of the vertices of A XYZ. The point A and B lie on XY and XZ respectively such that XA : AY = XB: BZ =1:2. Calculate the co – ordinates of X and Y.

B(X2 y2)= [3, 6]
∴ Co – ordinates of A = (1,4)
Co – ordinates of B = (3, 6)

OR

Find the Area of the triangle formed by joining the mid points of the sides of the triangle whose vertices are (0, -1) (2, 1) and

Co – ordinates of A
$$\frac{0+2}{2}, \frac{-1+1}{2}$$
= [1,0]
Co – ordinates of B
$$\frac{0+0}{2}, \frac{-1+3}{2}$$
= [0,1]
Co – ordinates of B
$$\frac{2+0}{2}, \frac{1+3}{2}$$
= [1,2]
∴ Area of A ABC = $$\frac{1}{2}$$ Σ x1 (y2 – y3)

= $$\frac{1}{2}$$ [1(1-2) + 0(2 – 0) + 1(0-1)]
= $$\frac{1}{2}$$ [-1 + 0 + 1]
∴ Area of ∆ ABC = 1 sq. unit

Question 28.
Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contacts.
Data : AB is the tangent drawn to a circle centered at O
‘C’ is the point of contact.

To prove that :
OC ⊥AB
Construction : Mark D .on AB. Join OD.
Let it cut the circle at E
Proof: OE < OD
OE is the radius of the circle.
OE = OC (radii of the same circle)
⇒ OC < OD
⇒ OC is the shortest distance between centre of the circle and tangent AB
∴ OC ⊥ AB
Hence proved.

Question 29.
A hollow sphere of internal and external radii are 6cm and Scm respectively is melted and recast into small cones of base radius 2cm and height 8cm. Find the number of cones.
Inner radius of a hollow sphere (r) = 6cm
and outer radius (R) = 8cms.

Radius of one small cone (r,) 2cm and
heigh (h) = 8 cm.

OR

A medicine capsule is in the shape of a cylinder with two hemispheres stuch to each of its ends. The length of the entire capsule is 14mm and the diameter of the capsule is 5 mm. Find its surface area.

Let r mm be the radius and h mm be the height of the cylinder.
r = $$\frac{5}{2}$$ mm = 2.5mm
h = (14 – 2 x $$\frac{5}{2}$$ mm
= (14-5) mm
= 9 mm
The radius of hemisphere = r = $$\frac{5}{2}$$ mm
Now, the surface area of the capsule. = Curved surface Area of the cylinder surface Area of two hemispheres.
= 2πrh + 2 x 2 πr²
= 2πr² (h + 2r)

Question 30.
Solve graphically
y = $$\frac{1}{2}$$ x and 3x + 4y – 20 = 0
y = $$\frac{1}{2}$$ x

3x + 4y – 20 = 0
3x + 4y = 20
4y = 20 – 3x

Question 31.
The following distribution gives the daily income of 65 workers of a factory. Convert this into less than type cumulative frequency distribution and draw its ogive.

Question 32.
The fourth term of an AP is 11, and 8th term exceeds twice the fourth term by 5. Find AP and find the sum of first 100 terms.
Given : T4 = -11 and T8 = 2T4 + 5
T8 = 2(11) + 5
a + 7d = 22 + 5 = 27

d = 4
a + 7d = 27
a + 7(4) = 27
a+ 28 = 27
a = 27-28 = -1
a = -1
A.P = -1,3, 7, …..
Sn = $$\frac{n}{2}$$ [2a + (n-1)d] = $$\frac{100}{2}$$ [2(-1) + (99)4]
= 50 [-2 + 396]
= 50 [394]
= 19700

Question 33.
A person on tour has ₹ 360 for his expenses. If he extends his tour for 4 days, he has to cut down his daily expenses by ₹ 3. Find the original duration of tour.
Let the original duration of the tour be x days.
Total expenditure on four = 360 ₹
Expenditure per day = $$\frac{360}{x}$$
Duration of the extended tour = (x +4) days
∴ Expenditure per day according to new schedule = $$\frac{360}{x+4}$$
It is given that daily expenses are cut down by 3₹
$$\frac{360}{x}-\frac{360}{x+4}$$ = 3
360 (x + 4) – 360x = 3(x) (x + 4)
360x + 1440 – 360x = 3x (x + 4)
1440 = 3x2+ 12x
3x2 + 12x – 1440 = 0
x2 + 4x – 480 = 0
x2 + 24x – 20x – 480 = 0
x(x + 24) – 20(x + 24) = 0
(x + 24) (x – 20) = 0
x + 24 = 0 and x – 20 = 0
x = -24 and x = 20
∴ Duration of tour = 20 days.

OR

Two pipes running together can fill a cistern in 3$$\frac{1}{13}$$ minutes. If one pipe
takes 3 minutes more than the other to fill it, find the time in which each pipe would fill the cistern.
Let the faster pipe takes x minutes.
Let the smaller pipe take (x + 3) minutes to fill.
Portion of the cistern filled by the faster 1
pipe in one minute = $$\frac{1}{x}$$
⇒ Portion of the cisterin filled by the faster pipe in $$\frac{40}{13}$$ minutes

|||ly portion of the cistern filled by the slower pipe in $$\frac{40}{13}$$ minutes.

It is given that the cistern is filled in $$\frac{40}{13}$$ min.

[40(x + 3) + 40x] = [(x) (x +3)] 13
40x + 120 + 40x = (x2 + 3x) 13
80x + 120= 13x2 + 39x
13x2 + 39x – 80x – 120 = 0
13x2 – 41x-120 = 0
13x2 – 65x + 24x – 120 = 0
13x (x – 5) + 24(x – 5) = 0
x – 5 = 0, 13x + 24 = 0
x = 5, 13x = -24, x = $$\frac{-24}{13}$$
Faster pipe fills the cistern in 5 minutes and the slower pipe takes 8 min to fill the cistern.

V. Answer the following : ( 4 x 4 = 16 )

Question 34.
In an AP whose first term is 2, the sum of first five terms is one fourth the sum of the next five terms. Show that T20 = -112. Find S20.
a = 2, T1 + T2 + T3 + T4+ T5
= $$\frac{1}{4}$$ (T6 + T7 + T8 + T9 + T10)
a + a + d + a + 2d + a + 3d + a + 4d
= $$\frac{1}{4}$$(a + 5d + a + 6d + a + 7d + a + 8d + a + 9d)
5a+ 10d = $$\frac{1}{4}$$ (5a + 35d)
5(a + 2d) = $$\frac{1}{4}$$ x 5( a + 7d)
a + 2d = $$\frac{1}{4}$$ (a + 7d)
4(a+2d) = a+7d
4a + 8d – a – 7d = 0
3a + d = 0
3(2) + d = 0
d = -6
T20 = a + 19d
= 2 +19(-6)
= 2 – 114
T20 = -112
Sn = $$\frac{n}{2}$$ [2a + (n-1)d]
Sn = $$\frac{20}{2}$$ [2(2) + (20-1)(-6)]
= 10 [4- 114]
= (10) (-110)
S20 = -1100.

OR

A man repays a loan of ₹ 3250 by paying 7 20 in the first month and then increases the payment by ₹ 15 every month. How long will it take to clear his loan?
Here, a = 20, c.d = 15, n = ?
Sn = $$\frac{n}{2}$$[2a + (n-1)d]
3250 = $$\frac{n}{2}$$ [2 x 20 + (n-1)15]
3250 x 2 = n(40 + 15n- 15)
6500 = n (25 +15n)
6500 = 25n + 15n2
(15n2 + 25n – 6500 = 0)÷ 5
3n2 + 5n – 1300 = 0
3n2-60n + 65n- 1300 = 0
3n(n – 20) + 65 (n – 20) = 0
(n – 20) (3n + 65) = 0
(n – 20) = 0, 3n + 65 = 0
n = 20, 3n = -65
n = $$\frac{-65}{3}$$
∴ Total amount will be paid in 20 months.

Question 35.
Find the mean, median and mode for * the following frequency distribution.

Question 36.
P.T $$\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}$$ = 1+ secθ cosecθ

Question 37.
Draw a right triangle in which the sides (other than hypotenuse) arc of lengths 8cm and 6cm, then construct another triangle whose sides are 5/3 times the corresponding sides of given triangle.

Verify
BC = 8cm
BC’= $$\frac{5}{3} \times 8=\frac{40}{3}$$
AB =6cms
A’B= $$\frac{5}{3} \times 6$$
=10cms.

VI. Answer the following : (5 x 1 = 5 )

Question 38.
State and prove pythagoras theorem.
In a right angled triangle, square on the hypotenuse is equal to sum of A the squares on the other sides.

Data : ∆ ABC, ∠B = 90°
T.P.T : AC2 = AB2 + BC2
Construction : Draw BD ⊥ AC
Proof: In ∆ ABC and ∆ ABD
∠ ABC = ∠ADB = 90 [∵ Data & construction]
∠A = ∠A (∵ common Angle)
∠ACB =∠ ABD [ ∵ Re maining angle]
∴ ∆ ABC and ∆ ABD are equiangular
=> ∆ ABC ~ ∆ABD

AB2= AC x AD ……… (1)
In ∆ABC and ∆ BDC
∠ ABC = ∠ BDC = 90 [∵ Data & construction]
∠C = ∠C (∵ common Angle)
∠BAC =∠ DBC [ ∵ Re maining angle]
∴ ∆ ABC and ∆ DBC are equiangular
∆ABC ~ ∆BDC

AB2 + BC2 = AC (AD + DC)
AB2 + BC2 = AC(AC)
AB2 + BC2=AC2
AC2 = AB2+ BC2

## Karnataka SSLC Maths Model Question Paper 4 with Answers

Students can Download Karnataka SSLC Maths Model Question Paper 4 with Answers, Karnataka SSLC Maths Model Question Papers with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

## Karnataka State Syllabus SSLC Maths Model Question Paper 4 with Answers

Time: 3 Hours
Max Marks: 80

I. In the following questions, four choices are given for each question, choose and write the correct answer along with its alphabet: ( 1 × 8 = 8 )

Question 1.
If the H.C.F of 65 and 117 is expressible in the form of 65m – 117, then the value of m is
a) 4
b) 3
c) 11
d) 2
Solution:
The H.C.F. of 65 and 117 = 65m- 117
Now 13 = 65m- 117
65m 117+13
65rn = 130
m = 130/65 = 2
∴ The value of m = 2
∴ HCF(65, 117)
= 13

Question 2.
If sinx = sin 60° cos 30° – cos 60 sin 30!l, then the value of x is
a) 0°
b) 30°
c) 45°
d) 60°
b) 30°
Solution:
Sinx = sin 60°cos30 – cos 60 sin 30

Question 3.
The angle between the radius of a circle and the tangent drawn at the point of contact is
a) 0°
b) 60°
c) 90°
d) 30°
c) 90°

Question 4.
The T.S.A. of a cuboid of dimension, l = 30cm, b = 20cm, c = 10cm, is ______
a) 600cm2
b) 60cm2
c) 6000cm2
d) 2200 cm2

Question 5.
Which of the following is a polynomial
a) x2 – 5x + 3√x
b) x1/2 + x1/2 – x +1
c) $$\sqrt{x}-\frac{1}{\sqrt{x}}$$
d) x2 – 4x + √2
x2 – 4x + √2
Solution:
x2 – 4x + √2 because variable has no negative or decimal powers.

Question 6.
The value of p if x, 2x + p and 3x + 6 are in A.P.
a) p = 3
b) p = 2
c) p = 1
d) p = 0
Solution:
x, 2x + p, 3x + 6 are inAP a, A, b are A.P.

2(2x + p) = 2(2x + 3)
2x + p = 2x + 3
p = 2x + 3 – 2k
P = 3

Question 7.
In triangle PQR, The value of y is
a) 4√3
b) 6√3
c) 5√3
d) √3
b) 6√3
Solution:

QS2 = PS x SR
QS2 = $$\sqrt{9 \times 12}$$
QS = $$\sqrt{9 \times 4 \times 3}$$
QS = 3 x 2√3
QS = 6√3

Question 8.
When 2 unbiased coins are tossed at ‘ a time, the probability of getting 2 heads is _____
a) 1/4
b) 1/2
c) 1
d) 0
a) 1/4
Solution:
S = {(H, H) (H, T) (T, H) (T, T)}
n(s) = 4
An event of getting 2 heads = A={(H, H)}
∴ P(A) = $$\frac{n(A)}{n(S)}=\frac{1}{4}$$

II. Answer the following Questions : ( 1 x 8 = 8 )

Question 9.
If the product of zeros of polynomial f(y) = ay3 – 6y2 + 11y – 6 is 4 then find the value of ‘a’.
α β γ = $$-\frac{\mathrm{d}}{\mathrm{a}}$$
4 = $$\frac{-(-6)}{a}$$
4a = 6
a = $$\frac{6}{4}$$
a = $$\frac{3}{2}$$

Question 10.
What is the value of C, if ax2 + bx + c = 0 has equal roots?
If ax2 + bx + c = 0 has equal roots then,
b2 – 4ac = 0
b2 = 4ac
4ac = b2
c = $$\frac{b^{2}}{4 a}$$

Question 11.
Find the second term if sum of the ‘n’ term of an AP is 2n2 + 1.
Sn = 2n2 + 1
S1 = 2(1)2 + l= 2xl + l= 2+ l= 3
∴ S1 = T1 = a = 3
S2 = 2(2)2 + 1 =8+ 1 =9
a + T2 = 9
T2 = 9 – 3 = 6

Question 12.
State converse of Pythagoras Theorem.
“The square on the greatest side of a triangle is equal to the sum of the squares on the other two sides, then the other two sides contian the right angle”.

Question 13.
What is the form of $$\frac{p}{q}$$ (p,q ∈ z q ≠ 0) form of 0.57 ?
$$0.5 \overline{7}=\frac{57-5}{90}=\frac{52}{90}=\frac{26}{45}$$

Question 14.
If sinθ = $$\frac{1}{3}$$ then find the value of (cot2θ + 2)
AC2 = AB2 + BC2
BC2 = AC2 – AB2
= 32 – 12
= 9 – 1 = 8
BC = √8 = 2√2
cotθ = $$\frac{2 \sqrt{2}}{1}$$ = 2√2
2cot2θ + 2 = 2(2√2)2 + 2
= 2 x 8 + 2
= 16 + 2
= 18

Question 15.
In sin(A + B) = $$\frac{\sqrt{3}}{2}$$ and cos (A – B) = 1, 0< A + B < 90°, A ≥ B
sin(A + B) = $$\frac{\sqrt{3}}{2}$$ = sin60°
A + B = 60° …. (1)
cos (A – B) = 1 = cos0°
A – B = 0 ….(2)
Solve (1) and (2)
A + B = 60
A – B = 0
2A = 60 or A = 30°
A + B = 60
30 + B = 60
B = 60 – 30° = 30°
∴ A = 30°, B = 30°

Question 16.
The surface area of a sphere is same as the C.S.A. of a right circular cylinder whose height and diameter are 4 cm each. Find the radius of the sphere.
Soln:
4πr² = 2πrh
4R² = dh .
4R²= 4 x 4
R² = $$\frac{4 \times 4}{4}$$ = 4
R = √4 = 2cm
A = 30°, B = 30°
∴ Radius of the sphere = 2cm.

III. Answer the following : ( 2 x 8 = 16 )

Question 17.
By Euclid’s division lemma, show that thte square of any positive integer is either of the form 3m or 3m + 1 for some integer m.
Let x be any positive integer. Then it is of the form 3q, 3q+1 or 3q + 2.
Now, (3q)2 = 9q2 = 3(3q2) = 3m where m = 3q2
(3q + 1)2 = 9q2 + 6q+ 1 = 3(3q2 + 2q) + 1
= 3m + 1, where m = 3q2 + 2q and (3q+2)2 = 9q2 + 12q + 4
= 3(3q2 + 4q + 1) +1
=3m +1, where m = 3q2 + 4q + 1
Hence, it is proved.

Question 18.
Solve : $$\frac{x+y}{x y}$$ = 2 and $$\frac{x-y}{x y}$$ = 6
x + y = 2xy
x – y = 6xy
2x = 8xy
2 = 8y

Question 19.
Solve : y2 – (√3 + 1)y + √3 = 0
y2 – √3y – y + √3 = 0
y(y – √3) – 1(y – √3) = 0
(y – √3)(y – 1) = 0
y – √3 = 0, y – 1 = 0
y = √3 and y = 1

Question 20.
Show that the points (3, 2) (-2, -3) and (2, 3) are collinear Or non – collinear.
Let A= (3, 2) B = (-2, -3) C = (2, 3)

AC ≠ AB + BC
∴ They are non – collinear.

Question 21.
In the given fig ∆DGH ~ ∆DEF, DH = 8cm, DF = 12cm, DG = (3x – l)cm and DE = (4x +. 2)cm, Find the lengths of DG and DE.
∆DGH ~ ∆DEF
$$\frac{\mathrm{DG}}{\mathrm{DE}}=\frac{\mathrm{DH}}{\mathrm{DF}}$$
$$\frac{3 x-1}{4 x+2}=\frac{8}{12}$$

12 (3x – 1) = 8(4x + 2)
36x – 12 = 32x + 16
36x – 32x = 16 + 12
4x = 28
x = 28/4
x = 7
∴ DG = 3x – 1 = 3 x 7 – 1 = 21 – 1 = 20
DE = 4x + 2 = 4 x 7 + 2 = 28 + 2 = 30

OR

D is a point on the side BC of A ABC such that ∠ADC = ∠BAC. Prove that
$$\frac{\mathbf{C A}}{\mathbf{C D}}=\frac{\mathbf{C B}}{\mathbf{C A}}$$
In ∆ ABC and ∆ DAC,
∠C = ∠C
∆ABC ~ ∆DAC

Question 22.
A card is drawn at random from a box containing 21 cards numbered 1 to 21. Find the probability that the card drawn is
a) Prime number b) Divisible by 3.
n(s) = 21
A= {2,3,5,7,11,13,17,19}
n(A) = 8
P(A) = $$\frac{n(A)}{n(S)}=\frac{8}{21}$$

b) n(S) = 21
B= {3,6, 9,12,15,18}
n(B) = 6
∴ P(B) = $$\frac{n(B)}{n(S)}=\frac{6}{21}=\frac{2}{7}$$

Question 23.
Draw a circle of radius 3cms. Construct a pair of tangents to it, from a point 5cm away from the circle.
r = 3cm d = 3 + 5 = 8cm

Question 24.
Express sinA and sec A in terms of cot A.
W.K.T cosec2A = 1 + cot2A

OR

If tanθ + sinθ = m and tan θ – sin θ= n, S.T m2 – n2 = 4$$\sqrt{\mathbf{m} \mathbf{n}}$$
L.H.S = m2 – n2 = (m +n) (m – n)
= {(tan θ + sin θ ) + (tan θ – sin θ )x(tan θ +sin θ) – (tan θ – sin θ)}
= {(2tanθ) x (2sinθ) = 4tanθsinθ

= 4$$\sqrt{\mathbf{m} \mathbf{n}}$$
= RHS.

IV. Answer the following : ( 3 x 9 = 27 )

Question 25.
The sum of the numerator and denominator of a fraction is 24. If 4 is subtracted from the numerator and 5 from its denominator, then it reduces to 1/4. Find the fraction.
Let the fraction be $$\frac{\mathrm{x}}{\mathrm{y}}$$
sum of its numerator and denominator = 24.
x + y = 24 ……..(1)
If 4 is subtracted from the numerator and
5 from its denominator we get = 1/4
⇒ $$\frac{x-4}{y-5}=\frac{1}{4}$$
4(x – 4) = 1(y – 5)
4x – 16 = y – 5
4x – y = -5 + 16
4x – y = 11 ……..(2)
From (1) and (2)

Consider,
x + y = 24
7 + y = 24
y = 24 – 7
y = 17
∴ The fraction is $$\frac{x}{y}=\frac{7}{17}$$

OR

Two women and five men can together finish an embroidary work in 4 days. While three women and 6 men can finish in 3 days. Find the time taken by one women alone and also that taken by one man alone.
Let 1 woman can finish the embroidary in x days and 1 man can finish the embroidary in y days.
1 woman’s 1 day’s work = $$\frac{1}{x}$$
1 man’s 1 day’s work = $$\frac{1}{y}$$

Let $$\frac{1}{x}$$ = u, $$\frac{1}{y}$$ = v
2u + 5v = $$\frac{1}{4}$$
3u + 6v = $$\frac{1}{3}$$
solve (1) and (2)
(8u + 20v = 1) x 9
(9u+ 18v= 1) x 8

v = $$\frac{1}{36}$$
Put v = $$\frac{1}{36}$$…….(1)
2u + 5v = $$\frac{1}{4}$$
2u + 5$$\frac{1}{36}$$ = $$\frac{1}{4}$$

Thus, 1 woman alone can finish the embroidery in 18 days and 1 man alone can finish it in 36 days.

Question 26.
If two zeroes of the polynomial p(x) = x4 – 6x3 – 26x2 + 138x – 35 are 2 – √3 and 2 + √3. Find the other two zeroes.
The two zeroes of p(x) are,
(2 – √3) and (2 + √3)
⇒ [x – (2 – √3] and [x – [2 + √3] are the factors p(x).
⇒ [ x – 2 + √3] [x – 2 – √3]
(x – 2)2 – (√3)2
x2 – 4x + 4 – 3
⇒ x2 – 4x + 1 is a factor of p(x)
Divide p(x) by x2 – 4x + 1 to get quotient

Consider the quotient, x2 – 2x – 35 = 0
x2 – 7x + 5x – 35 = 0
x(x – 7) + 5(x – 7) = 0
(x – 7) (x + 5) = 0
x-7 = 0 orx + 5 = 0
x = 7 or x = -5
∴ The other two zeroes are 7, -5.

Question 27.
A two digit number is such that the product of its digits is 18. when 63 is subtracted from the number, the digits interchange their places. Find the number.
Let the two digits be x and y.
⇒ The two digit number is = 10x + y
The product of three digits = xy = 18
When 63 is subtracted from the number, the digits interchange their places.
10x + y – 63 = 10y + x
10x – x – 10y + y = 63
9x – 9y = 63
9(x – y) = 63
x – y = 63/9 = 7
x – y = 7
x = 7 + y
Consider xy = 18 and substitute x = (y + 7) in it.
y(y + 7) = is . .
y2 + 7y – 18 = 0
y2 + 9y – 2y – 18 = 0
y (y + 9) – 2(y + 9) = 0 –
(y + 9) (y – 2) = 0 .
y + 9 = 0, y -2 = 0
y = -9 or y = 2
xy= 18
x(2) = 18
x = 18/2 = 9
x = 2
But negative digit is not considered.
∴ The two digit number is
= 10x + y
= 10(9) + 2 = 92

OR

A plane left 30 minutes later than the scheduled time and in order to reach its destination 1500 km away in time it has to increase its speed by 250 km / hr from its usual speed. Find its usual speed.
Let the usual speed of the plane be x km / hr. Time taken to cover 1500 km with the usual speed.
$$\frac{1500}{x}$$ hrs
Time taken to cover 1500 km with the speed of (x + 250) km / hr = $$\frac{1500}{x+250}$$

2(1500 x 250) = x2 + 250x
x2 + 250x-750000 = 0
x2 + 1000x – 750x – 750000 = 0
x(x + 1000) – 750(x + 1000) = 0
(x+ 1000)(x-750) = 0
x + 1000 = 0 or x – 750 = 0
x = -1000 or x= 750 ,
x = 750
Hence, the usual speed of the plane = 750 km/hr.

Question 28.
If the co – ordinates of the mid points of ∆ ABC are D(1, 2) E(0, -1) and (2, -1). Find the respective co – ordinates of ∆ ABC.

D is the midpoint of BC.

x1 + x2 = 4 and y1 + y2 = -2 ………(1)
E is the mid point of BC

x2 + x3 = 0 and y2 + y3 = -2 …….(2)
F is the mid point of BC

x1 + x3 = 4 and y2 + y3 = -2 ….(3)
Solve (1) and (3)

x2 = 0
x2 = 0
0 + x3 = 0
x3 = 0
x1 + x3 = 4
x1 + 0 = 4
x1 = 4 – 0 = 4
x1 = 4

y2 = -i
y2 + y3 = -2
-1 + y3 = -2
y3 = -2 +1
y3 = -1
y1 + y3 = -2
y1 + (-1) = -2
y1 -1 = -2
y1 = -2 + 1 = -1
y1 = -1
A(x1,y1) = A(4,-1)
B (x2 y2) = B(0, -1)
C(x3, y3) = C(0,-1)

OR

Find the length of the median through the vertex A(5,1) drawn to the triangle ABC where other two vertices are B(1, 5) andiC(-3, -1)

Question 29.
Prove that the tangents drawn from an external point are equal.

Data : ‘O is the centre of the circle PA and PB are the two tangents drawn from an external point P. OA and OB are radii of the circle.
To prove that : ∠APB + ∠AOB = 180°
∠OAP + ∠OBP = 90°
(∵ Angle between the radius and tangent at the point of contact is 90°)
OP = OP (Common side)
OA = OB (Radii of the same circle)
According to RHS postulate
∆APO = ∆BPO
∠ OAP = ∠OBP = 90°
∠OAP + ∠OBP = 90° +90° =180°
⇒ Opposite angles of OAPB quadrilateral are supplementary.
⇒ ∠ APB + ∠ AOB = 180°

Question 30.
If a chord of circle of radius 10cm subtend an angle of 60° at the centre of the circle. Find the area of the corresponding segment of the circle.
(Take π =3.14, √3 = 1.7)
O is the centre of the circle having radius 10 cm.
∠ AOB = 60° is angle subtended at the centre O of the circle by a chord AB.
Draw BL ⊥ OA
In right angled ∆ OLB,

Area of segment ∆ APB
= The area of sector OAPB – Area of ∆ OAB

OR

Find the area of the shaded region where PQRS is a square of side lOcms and semicircles are drawn with each side of square as diameter.
The four shaded regions are marked as I, II, III, IV.
All the four regions have equal areas due to § symmetry.
Area of I + Area of III
= Area of square ABCD – Sum of the areas of two semicircles of each of radius 5cm.

{10 x 10 – 2 x $$\frac{1}{2}$$ π x 52}cm2
= {100 – 25 x 3.14} cm2
= {100 – 78.5} cm2
= 21.5 cm2
Similarly Area of II + Area of IV = 21.5 cm2. Total Area of shaded region = 21.5 + 21.5
= 43 cm2

Question 31.
The following table gives the production yield per hectare of paddy of 50 farms of a village.

Question 32.
Find the mean of the following frequency distribution.

A = 50, h = 20, N = 100 and Σ fiui = 15
Mean = A + h$$\left\{\frac{1}{\mathrm{N}} \sum \mathrm{f}_{i} \mathrm{u}_{\mathrm{i}}\right\}$$
Mean = 50 + 20 x $$\frac{15}{100}$$
Mean = 50 + 3 = 53

Question 33.
Construct a triangle of sides 4cm, 5cm and 6cm and then a triangle similar to it whose Sides are 2/3 of the corresponding sides of it.

V. Answer the following : ( 4 x 4 = 16 )

Question 34.
Solve the pair of equations graphically.
4x – 3y + 4 = 0
4x + 3y – 20 = 0
4x – 3y + 4 = 0 ….. (1)
4x + 4 = 3y
y = $$\frac{4 x+4}{3}$$

4x + 3y – 20 = 0
3y = -4x + 20
y = $$\frac{-4 x+20}{3}$$

Question 35.
How many terms of the series 93 + 90 + 87 + ……….. amounts to 975. Find also the last term.
93 + 90 + 87 + …………
a = 93, d = 90 – 93 = -3, Sn = 975, n = ?,
T = ?
S = $$\frac{n}{2}$$[2a + (n – 1)d]
975 = $$\frac{n}{2}$$[2(93) + (n-1)(-3)]
975 x 2 = n[186 – 3n + 3]
1950 = n [189 – 3n]
1950 = n[189 – 3n]
1950 = 189n – 3n2
(3n2-189n+ 1950-0) H-3
n2 – 63n + 650 = 0
n2 – 50n-13n +650 = 0
n(n – 50) – 13(n – 50) = 0
(n – 50) (n -13) = 0
n – 50 = 0 or n – 13 = 0
n= 50 or n= 13
When n = 50, the last term
Tn = a + (n – 1)d
= 93 + (49) (-3)
= 93 – 147
T50 =-54
When n = 13, the last term
Tn = a + (n – 1)d ”
=93 + (13-1) (-3)
= 93 – 36
T13 = 57

OR

If m times the mth term of an A.P is equal to n times its n,h term, show that (m + n)th term is zero.
Soln: Let T1 = a and c.d = d
Given : m times m,h term = n times nthterm
=> mTm = nTn
m[a + (m – 1)d] = n[a + (n – 1)d]
m[a +(m – l)d] – n[a+(n -l)d] = 0
m[a + md – d] – n[a + nd – d] = 0
ma + m2d – md – na – n2d + nd = 0
a(m – n) + d(m – n) (m + n) – d(m – n) = 0
(m – n) {a + d(m +n) – d} = 0 . (m – n) {a + (m + n – l)d} = 0
m-n = 0, a + (m + n-l)d = 0
m= n Tm+n , = 0
But m ≠ n

Question 36.
A tower is 50m high. Its shadow is x mtrs shorter when the suns altitude is 45° than when it is 30°. Find the value of x.

Let PQ be the tower and let PA and PB be its shadows when the altitudes of the sum are 45° and 30° respectively.
∠ PAQ = 45°, ∠ PBQ = 30°
∠ BPQ = 90° and PQ = 50m ,
Let AB = x metres
From right angled ∆ APQ, we have
$$\frac{\mathrm{AP}}{PQ}$$ = cot 45°=1
$$\frac{\mathrm{AP}}{50}$$ = 1 => AP = 50m
From right angled ∆ BPQ,
$$\frac{\mathrm{BP}}{\mathrm{PQ}}$$ = cot30° = √3
⇒ $$\frac{x+50}{50}$$ = √3
x = 50(√3 – 1)m
= 50 (1.73 – l)m
= 50 x 0.732
= 36.6 m

Question 37.
Prove that “In a right angled triangle, square on the hypotenuse is equal to sum of the squares on the other sides.”

Verify
BC = 8cm
BC’= $$\frac{5}{3} \times 8=\frac{40}{3}$$
AB =6cms
A’B= $$\frac{5}{3} \times 6$$
=10cms.

VI. Answer the following : ( 5 x 1 = 5 )

Question 38.
The height of cone is 20m. A small cone is cut off from it at its top by the plane parallel to the base. If the volume of small cone 1/1000 th of the volume of given cone, at what height above the base the section is made.
In A OPB and A OQC A OPB ~aOQC OP _ PB OQ ~ QC

## Karnataka SSLC Maths Model Question Paper 3 With Answers

Students can Download Karnataka SSLC Maths Model Question Paper 3 with Answers, Karnataka SSLC Maths Model Question Papers with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

## Karnataka State Syllabus SSLC Maths Model Question Paper 3 with Answers

Time: 3 Hours
Max Marks: 80

I. In the following questions, four choices are given for each question, choose and write the correct answer along with its alphabet: ( 1 × 8 = 8 )

Question 1.
In the following numbers, irrational number is
a) 0.232332333……
b) 0.23233
c) 0.232323
d) 0.2323
a) 0.232332333……

Question 2.
10 sec2A – 10tan2A =
a) sec2A
b) 10
c) 1
d) 0
b) 10
Solution:
10 sec2A – 10tan2A
10(sec2A – tan2A) = 10(1)= 10

Question 3.
The length of the tangent drawn to a circle of radius 3cm from 5cm away from the centre is
a) 4cm
b) 5cm
c) 3cm
d) 2cm 4
a) 4cm
Solution:
d2 = r2 + t2
t2 = d2 – r2
52 – 32
= 25 – 9
∴ t = 16 = 4cm

Question 4.
A solid piece of copper of dimension 24 × 49 × 33 cms is moulded and recast into a sphere. The radius of the
sphere formed is ________
a) 49 cm
b) 24 cm
c) 21 cm
d)33 cm
e) 21 cm
l × b × h = $$\frac{4}{3} \pi r^{3}$$

Question 5.
The degree of the polynomial in the graph given below is

a) 4
b) 3
c) 1
d) 2
a) 4
Solution:
4 since it is intersecting the x – axis at 4 points.

Question 6.
The sum of the n terms of an AP is 2n2 + 5n and its common difference is 6, then its first term is
a) 0
b) 5
c) 2
d) 7
d) 7
Solution:
Sn = 2n2 + 5n
S1 = 2(1)2 + 5(1)
= 2 + 5 = 7

Question 7.
In ∆ PQR, PR = 12cm, QR = 6√3 cm, PQ = 6cm. The angle Q is
a) 45°
b) 90°
c) 30°
d) 120°
b) 90°
Solution:
PR2 = PQ2 + QR2
122= (6√3 )2 + 62
144 = 108 – 36
144=144
∴ ∠Q = 90°

Question 8.
A cube numbered 1 to 6 is thrown once, the probability of getting a number divisible by 3 is
a) 2/3
b) 0
c) 1/3
d) 1
Event A = {3,6}
n(s) = 6, n(A) =2
P(A) = $$\frac{n(A)}{n(S)}=\frac{2}{6}=\frac{1}{3}$$

II. Answer the following Questions : (1 x 8 = 8)

Question 9.
Given g(x) = 2x + 1, q(x) = (x3 + 3x2 – + 1), r(x) = 0, Find p(x)
p(x) = g(x) q(x) + r(x) 1 650 1170
= (2x + 1) (x3 + 3x2 – x + 1) + 0
p(x) = 2x (x3 + 3x2 -x + 1) + 1(x3 + 3x2 – x + 1)
= 2x4 + 6x3 – 2x2+ 2x + x3 + 3x2 – x + 1
= 2x4 + 7x3 + x2 + x + 1

Question 10.
If the sum of first n odd natural number is 1225, find the value of n.
Sum of “n” odd natural number = n2 = 1225
n = √1225
n = 35

Question 11.
In the fig. ∠AOD is divided into 2 parts which are in A.P. the smallest angle ∠AOB = 20° . Find the common difference between each angle.
∠AOB +∠BOC + ∠COD = 180°
a + a + d + a + 2d= 180°
3a + 3d= 180°
3(20) + 3d = 180°
60 + 3d = 180°
3d = 180 – 60
d = 120/3 = 40°

Question 12.
In ∆ ABC, if DE || BC, then $$\frac{\mathbf{A B}}{\mathbf{A D}}=\frac{\mathbf{A C}}{\mathbf{A} \mathbf{E}}=\frac{\mathbf{B C}}{\mathbf{D E}}$$ , state the theorem to justify this.
In a triangle, if a line is parallel to one of the sides then the sides of given triangle are propotional to sides of the intercepted triangle.

Question 13.
Find the largest number which divides 650 and 1170.
H.C.F of (650, 1170) = 130

Question 14.
If sin θ = 7/25, cos θ = 24/25 find the value of sin2 θ + cos2 θ

Question 15.
Find the value of cos60° cos 30° sin60° sin 30°
cos 60° cos30 – sin 60 sin 30

OR

This is in the form of
cos A cos B – sin A sin B = cos (A + B)
cos 60 cos 30 – sin 60 sin 30
= cos (60 + 30)
= cos (90)
– =0

Question 16.
The T.S.A of a solid hemisphere of radius 21 mm.
T.S.A of hemisphere = 3 πr²
= 3 × $$\frac{22}{7}$$ × 21 × 21 = 4158 mm2

III. Answer the following : ( 2 x 8 = 16 )

Question 17.
Prove that if x and y are odd positive integers, then x2 + y2 is even but not divisible by 4.
Let x = 2m + 1 and y = 2n + 1 for some integers m and n.
x2 + y2 = (2m + 1)2 + (2n + l)2
x2 + y2 = 4m2 + 4m + 1 + 4n2 + 1 + 4n
= 4m2 + 4n2 + 4m + 4n + 2.
x2 + y2 = 4(m2 + n2) + 4 (m + n) + 2
x2 + y2 = 4 {(m2 + n2) + (m + n)} +2
x2 + y2 = 4q + 2, where q = (m2 + n2 ) + (m+n)
⇒ x2 + y2 is even and leaves the remainder 2 when divided by 4.
⇒ x2 + y2 is even but not divisible by 4.

Question 18.
Solve : 100x + 200y = 700
200x + 100y = 800
Consider 100x + 200y = 700 ….(1)
200x + 100y = 800 ….(2)
Add (1) & (2) 300x + 300y= 1500
Divide by 300, x + y = 5 …… (3)
Subtract (1) and (2)
-100x + 100y= -100 Divide by 100.
-x + y = -1 …(4)
Solve (3) and (4)

y = 2
Substitute the value of y in equation (3)
x + y = 5
x + 2 = 5
x = 5 – 2 = 3
x = 3

Question 19.
Find the roots of the quadratic
equation 3x2 – 2√6x + 2 = 0 by formula method.

Question 20.
Find the value of x in which the points (1, -1) (x, 1) and (4, 5) are collinear.
A(1, -1) B(x, 1)C(4,5)
Area of the triangle = 0, when points are collinear.
0 = x1(y2 – y3) + x2(y3 – y1) + x3 (y1 – y2)
0 = (1) (1 – 5) + x(5 + 1) + 4(-1 – 1)
0 = -4 + 6x – 8
6x – 12 = 0
x = 12/6
x = 2

Question 21.
ABC is a right angled triangle, having [B = 90°. If BD = DC, Show that AC2 = 4AD2 – 3AB2

AC2 = AB2 + BC2 (Pythagoras)
AC2 = AB2 + 2(BD)2
AC2 = AB2 + 4BD2
In right angled ∆ ABD,
Consider AC2 = AB2 + 4BD2
= AB2 + 4(AD2 – AB2)
= AB2 + 4AD2 – 4AB2
AC2 = 4 AD2 – 3AB2

OR

Prove that area of the equilateral triangle described on the side of a square is half the area of the equilateral triangle described on its diagonal.

Data : ABCD is a square.
Equilateral triangles ∆ BCE and ∆ ACF have been described on side BC and diagonal AC respectively.
T.P.T. : Area of ( ∆ BCE) = 1/2 Area of ∆ ACF
Proof : Since ∆ BCE and ∆ ACF are equilateral.
∴ They are equiangular.
∆ BCE ~ ∆ ACF

Question 22.
A box contains 90 dices which are numbered from 1 to 90. If one dise is drawn at random from the box, find the probability that it bears
(i) two digit number
(ii) a perfect square number.
The numbers in the dise form the sample space
S = {1,2, 3,4,…….. 90}
One dise can be drawn out of 90 in 90 ways. .
n(s) = 90

i) There are 90 -9 = 81, two digit numbers, out of which one dise can be drawn in 81 ways.
∴ n(A) = 81
P(A) = $$\frac{n(A)}{n(S)}=\frac{81}{90}=\frac{9}{10}$$

ii) The perfect square numbers are B= {1,4,9, 16,25,36,49,64,81}
∴ n(B) = 9
P(B) = $$\frac{n(B)}{n(S)}=\frac{9}{90}=\frac{1}{10}$$

Question 23.
Draw a pair of tangents to a circle of radius 5cm which are inclined to each
other at an angle of 60°.

Question 24.
Prove that $$\frac{\tan \theta+\sin \theta}{\tan \theta-\sin \theta}=\frac{\sec \theta+1}{\sec \theta-1}$$

OR

Prove that (cosecθ – cotθ)2 = $$\frac{1-\cos \theta}{1+\cos \theta}$$
Consider (cosecθ – cotθ)2

IV. Answer the following : ( 3 × 9 = 27 )

Question 25.
The sum of the ages of A and B is 85 years. 5 years ago, the age of A was twice that of B. Find the present ages.
Let the present age of A be x years.
Let the present age of B be y years.
Sum of their ages = 85 x + y = 85 …….. (1)
Five years back, age of A was (x – 5) and that of B was (y – 5).
The age A is twice that of B.
∴ x – 5 = 2(y – 5) ……(2)
x – 5 = 2y – 10
x – 2y = -10 + 5
x – 2y = -5 ……(2)
Solve (1) and (2)

Substitute y in (1)
x + y=85
x + 30 = 85
x = 85 – 30 = 55
x = 55
∴ Present age of A is 55 years and B = 30 years.

OR

A piece of work can be done by 2 men and 7 boys in 4 days. The same piece of work can be done by 4 men and 4 boys in 3 days. How long it would take to do the same work by one man or one boy?
Let x and y be the number of days in which one man can complete the work.
∴ In 1 day a man can do 1/x th work and
In 1 day a man can do 1/y th work.
2 men and 7 boys can complete the work in 4 days.
∴ They can complete in 1/4 th of the work in one day.
∴ $$\frac{2}{x}+\frac{7}{y}=\frac{1}{4}$$ …… (1)
4 men and 4 boys together complete the work in 3days, and they can complete it in 1/3 rd of work.

4(2y + 7x) = xy
8y + 28x = xy …….(3)
3(4y + 4x) = xy
12y + 12x = xy ……..(4)
Solve (3) and (4)
(8y + 2x = xy) x 12
(l2y+ 12x = xy) x 18

4 y = 240
y = 240/4
y = 60

3(4y + 4x) xy
12y+12x = xy ………..(4)
4y = 240
y = 240/4 = 60
Substitute y in equation (3)
8y + 28x = xy
8(60) + 28x = xy
480 + 28x = x(60)
480 + 28x = 60x
480 = 60x – 28x
480 = 32x
x = $$\frac{480}{32}=\frac{120}{8}=\frac{30}{2}$$ = 15
x = 15
Thus, one man can complete the work in 15 days and one boy can do the work in 60 days.

Question 26.
Find the zeroes of the polynomial p(y) = y3 – 5y2 – 2y + 24 if it is given that Sum of the two zeros is Answer:
Let α, β,γ be the zeros of the given polynomial.
p(y) = y3 – 5y2 – 2y + 24
α + β = 7
Sum of the zeros

α + β + γ = 5
7 + r = 5
r = 5 – 7 = -2
r = -2
Sum of product of zeroes taken two at a

αβ + βr + rα = -2/1 = – 2
αβ + r(β + α) = -2
αβ + (7) (-2) = -2
αβ-14 = -2
αβ = -2+ 14
αβ = 12
(α – β )2 = (α + β)2– 4αβ
= (7)2 – 4(12)
(α – β)2 = 49 – 48
α – β = ±1
Solve for α and β. When α – β = +1
α + β = 7
α – β = 1
2α =8
α = 8/2 = 4
α = 4
Substitute the value of a in α + β =7
4+p =7
β =7-4 = 3
β = 3
Solve for α and β when α – β = -1

α = 6/2 = 3
α = 3
α + β = 7
3 + β=7
β = 7 – 3
a + β = 7
β = 4
When α – β = 1,
the values are α = 4, β = 3, r = -2.
When α – β = -1,
The values are α = 3, β = 4, r = -2.

Question 27.
The diagonal of a rectangular field is 60 metres more than the shorter side, if the longer side is 30 metres more than the shorter side, find the sides of the field.

Let the shorter D side be x mtr.
Longer side is 30m more than the shorter side. A Longer side = (x + 30)m.
Diagonal is 60m more than the shorter side.
Diagonal = (x + 60) m
AC2 = AB2 + BC2 (Pythagoras)
(x + 60)2 = (x + 30)2 + x2
x2+(60)2 + 2x (60) = x2 + (30)2 + 2(x)(30)+ x2
x2 + 3600 + 120x = x2 + 900 + 60x + x2
-x2 + 60x + 2700 = 0
x2 – 60x – 2700 = 0
x2 – 90x + 30x – 2700 = 0
x(x – 90) + 30(x – 90) = 0
(x – 90) (x + 30) = 0
x – 90 = 0 or x + 30 = 0
x = 90 or x = – 30
∴ Shorter side = 90m = BC = x
Longer side = x + 30 = 90 + 30 = 120m
Diagonal = x + 60 = 90 + 60 = 150m

OR

Sum of the areas of two squares is 468m2. If the difference of their perimeters is 24m, find the sides of two squares.
Let the sides of the two squares be x&y. Sum of the areas of two squares = 468
x2 + y2 = 468 ……(1)
Difference of their perimeters = 24m.
4x – 4y = 24 x – y = 6
x = 6 + y …….(2)
Substitute the value of x in (1)
x2 + y2 = 468
(6 + y)2 + y2 = 468
36 + y2+ 12y + y2 = 468
2y2 + 12y + 36 = 468
Divide by 2
y2 +6y+ 18 = 234 y2 + 6y = 234 – 18
y2 + 6y – 216 = 0
y2+18y – 12y -216 = 0
y(y + 18)-12(y + 18) = 0
(y + 18) (y – 12) = 0
y+18 = 0, y -12 = 0
y = -18,y = 12
∴ x = 6+y = 6 + 12 = 18m
The sides of two squares are 18m and 12m.

Question 28.
Show that the points x(2, -2) y(-2, 1) and z(5, 2) are the vertices of a right angled triangle XYZ and also calculate its area.

⇒ XYZ is an isosceles right angled triangle.
Area of a right angled triangle = $$\frac{1}{2}$$ x base x height
$$\frac{1}{2}$$ × 5 × 5 = $$\frac{25}{2}$$ = 12.5 cm

OR

Find the values of k for which the points A(k + 1, 2k) B(3k, 2k+3) and C(5k – 1, 5k) are collinear.
A(k + 1, 2k) = (x1, y1)
B(3k, 2k + 3) = (x2, y2)
C(5k-1, 5k) = (x3, y3)
Area of the triangle = 0, for the points to be collinear.
Area =[x1(y2 – y3) + x2(y3-y1) + x3(y1-y2)]
0 = $$\frac{1}{2}$${(k + 1) (2k+3 – 5k)+3k(5k- 2k) + (5k-1) [2k-2k-3]}
0= $$\frac{1}{2}$$]{(k+ 1) (-3k+ 3) + 3k (3k) + (5k- I)(-3)i
0 =$$\frac{1}{2}$${-3k2-3k + 3k + 3 + 9k2 – 15k+3}
0= $$\frac{1}{2}$${6k2– 15k + 6}
0 = 6k2 – 15k + 6
Divide by 3
2k2 – 5k + 2= 0
2k2 – 4k – 1k + 2 = 0
2k(k – 2) -1(k – 2) = 0
(k – 2) (2k -1) = 0
k – 2 = 0, or 2k- 1=0
k = 2 or k = 1/2

Question 29.
Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.

Data: ‘O’ is the centre of the circle PA and PB are the two tangents drawn from an external point P. OA and OB are radii of the circle.
To prove that: ∠APB+ ∠AOB = 180°
Proof: In AAPO and ABPO
∠OAP = ∠OBP = 90°
(∵ Angle between the radius and tangent at the point of contact is 90°)
OP = OP (∵ Common side)
OA = OB (∵ Radii of the same circle) According to RHS postulate AAPO ABPO
∠OAP = ∠OBP = 90°
∠OAP + ∠OBP = 90° + 90° = 180°
⇒ Opposite angles of OAPB quadrilateral are supplementary
∴ OAPB is a cylic quadrilateral
⇒ ∠APB + ∠AOB = 180°

Question 30.
Find the area of the shaded region where a circular arc of Radius 6 cm has been drawn with the vertex ‘O’ of an equilateral Triangle OAB of side 12cm as centre.
Area of the shaded Region = Area of circle area of sector OCDE + Area of equilateral
∆ le OAB

OR

From each corner of a square of side 4cm a quadrant of a circle of radius 1cm is cut and also a circle of diameter 2cm is cut as shown in Fig, Find the area of the remaining portion of the square.

Area of shaded region = Area of square – (Area of circle + Area of 4 quadrants)
The following table gives production yield per hectare of wheat of 100 farms of a

Question 31.
The following table gives production yield per hectare of wheat of 100 farms of a
village.

Change the distribution to a more than type distribution and draw its ogive.

Question 32.
If the median of the distribution given below is 28.5. Find the valucs of x and y.

Meadian = 28.5, lies in the C.I , 20 – 30
The median class = 20 – 30
l = 20, f = 20, c.f = 5 + h, h = 10, n = 60

Question 33.
Construct a triangle ABC with side BC = 7cm, ∠B = 45°, ∠C = 105°. Then construct a triangle whose sides are 4/3 times the corresponding sides of ∆ ABC.

V. Answer the following ( 4 × 4 = 16 )

Question 34.
Solve the pair of equations graphically.
x+y=3 and 3x-2y=4
x + y = 3
y = 3- x

3x – 2y = 4
-2y = 4 – 3x .

Question 35.
If the sum of first 8 terms of an Arithmetic progression is 136 and that of first 15 terms is 465, then find the sum of first 25 terms.
Given S8 = 136, S15 = 465, S25 = ?
Use the formula Sn = $$\frac{n}{2}$$ [2a + (n -l)d]
s8 = $$\frac{8}{2}$$[2a + (s-l)d]
136 = 4 (2a + 7d)
∴ 2a + 7d = 136/4
2a + 7d= 34…….(1)
S15 = $$\frac{15}{2}$$[2a + (15-l)d]
465 = $$\frac{15}{2}$$[2a + 14d]
465= $$\frac{15}{2}$$ x 2(a + 7d)
∴ a + 7d = 465/15
a + 7d = 31 …….(2)
From (1) and (2)

Consider a + 7d = 31
3 + 7d = 31
7d = 31 – 3
d = 28/7
d = 4
∴S25 = $$\frac{25}{2}$$[2a + (25-1)d]
S25 = $$\frac{25}{2}$$[2a + (25-1)d]
S25 = $$\frac{25}{2}$$[2a + (24)d]
S25 = $$\frac{25}{2}$$[2(3) + 24(4)d]
S25 = $$\frac{25}{2}$$ × 102
S25 = 1275
‘5 ‘ 9

OR

The sum of the 5th and 9th terms of an A.P is 40 and the sum of the 8th and 14th term is 64. Find the sum of the first 20 terms.
Given T5+ T9 = 40 and T8 + T14 = 64
a + 4d + a + 8d=40
2a + 12d = 40….(1)
a + 7d + a + 13d = 64
2a + 20d = 64…. (2)

Consider 2a + 12d = 40
2a + 12(3) = 40
2a + 36 = 40
2a = 40 – 36
a = 4/2
a = 2
Sn = $$\frac{n}{2}$$[2a + (n-1)d]
S20 = $$\frac{20}{2}$$[2(2) + (20-1)3]
= 104(4 + 57)
= 10 × 61
= 610

Question 36.
Two poles of equal heights are standing opposite to each other on either side of the road, which is 80m wide. From a point between them on to the road, the angles of elevation of the top of the poles re 60 and 30’, respectively. Find the height of the poles and distances of the point from the poles.

In ∆ ABE.
tan60 = $$\frac{\mathrm{AB}}{\mathrm{BE}}$$
√3 = x/BE
x = √3BE ….. (1)
tan30 = DC/EC
$$\frac{1}{\sqrt{3}}=\frac{x}{E C}$$
EC = x√3
EC = BE × √3 .√3 [∵ x = √3.BE]
FC = 3BE
∴ EC=3BF
We know that
BE + EC = 80 rn
BE 3BE 80m
4BE = 80
BE = 80/4 = 20
⇒ EC = 3BE 3(20)= 60m
∴ x = BE.
x = 20√3
⇒ AB = CD = 20√3 m
∴ The distance from the point to the pole arc 20m towards left and 60m towards the right
poles
Height of the poles 20√3m

Question 37.
Prove areas of similar triangles.
Areas of similar triangles are proportional
to the squares on the corresponding sides.

In ∆ABM and ∆DEN
∠AMB = ∠DNE = 90 [construction]
∠B =∠E (Data)
∠BAM = ∠EDN (Remainingang1e)
∆ABM ∼ ∆ DEN
∆ABM & ∆DEN are equiangular

VI. Answer the following : ( 5 × 1 = 5 )

Question 38.
A circus tent is made of canvas and is in the form of a right circular cylinder and a right circular cone above it. The diameter and height of the cylindrical part of the tent are 126m and 5m respectively. The total height of the tent is 21m. Find the total cost of the canvas used to make the tent when the cost per m2 of the canvas is ?15.

Total canvas used=CS A of cylindrical part -1- CS A of conical part
=2πrh + πrl
= 2 × $$\frac{22}{7}$$ × 63 × 5 + $$\frac{22}{7}$$ x 63 x 65
= 1980 m2 + 12870 m2
= 14850 m2
Total canvas used = 14850 m2
* Cost of canvas at the rate of ₹16 per m2
= 14850 × 15= ₹ 2,22.750.
Cost of canvas = ₹ 2,22,750

## Karnataka SSLC Maths Model Question Paper 2 with Answers

Students can Download Karnataka SSLC Maths Model Question Paper 2 with Answers, Karnataka SSLC Maths Model Question Papers with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

## Karnataka State Syllabus SSLC Maths Model Question Paper 2 with Answers

Time: 3 Hours
Max Marks: 80

I. In the following questions, four choices are given for each question, choose and write the correct answer along with its alphabet: ( 1 × 8 = 8 )

Question 1.
For some integer n every odd integer is of the form
a) 2n + 1
b) n + 1
c) 2n
d) n
2n + 1

Question 2.
The value of sin215° + sin225° + sin265° + sin275° is
a) 0
b) 1
c) 2
d) 3
2

Question 3.
If chord AB subtends an angle 50° at the centre of a circle then the angle between the tangents at A and B is
a) 40°
b) 100°
c) 130°
d) 120°
130°

Question 4.
The formula used to find the volume of a sphere
a) $$\frac{4}{3} \pi r^{3}$$
b) $$\frac{2}{3} \pi r^{3}$$
c) $$\frac{1}{3} \pi r^{3}$$
d) πr3
a) $$\frac{4}{3} \pi r^{3}$$

Question 5.
α + β are the zeroes of the polynomial x2 – 6x + 4, then the value of $$\frac{(\alpha+\beta)^{2}-2 \alpha \beta}{\alpha \beta}$$ is
a) 7
b) 8
c) -7
d) -8
a) 7
Solution:

Question 6.
If 29th term of an A.P is twice its 19th term, then the 9th term is
a) -1
b) 0
c) 1
d) 2
b) 0
Solution:
T29 = 2T19
a + 28d= 2(a + 18d)
a + 28d = 2a + 36d
0 = 2a + 36d-a-28d
0 = a + 8d
T9 = 0

Question 7.
In ∆ ABC, AB = 6√3 cm, AC = 12 cm, BC = 6cm, The angle B is
a) 45°
b) 90°
c) 60°
d) 30°
Solution:
AC2 = AB2 + BC2
122= (6√3 )2 + 62
144= 108 + 36 144=144
∴ ∠B = 90°

Question 8.
If the probability of an event is P(A) then the probability of its complimentary event will be
a) 1+P(A)
b) 1 – P(A)
c) P(A) – 1
d) 1 / P(A)
Solution:
We know that for any two complimentary events A and Ā
P(A) + P(Ā) = 1 → P(Ā) = 1 – P(A)

II. Answer the following questions : ( 1 × 8 = 8 )

Question 9.
If α and β are the zeroes of the quadratic polynomial 2 – 3x – x2 then what is the value of α + β + αβ ?
α + β = -b/a , αβ = c/a
= $$\frac{-(-3)}{-1}$$ αβ = $$\frac{2}{-1}$$ = – 2
= – 3
α + β + αβ = -3 -2 = -5

Question 10.
What are the roots of the quadratic equation x2 + (√3 + 1)x + √3 = 0?
2 + (√3 + 1)x + √3 = 0 = 0
x2 + x + √3x + √3 = 0
x(x + 1)+ √3 (x + 1) = 0
(x + 1)(x + √3) = 0
x + 1 = 0, x + √3 = 0
x = -1, x = -√3

Question 11.
If the nth terms of the two AP 9, 7, 5, ….. and 24, 21, 18, ……… are same. Find n.
9, 7, 5, …….
a = 9, d = 7-9 = -2 T = a + (n – 1)d
Tn = 9 + (n – 1) (-2)
Tn = 9 + (-2n + 2) ‘
Tn = 9 – 2n + 2
Tn = 11 – 2n
24,21,18,
a = 24, d = 21 – 24 = -3
Tn = a + (n – 1) d T=24 + (n – 1)(-3)
Tn = 24 – 3n + 3
Tn = 27 – 3n
Given : nth term of both A. P. is same.
∴ 11 – 2n = 27 – 3n
-2n +3n = 27 -11
n = 16

Question 12.
State A – A criterian theorem.
“If two triangles are equiangular, then the corresponding sides are proportional.”

Question 13.
Find the H.C.F of 455 and 42 with the help of Euclid’s division algorithm.
Step (1): a = 455 b = 42
a = bq + r
455 = 42 × 1 + 35

Step (2): a = 42, b = 35
a = bq + r
42 = 35 × 1 + 7

Step (3): a = 35, b = 7
a = bq + r
35 = 7 × 5 + 0
∴ H.C.F. (455, 42) = 7

Question 14.
Find θ if sin (θ + 56) = cosθ, where θ and ( θ + 56) are less than 90°.
sin (θ + 56) = cos θ
sin (θ + 56) = sin (90 – θ)
θ + 56 = 90 – θ
θ + θ = 90 – 56
2θ = 34
θ = 34/2
θ = 17

Question 15.
If x = a sin θ and y = b tan θ, then find the value of $$\frac{a^{2}}{x^{2}}-\frac{b^{2}}{y^{2}}$$
Consider

Question 16.
Calculate the height of a right circular cone where C.S.A. and base radius are 12320 cm2 and 56 cms. respectively.

C.S.A of a cone = π rl
12320 = $$\frac{22}{7}$$ x 56 x 1
1 = $$\frac{12320 \times 7}{22 \times 56}$$
1 = 70 cm
But 12 = r2 + h2
702 = 562 + h2
h22 = 702 – 562
h2 = (70 + 56) (70 – 56)
h2 =(126) (14)

III. Answer the following ( 2 × 8 = 16 )

Question 17.
Show that any positive odd integer is of the form 6q + 1 or 6q + 3 or 6q + 5 where q is an integer.
Let nbe any positive, integer using division algorithm,
a = bq + r
Taking a = n, b = 6
n = 6q + r
If any number is divided by 6, then the possible remainders are 0, 1, 2, 3, 4 or 5.
∴If n is odd, then r = 1, 3, 5
⇒ 6q + 1, 6q + 3, 6q + 5 are the positive odd integers.

Question 18.
Solve : 2x + 3y = 9
3x + 4y = 5

Substitute y = 17 in 2x + 3y = 9
2x + 3(17) = 9
2x + 51 =9
2x = 9 – 51
x = $$\frac{-42}{2}$$
= -21

Question 19.
Solve : (x – 2)2 + 1 = 2x – 3
x2 + 4 – 4x + 1 = 2x – 3
x2 – 4x + 5 – 2x + 3 = 0
x2 – 6x + 8 = 0 x2 – 4x – 2x + 8 = 0
x(x – 4) -2(x – 4) = 0
(x – 4) (x – 2) = 0
x – 4 = 0, x -2 = 0
x = 4 or x = 2

Question 20.
Show that the points (-2, 1) (2, -2) and (5, 2) are the vertices of a right angled triangle.

∴ ∆ ABC is a right angled triangle at B.

Question 21.
The equilateral triangles are drawn on the sides of a right triangle. Show that the area of the triangle on the hypotenuse is equal to the sum of the areas of the triangles on the other two sides.

∴ Area of ∆ XAB + Area of ∆ YBC = Area of ∆ ZAC .

OR

In the given figure, PA, QB and RC are each perpendicular to AC. Prove $$\frac{1}{x}+\frac{1}{z}=\frac{1}{y}$$

Question 22.
Two dice are thrown simultaneously. Find the probability that the sum of the numbers on the faces is neither divisible by 4 nor by 5.
S = {(1, 1) (1,2) (1, 3) (1,4) (1, 5) (1, 6) (2, 1) (2, 2) (2,3) (2,4) (2, 5) (2,6) (3.1) (3, 2) (3, 3) (3, 4) (3, 5) (3,6) (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4,6) (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5,6) (6.1) (6,2) (6, 3) (6,4) (6, 5) (6,6)}
n(S) = 36.
An event of getting sum of the numbers neither divisible by 4 nor by 5.
A={ 1,1) (1,2) (1,5) (1,6) (2,1) (2, 4)
(2, 5) (3, 3) (3, 4) (3, 6) (4, 2) (4,3) (4, 5) (5,1) (5, 2) (5, 4) (5, 6) (6,1) (6, 3) (6, 5)} n(A) = 20
n(A) = 20
P(A) = $$\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{20}{36}$$

Question 23.
Draw a circle of radius 3 cm. Take a point P outside the circle without using the centre of the circle, draw to tangents to the circle from an external point P.

Question 24.
Prove that (cosecθ-cotθ)2 = $$\frac{1-\cos \theta}{1+\cos \theta}$$
Consider LHS
(cosecθ-cotθ)2 = cosec2 θ +cot2 θ – 2cosecθ.cotθ

OR

If sinθ + cos θ = √2 sin (90 -θ) determine cot θ
sin θ + cos θ = √2 sin (90 – θ)
sin θ + cos θ = √2 cos θ
sin θ = ypi cos θ – cos θ
sin θ = cos θ (√2 -1)
sinθ /cosθ = √2 -1

IV. Answer the following : ( 3 × 9 = 27 )

Question 25.
Asha is 5 times as old as her daughter Usha, 5 years later Asha will be 3 times as old as her daughter Usha. Find the present ages of Asha and Usha.
Let the age of Asha be x years
Let the age of Usha be y years
According to question x=5y
x – 5y = 0 …….(1)
Five years later,
(x + 5) = 3(y + 5)
x + 5 = 3y + 15
x – 3y = 15 – 5
x – 3y = 15 – 5
x – 3y = 10 (2)
Substitute x = 5y in equation (2)
5y – 3y = 10
2y = 10
y = 10/2
y = 5
∴ x = 5y
= 5 × 5
x = 25
∴ The present age of Asha is 25 years.
The present age of Usha is 5 years.

OR

The sum of 2 digits of a 2 digit number is 12 the number obtained by interchanging the digits exceeds by the given number by 18. Find the number.
Let the two digits be x and y
The 2 digit number will be 10x + y
The sum of digits = x + y – 12 ……..(i)
The number obtained by interchanging the digits =10y + x
= 10x + y + 18
10y + x – 10x – y = 18
9y – 9x = 18
Divide by 9
y – x = 2
y = x + 2 ….(ii)
Consider equation (1)
x + y = 12
Substitute y = x + 2
x + x + 2 = 12 .
2x =12 – 2
2x= 10
x= 5
y = x + 2
y = 5 +2
y = 7
∴ The number is : 10x +y
= 10(5)+7 = 50+ 7
= 57

Question 26.
Find the other two zeroes of the polynomial y4 + y3 – 9y2 – 3y + 18 if the
zeroes are √3 and -√3
y = √3 and y = -√3
(y + √3) = o (y-V3) = o
(y + √3 )(y – √3 ) = o
y(y – √3 ) + √3 (y – √3 ) = 0
y2 – √3y + √3 y – 3 = o
y2 – 3 = 0

y2 + y – 6 = 0
y2 + 3y – 2y – 6 = 0
y(y + 3)-2(y + 3) = 0
(y + 3) = 0 & (y – 2) = 0
y=-3 & y = 2
The four zeros of polynomial are √3 , -√3 , -3 & 2

Question 27.
Solve for x.
$$\frac{1}{a+b+x}=\frac{1}{a}+\frac{1}{b}+\frac{1}{x}$$ (Where a ≠ 0, b ≠ 0 x ≠ 0, x ≠ -a,-b

(-a – b)ab = (a + b)x (a + b + x)
-a2b – ab2 = (a + b) (ax + bx + x2)
-a2b – ab2 = a2x + abx + ax2 + abx + b2x + x2b
-a2b – ab2= a2x + 2abx + ax2 + b2x + x2b
-ab(a+b) = ax(a+x) + xb(b+x)+2abx
=x[a(a+x)+b(b+x)+2ab]
=x[a2 + ax+b2 +bx + 2ab]
= x[(a+b)2 + x(a+b)]
-ab(a + b) = x(a+b)[a+b+x]
-ab = x(a + b + x)
-ab = ax + bx + x2
x2 + x (a + b) + ab = 0
x2 + ax + bx + ab = 0
x(x+a) + b(x + a) = 0
(x + a) (x +b) = 0
=>x + a = 0, x + b = 0
=> x = -a, x = -b

OR

The diagonal of a rectangular field is 60m more than the shorter side. If the larger side is 30m more than the shorter side, find the sides of the field.
Let the shorter side of the rectangular field be x m.
Longer side is x + 30 and the diagonal is x+60

In ∆ ABC,
AC2 = AB2 + BC2
(x + 60)2 = ( x + 30)2+ x2
x2 + 3600 + 120x = x2 + 900 + 60x + x2
x2 + 900 + 60x – 3600 – 120x = 0
x2 – 60x – 2700 = 0
x2 – 90x + 30x – 2700 = 0
x(x – 90) + 30(x – 90) = 0
(x – 90) (x + 30) = 0
x – 90 = 0, x + 30 = 0
x = 90, x = – 30.
∴ The shorter side = x = 90m
The longer side = x + 30 = 90+30 = 120m.

Question 28.
If the points (7, -2) (5, 1) and (3, 5) are collinear. Find the value of k.
Since the give points are collinear, the area of the triangle formed by them must be O i.e.
= $$\frac { 1 }{ 2 }$$ [x1(y2-y3) + x2(y3-y2) + x3(y1-y2)]
x1 = 7, x2 = 5, x3 = 3
y1 = – 2, y2 = 1, y3 = k
= $$\frac { 1 }{ 2 }$$ [7(1 – k) + 5 (k + 2) + 3(-2-1)] = 0
= $$\frac { 1 }{ 2 }$$ [7 – 7k + 5k + 10 – 9]-0
= $$\frac { 1 }{ 2 }$$ [7 – 2k + 1] = 0
= $$\frac { 1 }{ 2 }$$ [8 – 2k] = 0
8-2k = 0 or 2k=8
k = 8/2 = 4
k = 4

OR

Find the area of Rhombus if its vertices are (3, 0) (4, 5) (-1, 4) and (-2, -1) taken in order.

Area of rhombus = $$\frac { 1 }{ 2 }$$ d1 × d2
Let A = (3, 0) B = (4, 5)
C = (-1, 4) D = (-2, -1)
Diagonal

Question 29.
Prove the tangents drawn from an external point to a circle are equal.

Data: A is the centre of the circle B is an external point.
BP and BQ are the tangents.
AP, AQ and AB are joined.
To prove that: a) BP = BQ
b) ∠PAB-∠QAB.
c) ∠PBA = ∠QBA
Proof: In ∆ APB and ∆ AQB
AP = AQ [Radii of same circle]
∠ APB – ∠AQB = 90°
AB = AB [common side]
∴  ∆ APB ≅ ∆ AQB [RHS]
a) PB = QB
b) ∠PAB = ∠QAB
c) ∠PBA = ∠QBA

Question 30.
Find the area of the shaded region in the figure, where ABCD is a square of side 14 cm

Area of square ABCD
= 14 × 14=196 cm2

Hence, area of the shaded region
= Area of square – Area of four circles
= 196 – 154
= 42cm2

OR

Find the area of the shaded regions. Given PQRS a square of sides 14cm. Soln: S R

Question 31.
The following table gives the weight of 120 articles :

Change the distribution to a “more than type” distribution and draw its ogive.

Question 32.
The distribution below gives the weights of 30 students of a class. Find the median weight of the students.
Weight (in kg)

Question 33.
Construct a right triangle whose hypotenuse and one side measures 10cm and 8cm respectively. Then construct another triangle whose sides are 4/5 times the corresponding sides of the triangle .

V. Answer the following ( 4 × 4 = 16 )

Question 34.
Solve the pair of equations graphically.
x + y = 8 and x – y = -2
Answe:
x + y = 8
y = 8 – x

x – y = -2
x = -2 + y

x = 6/2 = 3
x + y = 8
3 + y = 8
y = 8 – 3
y = 5

Question 35.
Devide 20 into four parts which are in arithmetic progression and such that the product of first and fourth is to the product of second and third in the ratio 2:3.
Let the four parts in A.P be :
a – 3d, a – d, a +d, a + 3d
Their sum = (a – 3d) + (a – d) + (a + d) + (a + 3d) = 20
⇒ 4a = 20 or a = 5
Also, it is given that

or 75 -27d2 = 50 – 2d2
25d2 = 25 .
⇒ d2 = 1 or d = +√1 = ±1
When a = 5, d = 1, the four parts are 2, 4, 6,8
When a = 5, d = -1; the four parts are 8, 6, 4,2
Hence the four parts are (2, 4, 6, 8) or (8, . 6,4,2)

OR

The angles of a quadrilateral are in AP such that the greatest is double the least calculate all the angles of the quadrilateral.
The angles of the quadrilateral are in AP.
∴ The four angles are A, B, C and D which are a -3d, a – d, a + d, a + 3d respectively.
⇒ The least angle is a – 3d and the greatest angle is a + 3d.
It is given that
a + 3d = 2(a – 3d)
a + 3d = 2a – 6d
2a – a – 6d – 3d = 0
a – 9d ……(1)
9d = a
But ∠A + ∠B+∠C + ∠D = 360°
a – 3d + a – d + a + d + a +.3d = 360°
4a = 360 / 4 = 90°
But 9d = a
d = $$\frac{a}{9}=\frac{90}{9}$$ = 10°
∴ ∠A=a- 3d = 90 – 3(10) = 90 -30 = 60°
∠B = a – d = 90 – 10 = 80°
∠C = a + d = 90 + 10= 100
∠D = a + 3d = 9 + 3(10) = 90 + 30 = 120

Question 36.
A person on the lighhouse of height 100m above the sea level observes that the angle of depression of a ship sailing towards the light house changes from 30° to 45°. Calculate the distance travelled by the ship during the period of observation. (Take √3 ≈ 1.73)

P is the top of the light house PQ. We are given that its height PQ = 100m PX is horizontal line through P
∠APX = 30° and ∠BPX =45°
Then, ∠PAQ =30° and ∠PBQ = 45°
Suppose, AB = x metre and BQ = y metre
Karnataka SSLC Maths Model Question Paper 2 with Answers – 40
⇒ x + y =100√3 = 100√3 = 100× 1.732
⇒ x +y = 173.2
⇒ x + 100 = 173.2
⇒ x = 73.2
Therefore, the distance travelled = 73.2m

Question 37.
Prove that “the ratio of areas of two similar triangles is equal to the square of the ratio of their altitudes.

Data : ∆ABC ∼ ∆DEF
∠A = ∠D
∠B = ∠E
∠C = ∠F
$$\frac{A B}{D E}=\frac{A C}{D F}=\frac{B C}{E F}$$
T.P.T : $$\frac{\text { Area of } \Delta \mathrm{ABC}}{\text { Area of } \Delta \mathrm{DEF}}=\frac{\mathrm{AB}^{2}}{\mathrm{DE}^{2}}=\frac{\mathrm{AC}^{2}}{\mathrm{DF}^{2}}=\frac{\mathrm{BC}^{2}}{\mathrm{EF}^{2}}$$
Construction:Draw AM ⊥ BC and AN ⊥ EFim

VI. Answer the fallowing : ( 5 × 1 = 5 )

Question 38.
The radii of the circular ends of the frustrum of height – 6cm are 14 cm and 6cm respectively. Find the lateral surface area and total surface area of frustrum.
R = 14cm, r = 6cm and h =6cm
Now, let 1 be the slant height of the frustrum;

Now, lateral surface area of the frustrum
= π(R + r)l
= π (14 + 6) 10
= 628.57 cm2
And total surface area of the frustrum = π[(R+r)l + R2+ r2]
= π [(14 + 6)10 + (14)2 + (6)2]
= π(200 + 196 + 36)
= π(432)
= 1357.71 cm2.

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