## KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2

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## Karnataka State Syllabus Class 7 Maths Chapter 8 Comparing Quantities Ex 8.2

Question 1.
Convert the given fractional numbers to percents.
a) $$\frac{1}{8}$$
Solution:
Percent means per hundred. So multi-plied by 100.

b) $$\frac{5}{4}$$
Solution:

c) $$\frac{3}{40}$$
Solution:

d) $$\frac{2}{7}$$
Solution:

Question 2.
Convert the given decimal fractions to per cents.
a) 0.65
Solution:

b) 2.1
Solution:

c) 0.02
Solution:

d) 12.35
Solution:

Question 3.
Estimate what part of the figures is coloured and hence find the per cent which is coloured.

Solution:

∴ In the first figure 25% is coloured.

∴ in the second figur e 50% Is coloured,.

∴ In the third figure 37.5 % is coloured.

Question 4.
Find :
a) 15% of 250
Solution:

b) 1% of 1 hour
Solution:

c) 20% of ₹ 2500
Solution:

d) 75% of ₹ 1 kg
Solution:

Question 5.
Find the whole quantity if
a) 5% of it is 600.
Solution:
Let the whole quantity be ‘m’
∴ 5% of m = 600

∴ The whole quantity is 12,000.

b) 12% of its 1080.
Solution:
Let the whole quantity be ‘m’
∴ 12% of m = Rs. 1080

∴ The whole amount be Rs. 9.000

c) 40 % of it is 500 km
Solution:
Let the whole quantity be ‘m’
∴ 40% of m = 500 km

∴ The whole amount is 1250 kms

d) 70% of it is 14 minutes
Let the whole quantity be ‘m’

∴ The whole amount is 20 minutes

e) 8% of it is 40 litres.
Let the whole quantity be ‘m’

∴ The whole amount be 500 liters

Question 6.
Convert given per cents to decimal fractions and also to fractions in simplest forms:
Solution:

Question 7.
In a city, 30% are females, 40% are males and the remaining are children. What percent are children?
Solution:
females = 30%
males = 40%
children = 100 – (30 + 40)
remaining are children = 100 – 70 = 30%
∴ Percentage of children = 30%

Question 8.
Out of 15,000 votes in a constituency, 60% voted. Find the percentage of voters who did not vote. Can you now find how many actually did not vote?
Solution:
Total no. of voters in a constituency } = 15,000
Percentage of voted = 60%
∴ Percentage of voters who did not vote
= 100 – 60 = 40%
Actual number of voters who did not vote} = 40% of 15,000

6000 voters who did not vote.

Question 9.
Meeta saves ₹ 400 from her salary. If this is 10% of her salary. What is her salary?
Solution:
Let Meeta’s salary be ‘M’
10% of M is = Rs. 400
∴ M =?

∴ Her salary be Rs. 4,000/-

Question 10.
A local cricket team played 20 matches in one season. It won 25% of them. How many matches did they win?
Let the total matches won by them be M 25% of 20 = M

∴ They win 5 matches.

## KSEEB Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.4

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## Karnataka State Syllabus Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.4

Question 1.
Observe the patterns of digits made from line segments of equal length. You will find such segmented digits on the display of electronic watches or calculators.
If the number of digits framed is taken to be n, the number of segments required to form n digits is given by the algebraic expression appearing on the right of each pattern.
How many segments are required to form 5, 10, 100 digits of the kind 6, 4, 8.

Solution:
i) The number of segments required to form
5 digits of this kind
5n + 1 = 5 × 5 + 1 = 25 + 1 = 26

ii) The number of segments required to form 10 digits of this kind
5n + 1 = 5 × 10 + 1 = 50 + 1 = 51

iii) The number of segments required to form 100 digits of this kind
5n + 1 = 5 × 100 + 1 = 501

Let the number of digits formed be ‘n’.
Therefore the number of segments required to form ‘n’ digits is given by the algebraic expression 3n + 1.
i) The number of segments required to form 5 digits of this kind.
3n + 1 = 3 × 5 + 1 = 15 + 1 = 16

ii) The number of segments required to form 10 digits of this kind
3n + 1 = 3 × 10 + 1 = 30 + 1 = 31

iii) The number of segments required to form 100 digits of this kind
3n + 1 = 3 × 100 + 1 = 300 + 1 = 301

Let the number of digits formed be ‘n’. Therefore the number of segments required to form ‘n’ digits is given by the algebraic expression 5n + 2.
i) The number of segments required to form 5 digits of this kind.
5n + 2 = 5 × 5 + 2 = 25 + 2 = 27
ii) The number of segments required to form 10 digits of this kind
5n + 2 = 5 × 10 + 2 = 50 + 2 = 52

iii) The number of segments required to form 100 digits of this kind
5n + 2 = 5 × 100 + 2 = 500 + 2 = 502

Question 2.
Use the given algebraic expression to complete the table of number patterns.
Solution:

## KSEEB Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.3

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## Karnataka State Syllabus Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.3

Question 1.
If m = 2, find the value of:
i) m – 2
2 – 2 = 0

ii) 3m – 5
3 × 2 – 5 = 6 – 5 = 1

iii) 9 – 5m
9 – 5 × 2 ÷ 9 – 10 = -1

iv) 3m2 – 2m – 7
= 3 × 22 – 2 × 2 – 7
= 3 × 4 – 4 – 7
= 12 – 11 = 1

v)

Solution:

Question 2.
If p = -2, find the value of :
Solution:
i) 4p + 7
= 4x – 2 + 7
= -8 + 7 = -1

ii) -3p2 + 4p + 7
= -3 (-2)2 + 4 (-2) + 7
= 3 × 4 – 8 + 7
= 12 – 8 + 7
= -13

iii) -2p3 – 3p2 – 4p + 7
= -2 (-2)3 – 3(2)2 + 4 (-2) + 7
= -2(-8) – 3(4) – 8 + 7
= 16 – 12 – 8 + 7
= 16 + 7 – 12 – 8
= 23 – 20 = 3

Question 3.
Find the value of the following expressions, when x = -1:

i) 2x – 7
= 2(-1) – 7
= -2 – 7 = -9

ii) -x + 2
= -1 + 2 = 1

iii) x2 + 2x + 1
= 1 – 2 + 1 = 0

iv) 2x2 – x – 2
= 2(-1)2 – (-1) – 2
= 2 × 1 + 1 – 2
= 2 + 1 – 2 = 1

Question 4.
If a = 2, b = -2, find the value of:
i) a2 + b2
= 22 + (-2)2
= 4 + 4 = 8

ii) a2 + ab + b2
= 22 + 2 × -2 + (-2)
= 4 – 4 + 4 = 4

iii) a2 – b2
= (2)2 – (-2)2 = 4 – 4 = 0

Question 5.
When a = 0, b = -1, find the value of the given expressions :
i) 2a + 2b
= 2 × 0 + 2 × -1
= o – 2 = -2

ii) 2a2 + 2b2 + 1
= 2(0)2 + (-1)2 + 1
= 0 + 1 + 1 = 2

iii) 2a2b + 2ab2 + ab
= 2 × 02 × -1 + 2 × 0 × (-1)2 + 0 × -1
= 0 + 0 + 0 = 0

iv) a2 + ab + 2
= (0)2 + 0 × -1 + 2
= 0 + 0 + 2 = 2

Question 6.
Simplify the expressions and find the value if x is equal to 2
Solution:
i) x + 7 + 4 (x – 5)
= x + 7 + 4x – 20 (re-arranging the terms)
= 5x – 13 = 5(2) – 12
= 10 – 13 = -3

ii) 3(x + 2) + 5x – 7
= 3x + 6 + 5x – 1
= 8x – 1 = 8(2) – 1
= 16 – 1 = 15

iii) 6x + 5 + 5(x – 2)
= 6x+ 5x – 10
= 11x – 10 = 11(2) – 10
= 22 – 10 = 12

iv) 4(2x – 1) + 3x + 11
= 8x – 4 + 3x + 11
= 8x + 3x – 4 + 11 (re-arranging the terms)
= 11x + 7 = 11(2) + 7 = 22 + 7 = 29

Question 7.
Simplify these expressions and find their values if x = 3, a = -1, b = -2.
i) 3x – 5 – x + 9
= 3x – x + 9 – 5
= 2x + 4 = 2(3) + 4 = 6 + 4 = 10

ii) 2 – 8x + 4x + 4
= 2 + 4 – 4x
= -4x + 6
= -4 × 3 + 6
= -12 + 6 = -6

iii) 3a + 5 – 8a + 1
= 3a – 8a + 5 + 1
= -5a + 6
= -5(-1) + 6
= 5 + 6 = 11

iv) 10 – 3b – 4 – 5b
= -3b – 5b + 10 – 4
= -8b + 6
= -8 (-2) + 6
= 16 + 16 = 22

v) 2a – 2b – 4 – 5 + a
= 2a – 2b – 9
= 3a – 2b – 9
= 3(-1) – 2 (-2) – 9
= -3 + 4 – 9
= -12 + 4 = -8

Question 8.
i) If z = 10, find the value of z3 – 3(z – 10)
= (10)3 – 3 (10 – 10)
= 1000 – 3(0)
= 1000 – 0 = 1000

ii) If p = -10, find the value of p2 – 2p – 100
= (10)2 – 2(-10) – 100
= 100 + 20 – 100
= 120 – 100
= 20

Question 9.
What should be the value of a if the value of 2x2 + x – a equals to 5, when x = 0 ?
Solution:
= 2x2 + x – a
= 2(0)2 + 0 – a
= 5
= 0 + 0 – a = 5
∴ a = -5

Question 10.
Simplify the expression and find its value when a = 5 and b = -3, 2(a2 + ab) + 3 – ab
= 2a2 + 2ab + 3 – ab
= 2(5)2 + 2 × 5 × -3 + 3 – 5 × -3
= 2 × 25 – 30 + 3 + 15
= 50 – 30 + 3 + 15
= 50 + 3 + 15 – 30
= 68 – 30 = 38

## KSEEB Solutions for Class 7 Maths Chapter 8 Comparing Quantities Ex 8.1

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## Karnataka State Syllabus Class 7 Maths Chapter 8 Comparing Quantities Ex 8.1

Question 1.
Find the ratio of :
a) 5 to 5o paise
Solution:
Rs. 5 : 50 paise
5 × 100 : 50 paise
500 paise : 50 paise (Dividing by 50)
10 : 1

b) 15 kg to 210 g
15 kg: 210 g
15 × 1000: 210 g
15000 g : 210 g (Dividing by 30) 500 : 7

c) 9 m to 27 cm
9 m : 27 cm
9 × 100 : 27 cm
900 cm : 27 cm (Dividing by 9)
100 : 3

d) 30 days to 36 hours
30 days : 36 hours
30 × 24 : 36 hours
720 hours : 36 hours (Dividing by 36)
20 : 1

Question 2.
In a computer lab, there are 3 computers for every 6 students. How many computers will be needed for 24 students?
Solution:
Let ‘x’ computers to be needed for 24 students. Then the ratio will be 3 : x = 6 : 24

∴ 24 students need 12 computers.

Question 3.
Population of Rajasthan = 570 lakhs and population of UP = 1660 lakhs. Area of Rajasthan = 3 lakh km2 and area of UP= 2 lakh km2.
Solution:
Population of Rajasthan = 570 lakhs
Population of UP = 1660 lakhs
Area of Rajasthan = 3 lakh sq. kms
Area of U.P = 2 lakh sq. kms

i) How many people are there per km2 in both these States ?
Solution:

ii) Which State is less populated?
Solution:
Rajasthan State is less populated.

## KSEEB Solutions for Class 7 Maths Chapter 7 Congruence of Triangles Ex 7.2

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## Karnataka State Syllabus Class 7 Maths Chapter 7 Congruence of Triangles Ex 7.2

Question 1.
Which congruence criterion do you use in the following ?
a) Given : AC = D
AB = DE
BC = EF

So, ∆ ABC = ∆ DEF
SSS congruence criterion

b) Given : ZX = RP
RQ = ZY
∠PRQ = ∠XZY
So, ∆ PQR ≅ ∆ XYZ

SAS congruence criterion

c) Given: ∠MLN = ∠FGH
∠NML = ∠GFH
ML = FG
So, ∆LMN ≅ ∆GFH

ASA congruence criterion.

d) Given : EB = DB
AE = BC
∠A = ∠C = 90°
So, ∆ ABC ≅ ∆ CDB

RHS congruence criterion

Question 2.
You want to show that ∆ ART ≅ ∆ PEN,

a) If you have to use SSS criterion, then you need to show
i) AR =
AR = PE,
ii) RT =
RT = EN,
iii) AT =
AT = PN

b) If it is given that ∠T = ∠N and you are to use SAS criterion, you need to have

i) RT =
RT = EN and
ii) PN =
PN = AT
c) If it is given that AT = PN and you to use ASA criterion, you need to have
i) ?
∠ATR = ∠PNE and
ii) ?
∠TAR = NPE.

Question 3.
You have to show that ∆ AMP ≅ ∆ AMQ.
In the following proof, supply the missing reasons

Question 4.
In ∆ ABC, ∠A = 30°, ∠B = 40° and ∠C = 110°
In ∆ PQR,

A student says that ∆ ABC ≅ ∆ PQR by AAA congruence criterion. Is he justified? Why or why not?
Solution:
No, he is not justified because AAA is not a criterion for the congruence of triangles.

Question 5.
In the figure, the two triangles are congruent. The corresponding parts are marked. We can write ∆ RAT ≅ ?
Solution:

Question 6.
Complete the congruence statement:
Solution:

∆ BCA ≅ ? ∆ QRS ≅ ?
∆ BCA ≅ ? ∆ BTA ≅ ?
∆ QRS ≅ ? ∆ TPQ

Question 7.
In a squared sheet, draw two triangles of equal areas such that

i) The triangles are congruent.
Solution:
Consider the ∆S PQS and SQR

In ∆ PQS and ∆ SQR
PS = QR = 6 cms
∠SPQ = ∠QRS = 90°
QS = QS = common
By RHS congruence criterion
∆PQS ≅ ∆SQR.
Perimeter of the ∆ PQS = PQ + QS + PS
Perimeter of the ∆ SQR = SR + QS + QR
∴ Perimeter of the ∆ PQS = Perimeter of the ∆ SQR (∵ PQ = SR & PS = QR)

ii) the triangles are not congruent. What can you say about their perimeters?
Solution:

∴ area of ∆ PQS = Area of ∆ PQM.
By seeing the figure the ∆ PQS = PQ + PS + SQ = 8 + 6 + 10 = 24 cms.
Perimeter of the ∆ PQM = PQ + PM + QM
= 8 + 7.2 + 7.2
= 22.4 cms.
∴ Their perimeters are not equal.
PM = QM
(PM)2 = PN2 + MN2
= 42 + 62
= 16 + 36 = 52
PM = QM = $$\sqrt{52}$$ = 7.2

Question 8.
Draw a rough sketch of two triangles such that they have five pairs of congruent parts but still the triangles are not congruent.
Solution:

Considering the two triangles PQR and XYZ
In ∆ PQR and XYZ
PQ = XZ = 7 cms
PR = YZ = 6 cms
RQ = XY = 5 cms
∠PRQ = ∠XYZ
∠PQR = ∠XZY
In the above 2 triangles 5 pairs of the congruent present. Still, they are not congruent.

Question 9.
If ∆ ABC and ∆ PQR are to be congruent, name one additional pair of corresponding parts. What criterion did you use?

The additional corresponding part is BC = RQ by ASA congruence rules

Question 10.
Explain, why ∆ ABC ≅ ∆ FED

Solution:
∠ABC = ∠DEF = 90°
BC = DE ∠ACD = ∠EDF
( ∵ The sum of the measures of three angles of a triangle is 180°.)
∴ ∆ ABC ≅ ∆ DEF
(By ASA congruence criterion.)

## KSEEB Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.2

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## Karnataka State Syllabus Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.2

Question 1.
Simplify combining like terms :
i) 21b – 32 + 7b – 20b
(re-arranging the terms)
= 21b + 7b – 20b – 32
= 28b – 20b – 32
= 8b – 32

ii) z2 + 13z2 – 5z + 7Z2 – 15z
(re-arranging the terms)
= -z2 + 13z2 – 5z – 15z + 7z3
= 12z2 – 20z + 7z3

iii) p – (p – q) – q – (q – p)
= p – p + q – q – q + p
(re-arranging the terms)

= p – q

iv) 3a – 2b – ab – (a – b + ab) + 3ab + b – a
= 3a – 2b – ab – a + b – ab + 3ab + b – a
= 3a – a – a – 2b + b + b – ab – ab + 3ab

= a + ab

v) 5x2y – 5x2 + 3yx2 – 3y2 + x2 – y2 + 8xy2 – 3y2
(re-arranging the terms)
= 5x2y + 3yx2 – 5x2 + x2 – 3y2 – y2 – 3y2 + y2
= 8x2y – 4x2 – 7y2 + 8xy2

vi) (3y2 + 5y – 4) – (8y – y2 – 4) .
3y2 + 5y – 4 – 8y – y2 – 4
(re-arranging the terms)
3y2 + y2 + 5y – 8y – 4 + 4
= 14y2 – 3y

Question 2.
i) 3mn, -5mn, 8mn, -4mn
3mn + 8nm – 5mn – 4mn
(re-arranging the terms)
= 11mn – 9mn = 2mn

ii) t – 8tz, 3tz – z, z – t

(re-arranging the terms)
= -8tz + 3tz = -5tz

iii) -7mn + 5, 12mn + 2, 9mn – 8, – 2mn – 3
-7mn + 12 mn + 9mn – 2mn + 5 + 2 – 8 – 3
(re-arranging the terms)
= -7mn – 2mn + 12mn + 9mn + 7 – 11
= -9mn + 21 mn -4
= 12mn – 4

iv) a + b -3, b – a +3, a – b + 3

(re-arranging the terms)
= (2a – a + 2b – b + 6 – 3)
= a + b + 3

v) 14x + 10y – 12xy – 13, 18 – 7x – 10y + 8xy, 4xy
(re-arranging the terms)

= 17x + 51

vi) 5m – 7n, 3n – 4m + 2, 2m – 3mn – 5
(re-arranging the terms)
= 5m – 4m + 2m – 7n + 3n – 3mn + 2 – 5
= 7m – 4m – 7n + 3n – 3mn – 3
= 13m – 4n – 3mn – 3

vii) 4x2y, – 3xy2, 5y2, 5x2y
(Re-arranging the terms)
4x2y + 5x2y – 3xy2 – 5xy2
= 9x2y – 8xy2

viii) 3p2q2 – 4pq + 5, 10p2q2, 15 + 9pq + 7p2q2
3p2q2 – 10p2q2 + 7p2q2 – 4pq + 9pq + 5+ 15
(Re-arranging the terms)
= 3p2q2 + 7p2q2 – 10p2q2 – 4pq + 9pq + 5 + 15

= 5pq + 20

ix) ab – 4a, 4b – ab, 4a – 4b

(Re-arranging the terms)
(All terms are cancelling)
= 0

x) x2 – y2 – 1, y2 – 1 – x2, 1 – x2 – y2
(Re arranging the terms)

= -x2 – y2 – 1

Question 3.
Subtract :
i) -5y2 from y2
= y2 – (-5y2)
= y2 + 5y2 = 6y2

ii) 6xy from – 12xy
= -12xy – (6xy)
= -12xy – 6xy = -18xy

iii) (a – b) from (a + b)
= a + b – (a – b)

= 2b

iv) a(b – 5) from b(5 – a)
= ab – 5a from 5b – ab
= (5b – ab) – (ab – 5a)
= 5b – ab – ab + 5a
= 5a – 5b – 2ab

v) -m2 + 5mn from 4m2 – 3mn + 8
= (4m2 – 3mn + 8) – (-m2 + 5mn)
= 4m2 – 3mn + 8 + m2 – 5mn
= 4m2 + m2 – 3mn – 5mn + 8
= 5m2 – 8mn + 8

vi) -x2 + 10x – 5 from 5x – 10
= (5x – 10) – (-x2 + 10x – 5)
= 5x – 10 + x2 – 10x + 5
= 5x – 10x + x2 – 10 + 5
= x2 – 5x – 5

vii) 5a2 – 7 ab + 5b2 from 3ab – 2a2 – 2b2
= (3ab – 2a2 – 2b2) – (5a2 – 7ab + 5b2)
= 3ab – 2a2 – 2b2 – 5a2 + 7ab – 5b
= -2a2 – 5a2 – 2b2 – 5b2 + 7ab + 3ab
= -7a2 – 7b2 + 10ab

viii) 4pq – 5q2 – 3p2 from 5p2 + 3q2 – pq
= (5p2 + 3q2 – pq) – (4pq – 5q2 – 3p2)
= 5p2 + 3q2 – pq2 – 4pq + 5q2 + 3p2
= 5p2 + 3p2 + 3q2 + 5q2 – 4pq – pq
= 8p2 + 8q2 – 5pq

Question 4.
a) What should be added to x2 + xy + y2 to obtain 2x2 + 3xy ?
Solution:
= 2x2 + 3xy – (x2 + xy + y2)
= 2x2 + 3xy – x2 – xy – y2 – 2x2 – x2 + 3xy – xy – y2
(re-arranging the terms)
∴ x2 + 2xy – y2 should be added to x2 + xy + y2 to get 2x2 + 3xy

b) What should be subtracted from 2a + 8b + 10 to get -3a + 7b + 16 ?
Solution:
= (2a + 8b + 10) – (-3a + 7b + 16)
= 2a + 8b+ 10 + 3a – 7b – 16
= 2a + 3a + 8b – 7b + 10 – 16
(re-arranging the terms)
= 5a + b – 6
∴ (5a + b – 6) should be subtracted from 2a + 8b +10 to get -3a + 7b + 16.

Question 5.
What should be taken away from – 4y2 + 5xy + 20 to obtain -x2 – y2 + 6xy + 20?
Solution:
= (3x2 – 4y2 + 5xy + 20) – (-x2 – y2 + 6xy + 20)
= 3x2 – 4y2 + 5xy + 20 + x2 + y2 – 6xy – 20
= 3x2 + x2 – 4y2 + y2 + 5xy – 6xy + 20 – 20
(re – arranging the terms)
= 4x2 – 3y – xy
∴ 4x2 + 3y2 – xy should be taken away from 3x2 – 4y2 + 5xy + 20 to obtain -x2 – y2 + 6xy + 20

Question 6.
From the sum of 3x – y + 11 and -y – 11, subtract 3x – y – 11.
Solution:

b) From the sum of 4 + 3x and 5 – 4x + 2x2, subtract the sum of 3x2 – 5x and -x2 + 2x + 5.
Solution:

or
[(4 + 3x) + (5 – 4x + 2x2)] – [(3x2 – 5x) + (x2 + 2x + 5)]
= 4 + 3x + 5 – 4x + 2x2 – 3x2 + 5x + x2 – 2x – 5 = 4 + 3x – 4x + 2x2 + 5 – 3x2 + x2 + 5x – 2x – 5
(re-arranging the terms)

= 2x + 4

## KSEEB Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.5

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## Karnataka State Syllabus Class 7 Maths Chapter 10 Practical Geometry Ex 10.5

Question 1.
Construct the right angled ∆ PQR? where m∠Q = 90°, QR = 8 cm and PR = 10 cm.
Solution:

Solution:
Steps of Construction

1. Draw a line segment of length 8 cm and named as QR. At Q draw QM ⊥ QR.
2. With R as centre, draw an arc of radius 10 cm and cut the ⊥le line at P.
3. Join PR. Now we get the required PQR ∆

Question 2.
Construct a right-angled triangle whose hypotenuse is 6 cm long and one of the legs is 4
cm long.
Solution:

Steps of Construction

1. Draw a line segment of length 4 cm and named it AB. At A draw a ⊥ line AM.
2. With B as centre, draw an arc of 6 cms cut the x line at ‘C ’. Now we obtained the required triangle ABC.

Question 3.
Construct an isosceles right-angled triangle ABC, where m∠ACB = 90° and AC = 6 cm.
Solution:

Steps of Construction

1. Draw a line segment of 6 cm, ie., AC. At C draw CM ⊥ CA.
2. With ‘C’ as the centre draw an arc of radius 6 cm to intersect CM at ‘B’
3. Join AB. Now we get the required ∆ACB.

## KSEEB Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.4

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## Karnataka State Syllabus Class 7 Maths Chapter 10 Practical Geometry Ex 10.4

Question 1.
Construct ∆ ABC, given m∠A= 30° and m∠B = 30° and AB = 5.8 cm.
Solution:

Steps of Construction.

1. Draw a line segment AB of length 5.8 cm, at A draw a ray AM making an angle of 60° with AB.
2. At ‘B’ draw a ray ABN making an angle of 30° with BA.
3. Mark the point of intersection of two rays as ‘C’. Now we get the A ABC which is required.

Question 2.
Construct ∆ PQR if PQ = 5 cm, m∠PQR = 105° and m∠QRP = 40°. (Hint Recall angle-sum property of a triangle).
Solution:

By angle-sum property of a triangle ∠RPQ + ∠PQR + ∠QRP = 180°
∠RPQ + 105° + 40° = 180°
∠RPQ + 145° = 180°
∴ ∠RPQ = 180° – 145° = 35°,

Steps of Construction

1. Draw a line segment PQ of length 5 cm. At P draw a ray PM making an angle of 35° with PQ.
2. At Q draw a ray QN making an angle of 105° with QP.
3. Mark the point of intersection of two rays as R. We get the required P Q R triangle.

Question 3.
Examine whether you can construct ∆ DEF such that E F = 7.2 cm. m∠E = 110° and m∠F = 80°. Justify your answer.
Solution:
∠E + ∠F = 110°+ 80°= 190°
This triangle is not possible to construct because the measures of two angles exceed 180°.
The sum of three angles of a triangle is always equal to 180°.
So it is not possible.

## KSEEB Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.1

Students can Download Chapter 12 Algebraic Expressions Ex 12.1, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

## Karnataka State Syllabus Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.1

Question 1.
Get the algebraic expressions in the following cases using variables, constants and arithmetic operations.
i) Subtraction of z from y.
Solution:
y – z

ii) One-half of the sum of numbers x and
Solution:

iii) The number z multiplied by itself.
Solution:
Z × Z = Z2

iv) One- fourth of the product of numbers p and q.
Solution:

v) Numbers x and y both squared and added.
Solution:
x2 + y2

vi) Number 5 added to three times the product of numbers m and n.
Solution:
3mn + 5

vii) Product of numbers y and z subtracted from 10.
Solution:
10 – yz

viii) Sum of numbers a and b subtracted from their product.
Solution:
ab, – {a + b)

Question 2.
i) Identify the terms and their factors in the following expression show the terms and factors by tree diagrams.
Solution:

c) y – y3
Terms: y; -y3

Factors : y ; -1, y,

d) 5xy2 + 7x2y
Terms : 5xy2 ; 7x2y
Factors : 5, x, y, y; 7, x, x, y

e) -ab + 2b2 – 3a2
Terms : -ab ; 2b2; 3a2
Factors : 1, a, b ; 2, b, b ; -3, a, a

ii) Identify terms and factors in the expressions given below :

a) -4x + 5
Terms : -4x ; 5
Factors : -4, x; 5

b) -4x + 5y
Terms : -4x ; 5y
Factors :-4, x; 5, y

c) 5y + 3y2
Terms : 5y ; 3y2
Factors : 5, y ; 3, y, y

d) xy + 2x2y2
Terms : xy ; 2x2y2
Factors : x, y ; 2, x, x, y, y

e) pq + q
Terms : pq ; q
Factors : p, q ; q

f) 1.2 ab – 2.4b + 3.6a
Terms : 1.2 ab ; 2.4b ; 3.6a
Factors : 1.2, a, b ; -2.4, b ; 3.6, a

g)

Solution:

h) 0.1 p2 + 0.2 q2
Terms : 0.1p2; 0.2q2
Factors : 0.1, p, p ; 0.2, q, q

Question 3.
Identify the numerical coefficients of terms (other than constants) in the following expressions :
i) 5 – 3t2
Term other than constant = -3/2
numerical co-efficient = -3

ii) 1 + t + t2 + t3
Term other than constant = t, t2, t3
Numerical co-efficients = 1, 1, 1.

iii) x + 2xy + 3y
Terms other than constants = x, 2xy + 3y
Numerical co-efficients = 1, 2, 3

iv) 100m + 1000n
Terms other than constants = 100m, 1000n
Numerical co-efficients = 100, 1000.

v) -p2q2 + 7pq
Terms other than constants = -p2 q2, 7 pq
Numerical co-efficients = -1, 7

vi) 1.2a + 0.8b
Terms other than constants = 1.2a, 0.8b
Numerical co-efficients = 1.2, 0.8

vii) 3.14 r2
Terms other than constants = 3.14r2
Numerical co-efficients = 3.14

viii) 2 (l + b)
Terms other than constants = 2l, 2b
(∵ 2(l + b) = 2l + 2b)
Numerical co-efficients = 2, 2

ix) 0.1 y + 0.01 y2
Terms other than constants = 0.1 y, 0.01 y2
Numerical co-efficients = 0.1, 0.01

Question 4.
a) Identify terms which contain x and give the coefficient of x.
i) y2x + y
Terms which contains x — y2
co-efficient of x — y2

ii) 13y2 – 8yx
Terms which contains x — (x – 8yx)
co-efficient of x — (-8y)

iii) x + y + 2
Terms which contains x —- x
co-efficient of x —- 1

iv) 5 + z + zx
Terms which contains x — x
co-efficient of x — z

v) 1 + x + xy
Terms which contains x — x and xy
co-efficient of x — 1 y

vi) 12xy2 + 25 — 12xy2
Terms which contains x
co-efficient of x — 12y2

vii) 7x + xy2
7x Terms which contains x
co-efficient of x —- 7
Terms which contains x — xy2
y2 co-efficient of

b) Identify terms which contain y2 and give the coefficient of y’.
i) 8 – xy2
Terms which contains y2
xy2 co-efficient of y2 — x

ii) 5y2 + 7x
Terms which contains y2
co-efficient of y2 — 5

iii) 2x2 y – 15xy2 + 7y2
Terms which contains y2 — (-15xy2)
co-efficient of y2 — (-15x)
Terms which contain y2 — 7y2
co-efficient of y2 —- 7

Question 5.
Classify into monomials, binomials and trinomials.
i) 4y – 7z
This expression is binomial because it contains two terms. That is 4 y and -7z.
ii) y2
This is monomial expression because it contains only one term y2.
iii) x + y – xy
This is trinomial expression because it contains three terms x, y and -xy.
iv) 100
This is a monomial expression because it contains only one term 100.
v) ab – a – b
This is trinomial expression because it contains 3 terms ab, -a and -b.
vi) 5 – 3t
This is binomial expression because it contains 2 terms 5 and -3t.
vii) 4p2q – 4pq2
This is binomial expression because it contains two terms 4p2q and – 4 pq2
viii) 7mn
This is monomial expression because it contains only one term 7mn.
ix) z2 – 3z + 8
This is trinomial expression because it contains three terms z2, -3z and 8.
x) a2 + b2
This is binomial expression because it contains two terms a2 and b2.
xi) z2 + z
This is binomial expression because it contains 2 terms z2 and z.
xii) 1 + x + x2
This is trinomial expression because it contains 3 terms 1, x and x2.

Question 6.
State whether a given pair of terms is of like or unlike terms.
i) 1, 100
like terms
ii)
like terms
iii) -29x, -29y
unlike terms
iv) 14xy, 42yx
like terms
v) 4m2p, 4mp2
unlike terms
vi) 12xz, 12x2z2 unlike terms

Question 7.
Identify like terms in the following :
a) -xy2, -4yx2, 8x2, 2xy2, 7y, – 11x2, 100x, – 11yx, 20x2y, -6x2, y, 2xy, 3x
Solutions:
Like terms are
i) -xy2, 2xy2
ii) -4yx2, 20x2y
iii) 8x2, -11x2, -6x2
iv) 7y, y
v) -100x, 3x
vi) -11yx, 2xy

b) 10pq, 7p, 8q, -p2q2, -7qp, – 100q, -23, 12q2p2, -5p2, 41, 2405p, 75qp, 13p2q, qp2, 701p2
Solution:
like terms are
i) 10pq, – 7qp, 78qp,
ii) 7p, 2405p,
iii) 8q, -100q
iv) -23, 41,
v) 12q2p2, -p2q2,
vi) -5p2, 701p2,
vii) 13p2q, qp2

## KSEEB Solutions for Class 7 Maths Chapter 10 Practical Geometry Ex 10.3

Students can Download Chapter 10 Practical Geometry Ex 10.3, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

## Karnataka State Syllabus Class 7 Maths Chapter 10 Practical Geometry Ex 10.3

Question 1.
Construct ∆ DEF such that DE = 5 cm, DF = 3 cm and ∠EDF = 90°.
Solution:

Steps of Construction

1. Draw a line segment DE of length 5 cm. At D draw DM making 90° with DE.
2. With ‘D’ as centre, draw an arc of radius 3 cm, It cuts DM at the point F.
3. Join FE. The required triangle is obtained. That is ∆ DEF.

Question 2.
Construct an Isosceles triangle in which the lengths of each of its equal sides is 6.5 cm and the angle between them is 110°.

Solution:
Steps of Construction

1. Draw a line segment AB of length 6.5 cm. At A draw AM making 110° with AB using a protractor.
2. With ‘A’ as centre draw an arc of radius 6.5 cm and it cuts AM at C.
3. Join BC. The required triangle is ∆ ABC.

Question 3.
Construct ∆ ABC with BC = 7.5 cm. AC = 5 cm and ∠C = 60°.
Solution:

Steps of Construction

1. Draw a line segment BC of length 7.5 cm. At C draw CM making 60° with BC.
2. With ‘C’ as centre, draw an arc of radius 5cm, it cuts at CM at A.
3. Join AB. ∆ ABC is the required triangle.