KSEEB Solutions

KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4 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 3.4.

Karnataka SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.4

Question 1.
Solve the following pair of linear equations by the elimination method and the substitution method :
(i) x + y = 5 and 2x – 3y = 4
(ii) 3x + 4y =10 and 2x – 2y = 2
(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7
(iv) \(\frac{x}{2}+\frac{2 y}{3}=-1 \quad \text { and } \quad x-\frac{y}{3}=3\)
Solution:
(i) x + y = 5 and 2x – 3y = 4
x + y = 5 ………….. (i)
2x – 3y = 4 ………… (ii)
Multiplying eqn. (i) by 3
3x + 3y = 15 ………….. (iii)
By adding eqn. (ii) to eqn. (iii) ‘y’ is eliminated.

∴ \( x=\frac{19}{5}\)
Substituting the value of ‘x’ in eqn. (i),
x + y = 5
\(\frac{19}{5}+y=5\)
\(y=\frac{5}{1}-\frac{19}{5}\)
\(=\frac{25-19}{5}\)
\( \quad y=\frac{6}{5}\)
\( \quad x=\frac{19}{5} y=\frac{6}{5}\)

(ii) 3x + 4y = 10 and 2x – 2y = 2
3x + 4y = 10 ……….. (i)
2x – 2y = 2 …………. (ii)
Multiplying eqn. (ii) by 2,
4x – 4y = 4 ………….. (iii)
Adding eqn. (i) to eqn. (ii),

∴ \(\quad x=\frac{14}{7}=2\)
Substituting the value of ‘x’ in eqn. (i),
3x + 4y = 10
3 × 2 + 4y = 10
6 + 4y = 10
4y = 10 – 6
4y = 4
∴ \(\mathrm{y}=\frac{4}{4}=1\)
∴ x = 2, y = 1.

(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7
3x – 5y – 4 = 0 ⇒ 3x – 5y = 4 ……… (i)
9x = 2y + 7 ⇒ 9x – 2y = 7 ………. (ii)
Multiplying eqn. (i) by 3,
9x – 15y = 12 …………. (iii)
Subtracting eqn. (iii) from eqn. (ii)

∴ \(\quad y=\frac{-5}{13}\)
Substituting the value of ‘y’ in eqn. (i),
3x – 5y = 4
\(3 x-5\left(\frac{-5}{13}\right)=4\)
\(3 x+\frac{25}{13}=4 \quad \text { OR } \quad 3 x=4-\frac{25}{13}\)
\(3 x=\frac{52-25}{13}=\frac{27}{13}\)
\(3 x=\frac{27}{13}\)
∴ \(\quad x=\frac{27}{13} \times \frac{1}{3}\)
∴ \(\quad x=\frac{9}{13}\)
∴ \(\quad x=\frac{9}{13} \quad y=\frac{-5}{13}\)

(iv) \(\frac{x}{2}+\frac{2 y}{3}=-1 \text { and } x-\frac{y}{3}=3\)
\(\frac{x}{2}+\frac{2 y}{3}=-1 \quad \frac{x}{1}-\frac{y}{3}=3\)
\(\frac{3 x+4 y}{6}=-1 \quad \frac{3 x-y}{3}=3\)
3x + 4y = -6 ……….. (i) 3x – y = 9 …………… (ii)
Subtracting eqn. (ii) from eqn. (i),

\(y=\frac{-15}{3}=-3\)
y = -3
Substituting the value of ‘y’ in eqn. (i),
3x + 4y = -6
3x + 4 (-3) = -6
3x – 12 = -6
3x = -6 + 12
3x = 6
∴ \(\quad x=\frac{6}{3} \quad=2\)
∴ x = 2, y = 3

Question 2.
Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method:

(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes \(\frac{1}{2}\) if we only add 1 to the denominator. What is the fraction ?
Solution:
Let the fraction be \(\frac{x}{y}\), Here numerator is x and denominator is y.
∴ \(\quad \frac{\mathrm{x}+1}{\mathrm{y}-1}=1\)
x + 1 = y – 1
x – y = -1 – 1
x – y = -2 ……….. (i)
\(\frac{x}{y+1}=\frac{1}{2}\)
2 × x = y + 1
2x = y + 1
2x – y = 1 ………. (ii)
Subtracting eqn. (ii) from eqn. (i),

∴ x = 3
Substituting the value of ‘x’ in eqn. (i),
x – y = -2
3 – y = -2
-y = -2 -3
-y = -5
∴ y = 5
∴ Fraction is \(\frac{x}{y}=\frac{3}{5}\)

(ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu ?
Solution:
Present age of Nuri is ‘x’
Present age of Sonu is ‘y’
Five years ago, age of Nuri is x – 5
Five years ago, age of Sonu is y – 5
then, x – 5 = 3(y – 5)
x – 5 = 3y – 15
x – 3y = -15 + 5
x – 3y = -10 ……….. (i)
After 10 years, age of Nuri will be x + 10
After 10 years, age of Sonu will be y + 10
then, x +10 = 2(y + 10)
x + 10 = 2y + 20
x – 2y = 20 – 10
x – 2y = 10 …………. (ii)
Subtracting eqn. (ii) in eqn. (i),

∴ y = 20
Substituting the value of ‘y’ in eqn. (ii),
x – 2y = 10
x – 2(20) = 10
x – 40 = 10
x = 10 + 40
x = 50
∴ Present age of Nuri, x = 50
Present age of Sonu, y = 20.

(iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.
Solution:
Let the two digit number is 10x + y.
Sum of the digits of to digit number is
x + y = 9 ………….. (i)
Number obtained by reversing the order of the digits, we get 10y + x.
Twicing the 10y + x, it is nine times of first number.
∴ 2(10y + x) = 9(10x + 4)
20y + 2x = 90x + 9y
2x – 90x + 20y – 9y = 0
-88x + 11y = 0
88x – 11y = 0
8x – y = 0 ………….. (ii)
By adding eqn. (i) to eqn. (ii),

∴ \(x=\frac{9}{9}=1\)
Substituting the value of ‘x’ in eqn. (i),
x + y = 9
1 + y = 9
y = 9 – 1
y = 8
∴ x = 1, y = 8
∴ Two digit number = 10x + y
= 10 × 1 + 8
= 10 + 8
=18.

(iv) Meena went to a bank to withdraw Rs. 2000. She asked the cashier to give her Rs. 50 and Rs. 100 notes only. Meena got 25 notes in all. Find how many notes of Rs. 50 and Rs. 100 she received.
Solution:
Let the Rs. 50 notes be ‘x’
Rs. 100 notes b ‘y’
x + y = 25 …………. (i)
x + 2y = 40 …………. (ii)
Subtracting eqn. (ii) from eqn. (i),

∴ y = 15
Substituting the value of ‘y’ in eqn. (i)
x + y = 25
x+ 15 = 25
x = 25 – 15
∴ x = 10
∴ x = 10, y = 15
∴ Rs. 50’s notes are 10 and
Rs. 100’s notes are 15.

(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs. 27 for a book kept for seven days, while Susy paid Rs. 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.
Solution:
Fixed charge be Rs. x.
Additional charge be Rs. y.
x + 4y = 27 ………. (i)
x + 2y = 21 ………. (ii)
Subtracting Eqn. (ii) from eqn. (i),

∴ \(\quad y=\frac{6}{2} \quad=3\)
Substituting the value of ‘y’ in eqn. (i)
x + 4y = 27
x + 4(3) = 27
x + 12 = 27
∴ x = 27 – 12
∴ x = 15
∴ Fixed charge, x = Rs. 15.
Additional charge, y = Rs. 3.

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4 will help you. If you have any query regarding Karnataka Board SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.4, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Maths Solutions Chapter 2 Triangles Ex 2.3 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 2 Triangles Exercise 2.3.

Karnataka SSLC Class 10 Maths Solutions Chapter 2 Triangles Exercise 2.3

Question 1.
State which pairs of triangles In the following figures are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form:

Solution:
(i)

In ∆ABC and ∆PQR,
∠A = ∠P = 60°
∠B=∠Q = 80°
∠C =∠R = 40°
∴ SimilarIty criterion of two triangles is A.A.A.
∴ ∆ABC ~ ∆PQR.

(ii)

In ∆ABC and PQR, we have
\(\frac{A B}{Q R}=\frac{B C}{P R}=\frac{A C}{P Q}=\frac{1}{2}\)
i.e., sides of one triangle are proportional to the sides of the other triangle.
∴SImilarity criterion is S.S.S.
∴ ∆ABC ~ ∆PQR.

(iii)

∆LMP and ∆DEF are not similar triangles because one pair of sides are not proportional to each other.
\(\quad \frac{\mathrm{MP}}{\mathrm{DE}}=\frac{\mathrm{LP}}{\mathrm{DF}}=\frac{2}{4}=\frac{3}{6}=\frac{1}{2}\)
But, \(\frac{\mathrm{LM}}{\mathrm{EF}}=\frac{2.7}{5} \neq \frac{1}{2}\)
∴ These are not similar triangles

(iv)

In ∆MNL and ∆PQR, we have
∠M = ∠Q = 70°
But, \(\frac{\mathrm{MN}}{\mathrm{PQ}} \neq \frac{\mathrm{ML}}{\mathrm{QR}}\)
∴ These are not similar triangles

(v)

In ∆ABC and ∆DEF, we have
\(\frac{A B}{D F}=\frac{B C}{E F}=\frac{1}{2}\)
∠A = ∠F = 80°
∴ Similarly criterion for these triangles is S.A.S
∴ ∆ABC ~ ∆DEF

(vi)

Sum of three angles of is 180°, but in ∆DEF. sum of two angles. ∠F= 30°. In ∆PQR, ∠P = 70°.
In ∆DEF and ∆PQR. we have
∠D = ∠P = 70°
∠E =∠R = 80°
∠F = ∠R = 30°
∴ Three angles of these triangles are equal to each other.
∴ SimIlarity criterion here is A.A.A.
∴ ∆DEF ~ ∆PQR.

Question 2.
In the following figure. ∆OBA ~ ∆ODC, ∠BOC = 125° and ∠CDO = 70°. Find ∠DOC, ∠DCO and ∠OAB.

Solution:
∆OBA ~ ∆ODC (data)
In ∆OBA and ∆ODC, we have
∠ODC = ∠OBA = 70° (. Alternate angle)
∠AOB = 180°- 125° = 55° ( Adjacent angle)
∴ ∠AOB = ∠DOC = 55° (: VertIcally opposite angles)
∠OAB = ∠OCD = 55° (. Alternate angles)
∴ ∠DOC=55°
∠DCO = 55%
∠OAB= 55%.

Question 3.
Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that \(\frac{\mathrm{OA}}{\mathrm{OC}}=\frac{\mathrm{OB}}{\mathrm{OD}}\)

Solution:
Data: Diagonals AC and BD of a trapezium ABCD with AB || DC. intersect each other at the point O.
To Prove: \(\frac{\mathrm{OA}}{\mathrm{OC}}=\frac{\mathrm{OB}}{\mathrm{OD}}\)
In the trapezium ABCD AB || DC.
∴ ∠OCD = ∠OAB (Alternate angles)
∠ODC = ∠OBA (Alternate angles)
∠DOC = ∠AOB (Vertically opposite angles)
∴ These are equlangular triangles.
∴ These triangles are similar.
∴ ∆ODC ~ ∆OAB
\(\quad \frac{\mathrm{OA}}{\mathrm{OC}}=\frac{\mathrm{OB}}{\mathrm{OD}}\)
∵ Corresponding sides are in proportional.

Question 4.
In the following figure. \(\frac{\mathrm{QR}}{\mathrm{QS}}=\frac{\mathrm{QT}}{\mathrm{PR}}\) and ∠1 = ∠2. Show that ∆PQS ~ ∆TQR.

Solution:
Data: Here \(\frac{\mathrm{QR}}{\mathrm{QS}}=\frac{\mathrm{QT}}{\mathrm{PR}}\)and ∠1 = ∠2
To prove: ∆PQS ~ ∆TQR.
In ∆PQS and ∆TQR, we have
\(\frac{\mathrm{Q} \mathrm{R}}{\mathrm{QS}}=\frac{\mathrm{QT}}{\mathrm{PR}}\)
∠PQR = ∠PRQ (∵ ∠1 = ∠2)
Here, Similarity criterion used here is side, angle, side (SAS).
∴ ∆PQS ~ ∆TQR.

Question 5.
S and T are points on sides PR and QR of ∆PQR such that ∠P = ∠RTS. Show that ∆RPQ ~ ∆RTS.

Solution:
Data: S and T are points on sides PR and QR of ∆PQR such that ∠P = ∠RTS.
To Prove: ∆RPQ ~ ∆RTS.
In ∆RPQ and ∆RTS,
∠P = ∠RTS (Data)
∠PRQ = ∠SRT (Common)
∴ 3rd angle ∠PRQ = ∠SRT
∴ These are equiangularangular triangles.
∴ Here A.A.A. similarity criterion.
∴ ∆RPQ ~ ∆RTS.

Question 6.
In the given fIgure. ∆ABE ≅ ∆ACD. show that ∆ADE ~ ∆ABC.

Solution:
Data: ∆ABE ≅ ∆ACD.
To Prove: ∆ADE ~ ∆ABC.
∆ABE ≅ ∆ACD
AB = AC
AD = AE.
Then DC = BE
AB = AC
AD + DB = AE + EC
∴ DB = EC
(∵ DA = AE)
In ∆ADE and ∆ABC,
\(\frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{AE}}{\mathrm{AC}}=\frac{\mathrm{DE}}{\mathrm{BC}}\)
Here, corresponding sides are in proportion.
∴ Similarity criterion for ∆ is side, side, side
∴ ∆ADE ~ ∆ABC

Question 7.
In following figure. altitudes AD and CE of ∆ABC Intersect each other at the point P. Show that:

i) ∆AEP ~ ∆CDP
ii) ∆ABD ~ ∆CBE
iii) ∆AEP ~ ∆AÐB
iv) ∆PDC ~ ∆BEC.
Solution:
Data: altitudes AD and CE of ABC
Intersect each other at the point P.
To Prove: i) ∆AEP ~ ∆CDP
ii) ∆ABD ~ ∆CBE
iii) ∆AEP ~ ∆ADB
iv) ∆PDC ~ ∆BEC

i) In ∆AEP and ∆CDP.
∠AEP = ∠CDP = 90° (data)
∠APE = ∠CPD (Vertically opposite angle)
∴ ∠PAE = ∠PCD
These are equiangular triangles.
Similarity criterion for ∆ is A.A.A
∴ ∆AEP ~ ∆CDP

(ii) In ∆ABD and ∆CBE.
∠ADB = ∠CEB = 90° (data)
∠ABD = ∠CBE (common)
∴ ∠DAB = ∠BCE
These are equiangular triangles.
∴ Similarity criterion for is A.A.A.
∴ ∆ABD ~ ∆CBE

(iii) In ∆AEP and ∆ADB.
∠AEB = ∠ADB = 90° (data)
∠PAE = ∠DAB (common)
∴ ∠APE = ∠ABD.
∴ These are equiangular triangles.
∴ Similarity criterion for is A.A.A.
∴ ∆AEP ~ ∆ADB

(iv) In ∆PDC and ∆BEC.
∠PDC = ∠BEC = 90° (data)
∠PCD = ∠BCE (common)
∴ ∠CPD = ∠CBE
∴ These are equiangularangular triangles.
Similarity criterion for ∆ is A.A.A.
∴ ∆PDC ~ ∆BEC.

Question 8.
E is a point on the side AD produced of a parallelogram ABCD and BE Intersects CD at F. Show that ∆ABE ~ ∆CFB.

Solution:
Data: E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F.
To Prove: ∆ABE ~ ∆CFB
In \(\) ABCD Adjacent angles are equal.
Let ∠DAB = ∠BCD = 70°
∠DAB = ∠EAF = 70° (∵ Corresponding angle)
In ∆EDF, ∠DEF = 30° then,
∠EFD = 80°.
∠EFD = ∠BFC = 80° (vertIcally opposite angles)
In ∆FBC, ∠FBC = 30°. ,
Now in ∆ABE and ∆CFB,
∆EAB = ∆BCF = 70°
∆AEB = ∆FBC = 30°
∆ABE = ∆BFC = 80°
∴ Similarity criterion for ∆ is A.A.A.
∴ ∆ABE ~ ∆CFB

Question 9.
In the given figure. ∆ABC and ∆AMP are two right triangles, right angled at B and M respectively. Prove that:
i) ∆ABC ~ ∆AMP
ii) \(\frac{\mathrm{CA}}{\mathrm{PA}}=\frac{\mathrm{BC}}{\mathrm{MP}}\)

Solution:
Data: ∆ABC and ∆AMP are two right triangles, right angled at B and M respectvely.
To Prove: i) ∆ABC ~ ∆AMP
ii) \(\frac{\mathrm{CA}}{\mathrm{PA}}=\frac{\mathrm{BC}}{\mathrm{MP}}\)
(i) In ∆ABC and ∆AMP,
∠ABC = ∠AMP = 90° (data)
∠CAB = ∠MAP (common)
∴ ∠ACB = ∠MPA
∴ These are equiangular triangles.
Similarity criterion for ∆ is A.A.A.
∴ ∆ABC ~ ∆AMP

(ii) ∆ABC ~ ∆AMP(Proved)
∴ Corresponding sides are in proportion.
LC and LP are corresponding angles.
∴ Adjacent sides are CA, PA.
Similarly BC and MP are adjacent sides.
\(\quad \frac{\mathrm{CA}}{\mathrm{PA}}=\frac{\mathrm{BC}}{\mathrm{MP}}\)

Question 10.
CD and OH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ∆ABC and ∆EFG respectively.

If ∆ABC ~ ∆EFG, show that
i) \(\frac{C D}{G H}=\frac{A C}{F G}\)
ii) ∆DCB ~ ∆HGE
iii) ∆DCA ~ ∆HGF
Solution:
Data : CD and OH are respectively the bisectors of ∠ACB and ∠EGF such that D and H 11e on sides AB and FE of ∆ABC and ∆EFG respectively.
∆ABC ~ ∆EFG
To Prove: i) \(\frac{C D}{G H}=\frac{A C}{F G}\)
ii) ∆DCB ~ ∆HGE
iii) ∆DCA ~ ∆HGF
Proof: ∆ABC ~ ∆EFG (data given)
∴ Their corresponding sides are in proportion.
\(\quad \frac{\mathrm{AB}}{\mathrm{EF}}=\frac{\mathrm{BC}}{\mathrm{FG}}=\frac{\mathrm{AC}}{\mathrm{EG}}\)
∠B = ∠F, ∠A = ∠E, ∠C = ∠G.

(i) In ∆ADC and ∆EHG,
∠A = ∠E .
∠ACD = ∠EGH
∴ Their sides are in proportion.
\(\frac{\mathrm{CD}}{\mathrm{GH}}=\frac{\mathrm{AC}}{\mathrm{FG}}\)

(ii) In ∆DCB and ∆HGE.
\(\frac{\mathrm{CD}}{\mathrm{GH}}=\frac{\mathrm{BC}}{\mathrm{GF}}\)
∴ ∆DCB ~ ∆HGE

(iii) In ∆DCA and ∆HGE,
\(\frac{\mathrm{DC}}{\mathrm{GH}}=\frac{\mathrm{AD}}{\mathrm{EH}}=\frac{\mathrm{AC}}{\mathrm{EG}}\)
∴ ∆DCA ~ ∆HGE

Question 11.
In the following figure. E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that ∆ABD ~ ∆ECF

Solution:
Data: E Is a point on side CB produced of an isosceles ∆ABC with AB = AC. AD ⊥ BC and EF ⊥ AC,
To Prove: ∆ABD ~ ∆ECF
In ∆ABD and ∆ECF.
∠ADB = ∠EFC = 90° (data)
∠ABD = ∠FCE (∵ ∠B = ∠C)
∠BAD = ∠FEC
∴ Equiangular triangles.
∴ Similarity criterion for triangles is A.A.A.
∴ ∆ABD ~ ∆ECF.

Question 12.
Sides AB and BC and median AD of a ¿ABC are respectively proportional to sides PQ and QR and median PM of ∆PQR. (Sec figure gIven below) Show that ∆ABC ~ ∆PQR

Solution:
Data: Sides AB and BC and median AD of a ∆ABC are respectively proportional to sides P9 and QR and medIan PM of ∆PQR.
To Prove: ∆ABC ~ ∆PQR
In ∆ABC and ∆PQR,
Side \(\frac{\mathrm{AB}}{\mathrm{PG}}=\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{\mathrm{AD}}{\mathrm{PM}}\)
D is the mid-point of BC
∴ BD = DC
M is the mid-point of QR.
∴ QM = MR.
In ∆ADC and ∆PMR,
\(\frac{\mathrm{AD}}{\mathrm{PM}}=\frac{\mathrm{DC}}{\mathrm{MR}}=\frac{\mathrm{AC}}{\mathrm{PR}} \quad \left(\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{\frac{1}{2} \mathrm{BC}}{\frac{1}{2} \mathrm{QR}}\right)\)
Similarity criterion is S.S.S.
∴ ∆AÐC ~ ∆PMR
Now, In ∆ABC and ∆PQR.
\(\frac{A B}{P Q}=\frac{B C}{Q R}=\frac{A C}{P R}\)
∴ SimilarIty criterion is S.S.S.
∴ ∆ABC ~ ∆PQR.

Question 13.
D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that CA² = CB.CD.

Solution:
Data: D is a point on the side BC of a triangle ∆ABC such that ∠ADC = ∠BAC.
To Prove: CA² = CB × CD
Let ∠ADC = ∠BAC = 100°
In ∆ABC, If ∠B = 50°,then ∠C = 30°
In ∆ADC, If ∠C = 30°. then ∠DAC = 50°
In ∆BCP, ∠A= 100°, ∠B= 50°, ∠C= 30°
In ∆ADC, ∠ADC = 100. ∠DAC = 50°.
∠ACD = 30°
Similarity criterion of ∆ is A.A.A.
∴ In ∆ABC and ∆ADC,
\(\frac{\mathrm{CA}}{\mathrm{BC}}=\frac{\mathrm{DC}}{\mathrm{CA}}\)
∴ CA × CA = BC × DC
∴ CA² = BC × DC.

Question 14.
Sides AB and AC and median AD of a triangle ∆ABC are respectively proportional to sides PQ and PR and mediam PM of another triangle ∆PQR. Show that ∆ABC ~ ∆PQR.

Solution:
Data: Sides AB and AC and median AD of a triangle ∆ABC are respectively proportional to sides PQ and PR and medIan PM of another triangle ∆PQR.
To Prove: ∆ABC ~ ∆PQR
In ∆ABC and ∆PQR
\(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AC}}{\mathrm{PR}}=\frac{\mathrm{AD}}{\mathrm{PM}}\)
Corresponding sides are proportional.
Similarity criterion for ∆ is S.S.S
∴ ∆ABC ~ ∆PQR.

Question 15.
A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

Solution:
Height of PQ =?
In ∆ABC and ∆PQR,
∠B = ∠Q = 90°
∴ Hypotenuse AC is proportional to PQ
\(\quad \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{MR}}=\frac{\mathrm{AC}}{\mathrm{PR}}\)
\(\frac{6}{\mathrm{PQ}}=\frac{4}{28}\)
∴ 4 × PQ = 6 × 28
\(\quad \mathrm{PQ}=\frac{6 \times 28}{4} \quad \quad \mathrm{PQ}=42 \mathrm{m}\)

Question 16.
If AD and PM are medians of triangles ∆ABC and ∆PQR respectively, where ∆ABC ~ ∆PQR. prove that \(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AD}}{\mathrm{PM}}\)

Solution:
Data : AD and PM are medians of triangles ∆ABC and ∆PQR respectively. ∆ABC ~ ∆PQR.
To Prove: \(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AD}}{\mathrm{PM}}\)
∆ABC ~ ∆PQR.(data)
∴ we have \(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AC}}{\mathrm{PR}}=\frac{\mathrm{BC}}{\mathrm{QR}}\)
D is the mid-point of BC. BD = DC.
M Is the mid-point of QR. QM = MR
\(\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{\frac{1}{2} \mathrm{BC}}{\frac{1}{2} \mathrm{QR}}=\frac{\mathrm{BD}}{\mathrm{QM}}\)
\(\quad \frac{\mathrm{BC}}{\mathrm{QR}}=\frac{\mathrm{BD}}{\mathrm{QM}}\)
In ∆ABD and ∆PQM, we have
\(\quad \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BD}}{\mathrm{QM}}=\frac{\mathrm{AD}}{\mathrm{PM}}\)
\(\quad \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AD}}{\mathrm{PM}}\)

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 2 Triangles Ex 2.3 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 2 Triangles Exercise 2.3, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.2 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Exercise 5.2.

Karnataka SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Exercise 5.2

{Unless stated otherwise, use \(\pi=\frac{22}{7}\)}

Question 1.
Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.
Solution:
Radius of a sector of a circle, r = 6 cm.
Angle of a sector of a circle, θ = 60°
Area of a sector of a circle = ?
Area of a sector of a circle \(=\frac{\theta}{360} \times \pi r^{2}\)

= 18.8 sq.cm.

Question 2.
Find the area of a quadrant of a circle whose circumference is 22 cm.
Solution:
Circumference of a circle, 2πr = 22 cm.
Area of a quadrant of a circle = ?
2πr = 22

∴ Area of a quadrant of a circle

\(=\frac{77}{8}\)
= 9.62 sq. cm.

Question 3.
The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.
Solution:
Length of the minute hand of a clock, r = 14 cm.
Area swept by the minute hand in 5 minutes \(=360 \times \frac{5}{60}\)
= 30°
Angle of segment of a circle, 0 = 30°
∴ Area swept by the minute hand in 5 minute,

Question 4.
A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding:
(i) minor segment, (ii) major sector. (Use π = 3.14)
Solution:
Radius of the circle, r = 10 cm.
Angle at the centre of circle, θ = 90°
1) Area of minor segment = ?
2) Area of Major segment = ?
(i) Area of minor segment OAPB

ii) Area of Minor Segment, APB :
= Area of OAPB – Area of AOAB
\(=78.5 \mathrm{cm}^{2} \cdot-\frac{1}{2} \times \mathrm{OA} \times \mathrm{OB}\)
\(=78.5-\frac{1}{2} \times 10 \times 10\)
= 78.5 – 50
= 28.5 sq. cm.

(iii) Area of the Major Segment (AQBO):

Question 5.
In a circle of radius 21 cm, an arc subtends an angle of 60° at the Centre. Find:
(i) the length of the arc,
(ii) area of the sector formed by the arc,
(iii) area of the segment formed by the corresponding chord.
Solution:
Radius of the circle, r = 21 cm.
Angle at the centre of the circle, θ = 60°
1) the length of the arc?
2) Area of the sector formed by the arc?
3) Area of the segment.

1) The length of arc OAPB

= 22 cm

2) Area of the sector OAPB

= 231 sq. cm.

3) Area of the segment APB:
= Area of the sector OAPB – Area of the equilateral ∆OAB

= 231 – 190.7
= 40.3 sq. cm

Question 6.
A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments of the circle.
(Use π = 3.14 and \(\sqrt{3}=1.73\)).
Solution:
Radius of circle, r = 15 cm.
Angle at the centre, θ = 60°
1) Area of the minor segment =?
2) Area of the Major segment =?

(i) Area of Minor segment, APB :
= Area of sector OAPB – Area of ∆OAB

= 117.75 sq. cm. – 97.31
= 20.44 sq. cm.

ii) Area of the Major segment:
= Area of Circle – Area of a segment of circle OAPB

= 3.14 × 225 – 20.44
= 707.14 – 20.44
= 686.06 cm2.

Question 7.
A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment of the circle.
(Use π = 3.14 and \(\sqrt{3}=1.73\)).
Solution:
Radius of the circle, r = 12 cm.
Angle at the centre, θ = 120°
Area of segment APB =?
(i) Area of sector OAPB:

(ii) Area of segment APB:

∴ Area of segment APB:
= Area of sector OAPB – Area of ∆AOB
= 150.72 – 62.68
= 88.44 sq. cm.

Question 8.
A horse is tied to a peg at one corner of a square-shaped grass field of side 15 m by means of a 5 m. long rope (see Figure). Find
i) the area of that part of the field in which the horse can graze.
ii) the increase in the grazing area if the rope were 10 m long instead of 5 m. (use π = 3.14)

Solution:
Length of the rope, r = 5 m, θ = 90°
Area of segment =?

(ii) Length of the rope, r = 10 cm, θ = 90°
Area of segment =?

∴ Area of the field which is increased in the grazing
= Equation (2) – Equation (1)
= 78.5 – 19.63
= 58.87 sq. cm.

Question 9.
A brooch is made with silver wire in the form of a circle with a diameter of 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in the Figure. Find
i) the total length of the silver wire required,
ii) the area of each sector of the brooch.

Solution:
Diameter, d = 35mm θ = \(\frac{360^{\circ}}{10}\) = 36°
Radius, r = \(\frac{35}{2}\) mm
(i) Length of the silver wire required :
Length of the wire used in making 5 diameter + Circumference of Silver, wire required.

(ii) Area of each sector of the brooch = ?
Diameter of the brooch, d = 35mm
Radius, r = \(\frac{35}{2}\) mm
Angle at the centre, θ = \(\frac{360^{\circ}}{10}\) = 36°
The Area of each sector of the Brooch:

Question 10.
An umbrella has 8 ribs which are equally spaced (see the figure). Assuming the umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.

Solution:
Radius of circle, r = 45 cm.
Angle at the centre, θ = \(\frac{360^{\circ}}{8}\) = 45°
Area between the two consecutive ribs of the umbrella =?
Area of each segment of the circle,

Question 11.
A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades.
Solution:
Radius of the wiper of the car, r = 25 cm.
Angle at the centre, θ = 115°
Area of the sector =?
Area of the blade when it is swept once

Area of two blades when swept:

Question 12.
To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle 80° to a distance of 16.5 km. Find the area of the sea over which the ships are warned. (use π = 3.14)
Solution:
Radius, r = 16.5 km. = \(\frac{33}{2}\) km.
Angle, θ = 80°
Area of the Sector =?
Area of the sea over which the ships are warned

Question 13.
A round table cover has six equal designs as shown in Figure. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of Rs. 0.35 per cm².
(Use \(\sqrt{3}=\) = 1.7)

Solution:
Each angle at the centre,


Area of segment APB design 1 :
= Area of OAPB – Area of ∆AOB
= 410.6 – 333.2
= 77.46 sq.cm.
∴ Total are of SIX designs = 6 × 77.46
= 464.8 sq.cm.
Cost of designing for 1 sq.cm, is Rs. 0.35
∴ Cost of designing for 464.8 sq.cm ………….
= 464.8 × 0.35
= Rs. 162.68

Question 14.
Tick the correct answer in the following:
Area of a sector of angle p (in degrees) of a circle with radius R is

Solution:
D) \(\frac{p}{720}\) × 2πR²

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.2 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Exercise 5.2, 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 3.3 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 3.3.

Karnataka SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.3

ex 3.3 class 10 Question 1.
Solve the following pair of linear equations by the substitution method.
(i) x + y = 14
x – y = 14
(ii) s – t = 3
\(\frac{s}{3}+\frac{t}{2}=6\)
(iii) 3x – y = 3
9x – 3y = 9
(iv) 0.2x + 0.3y = 1.3
0.4x + 0.5y = 2.3
(v) \(\sqrt{2} x+\sqrt{3} y=0\)
\(\sqrt{3} x-\sqrt{8} y=0\)
(vi) \(\frac{3 x}{2}-\frac{5 y}{2}=-2\)
\(\frac{x}{3}+\frac{y}{2}=\frac{13}{6}\)
Solution:
(i) x + y = 14 ………. (i)
x – y = 4 …………. (ii)
In eqn. (i), x + 4 = 14
∴ x = 14 – y
Substituting the value of ‘x’ in equation (ii), we have
x – y = 4
(14 – y) – y = 4
14 – y – y = 4
14- 2y = 4
-2y = 4 – 14
-2y = -10
2y = 10
∴ x = 5
∴ Substituting y = 5 in x = 14 – y,
x = 14 – y
= 14 – 5
x = 9
∴ x = 9, y = 5.

(ii) s – t = 3
\(\frac{s}{3}+\frac{t}{2}=6\)
s – t = 3 …………. (i)
\(\frac{s}{3}+\frac{t}{2}=6\) ………… (ii)
From eqn. (i), s – t = 3
-t = 3 – s
t = -3 + s
Substituting the value of ‘t’ in eqn. (ii),
\(\frac{s}{3}+\frac{-3+s}{2}=6\)
\(\frac{2 s-9+3 s}{6}=6\)
5s – 9 = 6 × 6
5s – 9 = 36
5s = 36 + 9
∴ 5s = 45
∴ s=\(\frac{45}{5}\)
∴ s = 9
Substituting the value of s = 9 in t = -3 + s
t = -3 + s
= -3 + 9
t = 6
∴ s = 9, t = 6

(iii) 3x – y = 3 …………. (i)
9x – 3y = 9 ………… (ii)
From eqn. (i), 3x – y = 3
-y = 3 – 3x
y = – 3 + 3x
Substituting the value of ‘y’ in eqn. (ii),
9x – 3y = 9
9x – 3(-3 + 3x) = 9
9x + 9 – 9x = 9
Here we can give any value for ‘x’ i.e., infinite solutions are there,
y = 3x – 3 Means x = 0 then y = -3
x = 1 then y = 0
x = 2 then y = 3
etc.

(iv) 0.2x + 0.3y = 1.3 …………. (i)
0.4x + 0.5y = 2.3 …………. (ii)
From eqn. (i)„
0.2x + 0.3y = 1.3
0.2x = 1.3 – 0.3y
\(x=\frac{1.3-0.37}{0.2}\)
Substituting the value of ‘x’ in eqn (ii)
0.4x + 0.5y = 2.3
\(0.4\left(\frac{1.3-0.37}{0.2}\right)+0.5 y=2.3\)
0.2 (1.3 – 0.3y) + 0.5y = 2.3
2.6 – 0.6y + 0.5y = 2.3
2.6 – 0.1y = 2.3
-0.1y = 2.3 – 2.6
-0.1y = -0.3
0.1y = 0.3
\(y=\frac{0.3}{0.1}=\frac{3}{1}=3\)
Substituting the vlue of ‘y’ in
\(x=\frac{1.3-0.37}{0.2}\)
\(=\frac{1.3-0.3(3)}{0.2}\)
\(=\frac{1.3-0.9}{0.2}\)
\(=\frac{0.4}{0.2}\)
\(=\frac{4}{2}\)
∴ x = 2
∴ x = 2, y = 3

(v) \(\sqrt{2} x+\sqrt{3} y=0\) ………….. (i)
\(\sqrt{3} x-\sqrt{8} y=0\) ………….. (ii)
From eqn (i)
\(\sqrt{2} x+\sqrt{3} y=0\)
\(\sqrt{2} x=-\sqrt{3} y\)
\(\mathbf{x}=-\frac{\sqrt{3}}{\sqrt{2}} \mathbf{y}\)
Substituting the value of ‘x’ in eqn (ii)

substituting the value of ‘y’ in
\(\mathrm{x}=-\frac{\sqrt{3}}{\sqrt{2}} \mathrm{y}\)
\(=\frac{-\sqrt{3}}{\sqrt{2}} \times 0\)
∴ x = 0
∴ x = 0, y = 0

(vi) \(\frac{3 x}{2}-\frac{5 y}{2}=-2\) ………… (i)
\(\frac{x}{3}+\frac{y}{2}=\frac{13}{6}\) …………….. (ii)
From eqn (ii)
\(\frac{x}{3}+\frac{y}{2}=\frac{13}{6}\)
\(\frac{x}{3}=\frac{13}{6}-\frac{y}{2}\)
\(\frac{x}{3}=\frac{13-3 y}{6}\)
\(x=\frac{3(13-3 y)}{6}\)
\(x=\frac{13-3 y}{2}\)
Substituting the value of ‘x’ in eqn (i)
\(\frac{3 x}{2}-\frac{5 y}{2}=-2\)
\(\frac{3}{2}\left(\frac{13-3 y}{2}\right)-\frac{5 y}{2}=-2\)
\(\frac{39-9 y}{4}-\frac{5 y}{2}=-2\)
\(\frac{39-9 y-10 y}{4}=-2\)
\(\frac{39-19 y}{4}=-2\)
39 – 19y = -2 × 4
39 – 19y = -8
-19y = -8 + 39
-19y = -47
Or 19y = 47
\(y=\frac{47}{19}\)
∴ y = 3
Substituting the value of ‘y’ in
\(x=\frac{13-3 y}{2}\)
\(=\frac{13-3(3)}{2}\)
\(=\frac{13-9}{2}=\frac{4}{2}\)
∴ x = 2
∴ x = 2, y = 3

Question 2.
Solve 2x + 3y = 11 and 2x – 4y = -24 and hence find the value of ‘m’ for which y = mx + c
Solution:

ex 3.3 class 10 Question 2.
Solve 2x + 3y = 11 and 2x – 4y = -24 and hence find the value of ‘m’ for which y = mx + 3.
Solution:
2x + 3y = 11 …………… (i)
2x – 4y = -24 ………….. (ii)
From eqn. (i),
2x = 3y = 11
2x = 11 – 3y
\(x=\frac{11-3 y}{2}\)
Substituting the value of ‘x’ in eqn. (ii),
2x – 4y = -24
\(2\left(\frac{11-3 y}{2}\right)-4 y=-24\)
11 – 3y – 4y = -24 – 11
-7y = – 35
7y = 35
\(y=\frac{35}{7}\)
∴ y = 5.
Substituting the value of ‘y’ in
\(x=\frac{11-3 y}{2}\)
\(=\frac{11-3(5)}{2}\)
\(=\frac{11-15}{2}\)
\(=\frac{-4}{2}\)
∴ x = -2
∴ x = -2, y = 5
Now, y = mx + 3
5 = m(-2) + 3
5 = -2m + 3
5 – 3 = -2m
-2m = 2
\(\mathrm{m}=\frac{2}{-2}\)
∴ m= -1.

ex 3.3 class 10 Question 3.
Form the pair of linear equations for the following problems and find their solution by substitution method :
(i) The difference between the two numbers is 26 and one number is three times the other. Find them.
Solution:
Let one number be ‘x’.
another number be ‘y’.
Their difference is 26.
∴ x – y = 26 ………. (i)
One number is three times the other,
∴ x = 3y …………. (ii)
Substituting x = 3y in eqn. (i),
x – y = 26
3y – y = 26
2y = 26
\(y=\frac{26}{2}\)
∴ y = 13
Substituting the value of ‘y’ in eqn. (ii),
x = 3y
∴ x = 3 × 13
∴ x = 39, y = 13

(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
Solution:
In two supplementary angles, let one angle be ‘x’.
another angle be ‘y’
Sum of these = 180°
∴ x + y = 180 ………… (i)
The larger of two supplementary angles exceeds the smaller by 18 degrees.
∴ x-y = 18 ………… (ii)
From eqn. (ii),
x – y = 18
x = 18 + y
Substituting the value of ‘x’ in eqn. (i),
x + y = 180 .
18 +y + y = 180
18 + 2y = 180
2y = 180 – 18
2y = 162
\(\mathrm{y}=\frac{162}{2}\)
∴ y = 81
Substituting the value of y’ in x = 18 + y
x = 18 + 81
x = 99
∴ x = 99°, y = 81°.

(iii) The coach of a cricket team buys 7 bats and 6 balls for Rs. 3800. Later, she buys 3 bats and 5 balls for Rs. 1750. Find the cost of each bat and each ball.
Solution:
Let the cost of each bat b Rs. x.
Let the Cost of each ball be Rs. y.
∴ 7x + 6y = 3800
3x + 5y= 1750
From eqn. (ii),
3x + 5y = 1750
3x = 1750 – 5y
\(x=\frac{1750-5 y}{3}\)
Substituting the value of ‘x’ in eqn. (i),
\(7\left(\frac{1750-5 y}{3}\right)+6 y=3800\)
\(\frac{12250-35 y}{3}+\frac{6 y}{1}=3800\)
\(\frac{12250-35 y+18 y}{3}=3800\)
12250 – 17y = 3800 × 3
12250 – 17y = 11400
-17y = 11400 – 12250
-17y = -850
\(y=\frac{850}{17}\)
∴ y = Rs. 50
Substituting the value of ‘y’ in
\(x=\frac{1750-5 y}{3}\)
\(=\frac{1750-5 \times 50}{3}\)
\(=\frac{1750-250}{3}\)
\(=\frac{1500}{3}\)
∴ x = Rs. 500
∴ x = Rs. 500, Rs. y = 50
∴ Cost of each bat is Rs. 500,
Cost of each ball is Rs. 50.

(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs. 105 and for a journey of 15 km, the charge paid is Rs. 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?
Solution:
Let the fixed charge of the taxi be Rs. x.
If the charge for each km is Rs. y, then
Fixed charge + Charge for 10 km
= Rs. 105
∴ x + 10y = 105 ………. (i)
Fixed charge + Charge for 15 km travelled
= Rs. 155
∴ x + 15y = 155 ……….. (ii)
From eqn. (i),
x + 10 y = 105
x = 105 – 10y
Substituting the value of ’x’ in eqn. (ii),
x + 15y = 155
105 – 10y + 15y = 155
105 + 5y = 155
5y = 155 – 105
5y = 50
∴ \( \mathrm{y}=\frac{50}{5}\)
∴ Rs. y = 10.
Substituting the value of y in
x = 105 – 10y,
= 105 – 10 × 10
= 105 – 100
∴ x = Rs. 5
∴ Fixed Charge of Taxi is Rs. 5.
Charge for each km is Rs. 10
Charge for 1 km is Rs. 10
Charge for 25 km = 25 × 10 = Rs. 250.

(v) A fraction becomes \(\frac{9}{11}\). if 2 is added to both the numerator and the denominator. If 3 is added to both the numerator and the denominator it becomes \(\frac{5}{6}\). Find the fraction.
Solution:
If the Numerator is ‘x’ and denominator is ‘y’, then the Fraction is \(\frac{x}{y}\).
Adding 2 to both Numerator and denominator, then fraction is \(\frac{9}{11}\).
\(\frac{x+2}{y+2}=\frac{9}{11}\)
11 (x + 2) = 9(y + 2)
11x + 22 = 9y + 18
11x – 9y + 22 – 18 = 0
11x – 96 + 4 = 0 ……….. (i)
If 3 is added to both Numerator and denominator, then the fraction is \(\frac{5}{6}\).
\(\frac{x+3}{y+3}=\frac{5}{6}\)
6(x + 3) = 5 (y + 3)
6x + 18 = 5y + 15
6x – 5y + 18 – 15 = 0
6x – 5y + 3 = 0 ………… (ii)
From eqn. (i),
11x – 9y + 4 = 0
11x = 9y – 4
\(x=\frac{9 y-4}{11}\)
Substituting the value of ‘x’ in eqn. (ii),
6x – 5y + 3 = 0
\(6\left(\frac{9 y-4}{11}\right)-5 y+3=0\)
\(\frac{54 y-24}{11}-\frac{5 y}{1}+\frac{3}{1}=0\)
\(\frac{54 y-24-55 y+33}{11}=0\)
-y + 9 = 0
-y = -9
∴ y = 9
Substituting the value of ‘y’ in
\(x=\frac{9 y-4}{11}\)
\(=\frac{9 \times 9-4}{11}\)
\(=\frac{81-4}{11}\)
\(=\frac{77}{11}\)
∴ x = 7.
∴ Fraction is \(\frac{x}{y}=\frac{7}{9}\)

(vi) Let the present age of Jacob = x years
and the present age of his son = y years
∴ 5 years hence: Age of Jacob = (x + 5) years
Age of his son = (y + 5) years
According to given condition,
[Age of Jacob] = 3[Age of his son]
x + 5 = 3(y + 5) ⇒ x + 5 = 3y + 15
⇒ x – 3y -10 = 0 … (1)
5years ago : Age of Jacob = (x – 5) years,
Age of his son = (y – 5) years
According to given condition,
[Age of Jacob] = 7[Age of his son]
∴ (x – 5) = 7(y – 5) ⇒ x – 5 = 7y – 35
⇒ x – 7y + 30 = 0 … (2)
From (1) , x = [10 + 3y] … (3)
Substituting this value of x in (2) , we get
(10 + 3y) – 7y + 30 = 0
⇒ -4y = -40 ⇒ y = 10
Now, substituting y = 10 in (3) ,
we get x = 10 + 3(10)
⇒ x = 10 + 30 = 40
Thus, x = 40 and y = 10
⇒ Present age of Jacob = 40 years and present age of his son = 10 years

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3 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 3.3, drop a comment below and we will get back to you at the earliest

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

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Expert Teachers at KSEEBSolutions.com has created KSEEB Karnataka SSLC Social Science Model Question Papers 2021-2022 with Answers Pdf Download of KSEEB Class 10th Std Social Science Previous Year Model Question Papers, Sample Papers in Kannada Medium and English Medium are part of Karnataka SSLC Model Question Papers with Answers.

Here we have given Karnataka Secondary Education Examination Board KSEEB SSLC Social Science Model Question Papers for Class 10 State Syllabus Karnataka 2021-22 with Answers Pdf. Students can also read Karnataka SSLC KSEEB Solutions for Class 10 Social Science.

Social Science Model Question Papers for Class 10 Karnataka State Syllabus 2021-22

These SSLC 10th Social Science Model Question Papers 2021-2022 Karnataka State Board Kannada Medium and English Medium are useful to understand the pattern of questions asked in the board exam. Know about the important concepts to be prepared for Karnataka Board Exams and Score More marks.

These 10th Social Science Question Paper 2021-22 State Board Karnataka are designed according to the latest exam pattern, so it will help students to know the exact difficulty level of the question papers.

SSLC Social Science Model Question Papers for Class 10 State Syllabus Karnataka in English Medium

• Karnataka SSLC Social Science Model Question Paper 1
• Karnataka SSLC Social Science Model Question Paper 2
• Karnataka SSLC Social Science Model Question Paper 3
• Karnataka SSLC Social Science Model Question Paper 4
• Karnataka SSLC Social Science Model Question Paper 5
• Karnataka SSLC Social Science Model Question Paper 6

SSLC Social Science Model Question Papers for Class 10 State Syllabus Karnataka in Kannada Medium

• Karnataka SSLC Social Science Model Question Paper 1
• Karnataka SSLC Social Science Model Question Paper 2
• Karnataka SSLC Social Science Model Question Paper 3
• Karnataka SSLC Social Science Model Question Paper 4 (KSEEB Board Model Paper 2)
• Karnataka SSLC Social Science Model Question Paper 5 (KSEEB Board Model Paper 1)

Karnataka SSLC Social Science Model Question Paper Design

Instructions for preparing the SSLC Social Science Question Papers:

• In the 10th Social Science Model Question Papers Karnataka State Board, there are Four 4 marks questions (8 to 10 points), of which only one is given a choice, which should be selected from the History theme only.
• In SSLC Social Science Question Papers 2020 Kannada Medium, there are 9 three marks questions (6 to 8 points) out of which choice is given to four questions one from History, one from Geography, one from Economics, and one from Business Studies only.
• In SSLC Social Science Model Question Papers with Answers, there are 8 questions of 2 marks out of which only two of them will have choice selected from Political Science and Sociology only.
• In SSLC Model Question Papers Social Science English Medium, the question on map drawing should be asked only from the geography part and the locations, rivers, projects, natural divisions etc. Should be given in the textbook’s map or given in the explanation part in the textbook.
• While preparing the question paper, the Karnataka SSLC Social Science Model Question Papers with Answers should be prepared in accordance with the subject, category and objectives specified.

The basic objectives of using the new questionnaires are:

• This is done with the expectation that the children are just statistical and aiming at only marks not knowledge.
• Present days children and parents simply do not care about knowledge unless they are in the process of scoring.
• Children will not learn until what they have learned in the classroom is used in their lives.
• Learning should not be just memorizing. For all these reasons, a change in the testing system is essential.

Its purposes and uses

• To bring Innovation in children’s learning.
• Changes in the way of children being taught should be not just by memory but by logical thinking.
• Teaching all lessons with equal priority.
• Learning should not be a forced practice but learning should be by love.
• The main purpose is to bring cognitive development in the children
• Its mission is to develop language skills and writing skills in children.
• At first there were 47 questions and the short answer questions were more. But the questions were not encouraging the writing skills so long answer questions in the current model are more promising.
• There are totally 38 questions.
• Nine 3 mark questions and four 4 marks questions are increased along with one 5 mark question.

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If you have any queries regarding Karnataka State Syllabus KSEEB 10th Standard Social Science Model Question Papers 2021-22 with Answers Pdf, drop a comment below and we will get back to you at the earliest.

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