# Ncert Solutions For Class 10 Maths Ex 3.4

## Ncert Solutions For Class 10 Maths Chapter 3 Ex 3.4

Question 1:

Using the method of elimination and method of substitution, solve the given liner equations.

(1) a + b = 5 and 2a – 3b = 4

(2) 3a + 4b = 10 and 2a – 2b = 2

(3) 3a = 5b – 4 = 0 and 9a = 2b + 7

(4) a2+2b3=1$\frac{a}{2}+\frac{2b}{3}=\, -1$ and ab3=3$a-\frac{b}{3}=3$

Solution:

(1) By the method of elimination.

A + B = 5   – – – – – – – – (i)

2A – 3B = 4 – – – – – – – – (ii)

When the equation (i) and is multiplied by (ii), we get

2A + 2B = 10 – – – – – – – – (iii)

When the equation (ii) is subtracted from (iii) we get,

5B = 6

B = 65$\frac{6}{5}$ – – – – – – – (iv)

Substituting the values obtained in (i) we get,

A=565=195$A=5-\frac{6}{5}=\frac{19}{5}$ A=195,B=65$∴ A=\frac{19}{5},B=\frac{6}{5}$

By the method of substitution:

From the equation (i), we get:

A = 5 – B – – – – – – – (v)

When the value is put in equation (ii) we get,

2(5 – B) – 3B = 4

-5B = -6

B = 65$\frac{6}{5}$

When the values are substituted in equation (v), we get:

A=565=195$A=5-\frac{6}{5}=\frac{19}{5}$ A=195,B=65$∴ A=\frac{19}{5},B=\frac{6}{5}$

(2) By the method of elimination:

3A + 4B = 10 – – – – – – -(i)

2A – 2B = 2 – – – – — – – (ii)

When the equation (i) and (ii) is multiplied by 2, we get:

4A – 4B = 4 – – – – – – – (iii)

When the Equation (i) and (iii) are added, we get:

7A = 14

A=2 – – – – – – – – – (iv)

Substituting equation (iv) in (i) we get,

6 + 4B = 10

4B = 4

B = 1

Hence, A = 2 and B = 1

By the method of Substitution

From equation (ii) we get,

A = 1 + B – – – – – – – – (v)

Substituting equation (v) in equation (i) we get,

3(1 + B) + 4B = 10

7B = 7

B= 1

When B= 1 is substituted in equation (v) we get,

A = 1 + 1 = 2

Therefore, A =2 and B = 1

(3) By the method of elimination:

3A – 5B – 4 = 0 – – – – – – – (i)

9A = 2B + 7

9A – 2B – 7 = 0 – – – – – – – (ii)

When the equation (i) and (iii) is multiplied we get,

9A – 15B – 12 = 0 – – – – – (iii)

When the equation (iii) is subtracted from equation (ii) we get,

13B = -5

B=513$B=\frac{-5}{13}$ – – – – – – – – (iv)

When equation (iv) is substituted in equation (i) we get,

3A+25134=0$3A+\frac{25}{13}-4=0$ 3A=2713$3A=\frac{27}{13}$ A=913$A=\frac{9}{13}$ A=913,B=513$∴ A=\frac{9}{13},B=\frac{-5}{13}$

By the method of Substitution:

From the equation (i) we get,

A=5B+43$A=\frac{5B+4}{3}$ – – – – – – – – (v)

Putting the value (v) in equation (ii) we get,

9(5B+43)2B7=0$9(\frac{5B+4}{3})-2B-7=0$

13B = -5

B=513$B=-\frac{5}{13}$

Substituting this value in equation (v) we get,

A=5(513)+43$A=\frac{5(\frac{-5}{13})+4}{3}$ A=913$A=\frac{9}{13}$ A=913,B=513$∴ A=\frac{9}{13},B=\frac{-5}{13}$

(4) By the method of Elimination

A2=2B3=1$\frac{A}{2}=\frac{2B}{3}=-1$

3A + 4B = -6 – – – – – – – (i)

AB3=3$A-\frac{B}{3}=3$

3A – B = 9 – – – – – – – (ii)

When the equation (ii) is Subtracted from equation (i) we get,

5B = -15

B = 3 – – – – – – (iii)

When the equation (iii) is substituted in (i) we get,

3A – 12 = -6

3A = 6

A = 2

Hence, A = 2 , B = -3

By the method of Substitution:

From the equation (ii) we get,

A=B+93$A=\frac{B+9}{3}$ – – – – – – – – – (v)

Putting the value obtained from equation (v) in equation (i) we get,

3(B+93)+4B=6$3(\frac{B+9}{3})+4B=-6$

5B = -15

B= -3

When B= -3 is substituted in equation (v) we get,

A=3+93=2$A=\frac{-3+9}{3}=2$

Therefore, A = 2 and B = -3

Question 2:

Find the solutions for the given pair of linear equation by the method of elimination for the following questions:

(1) If 1 is added in the numerator and 1 subtracted from that of the denominator, a fraction is then reduced to 1. It becomes ½ if 1 is added only to the denominator. What is the fraction?

(2) Reuben was thrice as old as Alex, 5 years back. And Reuben will be twice as old as Alex, 10 years from now. What are the ages of Reuben and Alex?

(3) The sum of digits of a two digit number is 9. Also when this number is multiplied nine times, it is two times the number when obtained by reversing the order of the digits. Find the number.

(4) Manna went to the bank to withdraw a sum of Rs 2000. She had asked the cashier to give her Rs 50 and Rs 100 notes only. Manna got 25 notes in total. Find how many Rs 50 notes and Rs 100 notes she had received.

(5) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Sara paid a sum of Rs 27 for a book to keep it for seven days, while Rosy paid a sum of Rs 21 for the book that she had kept for five days. Find the fixed charge and the charge that will cost for each extra day.

Solution:

(1) Let the fraction be a/b

According to that of the given information,

A+1B1=1$\frac{A+1}{B-1}=1$ => A – B = -2 – – – – – – – – – – – (i)

AB+1=12$\frac{A}{B+1}=\frac{1}{2}$ => 2A – B = 1 – – – – – – – – (ii)

When equation (i) is Subtracted from equation (ii) we get A = 3 – – – – – – (iii)

When A = 3 is substituted in equation (i) we get,

3 – B = -2

-B = -5

B = 5

Hence, the fraction is 3/5

(2) Let the present age of Reuben = a

And the present age of Alex = b

According to the information that is given,

(A – 5 ) = 3 ( B – 5 )

A – 3B = -10 – – – – – – – – – (i)

(A + 10) = 2(B + 10)

A – 2B = 10 – – – – – – – – – – – (ii)

When the equation (i) is subtracted from equation (ii) we get,

B = 20 – – – – – – – – – (iii)

Substituting B = 20 in equation (i), we get:

A – 60 = -10

A = 50

Hence, the present age of Reuben is 50 yrs

And the present age of Alex is 20 yrs

(3) Let the unit digit and tens digit of a number be A and B respectively.

Then, Number (n) = 10B + A

N after reversing the digits = 10A + B

According to the given information, A + B = 9 – – – – – – – – – (i)

9(10B + A) = 2(10A + B)

88 B – 11 A = 0

-A + 8B = 0 – – – – – – – – (ii)

Adding the equations (i) and (ii) we get,

9B = 9B

= 1 (3)

Substituting this value in the equation (i) we get A= 8

Hence the number (N) is 10B + A = 10 x 1 +8 = 18

(4) Let the number of Rs 50 notes be A and the number of Rs 100 notes be B

According to the given information,

A + B = 25 – – – – – – – – – – – (i)

50A + 100B = 2000 – – – – – – – (ii)

When equation (i) is multiplied with (ii) we get,

50A + 50B = 1250 – – – – – – – – – (iii)

Subtracting the equation (iii) from the equation (ii) we get,

50B = 750

B = 15

Substituting in the equation (i) we get,

A = 10

Hence, Manna has 10 notes of Rs 50 and 15 notes of Rs 100.

(5) Let the fixed charge for the first three days be Rs A and the charge for each day extra be Rs B.

According to the information given,

A + 4B = 27 – – – – – (i)

A + 2B = 21 – – – – – – (ii)

When equation (ii) is subtracted from equation (i) we get,

2B = 6

B = 3 – – – – – – (iii)

Substituting B = 3 in equation (i) we get,

A + 12 = 27

A = 15

Hence, the fixed charge is Rs 15

And the Charge per day is Rs 3