# RS Aggarwal Solutions Class 10 Ex 3H

## RS Aggarwal Class 10 Ex 3H Chapter 3

Question 1: 2x – 3y = 7, (a + b)x + ( a + b – 3)y = (4a + b)

Solution:

2x – 3y – 7 = 0 …….. (1)

(a + b)x + ( a + b – 3)y – (4a + b) = 0 ……..(2)

a1 = 2, b1 = -3, c1 = -7

a2 = (a + b), b2 = -(a + b – 3), c2 = -(4a + b)

These holds only when,

$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$

$\frac{2}{a + b} = \frac{-3}{-(a + b – 3)} = \frac{-7}{-(4a + b)}$

$\frac{2}{a + b} = \frac{3}{(a + b – 3)} = \frac{7}{(4a + b)}$

$\Rightarrow \frac{2}{a + b} = \frac{7}{4a + b} or \frac{3}{(a + b – 3)} = \frac{7}{(4a + b)}$

8a + 2b = 7a + 7b and 12a + 3b = 7a + 7b – 21

a – 5b = 0 …….. (1)

5a – 4b = -21 …….. (2)

Putting a = 5b in (2), we get

(5)(5b) – 4b = -21

25b – 4b = -21

21b = -21

b = -1

Putting b = -1 in (1), we get

a – (5)(-1) = 0

a + 5 = 0

a = -5

Thus, a = -5, b = -1

Question 2: 2x + 3y = 7, a(x + y) – b(x – y) = 3a + b – 2

Solution:

2x + 3y = 7 ……..(1)

a(x + y) – b(x – y) = 3a + b – 2 ……..(2)

Equation (2) is

ax + ay – by + by = 3a  + b – 2

(a – b)x + (a + b)y = 3a + b -2

Comparing with the equations

a1x + b1y + c1 = 0, a2x + b2y + c2 = 0

Therefore, a1 = 2, b1 = (a + b), c2 = 3a + b = 2

There are infinitely many solution

If $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$

Or, $\frac{2}{a – b} = \frac{3}{a + b} = \frac{7}{3a + b – 2}$

Hence, $\frac{2}{a – b} = \frac{3}{a + b} and \frac{3}{a + b} = \frac{7}{3a + b – 2}$

2a + 2b = 3a – 3b and 3(3a + b -2) = 7(a + b)

-a = -5b and 9a – 7a + 3b – 7b = 6

Or 2a – 4b = 6

Or a – 2b = 3

Thus equation in a, b are,

a = 5b ……..(3)

a – 2b = 3 ……..(4)

Putting a = 5b in (4)

5b – 2b = 3 or 3b = 3 implies, b = 1

Putting b = 1 in (3)

a = 5 and b = 1

Question 3: 5x – 3y = 0, 2x + ky = 0

Solution:

We have 5x – 3y = 0 …….. (1)

2x + ky = 0 …….. (2)

Comparing the equation with

a1x + b1y + c1 = 0, a2x + b2y + c2 = 0

Therefore, a1 = 5, b1 = -3, a2 = 2, b2 = k

These equations have a non-zero solution if

$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}$

$\frac{5}{2} = \frac{-3}{k} \Rightarrow 5k = -6$

$k = \frac{-6}{5}$

Question 4: 5 chairs and 4 tables together cost Rs.2800, while 4 chairs and 3 tables together cost Rs.2170. Find the cost of a chair and that of a table.

Solution:

Let the cost of 1 chair be Rs. x and the cost of one table be Rs. y

The cost of 5 chairs and 4 tables = Rs(5x + 4y) = Rs. 2800

5x + 4y = 2800 …….. (1)

The cost of 4 chairs and 3 tables = Rs(4x + 3y) = Rs. 2170

4x + 3y = 2170 …….. (2)

Multiplying (1) by 3 and (2) by 4, we get

15x + 12y = 8400 …….. (3)

16x + 12y = 8680 …….. (4)

Subtracting (3) from (4), we get

x = 280

Putting the value of x in (1), we get

(5)(280 + 4y = 2800

Or 1400 + 4y = 2800

Or 4y = 1400

Hence, y = 350.

Question 5: 37 pens and 53 pencils together cost Rs.820, while 53 pens and 37 pencils together cost Rs.980. Find the cost of a pen and that of a pencil.

Solution:

Let the cost of a pen and a pencil be Rs. x and Rs. y respectively.

Cost of 37 pens and 53 pencils = Rs(37x + 53y) = Rs. 820

37x + 53y = 820 …….. (1)

Cost of 53 pens and 37 pencils = Rs(53x + 37y) = Rs. 980

53x + 37y = 980 …….. (2)

Adding (1) and (2), we get

90x + 90y = 1800

x + y = 20 …….. (3)

y = 20 – x

Putting the value of y in (1), we get

37x + 53(20 – x) = 820

37x + 1060 – 53x = 820

16x = 240

$\Rightarrow$ x = 15

From (3), y = 20 – x

Hence, substituting x we get, y = 5

Thus, the cost of a pen = Rs. 15 and cost of a pencil = Rs. 5

Question 6: A lady has 20-paisa and 25-paisa coins in her purse. If she has 50 coins in all totaling Rs.11.50, how many coins of each kind does she have?

Solution:

Let the number of 20 P and 25 P coins be x and y respectively

Total number of coins x + y = 50

i.e., x + y = 50 …….. (1)

Value of these coins = Rs. $\frac{x}{5} + \frac{y}{4}$ = Rs. 11.50 = $Rs. 11\frac{1}{2}$

Therefore, $\frac{x}{5} + \frac{y}{4} = \frac{23}{2}$

$\Rightarrow$ 4x + 5y = 230 ……..(2)

Multiplying (1) by 5 and (2) by 1, we get

5x + 5y = 250…….. (3)

4x + 5y = 230 …….. (4)

Subtracting (4) from (3), we get

x = 20

Putting x = 20 in (1),

y = 50 – x

= 50 – 20

= 30

Hence, number of 20 P coins = 20 and number of 25 P coins = 30

Question 7: The sum of two numbers is 137 and their difference is 43. Find the numbers.

Solution:

Let the two numbers be x and y respectively.

Given:

x + y = 137 …….. (1)

x – y = 43 ……..(2)

Adding (1) and (2), we get

2x = 180

x = 90

Putting x = 90 in (1), we get

90 + y = 137

y = 137 – 90

= 47

Hence, the two numbers are 90 and 47.

Question 8: Find two numbers such that the sum of twice the first and the thrice the second is 92, and four times the first exceeds seven times the second by 2.

Solution:

Let the first and second number be x and y respectively.

According to the question:

2x + 3y = 92 …….. (1)

4x – 7y = 2 …….. (2)

Multiplying (1) by 7 and (2) by 3, we get

14x + 21y = 644 …….. (3)

12x – 21y = 6 …….. (4)

Adding (3) and (4), we get

26x = 650

Thus, $x = \frac{650}{26} = 25$

Putting x = 25 in (1), we get

2(25) + 3y = 92

50 + 3y = 92

3y = 92 – 50

$y = \frac{42}{3} = 14$

Question 9: Find two numbers such that the sum of thrice the first and the second is 142, and four times the first exceeds the second by 138.

Solution:

Let the first and second numbers be x and y respectively.

According to the question:

3x + y = 142 …….. (1)

4x – y = 138 …….. (2)

Adding (1) and (2), we get

7x = 280

$x = \frac{280}{7} = 40$

Putting x = 40 in (1), we get

(3)(40) + y = 142

y = 142 – 120

y = 22

Hence, the first and second numbers are 40 and 22.

Question 10: If 45 is subtracted from twice the greater of two numbers, it results in the other number. If 21 is subtracted from twice the smaller number, it results in the greater number. Find the numbers.

Let the greatest number be x and the smaller be y respectively.

Solution:

According to the question:

2x – 45 = y

2x – y = 45 ……..(1)

And

2y – x = 21

-x + 2y = 21 ……..(2)

Multiplying (1) by 2 and (2) by 1, we get

4x – 2y = 90 ……..(3)

-x + 2y = 21 ……..(4)

Adding (3) and (4), we get

3x = 111

$x = \frac{111}{3} = 37$

Putting x = 37 in (1), we get

(2)(37) – y = 43

74 – y = 45

y = 29

Hence, the greater and smaller numbers are 37 and 29.

Question 11: If three times the larger of two numbers is divided by the smaller, we get 4 as the quotient and 8 as the remainder. If five times the smaller is divided by the larger, we get 3 as the quotient and 5 as the remainder. Find the numbers.

Solution:

Let the larger number be x and smaller by y respectively.

We know,

Dividend = Divisor × Quotient + Remainder

3x = y × 4 + 8

3x – 4y = 8 ……..(1)

And,

5y = x × 3 + 5

-3x + 5y = 5 …….(2)

Adding (!) and (2), we get

y = 13

Putting y = 13 in (1),

Value of these coins = Rs $\frac{x}{5} + \frac{y}{4} = Rs. 11.50 = rs 11\frac{1}{2}$

Therefore, $\frac{x}{5} + \frac{y}{4} = \frac{23}{2}$

$\Rightarrow$ 4x + 5y = 230 ……..(2)

Hence, the larger number is 20 and the smaller one is 13 respectively.

Question 12: If 2 is added to each of two given numbers, their ratio becomes 1:2. However, if 4 is subtracted from each of the given numbers, the ratio becomes 5:11. Find the numbers.

Solution:

Let the required numbers be x and y respectively.

Then,

$\frac{x + 2}{y + 2} = \frac{1}{2} \Rightarrow 2x + 4 = y + 2 \Rightarrow 2x – y = -2$

$\frac{x – 4}{y – 4} = \frac{5}{11} \Rightarrow 11x – 44 = 5y – 20 \Rightarrow 11x – 5y = 24$

Therefore,

2x – y = -2 ……..(1)

11x – 5y = 24 ……..(2)

Multiplying (1) by 5 and (2) by 1, we get

10x – 5y = -10 ……..(3)

11x – 5y = 24 ……..(4)

Subtracting (3) and (4), we get

x = 34

Putting x = 34 in (1), we get

2 × 34 – y = -2

68 – y = -2

Iy = – 2 – 68

y = 70

Hence, the required numbers are 34 and 70.

Question 13: The difference between two numbers is 14 and the difference between their squares is 448. Find the numbers.

Solution:

Let the numbers be x and y respectively.

According to the question:

x2 – y2 = 448 ……..(1)

x – y = 14 ……..(2)

From (2), we get

x = 14 + y …….. (3)

Putting x = 14 in (1), we get

(14 + y)2 – y2 = 448

196 + y2 + 28y – y2 = 448

196 + 28y = 448

28y = 448 – 196

y = 9

Putting y = 9 in (2), we get

x – 9 = 14

x = 14 + 9 = 23

Hence, the required numbers are 23 and 9.

Question 14: The sum of the digits of a two-digit number is 12. The number obtained by interchanging its digits exceeds the given number by 18. Find the number.

Solution:

Let the ten’s digit be x and unit’s digit be y respectively.

Then,

x + y = 12 …….. (1)

Let the ten’s digit of required number be x and its unit’s digit be y respectively

Required number = 10x + y

10x + y = 7(x + y)

10x + y = 7x + 7y

3x – 6y = 0 …….. (1)

Number found on reversing the digits = 10y + x

(10x + y) – 27 = 10y + x

10x – x + y – 10y = 27

9x – 9y = 27

(x – y) = 27

x – y = 3 ……..(2)

Multiplying (1) by 1 and (2) by 6

3x – 6y = 0 …….. (3)

6x – 6y = 18 …….. (4)

Subtracting (3) from (4), we get

3x = 18

X = 6

Putting x = 6 in (1), we get

3 × 6 – 6y = 0

18 -6y = 0

-6y = -18

y = 3

Number = 10x + y

= 10 × 6 + 3 = 60 + 3 = 63

Hence, the number is 63.

Question 15: The sum of the digits of a two-digit number is 12. The number obtained by interchanging its digits exceeds the given number by 18. Find the number.

Solution:

Let the ten’s digit and the unit’s digit of required number be x and y respectively.

Required number = 10x + y

Number obtained on reversing digits = 10y + x

According to the question:

10y + x(10x + y) = 18

10y + x- 10x – y = 18

9y – 9y = 18

y – x = 2 ……..(2)

Adding (1) and @), we get

2y = 14

y = 7

Putting y = 7 in (1), we get

x + 7 = 12

X = 5

Number = 10x + y

= 10 × 5 + 7

= 50 + 7

= 57

Hence, the number is 57.

Question 13: The sum of the digits of a two-digit number is 15. The number obtained by interchanging its digits exceeds the given number by 9. Find the number.

Solution:

Let the ten’s digit and the unit’s digit of required number be x and y respectively.

Then,

10x + y = 15 …….. (1)

Required number = 10x + y

Number obtained by interchanging the digits = 10y + x

10y + x(10y + x) = 9

10y + x – 10x – y = 9

9y – 9x = 9

9(y – x) = 9

y – x = 1

-x + y = 1 …….. (2)

Adding (1) and (2), we get

2y = 16

y = 8

Putting y = 8 in (1), we get

x + 8 = 15

x = 15 – 8 = 7

Required number = 10x + y

= 10(7) + 8 = 70 + 8 = 78

Hence, the required number is 78.

Question 14: A two-digit number is 3 more times the sum of its digits. If 18 is added to the number, the digits are reversed. Find the number.

Solution:

Let the ten’s digit and the unit’s digit of required number be x and y respectively.

Then,

Required number = 10x + y = 15

According to the given equation:

10x + y = 4(x + y) + 3

10x + y = 4x + 4y + 3

6x – 3y = 3

2x – y = 1 …….. (1)

And

10x + y + 18 = 10y + x

9x – 9y = -18

9(x – y) = -18

(x – y) = -2 …….. (2)

Subtracting (2) from (1), we get

x = 3

Putting x = 3, in (1) we get

(2)(3) – y = 1

y = 6 – 1 = 5

x = 3, y = 5

Required number = 10x + y

= 10(3) + 5 = 30 + 5 = 35

Hence, required number is 35.

#### Practise This Question

Express 512×729 in exponential form: