# Ncert Solutions For Class 10 Maths Ex 3.3

## Ncert Solutions For Class 10 Maths Chapter 3 Ex 3.3

Question 1:

Using substitution method, solve the given pair of linear equations.

1. i) x – y = 3

x3+y2=6$\frac{x}{3} + \frac{y}{2} = 6$

1. ii) a + b = 1

a – b = 4

iii)  0.2a + 0.3b = 1.3

0.4a+0.5b = 2.3

1. iv) 3a – b = 3

9a – 3b = 9

1. v) 3x25y3=2$\frac{3x}{2} – \frac{5y}{3} = -2$

x3+y2=136$\frac{x}{3} + \frac{y}{2} = \frac{13}{6}$

1. vi) 2x+3y=0$\sqrt{2}x + \sqrt{3}y = 0$

3x8y=0$\sqrt{3}x – \sqrt{8}y = 0$

Solution:

i) x – y = 3- – – – – – – – – – – (I)

x3+y2=6$\frac{x}{3} + \frac{y}{2} = 6$– – – – – – – – (II)

From (I), we get

x = y + 3- – – – – — – – – (III)

Substituting the value of x in equation (II), we get

y+33+y2=6$\frac{y+3}{3} + \frac{y}{2} = 6$

2y + 6 + 3y = 36

5y = 30

y = 6- – – – – – – – – (IV)

Substituting the value of y in equation (III), we get x = 9

∴ x = 9, y = 6

ii) a + b = 14- – – – – – – – – (I)

a – b = 4  — – – – – – – – – (II)

From (I), we get a = 14 – b- – – – – – – – – –  (III)

Substituting the value of a in equation (II), we obtain

(14 – b) – b = 4

14 – 2b = 4

10 = 2b

b = 5- – – – – – – – –  (IV)

Substituting (IV) in (III), we get

a = 14 – b

= 14 – 5

= 9

∴ a = 9, b = 5

iii)  0.2a + 0.3b = 1.3- – – – – – – (I)

0.4a + 0.5b = 2.3- – – – – – – –  (II)

From equation (I), we get

a=1.30.3b0.2$a=\frac{1.3-0.3b}{0.2}$ – – – – – – – – (III)

Substituting the value of a in equation (II), we get

0.4(1.30.3b0.2)+0.5b=2.3$0.4\left ( \frac{1.3-0.3b}{0.2} \right )+0.5b=2.3$

2.6 – 0.6b + 0.5b = 2.3

2.6 – 2.3 = 0.1b

0.3 = 0.1b

b = 3- – – – – – – – (IV)

Substituting the value of b in equation (III), we get

a=1.30.330.2$a=\frac{1.3-0.3*3}{0.2}$ =1.30.90.2=0.40.2=2$=\frac{1.3-0.9}{0.2}=\frac{0.4}{0.2}=2$

∴ a = 2, b = 3

iv) 3a – b = 3- – – – – – – – (I)

9a – 3b = 9  – – – – – – – – (II)

From (I), we get b = 3a – 3 – – – – – – – – (III)

Substituting the value of b in equation (II), we get

9a – 3(3a – 3) = 9

9a – 9a + 9 = 9

9 = 9

This is always true.

The given pair of equations has infinite solutions and the relation between these variables can be given by b = 3a – 3

Therefore, one of its possible solutions is a = 1, b = 0.

v) 3x25y3=2$\frac{3x}{2} – \frac{5y}{3} = -2$ – – – – – – – – (I)

x3+y2=136$\frac{x}{3} + \frac{y}{2} = \frac{13}{6}$ – – – – – – – – (II)

From equation (I), we get

9x – 10y = -12

x=12+10y9$x=\frac{-12+10y}{9}$ – – – – – – – – (III)

Substituting the value of x in equation (II), we get

12+10y93+y2=136$\frac{\frac{-12+10y}{9}}{3}+\frac{y}{2}=\frac{13}{6}$ 12+10y27+y2=136$\frac{-12+10y}{27}+\frac{y}{2}=\frac{13}{6}$ 24+20y+27y54=136$\frac{-24+20y+27y}{54}=\frac{13}{6}$

47y = 117 + 24

47y = 141

y = 3 – – – – – – – –  (IV)

Substituting the value of y in equation (III), we get

x=12+1039=189=2$x=\frac{-12+10*3}{9}=\frac{18}{9}=2$

Hence, x = 2, y = 3

vi) 2x+3y=0$\sqrt{2}x + \sqrt{3}y = 0$ – – – – – – – – (I)

3x8y=0$\sqrt{3}x – \sqrt{8}y = 0$ – – – – – – – – (II)

From equation (I), we get

x=3y2$x=\frac{-\sqrt{3}y}{\sqrt{2}}$ – – – – – – – –  (III)

Substituting the value of x in equation (II), we get

3(3y2)8y=0$\sqrt{3}\left ( -\frac{\sqrt{3}y}{\sqrt{2}} \right )-\sqrt{8}y=0$ 3y222y=0$-\frac{3y}{\sqrt{2}}-2\sqrt{2}y=0$ y(3222)=0$y\left ( -\frac{3}{\sqrt{2}}-2\sqrt{2} \right )=0$

y = 0- – – – – – – – (IV)

Substituting the value of y in equation (III), we get

x = 0

∴ x = 0, y = 0

Question 2:

Solve 2a + 3b = 11 and 2a – 4b = -24 and calculate the value of m in b = ma + 3.

Solution:

2a + 3b = 11                    (I)

2a – 4b = -24                   (II)

From equation (II), we get

a=113b2$a=\frac{11-3b}{2}$                   (III)

Substituting the value of a in equation (II), we get

2(113b2)4b=24$2\left(\frac{11-3b}{2}\right)-4b=-24$

11 – 3b – 4b = -24

-7b = -35

b = 5                  (IV)

Putting the value of b in equation (III), we get

a=11352=42=2$a=\frac{11-3*5}{2}=\frac{-4}{2}=-2$

Hence, a = -2, b = 5

Also,

b = ma + 3

5 = -2m +3

-2m = 2

m = -1

Question 3:

Using substitution method, find the solution by forming the pair of linear equations for the given problems.

i) The coach of a cricket team buys 7 bats and 6 balls. It cost her Rs 3800. Later, she buys 3 bats and 5 balls. It cost her Rs 1750. Find the cost of single bat and single

Solution:

Let the cost a bat be x and cost of a ball be y.

According to the question,

7x + 6y = 3800 …………….. (I)

3x + 5y = 1750 ……………… (II)

From (I), we get

y=38007x6$y=\frac{3800-7x}{6}$ ………………. (III)

Substituting (III) in (II). we get

3x+5(38007x6)=1750$3x+5\left ( \frac{3800-7x}{6} \right )=1750$ 3x+9500335x6=1750$3x+\frac{9500}{3}-\frac{35x}{6}=1750$ 3x35x6=175095003$3x-\frac{35x}{6}=1750-\frac{9500}{3}$ 18x35x6=525095003$\frac{18x-35x}{6}=\frac{5250-9500}{3}$ 17x6=42503$-\frac{17x}{6}=\frac{-4250}{3}$

-17x = -8500

x = 500 ……………….. (IV)

Substituting the value of x in (III), we get

y=380075006=3006=50$y=\frac{3800-7*500}{6}=\frac{300}{6}=50$

Hence, the cost of a bat is Rs 500 and cost of a ball is Rs 50.

ii) The difference between two numbers is 26 and one number is three times the older. Find the two numbers.

Solution:

Let the two numbers be x and y respectively, such that y > x.

According to the question,

y = 3x ……………… (1)

y – x = 26 …………..(2)

Substituting the value of (1) into (2), we get

3x – x = 26

x = 13 ……………. (3)

Substituting (3) in (1), we get y = 39

Hence, the numbers are 13 and 39.

iii) Five years hence, the age of Rahul will be three times that of his son. Five years ago, Rahul’s age was seven times that of his son. Calculate their present ages.

Solution:

Let the age of Rahul and his son be x and y respectively.

According to the question,

(x + 5) = 3(y + 5)

x – 3y = 10 ………………… (1)

(x – 5) = 7(y – 5)

x – 7y = -30 ……………….. (2)

From (1), we get x = 3y + 10 ……………………. (3)

Substituting the value of x in (2), we get

3y + 10 – 7y = -30

-4y = -40

y = 10 ………………… (4)

Substituting the value of y in (3), we get

x = 3 x 10 + 10

= 40

Hence, the present age of Rahul and his son is 40 years and 10 years respectively.

iv) If 2 is added to both the numerator and the denominator a fraction becomes 8/11. If 3 is added to both the numerator and the denominator it becomes 5/6. Find the fraction.

Solution:

Let the fraction be x/y.

According to the question,

x+2y+2=911$\frac{x+2}{y+2}=\frac{9}{11}$

11x + 22 = 9y + 18

11x – 9y = -4 …………….. (1)

x+3y+3=56$\frac{x+3}{y+3}=\frac{5}{6}$

6x + 18 = 5y +15

6x – 5y = -3 ………………. (2)

From (1), we get x=4+9y11$x=\frac{-4+9y}{11}$ …………….. (3)

Substituting the value of x in (2), we get

6(4+9y11)5y=3$6\left ( \frac{-4+9y}{11} \right )-5y=-3$

-24 + 54y – 55y = -33

-y = -9

y = 9 ………………… (4)

Substituting the value of y in (3), we get

x=4+8111=7$x=\frac{-4+81}{11}=7$

Hence the fraction is 7/9.

v) The larger of two supplementary angles exceeds the smaller by 18o. Find the two angles.

Solution:

Let the larger angle by xo and smaller angle be bo.

We know that the sum of two supplementary pair of angles is always 180o.

According to the question,

x + y = 180o……………. (1)

x – y = 18……………..(2)

From (1), we get x = 180o – y …………. (3)

Substituting (3) in (2), we get

180– y – y =18o

162o = 2y

y = 81o ………….. (4)

Using the value of y in (3), we get

x = 180o – 81o

= 99o

Hence, the angles are 99o and 81o.

vi) A taxi charge consists of a fixed charge + charge for the distance covered. For a distance of 10 km and 15 km, the charge paid is Rs 105 and Rs 155 respectively. Calculate the fixed charge and the charge per km. Also, calculate travel cost for 25 km.

Solution:

Let the fixed charge be Rs x and per km charge be Rs y.

According to the question,

x + 10y = 105 …………….. (1)

x + 15y = 155 …………….. (2)

From (1), we get x = 105 – 10y ………………. (3)

Substituting the value of x in (2), we get

105 – 10y + 15y = 155

5y = 50

y = 10 …………….. (4)

Putting the value of y in (3), we get

x = 105 – 10 * 10 = 5

Hence, fixed charge is Rs 5 and per km charge = Rs 10

Charge for 25 km = x + 25y = 5 + 250 = Rs 255