NCERT Solutions for Class 10 Maths Exercise 3.7 of the Chapter 3 Linear Equations in Two Variables - Future Study Point

# NCERT Solutions for Class 10 Maths Exercise 3.7 of the Chapter 3 Linear Equations in Two Variables

NCERT Solutions for Class 10 Maths Exercise 3.7 of the Chapter 3 Linear Equations in Two Variables are the NCERT solutions of an optional exercise of the chapter 3 Linear Equation in Two Variables which contains most important questions not only for the 10 class CBSE board but also important for competitive entrance exams. All NCERT solutions of the exercise 3.7 of the chapter 3 solved here by a step by step method.

Exercise 3.1- Linear Equation in Two Variables

Exercise 3.7 – Linear Equations in Two Variables

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## NCERT Solutions for Class 10 Maths Exercise 3.7 of the Chapter 3 Linear Equations in Two Variables

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Q1.The ages of two friends Ani and Biju differ by 3 years. Ani’s father Dharam is twice as old as Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Dharam differ by 30 years. Find the ages of Ani and Biju.

Ans. Let the age of Ani is = x and of Biju is y

It is given that the difference of both age is 3

If the age of Ani is > The age of Biju

x – y = 3….(i)

It is given to us that Ani’s father Dharam’s age = 2x

Also given

y = 2× Cathy’s age⇒ Cathy’s age is = y/2

It is given that the ages of Cathy and Dharam differ by 30 years

It is clear that Dharam’s age is(i.e 2x) > The age of Cathy(y/2)

2x – y/2 = 30

4x – y = 60……(ii)

Subtracting equation (i) from equation (ii),we get

3x  = 57 ⇒x = 19

Putting the value x = 19 in equation (i)

19 – y = 3 ⇒ y =16

If the age of Biju is more than the age of Ani

y – x = 3…..(iii)

In this case also the age of Dharam is more than the age of Cathy,so the equation reamins the same as (ii)

Adding both equation (iii) and second equation (ii)

3x = 63 ⇒ x = 21

Putting x =21 in equation (iii)

y -21 = 3 ⇒ y = 24

Hence the age of ani when she is older than Biju is 19 and of Biju is 16.and  the age of Ani when Biju is older than Ani is 21 and of Biju is 24

Q2.One says, “Give me a hundred, friend! I shall then become twice as rich as you”. The other replies, “If you give me ten, I shall be six times as rich as you”. Tell me what is the amount of their (respective) capital? [From the Bijaganita of Bhaskara II] [Hint : x + 100 = 2(y – 100), y + 10 = 6(x – 10)].

Ans.Let one has the capital of  Rs x  and the other has the capital of Rs y

According to first condtion

x +100 = 2(y -100)

x -2y = -300……(i)

According to second condition

y + 10 = 6(x -10)

-6x+y  = -70…….(ii)

Multiplying equation (i) by 6,we get equation (iii)

6x -12y = -1800…..(iii)

Adding both equation (ii) and equation (iii)

-11y = -1870

y = 170

Putting y = 170 in equation (i)

x – 2×170 = -300

x -340 = -300

x = 40

Hence one has the capital of Rs 40 and other has the capital of Rs 170

Q3.A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/h, it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.

Ans. Let the original speed of the train is x km/h and the scheduled time to cover the distance is y.

Distance = Original speed × Scheduled time = xy

Speed = distance/time

According to first condition

x + 10 = xy/(y-2)

(x + 10)(y -2) = xy

xy -2x + 10y -20 = xy

-2x + 10 y = 20…….(i)

According to second condition

x -10 = xy/(y + 3)

(x -10)( y+ 3) = xy

xy + 3x -10y -30 = xy

3x -10 y = 30……(ii)

Adding both equation (i) and the equation (ii)

x = 50

Putting x = 50 in equation (i)

-2×50 +10 y = 20

-100 + 10 y =20

10y =120

y = 12

Distance covered is = x y =50×12 = 600

Hence the distance covered by the train is 600 km

See the video for the solutions of Q1,Q2 and Q3

Q4.The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.

Ans. Let the number of rows are x and the number of students in a row is y

The number of students in the class = number of rows ×number of students in a row = xy

According to first condition

xy = (x -1)(y +3)

xy = xy +3x -y -3

3x -y = 3……..(i)

According to second condition

xy = (x +2)(y -3)

xy = xy – 3x + 2y -6

3x- 2y =-6…..(ii)

Subtracting equation (ii) from equation (i)

y = 9

Putting the value y = 9 in equation (i)

3x – 9 = 3

3x = 12⇒ x = 4

The number of students in the class = number of rows(x) ×number of students in a row(y) = 4×9= 36

Q5.In a ∆ABC, C = 3 B = 2 (A + B). Find the three angles.

Ans.We are given

In a ∆ABC, ∠ C = 3 ∠ B = 2 (∠A + ∠ B)

Let ∠ C = 3 ∠ B = 2 (∠A + ∠ B) = x

∠ C = x…..(i) ∠ B = x/3 ….(ii) 2 (∠A + ∠ B) = x⇒ ∠A + ∠ B = x/2…..(iii)

From equation (ii) and equation (iii)

∠A + x/3= x/2

∠A = (x/2) – (x/3)

∠A = x/6……(iv)

From equation (i),equation (ii) and equation (iv)

∠A : ∠ B : ∠ C = x/6  :  x/3  : x

∠A : ∠ B : ∠ C = 1 :  2 : 6

∠A = (1/9) ×180 = 20°

∠ B = (2/9) × 180 =40°

∠ C = (6/9) × 180 =120°

Hence three angles are 20°,40° and 120°

Q6. Draw the graphs of the equations 5x -y = 5 and 3x -y = 3. Determine the coordinates of the vertices of the triangle formed by these lines and the y axis.

Ans. Solutions of the equation 5x -y = 5

 x 0 1 2 y -5 0 5

Solutions of the 3x -y = 3

 x 0 1 2 y -3 0 3

As shown in the graph vertices of the triangle formed by these lines and the y axis are (0,-3),(0,-5) and (1,0)

See the video of Solutions of Question 4 , Question 5, and Question 6

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