NCERT Solutions Maths Class 10 Chapter 3

# NCERT Solutions Maths Class 10 Chapter 3

## NCERT Solutions Maths Class 10 Exercise 3.7

1. 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.

Solution: Let the ages of Ani and Biju be x years and y years, respectively.

Age of Dharam = 2x years and age of Cathy = y/2 years

Case 1: Let x > y
According to the question, we have

xy = 3 … (1)

And 2x y/2 = 30

4xy = 60 … (2)

Subtracting equation (1) from equation (2), we get
3x = 60 – 3 = 57

3x = 57

x = 19

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

19 – y = 3

y = 16

Thus, the age of Ani is 19 years and the age of Biju is 16 years.

Case 2: Let y > x

According to the question, we have

yx = 3 … (3)

And 2xy/2 = 30

4xy = 60 … (4)

Adding equations (3) and (4), we get

3x = 63

x = 21

Putting the value of x in equation (3), we get

y – 21 = 3

y = 24
Thus, the age of Ani is 21 years and the age of Biju is 24 years.

2. 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]

Solution:  Let the money with the first person be Rs x and the money with the second person be Rs y.
According to the first condition, we have

x + 100 = 2(y – 100)

x + 100 = 2y – 200

x – 2y = –300 … (1)

According to the second condition, we have

6(x – 10) = (y + 10)

6x – 60 = y + 10

6xy = 70 … (2)

Multiplying equation (2) by 2, we get

12x – 2y = 140 … (3)

Subtracting equation (1) from equation (3), we get

11x = 140 + 300

11x = 440

x = 40

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

40 – 2y = –300

40 + 300 = 2y

2y = 340

y = 170

Thus, the two friends have Rs 40 and Rs 170 with them.

3. 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.

Solution: Let the speed of the train be x km/h, the time taken by the train to cover the given distance be t hours and the distance to cover be d km.

Since speed = distance/time

x = d/t

d = xt … (1)
According to the first condition, we have

x + 10 = d/(t – 2)

(x + 10)(t – 2) = d

xt + 10t – 2x – 20 = d

10t – 2x = 20 … (2)                    [Using eq. (1)]

According to the second condition, we have

x – 10 = d/(t + 3)

(x – 10)(t + 3) = d

xt – 10t + 3x – 30 = d

3x – 10t = 30  … (3)                 [Using eq. (1)]

Adding equations (2) and (3), we get
x = 50
Putting the value of x in equation (2), we get
10t – 2 × 50 = 20

10t – 100 = 20

10t = 120

t = 12

From equation (1), we get

d = xt = 50 × 12 = 600

Thus, the distance covered by the train is 600 km.

4. 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.

Solution: Let the number of rows be x and the number of students in a row be y.
Total number of students in the class = Number of rows × Number of students in a row

= xy

According to the first condition, we have

Total number of students = (x – 1) (y + 3)

xy = (x – 1) (y + 3)

xy = xyy + 3x – 3

3xy – 3 = 0

3xy = 3 … (1)

According to the second condition, we have

Total number of students = (x + 2) (y – 3)

xy = xy + 2y – 3x – 6

3x – 2y = –6 … (2)
Subtracting equation (2) from equation (1), we get
y = 9
Putting the value of y in equation (1), we get
3x – 9 = 3

3x = 9 + 3 = 12

x = 4

Thus, the number of rows = 4

And the number of students in a row = 9
Hence, the total number of students in the class = xy = 4 × 9 = 36

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

Solution: C = 3B = 2(A + B)

Taking C = 3B  … (1)

And 3B = 2(A + B)

3B = 2A + 2B

B = 2A

2A – B = 0 … (2)

We know from the angle sum property that the sum of the measures of all the angles of a triangle is 180°.

A + B + C = 180°

A + B + 3B = 180°       [From eq. (1)]

A + 4B = 180°… (3)

Multiplying equation (2) by 4, we get

8A – 4B = 0 … (4)

Adding equations (3) and (4), we get

9A = 180°

A = 20°

Putting the value of A in equation (3), we get,

20° + 4B = 180°

B = 40°

And C = 3 × 40° = 120°

Hence, the measures of A, B and C are 20°, 40° and 120°, respectively.

6. Draw the graphs of the equations 5xy = 5 and 3xy = 3. Determine the co-ordinate of the vertices of the triangle formed by these lines and the y-axis.

Solution: We have the first equation: 5xy = 5

y = 5x – 5

For x = 0, 1, 2, we have y = –5, 0, 5 as shown in the following table.

 x 0 1 2 y –5 0 5

We have the second equation:

3xy = 3

y = 3x – 3

For x = 0, 1, 2, we have y = –3, 0, 3 as shown in the following table.

 x 0 1 2 Y –3 0 3

Let us draw these points on a graph paper to find the lines for the given equations.

It can be seen from the above graph that the required triangle is ACE.

The coordinates of its vertices are A(1, 0), C(0, –3), E(0, –5).

7. Solve the following pair of linear equations:

(i) px + qy = pq

qxpy = p + q

(ii) ax + bx = c

bx + ay = 1 + c

(iii) x/a y/b = 0

ax + by = a2 + b2

(iv)  (ab)x + (a + b)y = a2 – 2abb2

(a + b)(x + y) = a2 + b2

(v) 152x – 378y = –74

–378x + 152y = –604

Solution: (i) px + qy = pq …. (1)

qxpy = p + q …. (2)

Multiplying equation (1) by p and equation (2) by q, we get

p2x + pqy = p2 pq … (3)
q2xpqy = pq + q2 … (4)
Adding equations (3) and (4), we get

p2x + q2x = p2 + q2

(p 2 + q2)x = (p2 + q2)
x = 1

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

p(1) + qy = pq

qy = –q

y = –1

Hence, x = 1 and y = –1

(ii) ax + bx = c …. (1)

bx + ay = 1 + c  …. (2)

Multiplying equation (1) by a and equation (2) by b, we get

a2x + aby = ac … (3)
b2x + aby = b + bc … (4)
Subtracting equation (4) from equation (3), we get
(a2 b2)x = acbcb

(a2 b2)x = c(ab) – b

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

(iii) x/ay/b = 0

bx ay = 0 …… (1)

ax + by = a2 + b2  …… (2)

Multiplying equation (1) by b and equation (2) by a, we get

b2xaby = 0  ……. (3)

a2x + aby = a3 + ab2 ……. (4)

Adding equations (3) and (4), we get

b2x + a2x = a3 + ab2

x(b2 + a2) = a(a2 + b2)

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

b(a) – ay = 0

abay = 0

y = b

Thus, x = a and y = b

(iv)  (ab)x + (a + b)y = a2 – 2abb2 ….. (1)

(a + b)(x + y) = a2 + b2

(a + b)x + (a + b)y = a2 + b2 ….. (2)

Subtracting equation (2) from equation (1), we get

(ab)x – (a + b)x = (a2 – 2abb2) – (a2 + b2)
(abab)x = –2ab – 2b2
–2bx = –2b(a + b)
x = a + b
Putting the value of x in equation (1), we get

(ab)(a + b) + (a + b)y = a2 – 2abb2
a2 b2 + (a + b)y = a2 – 2abb2
(a + b)y = –2ab

y = –2ab/(a + b)
Thus, x = a + b and y = –2ab/(a + b)

(v) 152x – 378y = –74 ….. (1)

–378x + 152y = –604 ….. (2)

Adding the equations (1) and (2), we get

–226x – 226y = –678

x + y = 3 …… (3)

Subtracting equation (2) from equation (1), we get

530x – 530y = 530

xy = 1 ……. (4)
Adding equations (3) and (4), we get

2x = 4

x = 2
Putting the value of x in equation (3), we get

2 + y = 3

y = 1

Thus, x = 2 and y = 1

8. ABCD is a cyclic quadrilateral (see figure). Find the angles of the cyclic quadrilateral.

Solution: We know that the sum of the measures of opposite angles in a cyclic quadrilateral is 180°.

Therefore, A + C = 180°
4y + 20 – 4x = 180°
–4x + 4y = 160°
xy = –40° …… (1)
Also,
B + D = 180°
3y – 5 – 7x + 5 = 180°
–7x + 3y = 180°…… (2)
Multiplying equation (1) by 3, we get

3x – 3y = –120° ……… (3)
Adding equations (2) and (3), we get

–4x = 60°

x = –15°
Putting the value of x in equation (1), we get

–15 – y = –40°
y = –15 + 40

y = 25
Therefore,
A = 4y + 20 = 4 × 25 + 20 = 120°
B = 3y – 5 = 3 × 25 – 5 = 70°
C = –4x = –4 × (–15) = 60°
D = –7x + 5 = –7(–15) + 5 = 110°