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
x – y = 3 … (1)
And 2x – y/2
= 30
⇒ 4x – y
= 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
y – x = 3 … (3)
And 2x – y/2 = 30
⇒ 4x – y
= 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
⇒ 6x – y
= 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 = xy – y + 3x – 3
3x – y – 3 = 0
3x – y = 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 = 3∠B = 2(∠A + ∠B). Find the three angles.
Solution: ∠C = 3∠B = 2(∠A + ∠B)
Taking ∠C = 3∠B … (1)
And 3∠B = 2(∠A + ∠B)
3∠B = 2∠A + 2∠B
∠B = 2∠A
2∠A – ∠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 + 3∠B = 180° [From eq.
(1)]
∠A + 4∠B = 180°… (3)
Multiplying equation (2) by 4, we get
8∠A – 4∠B = 0 … (4)
Adding equations (3) and (4), we get
9∠A = 180°
∠A = 20°
Putting the value of ∠A in
equation (3), we get,
20° + 4∠B = 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 5x
– y = 5 and 3x – y
= 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: 5x
– y = 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:
3x – y = 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 = p – q
qx – py = p + q
(ii) ax + bx = c
bx + ay = 1 + c
(iii) x/a – y/b = 0
ax + by = a^{2 }+ b^{2}
(iv) (a – b)x
+ (a + b)y = a^{2 }– 2ab – b^{2}
(a + b)(x + y)
= a^{2 }+ b^{2}
(v) 152x – 378y = –74
–378x + 152y = –604
Solution: (i) px + qy = p
– q …. (1)
qx – py = p + q …. (2)
Multiplying equation (1) by p and equation (2) by q, we get
p^{2}x + pqy = p^{2 }– pq …
(3)
q^{2}x – pqy = pq + q^{2} … (4)
Adding equations (3) and (4), we get
p^{2}x + q^{2}x = p^{2
}+ q^{2}
(p^{ 2
}+ q^{2})x = (p^{2
}+ q^{2})
x =
1
Putting the value of
x in equation (1), we get
p(1) + qy = p – q
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
a^{2}x + aby = ac …
(3)
b^{2}x + aby = b + bc …
(4)
Subtracting equation (4) from equation (3), we get
(a^{2 }– b^{2})x = ac – bc
– b
(a^{2 }–
b^{2})x = c(a – b)
– b
Putting the value of x in equation (1), we get
(iii) x/a – y/b = 0
bx – ay = 0 ……
(1)
ax + by = a^{2 }+ b^{2 } …… (2)
Multiplying equation (1) by b and equation (2) by a,
we get
b^{2}x – aby =
0 ……. (3)
a^{2}x + aby = a^{3 }+ ab^{2} ……. (4)
Adding equations (3) and (4), we get
b^{2}x + a^{2}x = a^{3
}+ ab^{2}
x(b^{2 }+ a^{2}) = a(a^{2 }+ b^{2})
x = a
Putting the value of x in equation (1), we get
b(a) – ay = 0
ab –
ay = 0
y =
b
Thus, x =
a and y = b
(iv) (a – b)x
+ (a + b)y = a^{2 }– 2ab – b^{2 }….. (1)
(a + b)(x + y) = a^{2 }+ b^{2}
(a + b)x + (a
+ b)y = a^{2 }+ b^{2 }….. (2)
Subtracting equation (2) from equation (1), we get
(a – b)x
– (a + b)x = (a^{2 }– 2ab – b^{2}) – (a^{2 }+ b^{2})
(a – b – a – b)x
= –2ab – 2b^{2}
–2bx = –2b(a + b)
x = a + b
Putting the value of x in equation (1), we get
(a – b)(a
+ b) + (a + b)y = a^{2
}– 2ab – b^{2}
a^{2 }– b^{2 }+ (a + b)y
= a^{2 }– 2ab – b^{2}
(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
x – y = 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°
x – y = –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°