NCERT Solutions For Class 10 Maths Chapter – 3 Exercise – 3.7
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 dffer by 30 years Find the ages of Ani and Biju.
Solution:-
X – y = 3 (I)
X – y = -3 (ii)
Ani father age = 2x years
Biju sister Cathy age = y/2 years
2x – y/2 = 30 = 4x – y = 60 (iii)
Solving (I) and (iii) we get x = 19 , y = 16
Solving (ii) and (iii) we get x = 21 , y = 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]
Solution:-
X – 2y = -300 (I)
6x – y = 70 (ii)
Solving (I) and (ii) we get x = 40 , y = 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 10km/h It would have taken 3 hours more than the scheduled time. Find the distance covered by the train.
Solution:-
Y2 + 10y – 5x = 0 (I)
3y2 – 30y – 10x = 0 (ii)
On solving equations (I) and (ii) we will obtain x = 600 km.
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.
Solution:-
Total number of students = xy
(x+3)(y-1)=xy
-x + 3y = 3 (ii)
(x-3)(y+2)=xy
2x – 3y = 6 (iii)
Solving (ii) and (iii) we get
X = 9 , y = 4
5. In a ∆ABC, ∠ C = 3 ∠ B = 2 (∠A + ∠ B). Find the three angles.
Solution:
Given,
∠C = 3 ∠B = 2(∠B + ∠A)
∠B = 2 ∠A+2 ∠B
∠B=2 ∠A
∠A – ∠B= 0- – – – – – – – – – – – (i)
We know, the sum of all the interior angles of a triangle is 180O.
Thus, ∠ A +∠B+ ∠C = 180O
∠A + ∠B +3 ∠B = 180O
∠A + 4 ∠B = 180O– – – – – – – – – – – – – – -(ii)
Multiplying 4 to equation (i) , we get
8 ∠A – 4 ∠B = 0- – – – – – – – – – – – (iii)
Adding equations (iii) and (ii) we get
9 ∠A = 180O
∠A = 20O
Using this in equation (ii), we get
20O+ 4∠B = 180O
∠B = 40O
3∠B =∠C
∠C = 3 x 40 = 120O
Therefore, ∠A = 20O
∠B=40O
∠C = 120O
Q6. Draw the graphs of the equation5x – y = 5 and 3x – y = 3 Determine the co – ordinates of the vertices of the triangle formed by these lines and the y axis.
Solution:-
5x – y = 5
=> y = 5x – 5
Its solution table will be.
| x | 2 | 1 | 0 |
| y | 5 | 0 | -5 |
Also given , 3x – y = 3
Y = 3x – 3
| x | 2 | 1 | 0 |
| y | 5 | 0 | -5 |
The graphical representation of these lines will be as follows:
Q7. Solve the following pair of linear equations:
- Px + qy = p – q
Qx – py = p + q
- Ax + by = c
Bx + ay = 1 + c
- X/a – y/b = 0
Ax + by = a2 + b2
- (a – b)x + (a + b)y = a2– 2ab – b2
(a + b)(x + y) = a2 + b2
- 152x – 378y = -74
-378x + 152y = -604
Solution:-
- Px + qy = p – q (I)
qx – py = p + q (ii)
Multiplying p to equation (1) and q to equation (2) we get
P2x + pqy = p2 – pq (iii)
q2x – pqy = pq + q2 (iv)
Adding equation (iii) and equation (iv) we get
P2x + pqy = p2 + q2
(p2 + q2 )x = p2 + q2
X = (p2 + q2) / p2 +q2 = 1
From equation (I) we get
P(1) + qy = p – q
qy = p – q – p
Y = -1
- Ax + by = c (I)
Bx + ay = 1 + c (ii)
Multiplying a to equation (I) and b to equation (ii), we obtain
A2x + aby = ac (iii)
B2x + aby = b + bc (iv)
Subtracting equation (iv) from equation (iii)
(a2 – b2)x = ac – bc – b
X = (ac – bc – b)/ (a2 – b2)
X = c(a – b) – / (a2 + b2)
From equation (I) we obtain
Ax + by = c
A{c(a – b) – b)/(a2 – b2)} + by = c
By = abc – b2c + ab/a2 – b2
Y = c(a – b) + a/a2-b2
- X/a – y/b = 0 (I)
X + by = a2 + b2 (ii)
Multiplying a and b to equation (I) and (ii) respectively we get
B2x = aby = 0 (iii)
A2x + aby = a3 + ab3 (iv)
Adding equations (iii) and (iv) we get
B2x + a2x = a3 + ab2
X(b2 + a2) = a(a2 + b2)x = a
Using equation (I) we get
B(a) – ay = 0
Ab – ay = 0
Ay = ab,
Y = b
- (a – b)x + (a +b)y = a2-2ab – b2 (I)
(a + b)(x + y) = a2 + b2 (ii)
Subtracting equation (ii) from equation(I) we get
(a – b)x – (a +b)x = (a2 – 2ab – b2) – (a2 + b2)
X = b + a
Substituting this value in equation (I) we get
(a +b)(a-b) + y(a +b) = a2– 2ab – b2
Y = -2ab/(a +b)
- 152x – 378y = -74 (I)
76x – 189y = -37 (ii)
Using the value of x in equation (ii) we get
-189(189y – 37/76) + 76y = -302
-(189)2y + 189 x 37 + (76)2y = -302 x 76
Y = 1
Using equation (I) we get
X = (189-37)76
X = 2
8. ABCD is a cyclic quadrilateral (see Fig. 3.7). Find the angles of the cyclic quadrilateral.
Solution:
It is known that the sum of the opposite angles of a cyclic quadrilateral is 180o
Thus, we have
∠C +∠A = 180
4y + 20− 4x = 180
− 4x + 4y = 160
x − y = − 40 ……………(1)
And, ∠B + ∠D = 180
3y − 5 − 7x + 5 = 180
− 7x + 3y = 180 ………..(2)
Multiplying 3 to equation (1), we get
3x − 3y = − 120 ………(3)
Adding equation (2) to equation (3), we get
− 7x + 3x = 180 – 120
− 4x = 60
x = −15
Substituting this value in equation (i), we get
x − y = − 40
-y−15 = − 40
y = 40-15
= 25
∠A = 4y + 20 = 20+4(25) = 120°
∠B = 3y − 5 = − 5+3(25) = 70°
∠C = − 4x = − 4(− 15) = 60°
∠D = 5-7x
∠D= 5− 7(−15) = 110°
Hence, all the angles are measured.