﻿ Class 10 NCERT Math Solution
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Exercise - 3.7 (Optional)

Question-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 age of Ani and Biju is x years and y years respectively.
Therefore, Ani's father Dharam = 2x years
And Biju's sister Cathy = y/2 years
According to question :

Case I. When Ani is older than Biju by 3 years,

x - y = 3 .....(i)

2x - y/2 = 30
4x - y = 60 ....(ii)

By substracting eq (i)and eq (ii)
x - y - 4x + y = 3 - 60
-3x = -57
x = 57/3
x = 19

Put the value of x = 19 in eq (i)
19 - y = 3
-y = 3 - 19
-y = -16
y = 16

Hence, Ani's age is 19 years and Biju's age is 16 years.

Case II. When Biju is older than Ani

y - x = 3 .......(i)

2x - y/2 = 30
4x - y = 60 .....(ii)

By adding eq (i)and eq (ii)
y - x + 4x - y = 3 + 60
3x = 63
x = 63/3
x = 21

Put the value of x = 21 in eq (i)
y - 21 = 3
y = 3 +21
y = 24

Hence, Ani's age is 21 years and Biju's age is 24 years.
```

Question-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 those freinds having x and y respectively.
Acccording to question :

x + 100 = 2(y - 100)
x + 100 = 2y - 200
x - 2y = -200 - 100
x - 2y = -300 ......(i)

6(x - 10) = (y + 10)
6x - 60 = y + 10
6x - y = 10 + 60
6x - y = 70  ......(ii)

For equal of any one variable(x or y), mutiplied by their coefficients following below:
x - 2y = -300 ......X 6
6x - y = 70 ........X 1

Now,
6x - 12y = -1800  ......(iii)
6x - y = 70 .....(iv)

By substracting eq (iii) and eq (iv)
6x - 12y - 6x + y = -1800 - 70
-11y = -1870
y = 1870/11
y = 170

Put the value of y = 170 in eq (i)
x - 2 x 170 = -300
x = -300 + 340
x = 40

Hence, friends have Rs. 40 and Rs. 170 respectively.
```

Question-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 speed of train be x km/h, time taken by train is t and distance covered by train is d.
We know that, speed = Distance/time taken
so, x = d/t or d = tx
According to question :

(x + 10) = d/(t - 2)
(x + 10)(t - 2) = d
tx - 2x + 10t - 20 = tx
-2x + 10t = 20 .....(i)

(x - 10) = d/(t + 3)
(x - 10)(t + 3) = d
tx + 3x - 10t - 30 = tx
3x - 10t = 30 ......(ii)

By adding eq (i) and eq (ii)
-2x + 10t + 3x - 10t = 20 + 30
x = 50

Put the value of x= 50 in eq (i)
-2 x 50 + 10t = 20
-100 + 10t = 20
10t = 20 + 100
t = 120/10
t = 12

Hence, speed of train (x) = 50 km/h and time taken (t) = 12 hours
So, distance covered by train (d) = tx = 12 x 50 = 600 km
```

Question-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 no. of rows be x and no. of students in a row be y.
Now, total students in a class = no. of rows x no. of students in a row = xy
According to question :

Case I.
total no. of students = (x - 1)(y + 3)
xy = xy + 3x - y - 3
3x - y = 3 .....(i)

Case II.
total no. of students = (x + 2)(y - 3)
xy = xy - 3x + 2y - 6
-3x + 2y = 6 .....(ii)

By Adding eq (i) and eq (ii)
3x - y - 3x + 2y = 3 + 6
y = 9

Put the value of y = 9 in eq (i)
3x - 9 = 3
3x = 3 + 9
x = 12/3
x = 4

Hence, no. of rows (x) = 4 and no. of students in a row (y) = 9.
Now, total students in a class (xy) = 4 x 9 = 36
```

Question-5 :-  In a Δ ABC, ∠ C = 3 ∠ B = 2 (∠ A + ∠ B). Find the three angles.

Solution :-
```  Given that : ∠ C = 3 ∠ B = 2 (∠ A + ∠ B)
∠C = 3∠B = 2(∠A + ∠B)
3∠B = 2(∠A + ∠B)
3∠B = 2∠A + 2∠B
3∠B - 2∠B = 2∠A
∠B = 2∠A
-2∠A + ∠B = 0 .....(i)

We know that the sum of all angles in a triangle is 180°.
Therefore, ∠A + ∠B + ∠C = 180°
∠A + ∠B + 3∠B = 180°
∠A + 4∠B = 180° .....(ii)

Let us consider by eq (i)
∠B = 2∠A ......(iii)

Put the value of ∠B = 2∠A in eq (ii)
∠A + 4 x 2∠A = 180°
∠A + 8∠A = 180°
9∠A = 180°
∠A = 180°/9
∠A = 20°

Put the value of ∠A = 20° in eq (iii)
∠B = 2∠A = 2 x 20° = 40°

Put the value of ∠B = 40° in ∠C = 3∠B
∠C = 3 x 40° = 120°
```

Question-6 :-  Draw the graphs of the equations 5x – 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 and 3x – y = 3

Lets represent these equations in graphically:
5x – y = 5  .......(i)

x01

y-50

3x – y = 3......(ii)

x01

y-30

Plot the points A(0, -5), B(1, 0) and P(0, -3), Q(1, 0), corresponding to the solutions in Table.
Now draw the lines AB and PQ, representing the equations 5x – y = 5  and 3x – y = 3, as shown in following Figure: ```

Question-7 :-  Solve the following pair of linear equations:
(i) px + qy = p – q and qx – py = p + q
(ii) ax + by = c and bx + ay = 1 + c
(iii) x/a - y/b = 0 and ax + by = a² + b²
(iv) (a – b)x + (a + b) y = a² – 2ab – b² and (a + b)(x + y) = a² + b²
(v) 152x – 378y = – 74 and –378x + 152y = – 604

Solution :-
```(i)  px + qy = p – q .......(i)
qx – py = p + q ..........(ii)

For equal of any one variable(x or y), mutiplied by their coefficients following below:
px + qy = p – q .......X p
qx – py = p + q .......X q

Now,
p²x + pqy = p² - pq .....(iii)
q²x - pqy = pq + q² .....(iv)

By adding eq (i) and eq (ii)
p²x + pqy + q²x - pqy = p² - pq + pq + q²
x(p² + q²) = p² + q²
x = (p² + q²)/(p² + q²)
x = 1

Put the value of x = 1 in eq (i)
p x 1 + qy = p - q
p + qy = p - q
qy = -q
y = -q/q
y = -1

x = 1, y = -1
```
```(ii) ax + by = c .....(i)
bx + ay = 1 + c .....(ii)

For equal of any one variable(x or y), mutiplied by their coefficients following below:
ax + by = c ........X b
bx + ay = 1 + c ....X a

Now,
abx + b²y = bc ......(iii)
abx + a²y = a + ac ...(iv)

By substracting eq (iii) and eq (iv)
abx + b²y - abx - a²y = bc - a - ac
y(b² - a²) = bc - a - ac
y = (bc - a - ac)/(b² - a²)

Put the value of y = (bc - a - ac)/(b² - a²) in eq (i)
ax + b [(bc - a - ac)/(b² - a²)] = c
ax = c - (b²c - ab - abc)/(b² - a²)
ax = (cb² - ca² - b²c + ab + abc)/(b² - a²)
ax = a(b + bc - ca)/(b² - a²)
x = (b + bc - ca)/(b² - a²)

x = (b + bc - ca)/(b² - a²), y = (bc - a - ac)/(b² - a²)
```
```(iii) x/a - y/b = 0
bx - ay = 0 ......(i)
ax + by = a² + b² ......(ii)

For equal of any one variable(x or y), mutiplied by their coefficients following below:
bx - ay = 0 .........X a
ax + by = a² + b² ...X b

Now,
abx - a²y = 0 .....(iii)
abx + b²y = b(a² + b²) ......(iv)

By  substracting eq (iii) and eq (iv)
abx - a²y - abx - b²y = 0 - b(a² + b²)
-y(a² + b²) = - b(a² + b²)
y = b

Put the value of y = b in eq (i)
bx - ab = 0
b(x - a) = 0
x - a = 0
x = a

x = a, y = b

bx - ay = 0 ......(i)
```
```(iv) (a – b)x + (a + b) y = a² – 2ab – b² ....(i)
(a + b)(x + y) = a² + b²
(a + b)x + (a + b)y = a² + b² ......(ii)

By substracting eq (i) and eq (ii)
(a – b)x + (a + b)y - (a + b)x - (a + b)y = a² – 2ab – b² - a² - b²
x[(a - b - a - b)] = -2ab - 2b²
-2bx = -2b(a + b)
x = a + b

Put the value of x = a + b in eq (ii)
(a + b)(a + b) + (a + b)y = a² + b²
a² + b² + 2ab + (a + b)y = a² + b²
(a + b)y = -2b
y = -2b/(a + b)
x = a + b, y = -2b/(a + b)
```
```(v) 152x – 378y = – 74
76x - 189y = -37 .....(i)

–378x + 152y = – 604
-189x + 76y = -302 .....(ii)

Let us consider eq (i)
76x - 189y = -37
76x = -37 + 189y
x = (-37 + 189y)/76 ....(iii)

Put the value of x = (-37 + 189y)/76 in eq (ii)
-189[(-37 + 189y)/76] + 76y = -302
-189(-37 + 189y) + 5776y = -22952
6993 - 35721y + 5776y = -22952
-29945y = -22952 - 6993
-29945y = -29945
y = 1

Put the value of y = 1 in eq (iii)
x = (-37 + 189 x 1)/76
x = 152/76
x = 2

x = 2, y = 1
```

Question-7 :-  ABCD is a cyclic quadrilateral below figure. Find the angles of the cyclic quadrilateral. Solution :-
```  We know that sum of the measures of opposite angles in a cyclic quadrilateral is 180°.
Here, ∠A = 4y + 20, ∠B = 3y - 5, ∠C = -4x, ∠D = -7x + 5
Therefore, ∠A + ∠c = 180°
4y + 20 - 4x = 180°
-4x + 4y = 180° - 20
4(-x  + y) = 160°
-x + y = 40° ......(i)

Also, ∠B + ∠D = 180°
3y - 5 - 7x + 5 = 180°
-7x + 3y = 180° .....(ii)

For equal of any one variable(x or y), mutiplied by their coefficients following below:
-x + y = 40°  ..........X 7
-7x + 3y = 180° ........X 1

Now,
-7x + 7y = 280°  ......(iii)
-7x + 3y = 180° ......(iv)

By substracting eq (iii) and eq (iv)
-7x + 7y + 7x - 3y = 280° - 180°
4y = 100°
y = 100°/4
y = 25°

Put the value of y = 25° in eq (i)
-x + 25° = 40°
-x = 40° - 25°
-x = 15°
x = -15°

Hence, ∠A = 4y + 20 = 4 x 25 + 20 = 120°
∠B = 3y - 5 = 3 x 25 - 5 = 70°
∠C = -4x = (-4) x (-15) = 60°
∠D = -7x + 5 = (-7) x (-15) + 5 = 110°
```
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