### NCERT Solutions Class 10 Maths Chapter – 10 Exercise – 3.7

- by Dev Bhardwaj

*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

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:-**

** **

Y^{2} + 10y – 5x = 0 (I)

3y^{2} – 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 180^{O}.

Thus, ∠ A +∠B+ ∠C = 180^{O}

∠A + ∠B +3 ∠B = 180^{O}

∠A + 4 ∠B = 180^{O}– – – – – – – – – – – – – – -(ii)

Multiplying 4 to equation (i) , we get

8 ∠A – 4 ∠B = 0- – – – – – – – – – – – (iii)

Adding equations (iii) and (ii) we get

9 ∠A = 180^{O}

∠A = 20^{O}

Using this in equation (ii), we get

20^{O}+ 4∠B = 180^{O}

∠B = 40^{O}

3∠B =∠C

∠C = 3 x 40 = 120^{O}

Therefore, ∠A = 20^{O}

∠B=40^{O}

∠C = 120^{O}

**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:**

**P****x + qy = p – q**

**Q****x – py = p + q**

**A****x + by = c**

**B****x + ay = 1 + c**

**X****/a – y/b = 0**

**A****x + by = a**^{2}** + b**^{2}

**(a – b)x + (a + b)y = a**^{2}**– 2ab – b**^{2}

**(a + b)(x + y) = a**^{2}** + b**^{2}

**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

P^{2}x + pqy = p^{2} – pq (iii)

q^{2}x – pqy = pq + q^{2} (iv)

Adding equation (iii) and equation (iv) we get

P^{2}x + pqy = p^{2} + q^{2}

(p^{2} + q^{2} )x = p^{2} + q^{2}

X = (p^{2} + q^{2}) / p^{2} +q^{2} = 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

A^{2}x + aby = ac (iii)

B^{2}x + aby = b + bc (iv)

Subtracting equation (iv) from equation (iii)

(a^{2} – b^{2})x = ac – bc – b

X = (ac – bc – b)/ (a^{2} – b^{2})

X = c(a – b) – / (a^{2} + b^{2})

From equation (I) we obtain

Ax + by = c

A{c(a – b) – b)/(a^{2} – b^{2})} + by = c

By = abc – b^{2}c + ab/a^{2} – b^{2}

Y = c(a – b) + a/a^{2}-b^{2}

- X/a – y/b = 0 (I)

X + by = a^{2} + b^{2 }(ii)

Multiplying a and b to equation (I) and (ii) respectively we get

B^{2}x = aby = 0 (iii)

A^{2}x + aby = a^{3} + ab^{3} (iv)

Adding equations (iii) and (iv) we get

B^{2}x + a^{2}x = a^{3}_{ }+ ab^{2}

X(b^{2} + a^{2}) = a(a^{2} + b^{2})x = a

Using equation (I) we get

B(a) – ay = 0

Ab – ay = 0

Ay = ab,

Y = b

- (a – b)x + (a +b)y = a
^{2}-2ab – b^{2 }(I)

(a + b)(x + y) = a^{2 } + b^{2} (ii)

Subtracting equation (ii) from equation(I) we get

(a – b)x – (a +b)x = (a^{2} – 2ab – b^{2}) – (a^{2} + b^{2})

X = b + a

Substituting this value in equation (I) we get

(a +b)(a-b) + y(a +b) = a^{2}– 2ab – b^{2}

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)^{2}y + 189 x 37 + (76)^{2}y = -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 180^{o}

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.

NCERT Solutions For Class 10 Maths Chapter – 3 Exercise – 3.7 Q1. The ages of two friends Ani and…

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