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**Q.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:** The age difference between Ani and Biju is 3 yrs.

Either Biju is 3 years older than that of Ani or Ani is 3 years older than Biju. From both the cases we find out that Ani’s father’s age is 30 yrs more than that of Cathy’s age.

Let the ages of Ani and Biju be A and B respectively.

Therefore, the age of Dharam = 2 x A = 2A yrs.

And the age of Biju sister Ann B / 2 yrs

By using the information that is given,

When Ani is older than that of Biju by 3 yrs then A – B = 3.............(1)

2A − B / 2 = 30

4A – B = 60...........(2)

By subtracting the equations (1) and (2) we get,

3A = 60 – 3 = 57

A = 57 / 3 = 19

Therefore, the age of Ani = 19 yrs

And the age of Biju is 19 – 3 = 16 yrs.

When Biju is older than Ani, B – A = 3 ...........(1)

2A − B / 2 = 30

4A – B = 60..............(2)

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

3A = 63

A = 21

Therefore, the age of Ani is 21 yrs

And the age of Biju is 21 + 3 = 24 yrs.

Solution:

Using the information that is given we get,

A + 100 = 2(B – 100) ⇒ A + 100 = 2B – 200

Or A – 2B = -300.............(1)

And

6(A – 10) = ( B + 10 )

Or 6A – 60 = B + 10

Or 6A – B = 70............(2)

When equation (2) is multiplied by 2 we get,

12A – 2B = 140...............(3)

When equation (1) is subtracted from equation (3) we get,

11A = 140 + 300

11A = 440

⇒ A = 440 / 11 = 40

Using A = 40 in equation (1) we get,

40 – 2B = -300

40 + 300 = 2B

2B = 340 B = 170

Therefore, Sangam had Rs 40 and Reuben had Rs 170 with them.

Solution:

Speed of the train = Distance travelled by train / Time taken to travel that distance

A = N (distance) / X (time)

Or, N = AX.............(1)

Using the information that is given, we get:

(A + 10) = X / (N - 2)

(A + 10) (N – 2) = X

AN + 10N – 2A – 20 = X

By using the equation (1) we get,

– 2A + 10N = 20 .............(2)

(A - 10) = X / (N + 3)

(A - 10) (N + 3) = X

AN - 10N + 3A – 30 = X

By using the equation (1) we get,

3A – 10N = 30...........(3)

Adding equation (2) and equation (3) we get,

A = 50

Using the equation (2) we get,

(-2) x (50) + 10N = 20

-100 + 10N = 20

=> 10N = 120

N = 12hours

From the equation (1) we get, Distance travelled by the train, X = AN

= 50 x 12

= 600 km

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

Solution:

Total number of students = Number of rows x Number of students in a row

= AB

Using the information, that is given,

Total number of students = (A – 1) ( B + 3)

Or AB = ( A – 1 )(B + 3) = AB – B + 3A – 3

Or 3A – B – 3 = 0

Or 3A – Y = 3..................(1)

Total Number of students = (A + 2 ) ( B – 3 )

Or AB = AB + 2B – 3A – 6

Or 3A – 2B = -6.................(2)

When equation (2) is subtracted from (1)

(3A – B) – (3A – 2B) = 3 – (-6)

-B + 2B = 3 + 6B = 9

By using the equation (1) we get,

3A – 9 =3

3A = 9 + 3 = 12

A = 4

Number of rows, A = 4

Number of students in a row, B = 9

Number of total students in a class = AB = 4 x 9 = 36

Solution:

∠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

Solution:

5x – y = 5

=> y = 5x – 5

Its solution table will be.

Also given,3x – y = 3

y = 3x – 3

The graphical representation of these lines will be as follows:

From the above graph we can see that the triangle formed is ∆ABC by the lines and the y axis. Also the coordinates of the vertices are A(1, 0), C(0, -5) and B(0, -3).

(i) px + qy = p – q

qx – py = p + q

(ii) ax + by = c

bx + ay = 1 + c

(iii) x / a – y / b = 0

ax + by = a

(iv) (a – b)x + (a + b) y = a

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

(v) 152x – 378y = – 74

–378x + 152y = – 604

Solution:

qx – py = p + q……………….(ii)

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

p

q

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

p

(p

x = (p

From equation (i), we get

p(1) + qy = p – q

qy = p - q - p

qy = -q

y = -1

bx + ay = 1 + c………… ..(ii)

Multiplying a to equation (i) and b to equation (ii), we obtain

a

b

Subtracting equation (iv) from equation (iii),

(a

x = (ac − bc – b) / (a

x = c(a - b) – b / (a

From equation (i), we obtain

ax + by = c

a{c(a − b) − b) / (a

ac(a − b) − ab / (a

by = c – ac(a − b) − ab / (a

by = abc – b

y = c(a - b) + a / a

ax + by = a

x / a – y / b = 0

=> bx − ay = 0 ……. (i)

ax + by = a

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

b

a

Adding equations (iii) and (iv), we get

b

x (b

Using equation (i), we get

b(a) − ay = 0

ab − ay = 0 ay = ab,

y = b

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

(a + b) y + (a – b) x = a

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

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

Subtracting equation (ii) from equation (i), we get

(a − b) x − (a + b) x = (a

x(a − b − a − b) = − 2ab − 2b

− 2bx = − 2b (b + a) x = b + a

Substituting this value in equation (i), we get

(a + b)(a − b) +y (a + b) = a

a

(a + b) y = − 2ab

y = -2ab / (a + b)

76x − 189y = − 37

x = (189y - 137) / 76……………..…(i)

− 378x + 152y = − 604

− 189x + 76y = − 302 ………….. (ii)

Using the value of x in equation (ii), we get

−189(189y − 37/76) + 76y = − 302

− (189)

189 × 37 + 302 × 76 = (189)

6993 + 22952 = (189 − 76) (189 + 76) y

29945 = (113) (265) y

y = 1

Using equation (i), we get

x = (189 - 37) / 76

x = 152 / 76 = 2

Solution:

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.

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