2 Variable Equations

5m + 7n = 11         ...(1)

7m + 5n = 1         ...(2)

Adding equations (1) and (2) we get,  

12m + 12n = 12

∴ m + n = 1              ...(3)

Subtracting equation (1) from equation (2) we get,

2m − 2n = −10

∴ m − n = − 5           ...(4)

Solving equation (3) and equation (4) we get, 

m = −2 and n = 3  

Alternately,

(1) × 7 – (2) × 5, we get

24n = 72

⇒ n = 3

Substituting value of n in 1st equation, we get m = -2.

Hence, option (d).

Give, 2x – x/3 = 5.

⇒ 6x - x = 15.

⇒ 5x = 15

⇒ x = 3

Hence, 3.

Let the two numbers be x and y. 

⇒ x + y = 72    …(1)

and x - y = 8    …(2)

Solving (1) and (2) we get,

x = 40 and y = 32

Hence, option (c).

Let the costs of each pen and each eraser be Rs. p and Rs. e respectively

∴ 4p + 5e = 128    …(1)

∴ 5p + 4e = 124    …(2)

Adding (1) and (2),

⇒ 9 × (p + e) = 252

⇒ p + e = 28    …(3)

Subtracting (1) from (2), 

⇒ p - e = - 4    …(4)

Adding (3) and (4),

⇒ 2p = 24

⇒ p = 12.

Hence, option (b).

Let the number of hundred rupee notes be x and the number of fifty rupee notes be y.

x + y = 64                      ...(1)

100x + 50y = 5500        ...(2)

Equation (2) can be rewritten as,

50x + 50x + 50y = 5500

⇒ 50x + 50(x + y) = 5500

∴ 50x + 3200 = 5500

∴ x = 46

∴ The number of hundred rupee notes is 46. 

Hence, option (a).

Let the son's present age be x years. Then, man's present age = (x + 48) years.

In four years, his age will be twice the age of his son.

∴ (x + 48) + 4 = 2(x + 4)

⇒ x + 52 = 2x + 8

⇒ x = 44.

Hence, option (d).

Let the fraction be x/y.

From the first condition we have, 

$\frac{\mathrm{x}+1}{\mathrm{y}+1}=\frac{2}{3}$

∴ 3x – 2y = – 1      … (1)

From the second condition we have,

$\frac{\mathrm{x}-1}{\mathrm{y}-1}=\frac{5}{8}$

∴ 8x – 5y = 3                 ... (2)

Solving equation (1) and (2)

⇒ x = 11 and y = 17

∴ The original fraction = 11/17

Hence, option (c).

Let p = 1/x and q = 1/y

5p + 7q = 17         ...(1)

11p + 6q = 28         ...(2)

Now, we have got a system of linear simultaneous equations.

(1) × 11 - (2) × 5 we get,

⇒ 77q – 30q = 187 - 140

⇒ 47q = 47

⇒ q = 1

Substituting value of q in (1) we get p = 2

Now, p = 2 = 1/x ⇒ x = ½

and, q = 1 = 1/y ⇒ y = 1

∴ x + y = 3/2

Hence, option (a).

Let the present ages of son and father be x and (36 - x) years respectively.

Three years ago, father's age was five times the age of the son.

∴ (36 - x) - 3 = 5(x - 3)

⇒ 33 - x = 5x - 15

⇒ 6x = 48

⇒ x = 8.

∴ Son's age after 3 years = (x + 3) = 11 years.

Hence, option (d).

Let the number be ‘ab’.

Sum of the digits i.e., a + b = 7   …(1)

Difference of the digits is 1.

Now we don’t know out of a and b which is greater, hence, we’ll have to make two cases.

Case 1: a ≥ b

⇒ a – b = 1   …(2)

Solving equations (1) and (2) we get,

a = 4 and b = 3.

Hence, the number is 43.

Case 1: b ≥ a

⇒ b – a = 1   …(3)

Solving equations (1) and (3) we get,

a = 3 and b = 4.

Hence, the number is 34.

Now the smallest such number is clearly 34.

Hence, 34.