# 2 Variable Equations | Algebra - Simple Equations

**2 Variable Equations | Algebra - Simple Equations**

Find the value of m, from the following simultaneous equations.

5m + 7n = 11

7m + 5n = 1

- A.
3

- B.
-3

- C.
2

- D.
-2

- E.
-1

Answer: Option D

**Explanation** :

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).

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**2 Variable Equations | Algebra - Simple Equations**

Twice of a number is 5 more than one-third of it. Find the number.

Answer: 3

**Explanation** :

Give, 2x – x/3 = 5.

⇒ 6x - x = 15.

⇒ 5x = 15

⇒ x = 3

Hence, 3.

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**2 Variable Equations | Algebra - Simple Equations**

The sum of two numbers is 72. The difference of the numbers is 8. Find the numbers.

- A.
22, 14

- B.
24, 48

- C.
40, 32

- D.
16, 24

Answer: Option C

**Explanation** :

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).

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**2 Variable Equations | Algebra - Simple Equations**

Four pens and five erasers cost Rs. 128. Five pens and four erasers cost Rs. 124. Find the cost of each pen (in Rs.).

- A.
8

- B.
12

- C.
16

- D.
20

Answer: Option B

**Explanation** :

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).

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**2 Variable Equations | Algebra - Simple Equations**

Chetan has Rs. 5,500 in her purse in denominations of hundred and fifty. She has 64 notes in all counting both hundred and fifty. How many hundred rupee notes does she have in her purse?

- A.
46

- B.
18

- C.
48

- D.
16

- E.
50

Answer: Option A

**Explanation** :

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).

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**2 Variable Equations | Algebra - Simple Equations**

A man is 48 years older than his son. In four years, his age will be twice the age of his son. The present age of his son is:

- A.
28 years

- B.
36 years

- C.
40 years

- D.
44 years

Answer: Option D

**Explanation** :

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).

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**2 Variable Equations | Algebra - Simple Equations**

If 1 is added to the numerator and denominator of a certain fraction, its value becomes 2/3 and if 1 is subtracted from the numerator and denominator of the original fraction, its value becomes 5/8. Find the original fraction.

- A.
20/57

- B.
13/39

- C.
11/17

- D.
14/42

- E.
13/38

Answer: Option C

**Explanation** :

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).

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**2 Variable Equations | Algebra - Simple Equations**

Find the value of (x + y), from the given set of equations.

$\frac{5}{x}+\frac{7}{y}=17$

$\frac{11}{x}+\frac{6}{y}=28$

- A.
3/2

- B.
1/2

- C.
5/2

- D.
2

Answer: Option A

**Explanation** :

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).

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**2 Variable Equations | Algebra - Simple Equations**

The sum of the present ages of a father and his son is 36 years. Three years ago, father's age was five times the age of the son. After 3 years, son's age will be:

- A.
8 years

- B.
14 years

- C.
18 years

- D.
11 years

Answer: Option D

**Explanation** :

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).

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**2 Variable Equations | Algebra - Simple Equations**

In a two digit number sum of the digits is 7 and difference of the digits is 1. Find the smallest such number.

Answer: 34

**Explanation** :

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.

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