This is aptitude questions and answers section on Simple Equations with explanation for various interview, competitive examinations and entrance tests.

1. Solve the equation for x : 6x - 27 + 3x = 4 + 9 - x
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2. Solve the equation for x : 19(x + y) + 17 = 19(-x + y) - 21
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Answer: Option A

Explanation:

19x + 19y + 17 = -19x + 19y - 21

38x = -38 => x = -1

3. The cost of 2 chairs and 3 tables is Rs.1300. The cost of 3 chairs and 2 tables is Rs.1200. The cost of each table is more than that of each chair by?
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Answer: Option E

Explanation:

2C + 3T = 1300 --- (1)

3C + 3T = 1200 --- (2)

Subtracting 2nd from 1st, we get

-C + T = 100 => T - C = 100

4. The denominator of a fraction is 1 less than twice the numerator. If the numerator and denominator are both increased by 1, the fraction becomes 3/5. Find the fraction?
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Answer: Option D

Explanation:

Let the numerator and denominator of the fraction be 'n' and 'd' respectively.

d = 2n - 1

(n + 1)/(d + 1) = 3/5

5n + 5 = 3d + 3

5n + 5 = 3(2n - 1) + 3 => n = 5

d = 2n - 1 => d = 9

Hence the fraction is : 5/9.

5. The cost of 10 kg of apples is equal to the cost of 24 kg of rice. The cost of 6 kg of flour equals the cost of 2 kg of rice. The cost of each kg of flour is Rs.20.50. Find the total cost of 4 kg of apples, 3 kg of rice and 5 kg of flour?
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Answer: Option B

Explanation:

Let the costs of each kg of apples and each kg of rice be Rs.a and Rs.r respectively.

10a = 24r and 6 * 20.50 = 2r

a = 12/5 r and r = 61.5

a = 147.6

Required total cost = 4 * 147.6 + 3 * 61.5 + 5 * 20.5

= 590.4 + 184.5 + 102.5 = Rs.877.40