Basic Numeracy/Quantitative Aptitude

Detailed Solution

314. d: choice for the first digit. 4 choices for the last 3 digits that form the 4 digit number since repetition is allowed. Hence the number N of numbers that we may form is given by
N = 1 * 4 * 4 * 4 = 64

315. b: 1 choice for the first digit. 3 choices for the second digit of the number to be formed since repetition is allowed. 2 choices for the third digit of the number to be formed. 1 choice for the fourth digit of the number to be formed. Hence the number N of numbers that we may form is given by
N = 1 * 3 * 2 * 1 = 6

316. d: For the number to be formed to be divisible by two, the last digit must be 2, hence one choice for this digit. 4 choices for each of the other digits since repetition is allowed. Hence the number N of numbers that we may form is given by
N = 4 * 4 * 4 * 1 = 64

317. b: For the number to be formed to be divisible by two, the last digit must be 2, hence one choice for this digit. 3 choices for the first digit, 2 choices for the second digit and 1 choice for the third digit that form the number. Hence the number N of numbers that we may form is given by
N = 3 * 2 * 1 * 1 = 6

318. d: Let there by 'x' chickens and 'y' pigs.
Therefore, x + y = 200 --- (1)
Each chicken has 2 legs and each pig has 4 legs
Therefore, 2x + 4y = 540 --- (2)
Solving equations (1) and (2), we get x = 130 and y = 70. There were 130 chickens and 70 pigs in the farm.

319. d: Let the age of the son 3 years back be x years
Therefore, the age of the father 3 years back was x + 24
At present the age of the son is x + 3 and the father is 5 times as old as the son. i.e., x + 24 + 3 = 5(x + 3)
i.e., x + 27 = 5x + 15
or 4x = 12 or x = 3.
Therefore, the son was 3 years old 3 years back and he will be 9 years old three years from now.

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