Problems on Numbers Aptitude

Problems on Numbers form a crucial part of quantitative aptitude tests. They involve understanding and applying concepts like divisibility, properties of numbers, arithmetic operations, and number series. These problems require a good grasp of basic mathematical concepts and often involve finding patterns or applying logical reasoning to solve numerical challenges.

1. Find the smallest number which when divided by 20, 25, 35, and 40 leaves a remainder of 5, 10, 20, and 25 respectively.

LCM of 20, 25, 35, and 40 is 1400.
Required number = LCM + (remainder - divisor) = 1400 + (5 - 20) = 1415.

2. If a number is divided by 357, the remainder is 47. What will be the remainder if the same number is divided by 17?

Number = 357k + 47. When divided by 17, remainder = 47 mod 17 = 13.

3. The sum of two numbers is 15 and the sum of their squares is 113. Find the numbers.

Let the numbers be x and y. Then, x + y = 15 and x² + y² = 113.
(x + y)² = x² + y² + 2xy = 15² = 225.
2xy = 225 - 113 = 112.
xy = 56. Solving x + y = 15 and xy = 56 gives x = 7, y = 8.

4. What is the smallest number which when increased by 3 is exactly divisible by 27, 35, and 125?

LCM of 27, 35, and 125 is 4725.
Required number = (4725k - 3) that is divisible by 4725. k = 3 gives the smallest number, 10499.

5. A number when divided by 296 gives a remainder of 75. What will be the remainder when the same number is divided by 37?

Number = 296k + 75. When divided by 37, remainder = 75 mod 37 = 1.

6. The product of two numbers is 2028 and their HCF is 13. Find the numbers.

Let the numbers be 13a and 13b where a and b are co-prime. Then, 13a × 13b = 2028.
ab = 12. Possible pairs are (1, 12) and (3, 4). Co-prime pair is (3, 4).
Numbers are 13 × 3 and 13 × 4 = 39 and 52.

7. Find the largest 3-digit number which is exactly divisible by 88.

Largest 3-digit number = 999. Dividing 999 by 88 gives a remainder of 15. Required number = 999 - 15 = 984.

8. The difference between a number and its three-fifth is 50. What is the number?

Let the number be x. Then, x - 3x/5 = 50. Solving gives x = 125.

9. If the sum of a number and its square is 182, what is the number?

Let the number be x. Then, x + x² = 182. Solving x² + x - 182 = 0 gives x = 13.

10. The sum of three consecutive even numbers is 54. Find the numbers.

Let the numbers be x, x + 2, x + 4. Then, x + (x + 2) + (x + 4) = 54. Solving gives x = 18.

11. A number consists of two digits whose sum is 8. If 18 is added to the number, its digits are reversed. Find the number.

Let the number be 10x + y. Then, x + y = 8 and 10x + y + 18 = 10y + x. Solving gives x = 3, y = 5.

12. Find the smallest number which leaves a remainder of 5 when divided by 20, 25, 35, and 40.

LCM of 20, 25, 35, and 40 is 1400. Required number = LCM + remainder = 1400 + 5 = 1415.

13. If the HCF of two numbers is 17 and their LCM is 765, find the numbers.

Let the numbers be 17a and 17b. Then, 17a × 17b = 17 × 765.
ab = 45. The pairs of factors of 45 are (1, 45) and (3, 15).
Numbers are 17 × 3 and 17 × 15 = 51 and 255.

14. The sum of two numbers is 50 and their difference is 10. Find the numbers.

Let the numbers be x and y. Then, x + y = 50 and x - y = 10. Solving gives x = 30, y = 20.

15. The sum of a number and its reciprocal is 10/3. Find the number.

Let the number be x. Then, x + 1/x = 10/3. Solving x² - 10x/3 + 1 = 0 gives x = 3.

16. A number is increased by 20% and then decreased by 20%. What is the net change in the number?

Let the number be 100. After increasing by 20%, it becomes 120. After decreasing by 20%, it becomes 96. Net change = 96 - 100 = -4%.

17. The average of 5 consecutive odd numbers is 61. What is the middle number?

Let the numbers be x - 4, x - 2, x, x + 2, x + 4. Then, (x - 4 + x - 2 + x + x + 2 + x + 4)/5 = 61. Solving gives x = 61.

18. A number when divided by 342 gives a remainder of 47. What will be the remainder if the same number is divided by 19?

Number = 342k + 47. When divided by 19, remainder = 47 mod 19 = 9.

19. If the digits of a two-digit number are reversed, the number obtained is 27 more than the original number and the sum of the digits is 9. Find the original number.

Let the original number be 10x + y. Then, 10y + x = 10x + y + 27 and x + y = 9. Solving gives x = 5, y = 4. Original number = 54.

20. The product of two numbers is 120 and the sum of their squares is 289. Find the numbers.

Let the numbers be x and y. Then, xy = 120 and x² + y² = 289.
(x + y)² = x² + y² + 2xy = 289 + 2 × 120 = 529.
x + y = √529 = 23.
Solving xy = 120 and x + y = 23 gives x = 10, y = 12.