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## Problems on Numbers with Solutions

Applicants who want to practice Problems on Numbers are most welcome on this page because here candidates can get Problems on Numbers with Solutions. On this page we are describing full Explanation of questions. You can also get Shortcuts to solve question in some time and formulas, which is structured by team of recruitmentresult.com.

1. If one-third of one-fourth of a number is 15, then three-tenth of that number is:

Let the number be x.

Then,  1/3 of 1/4 of x = 15      x = 15 x 12 = 180.

So, required number = (3/10 x 180 ) = 54.

2. Three times the first of three consecutive odd integers is 3 more than twice the third. The third integer is:

Let the three integers be x, x + 2 and x + 4.

Then, 3x = 2(x + 4) + 3      x = 11.

Third integer = x + 4 = 15.

3. The difference between a two-digit number and the number obtained by interchanging the positions of its digits is 36. What is the difference between the two digits of that number?

Let the ten’s digit be x and unit’s digit be y.

Then, (10x + y) – (10y + x) = 36

9(x – y) = 36

x – y = 4.

4. The difference between a two-digit number and the number obtained by interchanging the digits is 36. What is the difference between the sum and the difference of the digits of the number if the ratio between the digits of the number is 1 : 2 ?

Since the number is greater than the number obtained on reversing the digits, so the ten’s digit is greater than the unit’s digit.

Let ten’s and unit’s digits be 2x and x respectively.

Then, (10 x 2x + x) – (10x + 2x) = 36

9x = 36

x = 4.

Required difference = (2x + x) – (2x – x) = 2x = 8.

5. A two-digit number is such that the product of the digits is 8. When 18 is added to the number, then the digits are reversed. The number is:

Let the ten’s and unit digit be x and 8/x respectively.

Then,  (10x + 8/ x) + 18 = 10 x 8/x + x

10x2 + 8 + 18x = 80 + x2

9×2 + 18x – 72 = 0

x2 + 2x – 8 = 0

(x + 4)(x – 2) = 0

x = 2.

6. The sum of the digits of a two-digit number is 15 and the difference between the digits is 3. What is the two-digit number?

Let the ten’s digit be x and unit’s digit be y.

Then, x + y = 15 and x – y = 3   or   y – x = 3.

Solving x + y = 15   and   x – y = 3, we get: x = 9, y = 6.

Solving x + y = 15   and   y – x = 3, we get: x = 6, y = 9.

So, the number is either 96 or 69.

Hence, the number cannot be determined.

7. The sum of the squares of three numbers is 138, while the sum of their products taken two at a time is 131. Their sum is:

Let the numbers be a, b and c.

Then, a2 + b2 + c2 = 138 and (ab + bc + ca) = 131.

(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca) = 138 + 2 x 131 = 400.

(a + b + c) = (400)1/2 = 20.

8. A number consists of two digits. If the digits interchange places and the new number is added to the original number, then the resulting number will be divisible by:

Let the ten’s digit be x and unit’s digit be y.

Then, number = 10x + y.

Number obtained by interchanging the digits = 10y + x.

(10x + y) + (10y + x) = 11(x + y), which is divisible by 11.

9. In a two-digit, if it is known that its unit’s digit exceeds its ten’s digit by 2 and that the product of the given number and the sum of its digits is equal to 144, then the number is:

Let the ten’s digit be x.

Then, unit’s digit = x + 2.

Number = 10x + (x + 2) = 11x + 2.

Sum of digits = x + (x + 2) = 2x + 2.

(11x + 2)(2x + 2) = 144

22×2 + 26x – 140 = 0

11×2 + 13x – 70 = 0

(x – 2)(11x + 35) = 0

x = 2.

Hence, required number = 11x + 2 = 24.

10. Find a positive number which when increased by 17 is equal to 60 times the reciprocal of the number.

Let the number be x.

Then, x + 17 = 60/x

x2 + 17x – 60 = 0

(x + 20)(x – 3) = 0

x = 3.

11. The product of two numbers is 9375 and the quotient, when the larger one is divided by the smaller, is 15. The sum of the numbers is:

Let the numbers be x and y.

Then, xy = 9375 and x/ y= 15.

Xy/ (x/y) = 9375/ 15

y2 = 625.

y = 25.

x = 15y = (15 x 25) = 375.

Sum of the numbers = x + y = 375 + 25 = 400.

12. The product of two numbers is 120 and the sum of their squares is 289. The sum of the number is:

Let the numbers be x and y.

Then, xy = 120 and x2 + y2 = 289.

(x + y)2 = x2 + y2 + 2xy = 289 + (2 x 120) = 529

x + y = (529)1/2 = 23.

13. A number consists of 3 digits whose sum is 10. The middle digit is equal to the sum of the other two and the number will be increased by 99 if its digits are reversed. The number is:

Let the middle digit be x.

Then, 2x = 10 or x = 5. So, the number is either 253 or 352.

Since the number increases on reversing the digits, so the hundred’s digits is smaller than the unit’s digit.

Hence, required number = 253.

14. The sum of two number is 25 and their difference is 13. Find their product.

Let the numbers be x and y.

Then, x + y = 25 and x – y = 13.

4xy = (x + y)2 – (x- y)2

= (25)2 – (13)2

= (625 – 169)

= 456

xy = 114.

15. What is the sum of two consecutive even numbers, the difference of whose squares is 84?

Let the numbers be x and x + 2.

Then, (x + 2)2 – x2 = 84

4x + 4 = 84

4x = 80

x = 20.

The required sum = x + (x + 2) = 2x + 2 = 42.

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