# Reasoning Question And Answer On Problem On Numbers For Upcoming Exams

**Reasoning Question And Answer On Problem On Numbers For Upcoming Exams**–Hello all friends and users you all know that we always share lot of informative regarding your competitive exams like E-Books ,R.S Aggarwal Series,Quantitative Aptitude like *Reasoning Question And Answer *or Problem On Numbers For Upcoming Exams.We have our whole series of quantitative Aptitude you can find here *Sample Reasoning Questions,**Reasoning Assessment Questions*,Analytical Reasoning Questions and Answers,Verbal Reasoning Questions and Answers and more….so lets start

**PROBLEMS ****ON NUMBERS**

** **

In this section, questions involving a set of numbers are put in the form of a puzzle. You have to analyse the given conditions, assume the unknown numbers and form equations accordingly, which on solving yield the unknown numbers.

__ SOLVED EXAMPLES __

__ __

**Ex.1. A number is as much greater than 36 as is less than 86. Find the number***.*

**Sol.** Let the number be x. Then, x – 36 = 86 – x => 2x = 86 + 36 = 122 => x = 61. Hence, the required number is 61.

Ex. 2. Find a number such that when 15 is subtracted from 7 times the number, the **Result is 10 more than twice the number***.* (Hotel Management, 2002)

**Sol.** Let the number be x. Then, 7x – 15 = 2x + 10 => 5x = 25 =>x = 5.

Hence, the required number is 5.

**Ex. 3. The sum of a rational number and its reciprocal is 13/6. Find the number.**

(S.S.C. 2000)

**Sol.** Let the number be x.

Then, x + (1/x) = 13/6 => (x2 + 1)/x = 13/6 => 6×2 – 13x + 6 = 0

=> 6×2 – 9x – 4x + 6 = 0 => (3x – 2) (2x – 3) = 0

- x = 2/3 or x = 3/2

Hence the required number is 2/3 or 3/2.

* *

**Ex. 4. The sum of two numbers is 184. If one-third of the one exceeds one-seventh**

**of the other by 8, find the smaller number.**

**Sol.** Let the numbers be x and (184 – *x). *Then,

(X/3) – ((184 – x)/7) = 8 => 7x – 3(184 – x) = 168 => 10x = 720 => x = 72.

So, the numbers are 72 and 112. Hence, smaller number = 72.

**Ex. 5. The difference of two numbers is 11 and one-fifth of their sum is 9. Find the numbers***.*

**Sol****.** Let the number be x and y. Then,

x – y = 11 —-(i) and 1/5 *(x *+ *y) *= 9 => *x *+ *y *= 45 —-(ii)

Adding (i) and (ii), we get: 2x = 56 or x = 28. Putting x = 28 in (i), we get: y = 17.

Hence, the numbers are 28 and 17.

**Ex. 6. If the sum of two numbers is 42 and their product is 437, then find the**

**absolute difference between the numbers***.* (S.S.C. 2003)

**Sol.** Let the numbers be x and y. Then, x + y = 42 and xy = 437

*x *– *y *= sqrt[(x + *y)2 *– 4xy] = sqrt[(42)2 – 4 x 437 ] = sqrt[1764 – 1748] = sqrt[16] = 4.

Required difference = 4.

**Ex. 7. The sum of two numbers is 16 and the sum of their squares is 113. Find the **

**numbers.**

** Sol**. Let the numbers be x and (15 – *x).*

Then, x2 + (15 – x)2 = 113 => x2 + 225 + X2 – 30x = 113

=> *2×2 *– 30x + 112 = 0 => x2 – *15x *+ 56 = 0

=> *(x *– 7) *(x *– 8) = 0 => x = 7 or x = 8.

* So, *the numbers are 7 and 8.

**Ex. 8. The average of four consecutive even numbers is 27. Find the largest of these**

**numbers.**

**Sol.** Let the four consecutive even numbers be x, x + 2, x + 4 and x + 6.

Then, sum of these numbers = (27 x 4) = 108.

* So, *x + *(x *+ 2) + *(x *+ 4) + *(x *+ 6) = 108 or *4x *= 96 or x = 24.

:. Largest number = *(x *+ 6) = 30.

**Ex. 9. The sum of the squares of three consecutive odd numbers is 2531.Find the**

**numbers. **

** Sol.** Let the numbers be x, x + 2 and x + 4.

Then, X2 + *(x *+ 2)2 + *(x *+ 4)2 = 2531 => *3×2 *+ *12x *– 2511 = 0

=> X2 + *4x *– 837 = 0 => * (x *– 27) *(x *+ 31) = 0 => x = 27.

Hence, the required numbers are 27, 29 and 31.

**Ex. 10. Of two numbers, 4 times the smaller one is less then 3 times the 1arger one by 5. If the sum of the numbers is larger than 6 times their difference by 6, find the two numbers**.

** Sol**. Let the numbers be x and *y, *such that x > y

Then, *3x *– 4y = 5 …(i) and *(x *+ *y) *– 6 *(x *– *y) *= 6 => –*5x *+ 7y = 6 …(ii)

Solving (i) and *(ii), *we get: x = 59 and *y *= 43.

Hence, the required numbers are 59 and 43.

**Ex. 11. The ratio between a two-digit number and the sum of the digits of that**

**number is 4 : 1.If the digit in the unit’s place is 3 more than the digit in the ten’s place, what is the number?**

**Sol.** Let the ten’s digit be x. Then, unit’s digit = *(x *+ 3).

Sum of the digits = x + *(x *+ 3) = *2x *+ 3. Number = l0x + *(x *+ 3) = llx + 3.

11x+3 / 2x + 3 = 4 / 1 => 1lx + 3 = 4 (2x + 3) => 3x = 9 => *x *= 3.

Hence, required number = *11x *+ 3 = 36.

**Ex. 12. A number consists of two digits. The sum of the digits is 9. If 63 is subtracted**

**from the number, its digits are interchanged. Find the number.**

**Sol.** Let the ten’s digit be x. Then, unit’s digit = (9 – *x).*

Number = l0x + (9 – x) = *9x *+ 9.

Number obtained by reversing the digits = 10 (9 – x) + x = 90 – 9x.

therefore, *(9x *+ 9) – 63 = 90 – *9x* =>* 18x *= 144 => x = 8.

* So, *ten’s digit = 8 and unit’s digit = 1.

Hence, the required number is 81.

**Ex. 13. A fraction becomes 2/3 when 1 is added to both, its numerator and denominator. **

**And ,it becomes 1/2 when 1 is subtracted from both the numerator and denominator. Find the fraction.**

**Sol**. Let the required fraction be x/y. Then,

x+1 / y+1 = 2 / 3 => 3x – 2y = – 1 …(i) and x – 1 / y – 1 = 1 / 2

- 2x – y = 1 …(ii)

Solving (i) and (ii), we get : x = 3 , y = 5

therefore, Required fraction= 3 / 5.

Ex. 14. 50 is divided into two parts such that the sum of their reciprocals is 1/ 12.Find the two parts.

**Sol.** Let the two parts be x and (50 – *x).*

Then, 1 / x + 1 / (50 – x) = 1 / 12 => (50 – x + x) / x ( 50 – x) = 1 / 12

=> x2 – 50x + 600 = 0 => (x – 30) ( x – 20) = 0 => x = 30 or x = 20.

So, the parts are 30 and 20.

**Ex. 15. If three numbers are added in pairs, the sums equal 10, 19 and 21. Find the**

numbers )

**Sol**. Let the numbers be x, *y *and z. Then,

x+ *y *= 10* …(i) y *+ z = 19* …(ii) * x + z = 21 …(iii)

Adding (i) *,(ii) *and *(iii), *we get: *2 (x *+ *y + z ) *= 50 or (x + *y *+ z) = 25.

Thus, x= (25 – 19) = 6; *y *= (25 – 21) = 4; z = (25 – 10) = 15.

Hence, the required numbers are 6, 4 and 15.

# Download PDF OF Reasoning Question And Answer On Problem On Numbers For Upcoming Exams

**This is More Interesting**

- Reasoning Question And Answer On Problem On Numbers For Upcoming Exams
- FREE VERBAL ABILITY EXERCISE @ FREEJOBAWARE.COM
- Dicussion On-Increasing Competition and Today’s Youth
- Average Reasoning Mathematics Question Answers For Various Exams
- Square Roots And Cube Roots Formulas and Questions For Competitive Exams

## Leave a Reply