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**PROBLEMS**

**ON NUMBERS**

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

__EXAMPLE Problems__**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 => (x

^{2}+ 1)/x = 13/6 => 6x^{2}– 13x + 6 = 0 => 6x

^{2 }– 9x – 4x + 6 = 0 => (3x – 2) (2x – 3) = 0ð 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)*– 4xy]

^{2}*= 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, x

^{2}+ (15 – x)^{2}= 113 => x^{2}+ 225 + X^{2}– 30x = 113 =>

*2x*– 30x + 112 = 0 => x^{2}^{2}–*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, X

^{2}+*(x*+ 2)^{2}+*(x*+ 4)^{2}= 2531 =>*3x*+^{2}*12x*– 2511 = 0 => X

^{2}+*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

ð 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

=> x

^{2}– 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.