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PROBLEMS ON NUMBERS

Aptitude

PROBLEMS ON NUMBERS -> DESCRIPTION

Types of Numbers:
Natural Numbers : Counting numbers 1,2,3,4,5,..... are called natural numbers.
Whole Numbers : All counting numbers together with zero from the set of whole numbers. Thus,
(i). 0 is the only whole number which is not a natural number.
(ii). Every natural number is a whole number.
Even Numbers : A number divisible by 2 is called an even number. e.g. 2,4,6,7,10,etc.
Odd Numbers : A number is not divisible by 2 is called an odd number. e.g. 1,3,5,6,7,9,11, etc.

PROBLEMS ON NUMBERS -> SOLVED EXAMPLES

1. 50 is divided into tow 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.
2. 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.
3. Find a number such that when 15 is subtracted from 7 times the number, the result is 10 more than twice the number.
  Sol.
Let the number be x. Then, 7x - 15 = 2x + 10 ⇔ 5x = 25 ⇔ x = 5.
Hence, the required number is 5.
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 ⇔ 7

PROBLEMS ON NUMBERS -> Exercise

16. If three numbers are added in pairs, the sums equal 10,19 and 21. Find the numbers?
 
  • A. 5, 10, 15
  • B. 6, 4, 15
  • C. 2, 4, 8
  • D. 15, 20, 25
Ans: B.
Sol.
Let the numbers be x, y and z.
Then, x + y = 10 ............(i) and
y + z = 19 ....................(II) and
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.
 
17. Find a positive number which when increased by 17 is equal to 60 times the reciprocal of the number.
 
  • A. 3
  • B. 5
  • C. 7
  • D. 12
Ans: A.
Sol.
Let the number be x.
Then, x+17 = 60/x
⇔ x2 + 17x - 60 = 0
⇔ (x+20)(x-3) = 0
⇔ x = 3.
 
 
18. The sum of two numbers is 40 and their difference is 4. The ratio of the numbers is
 
  • A. 11 : 9
  • B. 11 : 18
  • C. 21 : 19
  • D. 22 : 9
Ans: A.
Sol.
Let the numbers be x and y.
Then, x+y / x -y = 40/4 = 10
⇔ (x+y) = 10(x-y)
⇔ 9x = 11y
⇔ x/y = 11 / 9.