Problems On NumberPage 1
1.  A number is as much greater than 36 as is less than 86. Find the number?  (a)60  (b)61  (c)62  (d)63 

Answer is: BLet, the number be X. Then
X – 36 = 86 – X
2X = 122
X = 61
2.  Find a number such that when 15 is subtracted from 7 times the number, the result is 10 more than twice the number.  (a)5  (b)10  (c)15  (d)20 

Answer is: ALet, the number be X. Then
7X – 15 = 2X + 10
5X = 25
X = 5
3.  The sum of rational number and its reciprocal is 13/6 . Find the number?  (a)2/3 or 3/2  (b)4/5 or 5/4  (c)1/2 or 2/5  (d)5/2 or 4/3 

Answer is: ALet the number be X.
Then, X + 1/X = 13/6 ⇔(X² + 1)/X = 13/6 ⇔6X²  13X + 6 = 0
⇒ 6X²  9X – 4X +6 = 0
⇒ (3X  2)(2X  3) = 0
⇒ X = 2/3 or 3/2
Hence, the required number is 2/3 or 3/2
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?  (a)70  (b)71  (c)72  (d)73 

Answer is: CLet the numbers be X and (184  X). Then,
⇒ X/3 – (184  X)/7 = 8
⇒ 7X – 3(184  X) = 168
⇒ 10X = 720
⇒ X = 72
5.  The sum of two numbers is 15 and the sum of their squares is 113. Find the numbers?  (a)7 and 8  (b)8 and 9  (c)9 and 10  (d)10 and 11 

Answer is: ALet the numbers be X and (15  X)
Then, X² + (15  X)² = 113
⇒ X² + 225 + X²  30X = 113
⇒ 2X²  30X + 112 = 0
⇒ X² 15X + 56 = 0
⇒ (X  7)(X  8) = 0
⇒ X = 7 or X = 8
So, the numbers are 7 and 8.
6.  The average of four consecutive even numbers is 27. Find the largest of these numbers?  (a)25  (b)30  (c)35  (d)40 

Answer is: BLet, the 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
4X = 96
X = 24
∴ Largest number = (X + 6) = 30
7.  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 palace, what is the number?  (a)33  (b)34  (c)35  (d)36 

Answer is: DLet, the ten's digit be X. then unit's digit = (X + 3).
Sum of digits = X + (X + 3) = 2X + 3
Number = 10X + (X + 3) = 11X + 3
∴ (11X + 3) ÷ (2X + 3) = 4/1
⇒ 11 X + 3 = 4(2 X +3)
⇒ 3X =9
⇒ X = 3
Hence, Required number = 11 X + 3 = 36
8.  50 is divided in two parts such that the sum of their reciprocals is 1/12. Find the two parts?  (a)30 and 20  (b)40 and 30  (c)40 and 20  (d)20 and 10 

Answer is: ALet, the two parts be X and (50  X).
Then, 1/X + 1/(50X) = 1/12
⇒ (50 – X + X) ÷ X(50 – X) = 1/12
⇒ X²  50X + 600 = 0
⇒ X²  30X – 20X + 600 = 0
⇒ (X  30) (X  20) = 0
⇒ X = 30 or X = 20
So, the parts are 30 and 20.