Aptitude 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/(50-X) = 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.
Comments