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Number System : Aptitude Test


  • 1. What is the place value of 7 in the numeral 2734? 
A.) 70
B.) 7
C.) 700
D.) 7.00

Answer: Option 'C'

7 × 100 = 700

  • 2. What is the place value of 3 in the numeral 3259 
A.) 300
B.) 30
C.) 3
D.) 3000

Answer: Option 'D'

3 × 1000 = 3000

  • 3. What is the diffference between the place value of 2 in the numeral 7229? 
A.) 20
B.) 200
C.) 180
D.) 18

Answer: Option 'C'

200 - 20 = 180

  • 4. What is the place value of 0 in the numeral 2074?
A.) 100
B.) 70
C.) 7.0
D.) 0

Answer: Option 'D'

Note : The place value of zero (0) is always 0. It may hold any place in a number,
its value is always 0.

  • 5. What is the diffference between the place value and face value of 3 in the numeral 1375? 
A.) 300
B.) 3
C.) 297
D.) 303

Answer: Option 'C'

place value of 3 = 3 × 100 = 300 
face value of 3 = 3
300 - 3 = 297

  • 6. A number when divided by a divisor leaves a remainder of 24.
    When twice the original number is divided by the same divisor, the remainder is 11. What is the value of the divisor? 
A.) 73
B.) 37
C.) 64
D.) 53

Answer: Option 'B'

Let the original number be 'a'
Let the divisor be 'd'
Let the quotient of the division of aa by dd be 'x'
Therefore, we can write the relation as a/d = x and the remainder is 24.
i.e., a=dx+24 When twice the original number is divided by d, 2a is divided by d.
We know that a=dx+24. Therefore, 2a = 2dx + 48
The problem states that (2dx+48)/d leaves a remainder of 11.
2dx2dx is perfectly divisible by d and will therefore, not leave a remainder.
The remainder of 11 was obtained by dividing 48 by d.
When 48 is divided by 37, the remainder that one will obtain is 11.
Hence, the divisor is 37.

  • 7. The largest number amongst the following that will perfectly divide 101100 – 1 is: 
A.) 100
B.) 10000
C.) 100^100
D.) 10

Answer: Option 'C'

The easiest way to solve such problems for objective exam purposes is trial and error or by back
substituting answers in the choices given.
1012 = 10,201
1012 − 1 = 10,200. 
This is divisible by 100. 
Similarly try for 1013 − 1 = 1,030,301−1 = 1,030,300.
So you can safely conclude that (1011 − 1) to (1019 − 1) will be divisible by 100.
(10110 − 1) to (10199 − 1) will be divisible by 1000. 
Therefore, (101100 − 1) will be divisible by 10,000.

  • 8. In an election, candidate A got 75% of the total valid votes. If 15% of the total votes were declared invalid and the total numbers of votes is 560000, find the number of valid vote polled in favour of candidate. 
A.) 357600
B.) 356000
C.) 367000
D.) 357000

Answer: Option 'D'

Total number of invalid votes = 15 % of 560000
= 15/100 × 560000
= 8400000/100
= 84000
Total number of valid votes 560000 – 84000 = 476000
Percentage of votes polled in favour of candidate A = 75 %
Therefore, the number of valid votes polled in favour of candidate A = 75 % of 476000
= 75/100 × 476000
= 35700000/100
= 357000

  • 9. Aravind had $ 2100 left after spending 30 % of the money he took for shopping. How much money did he take along with him? 
A.) $ 3600
B.) $ 3300
C.) $ 3000
D.) $ 3100

Answer: Option 'C'

Let the money he took for shopping be m.
Money he spent = 30 % of m
= 30/100 × m
= 3/10 m
Money left with him = m – 3/10 m = (10m – 3m)/10 = 7m/10
But money left with him = $ 2100
Therefore 7m/10 = $ 2100 
m = $ 2100× 10/7
m = $ 21000/7
m = $ 3000
Therefore, the money he took for shopping is $ 3000.

  • 10. A shopkeeper bought 600 oranges and 400 bananas. He found 15% of oranges and 8% of bananas were rotten. Find the percentage of fruits in good condition. 
A.) 87.8%
B.) 86.8%
C.) 85.8%
D.) 84.8%

Answer: Option 'A'

Total number of fruits shopkeeper bought = 600 + 400 = 1000
Number of rotten oranges = 15% of 600
= 15/100 × 600 = 9000/100 = 90
Number of rotten bananas = 8% of 400
= 8/100 × 400 = 3200/100 =32
Therefore, total number of rotten fruits = 90 + 32 = 122
Therefore Number of fruits in good condition = 1000 - 122 = 878
Therefore Percentage of fruits in good condition = (878/1000 × 100)%
= (87800/1000)% = 87.8%



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