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#### Quantitative Aptitude Problems With Answers - Free Online Practice Test

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**Formulas:-****0, 1, 2, 3, ----9. are called digits.****10, 11, 12,----- are called Number.****Natural number (N) :-**Counting numbers are called natural numbers.

Example:- 1, 2, 3,---etc. are all natural numbers. minimum natural number 1 and maximum natural

number ∞**Whole numbers (W) :-**All counting numbers together with zero from the set of whole numbers

Example:- 0, 1, 2, 3, 4, ------ are whole number.**Integers (Z) :-**All counting numbers, 0 and -ve of counting numbers are called integers.

Example:- -∞---------, -3, -2, -1, 0, 1, 2, 3, -------∞**Rational Numbers (Q) :-**A Rational Number is a real number that can be written as a simple fraction

Example:- {p/q/p,q∈Z}**Irrational NUmbers :-**An Irrational Number is a real number that cannot be written as a

simple fraction.

Example:- √**Even numbers :-**A number divisible by 2 is called an even number.

Example:- 0, 2, 4, 6, - - - - - - - - -**Odd numbers :-**A number not divisible by 2 is called an odd number.

Example:- 1, 3, 5, 7, - - - - - -**Composite Numbers :-**Numbers greater than 1 which are not prime, are called composite numbers.

Example:- 4, 6, 8, 9, 10, - - - -. 6 -> 1,2,3,6.**Prime Numbers:-**A number greater than 1 having exactly two factors, namely 1 and itself is called

a prime number.

Upto 100 prime numbers are:

**2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 71, 73, 79, 83, 89, 97****Co-prime Numbers:-**Two natural numbers a and b are said to be co-prime if their HCF is 1.

Example:- (21, 44), (4, 9), (2, 3), - - - - -**Twin prime numbers :-**A pair of prime numbers (as 3 and 5 or 11 and 13) differing by two are

called twin prime number.

Example:- The twin pair primes between 1 and 100 are

**(3, 5), (5, 7), (11, 13), (17, 19), (29, 31), (41, 43), (59, 61), (71, 73).****Face value:-**Face value is the actual value of the digit.

Example:- In the number 7635, the "7" has a face value of 7, the face value of 3 is 3 and so on**Place value:-**The value of where the digit is in the number, such as units, tens, hundreds, etc.

Example:- In 352, the place value of the 5 is "tens"

Place value of 2 * 1 = 2;

Place value of 5 * 10 = 50;

Place value of 3 * 100 = 300.

- 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 101
^{100}– 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.

101^{2} = 10,201

101^{2} − 1 = 10,200.

This is divisible by 100.

Similarly try for 101^{3} − 1 = 1,030,301−1 = 1,030,300.

So you can safely conclude that (101^{1} − 1) to (101^{9} − 1) will be divisible by 100.

(101^{10} − 1) to (101^{99} − 1) will be divisible by 1000.

Therefore, (101^{100} − 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%

**To whom this Number System Question and Answers section is beneficial?**

Students can learn and improve on their skillset for using Number System effectively and can also prepare for competitive examinations like...

- All I.B.P.S and Public Sector Bank Competitive Exam
- Common Aptitude Test (CAT) Exams
- UPSC Paper-II or CSAT Exams
- SSC Competitive Exams
- Defence Competitive Exams
- L.I.C Assistant Administrative Officer (AAO)/ G.I.C AAO and Clerk Competitive Exams
- Railway Competitive Exam
- University Grants Commission (UGC)
- Career Aptitude Test (IT Companies) and etc.