#### 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.

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