model 2 divisibility, multiples, add and subtract based number system Section-Wise Topic Notes With Detailed Explanation And Example Questions

MOST IMPORTANT quantitative aptitude - 6 EXERCISES

Top 10,000+ Aptitude Memory Based Exercises

The following question based on number system topic of quantitative aptitude

Questions : Each member of a picnic party contributed twice as many rupees as the total number of members and the total collection was 3042. The number of members present in the party was ?

(a) 2

(b) 32

(c) 40

(d) 39

The correct answers to the above question in:

Answer: (d)

Let the total number of members be X.

Then, each member's contribution =Rs. 2X

Given, X × 2X = 3042

⇒ 2X= 3042

⇒ X= 1521

⇒ X = 39

Practice number system (model 2 divisibility, multiples, add and subtract based number system) Online Quiz

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Read more divide multiple add subtract Based Quantitative Aptitude Questions and Answers

Question : 1

The expression 26n – 42n, where n is a natural number is always divisible by

a) 15

b) 18

c) 36

d) 48

Answer: (d)

The expression 26n − 42n = (26)n − (42)n

= 64n − 16n

Which is divisible by 64 –16= 48

Therefore the required answer is 48.

 

Question : 2

A student was asked to divide a number by 6 and add 12 to the quotient. He, however, first added 12 to the number and then divided it by 6, getting 112 as the answer. The correct answer should have been ?

a) 124

b) 122

c) 118

d) 114

Answer: (b)

Let the number be p.

According to the question,

∴ $\text"p + 12"/6 = 112$

⇒   p + 12 = 672

⇒   p = 672 – 12 = 660

∴  Correct answer = $\text"600"/6 = 112$

 

Correct answer =  110 + 12 = 122

Question : 3

A six digit number is formed by repeating a three digit number; for example, 256, 256 or 678, 678 etc. Any number of this form is always exactly divisible by :

a) 11 only

b) 7 only

c) 1001

d) 13 only

Answer: (c)

The number (x y z x y z) can be written, after giving corresponding weightage of the places at which the digits occur, as 100000 x + 10000y + 1000z + 100x + 10y + z

= 100100x + 10010y + 1001z

= 1001 (100x + 10y + z)

Since 1001 is a factor, the number is divisible by 1001.

7 × 11 × 13 = 1001

As the number is divisible by 1001, it will also be divisible by all three namely, 7, 11 and 13 and not by only one of these because all three are factors of 1001.

So, the answer is 1001.

Question : 4

A number when divided by 91 gives a remainder 17. When the same number is divided by 13, the remainder will be ?

a) 0

b) 4

c) 6

d) 3

Answer: (b)

Here, the first divisor (91) is a multiple of the second divisor (13).

∴ Required remainder = Remainder obtained on dividing 17 by 13

⇒ 17 = ( 13 × 1 ) + 4

Hence Required remainder = 4

Question : 5

The greatest common divisor of 33333 + 1 and 33334 + 1 is ?

a) 2

b) 1

c) 33333 + 1

d) 20

Answer: (c)

Question : 6

If a and b are two odd positive integers, by which of the following integers is (a4 – b4) always divisible ?

a) 3

b) 6

c) 8

d) 12

Answer: (c)

$a^4 - b^4 = (a - b) (a + b) (a^2 + b^2)$,

Where a and b are odd positive integers.

If two positive integers are odd, then their sum, difference and sum of their squares are always even.

∴ (a - b) (a + b) and $(a^2 + b^2)$ are divisible by 2.

Hence (a - b) (a + b) x $(a^2 + b^2) = a^4 - b^4$ is always divisible by $2^3 = 8$

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