**Number System Questions For IIFT Set-2 PDF**

Download important IIFT Number System Questions PDF based on previously asked questions in IIFT and other MBA exams. Practice Number System questions and answers for IIFT exam.

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**Question 1: **Arrange the following letters to form a meaningful word.

a) 1 3 5 2 4 6 8 7

b) 8 6 7 1 4 2 3 5

c) 4 2 3 5 8 6 7 1

d) 5 3 7 1 8 4 2 6

**Question 2: **A rod is cut into 3 equal parts. The resulting portions are then cut into 12, 18 and 32 equal parts, respectively. If each of the resulting portions have integer length, the minimum length of the rod is

a) 6912 units

b) 864 units

c) 288 units

d) 240 units

**Question 3: **If $a, b, c,$ and $d$ are integers such that $a+b+c+d=30$ then the minimum possible value of $(a – b)^{2} + (a – c)^{2} + (a – d)^{2}$ is

**Question 4: **A number is interesting if on adding the sum of the digits of the number and the product of the digits of the number, the result is equal to the number. What fraction of numbers between 10 and 100 (both 10 and 100 included) is interesting?

a) 0.1

b) 0.11

c) 0.16

d) 0.22

e) None of the above

**Question 5: **Please read the following sentences carefully:

I — 103 and 7 are the only prime factors of 1000027

II — $\sqrt[6]{6!}>\sqrt[7]{7!}$

III — If I travel one half of my journey at an average speed of x km/h, it will be impossible for me to attain an average speed of 2x km/h for the entire journey.

a) All the statements are correct

b) Only Statement Il is correct

c) Only Statement Ill is correct

d) Both statements I and Il are correct

e) Both statements I and Ill are correct

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**Question 6: **An antique store has a collection of eight clocks. At a particular moment, the displayed times on seven of the eight clocks were as follows: 1:55 pm, 2:03 pm, 2:11 pm, 2:24 pm, 2:45 pm, 3:19 pm and 4:14 pm. If the displayed times of all eight clocks form a mathematical series, then what was the displayed time on the remaining clock?

a) 1:53 pm

b) 1:58 pm

c) 2:18 pm

d) 3:08 pm

e) 5:08 pm

**Question 7: **If a, b and c are 3 consecutive integers between -10 to +10 (both inclusive), how many integer values are possible for the expression?

$\frac{a^3+b^3+c^3+3abc}{(a+b+c)^2}$=?

a) 0

b) 1

c) 2

d) 3

e) 4

**Question 8: **If x and y are real numbers, then the minimum value of $x^{2}+ 4xy+ 6y^{2}-4y+ 4$ is

a) ‒4

b) 0

c) 2

d) 4

e) None of the above

**Question 9: **If $n^2 = 123456787654321$, what is $n$?

a) 12344321

b) 1235789

c) 11111111

d) 1111111

**Question 10: **A is the set of positive integers such that when divided by 2, 3, 4, 5, 6 leaves the remainders 1, 2, 3, 4, 5 respectively. How many integers between 0 and 100 belong to set A?

a) 0

b) 1

c) 2

d) None of these

**Answers & Solutions:**

**1) Answer (C)**

OVERHEAD is the word that can be formed from the given letters.

Hence, option C is the correct answer.

**2) Answer (B)**

The rod is cut into 3 equal parts thus the length of the rod will be a multiple of 3.

Each part is then cut into $12 = 2^2*3$

$18 = 2*3^2$ and $32 = 2^5$ parts and thus, each part of rod has to be a multiple of $2^5*3^2 = 288$

Thus, the rod will be a multiple of $288*3 = 864$

Thus, the minimum length of the rod is 864 units.

Hence, option B is the correct answer.

**3) Answer: 2**

For the value of given expression to be minimum, the values of $a, b, c$ and $d$ should be as close as possible. 30/4 = 7.5. Since each one of these are integers so values must be 8, 8, 7, 7. On putting these values in the given expression, we get

$(8 – 8)^{2} + (8 – 7)^{2} + (8 – 7)^{2}$

=> 1 + 1 = 2

**4) Answer (E)**

As the number is between 10 and 100 and 100 cannot be the number we are looking for, we can assume the number to be of two-digits.

Let the number be xy.

According to the question, for the number to be interesting

x + y + xy = 10x + y

On solving, we get

xy = 9x

or, x (9 – y) = 0

x cannot be 0, because we need a number greater than or equal to 10.

So, 9 – y = 0

=> y = 9

For all the numbers whose unit digit is 9 will be an interesting number.

So, the numbers are 19, 29, 39, 49, …..99

There are 9 such numbers out of 91 total numbers between 10 and 100 including both.

Required fraction = $\dfrac{9}{91}$ = 0.0989

As this is not given in any of the options, the answer will be “none of the above”.

Hence, option E is the correct answer.

**5) Answer (C)**

Let us evaluate the statements one by one:

I: 103 and 7 are the only prime factors of 1000027

On successively dividing 1000027 by 103 and 7, we get 1387 as the answer.

1387 is divisible by 19.

Therefore, statement I is false.

II: $\sqrt[6]{6!}>\sqrt[7]{7!}$

Raising the power to 42 on both sides, we get,

$[6!]^7>[7!]^6$

$6!*[6!]^6 > 7^6*[6!]^6$

$7^6$ is greater than $6!$.

Therefore, statement II is false.

III:It has been given that the person travels one-half of the journey at x kmph.

Let us assume the distance to be ‘2d’.

Let us assume the average speed to be 2d and check for the feasibility.

Let the speed at which the person travels the other half of the journey be y.

d/x + d/y = 2d/2x

d/x + d/y = d/x

=> d/y = 0 or y tends to infinity.

Therefore, such a scenario is not possible and hence, statement III is true.

Only statement III is true. Therefore, option C is the right answer.

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**6) Answer (B)**

Let us find out the difference between the times given to figure out the pattern.

The times given are 1:55 pm, 2:03 pm, 2:11 pm, 2:24 pm, 2:45 pm, 3:19 pm and 4:14 pm.

The difference between 2 consecutive times given are 8 minutes, 8 minutes, 13 minutes, 21 minutes, 34 minutes, and 55 minutes.

We can observe that the difference between the times are in the Fibonacci series.

8 + 13 = 21

21 + 13 = 34

34 + 21 = 55

The Fibonacci series is as follows:

1,1,2,3,5,8,13,21,34,55.

But the first difference in the times given is 8.

Therefore, the missing time must be such that it divides the interval of 8 minutes into 3 minutes and 5 minutes.

The missing time should be 1:58 pm and hence, option B is the right answer.

**7) Answer (C)**

Since a,b,c are consecutive integers

=> $a = b-1$ and $c = b+1$

Expression : $\frac{a^3+b^3+c^3+3abc}{(a+b+c)^2}$

= $\frac{(b – 1)^3 + b^3 + (b + 1)^3 + 3 (b – 1) b (b + 1)}{(b – 1 + b + b + 1)^2}$

= $\frac{b^3 + 3b + b^3 + b^3 + 3b + 3b^3 – 3b}{9 b^2}$

= $\frac{6 b^3 + 3 b}{9 b^2} = \frac{2 b^2 + 1}{3 b}$

Putting different values of b from – 10 to 10, we can verify that only – 1 and 1 satisfies to get integer values for the expression.

Ans – (C)

**8) Answer (C)**

Expression : $x^{2}+ 4xy+ 6y^{2}-4y+ 4$

= $(x^2 + 4xy + 4y^2) + (2y^2 – 4y + 4)$

= $(x + 2y)^2 + \frac{1}{2} (4y^2 – 8y + 8)$

= $(x + 2y)^2 + \frac{1}{2} [(2y)^2 – 2 (2y) (2) + (2)^2 + 4]$

= $(x + 2y)^2 + \frac{1}{2} [(2y – 2)^2 + 4]$

= $(x + 2y)^2 + \frac{1}{2} (2y – 2)^2 + 2$

Since, $x$ and $y$ are real, => Min value of $(x + 2y)^2 = 0$

Minimum value of $(2y – 2)^2 = 0$

$\therefore$ Minimum value of expression = $0 + 0 + 2 = 2$

**9) Answer (C)**

Observe the pattern given below.

$11^2 = 121$

$111^2 = 12321$

$1111^2 = 1234321$ and so on.

So, $11111111^2 = 123456787654321$

**10) Answer (B)**

Let the number ‘n’ belong to the set A.

Hence, the remainder when n is divided by 2 is 1

The remainder when n is divided by 3 is 2

The remainder when n is divided by 4 is 3

The remainder when n is divided by 5 is 4 and

The remainder when n is divided by 6 is 5

So, when (n+1) is divisible by 2,3,4,5 and 6.

Hence, (n+1) is of the form 60k for some natural number k.

And n is of the form 60k-1

Between numbers 0 and 100, only 59 is of the form above and hence the correct answer is 1

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