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Download Number System Questions for XAT PDF – XAT Number System questions pdf by Cracku. Top 10 very Important Number Systems Questions for XAT based on asked questions in previous exam papers.

Question 1: For a 4-digit number, the sum of its digits in the thousands, hundreds and tens places is 14, the sum of its digits in the hundreds, tens and units places is 15, and the tens place digit is 4 more than the units place digit. Then the highest possible 4-digit number satisfying the above conditions is

Question 2: For all possible integers n satisfying $2.25\leq2+2^{n+2}\leq202$, then the number of integer values of $3+3^{n+1}$ is:

Question 3: Zahir and Raman are at the entrance of a dark cave. To enter this cave, they need to open a number lock. Raman sees a note on a rock: “ … chest of pure diamonds kept for the smart one … number has six digits … second last digit is 2, third last is 4 … divisible by all prime numbers less than 15 …”. Excited, Zahir and Raman seek your help: which of these can be the first digit of the six-digit number that will help them open the lock?

a) 5

b) 3

c) 9

d) 1

e) 4

Question 4: An encryption system operates as follows:
Step 1. Fix a number k $(k \leq 26)$.
Step 2. For each word, swap the first k letters from the front with the last k letters from the end in reverse order. If a word contains less than 2k letters, write the entire word in reverse order.
Step 3. Replace each letter by a letter k spaces ahead in the alphabet. If you cross Z in the process to move k steps ahead, start again from A.
Example: k = 2: zebra –> arbez –> ctdgb.
If the word “flight” becomes “znmorl” after encryption, then the value of k:

a) 5

b) 4

c) 7

d) Cannot be determined uniquely from the given information

e) 6

Question 5: How many 4-digit numbers, each greater than 1000 and each having all four digits distinct, are there with 7 coming before 3?

Question 6: How many pairs(a, b) of positive integers are there such that $a\leq b$ and $ab=4^{2017}$ ?

a) 2018

b) 2019

c) 2017

d) 2020

Question 7: How many of the integers 1, 2, … , 120, are divisible by none of 2, 5 and 7?

a) 42

b) 41

c) 40

d) 43

Question 8: Let N, x and y be positive integers such that $N=x+y,2<x<10$ and $14<y<23$. If $N>25$, then how many distinct values are possible for N?

Question 9: How many integers in the set {100, 101, 102, …, 999} have at least one digit repeated?

Question 10: Let m and n be natural numbers such that n is even and $0.2<\frac{m}{20},\frac{n}{m},\frac{n}{11}<0.5$. Then $m-2n$ equals

a) 3

b) 1

c) 2

d) 4

Given the 4 digit number :

Considering the number in thousands digit is a number in the hundredth digit is b, number in tens digit is c, number in the units digit is d.

Let the number be abcd.

Given that a+b+c = 14. (1)

b+c+d = 15. (2)

c = d+4. (3).

In order to find the maximum number which satisfies the condition, we need to have abcd such that a is maximum which is the digit in thousands place in order to maximize the value of the number. b, c, and d are less than 9 each as they are single-digit numbers.

Substituting (3) in (2) we have b+d+4+d = 15, b+2*d = 11.  (4)

Subtracting (2) and (1) : (2) – (1) = d = a+1.   (5)

Since c cannot be greater than 9 considering c to be 9 the value of d is 5.

If d = 5, using d = a+1, a = 4.

Hence the maximum value of a = 4 when c = 9, d = 5.

Substituting b+2*d = 11. b = 1.

The highest four-digit number satisfying the condition is 4195

$2.25\leq2+2^{n+2}\leq202$

$2.25-2\le2+2^{n+2}-2\le202-2$

$0.25\le2^{n+2}\le200$

$\log_20.25\le n+2\le\log_2200$

$-2\le n+2\le7.xx$

$-4\le n\le7.xx-2$

$-4\le n\le5.xx$

Possible integers = -4, -3, -2, -1, 0, 1, 2, 3, 4, 5

If we see the second expression that is provided, i.e

$3+3^{n+1}$, it can be implied that n should be at least -1 for this expression to be an integer.

So, n = -1, 0, 1, 2, 3, 4, 5.

Hence, there are a total of 7 values.

Let the 6 digit number be  _ _ _ 42_

It is divisible by 2,3,5,7,11,13

Since the number is divisible by both 2 and 5 the last digit of the number must be 0.

The number is also divisible by 7, 11, and 13.

Hence the number must also be divisible by 7*11*13.

=7*11*13 = 1001.

A number which is a multiple of 1001 is of the form abcabc.

This is because abc*(1001) = abc*(1000+1) = abc000 + abc = abcabc.

Hence the number is 420420.

The first digit is 4.

Flight become znmorl

Let’s assume $k>3$

So flight will become thgilf -> znmrol. Hence the value of k will be 6

Here there are two cases possible

Case 1: When 7 is at the left extreme

In that case 3 can occupy any of the three remaining places and the remaining two places can be taken by (0,1,2,4,5,6,8,9)

So total ways 3(8)(7)= 168

Case 2: When 7 is not at the extremes

Here there are 3 cases possible. And the remaining two places can be filled in 7(7) ways.(Remember 0 can’t come on the extreme left)

Hence in total 3(7)(7)=147 ways

Total ways 168+147=315 ways

$ab\ =\ 4^{2017}=2^{4034}$

The total number of factors = 4035.

out of these 4035 factors, we can choose two numbers a,b such that a<b in [4035/2] = 2017.

And since the given number is a perfect square we have one set of two equal factors.

.’. many pairs(a, b) of positive integers are there such that $a\leq b$ and $ab=4^{2017}$ = 2018.

The number of multiples of 2 between 1 and 120 = 60

The number of multiples of 5 between 1 and 120 which are not multiples of 2 = 12

The number of multiples of 7 between 1 and 120 which are not multiples of 2 and 5 = 7

Hence, number of the integers 1, 2, … , 120, are divisible by none of 2, 5 and 7 = 120 – 60 – 12 – 7 = 41

Possible values of x = 3,4,5,6,7,8,9

When x = 3, there is no possible value of y

When x = 4, the possible values of y = 22

When x = 5, the possible values of y=21,22

When x = 6, the possible values of y = 20.21,22

When x = 7, the possible values of y = 19,20,21,22

When x = 8, the possible values of y=18,19,20,21,22

When x = 9, the possible values of y=17,18,19,20,21,22

The unique values of N = 26,27,28,29,30,31

Total number of numbers from 100 to 999 = 900

The number of three digits numbers with unique digits:

_ _ _

The hundredth’s place can be filled in 9 ways ( Number 0 cannot be selected)

Ten’s place can be filled in 9 ways

One’s place can be filled in 8 ways

Total number of numbers = 9*9*8 = 648

Number of integers in the set {100, 101, 102, …, 999} have at least one digit repeated = 900 – 648 = 252

$0.2<\frac{n}{11}<0.5$

=> 2.2<n<5.5

Since n is an even natural number, the value of n = 4

$0.2<\frac{m}{20}<0.5$  => 4< m<10. Possible values of m = 5,6,7,8,9

Since $0.2<\frac{n}{m}<0.5$, the only possible value of m is 9

Hence m-2n = 9-8 = 1

We hope these Number Systems questions for XAT pdf for the XAT exam will be highly useful for your preparation.