# CAT Number System Questions PDF [Most Important with Solutions]

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Number Systems is one of the key topics in the CAT Quantitative Ability (QA) section. Although the Number System concept is vast, there will be certain easy sums in the exam. If you’re weak in this concept, make sure you learn at least the concepts.  You can check out these Number System questions from the Practice a good number of questions in CAT Number System so that you can answer these questions with ease in the exam. In this post, we will look into some important Number System Questions for CAT Quants. These are a good source for practice; If you want to practice these questions, you can download this Important CAT Number System Questions PDF below, which is completely Free.

Question 1: If the sum of squares of two numbers is 97, then which one of the following cannot be their product?

a) -32

b) 16

c) 48

d) 64

Solution:

Let ‘a’ and ‘b’ are those two numbers.

$\Rightarrow$ $a^2+b^2 = 97$

$\Rightarrow$ $a^2+b^2-2ab = 97-2ab$

$\Rightarrow$ $(a-b)^2 = 97-2ab$

We know that $(a-b)^2$ $\geq$ 0

$\Rightarrow$ 97-2ab $\geq$ 0

$\Rightarrow$ ab $\leq$ 48.5

Hence, ab $\neq$ 64. Therefore, option D is the correct answer.

Question 2: The smallest integer n for which $4^{n} > 17^{19}$ holds, is closest to

a) 37

b) 35

c) 33

d) 39

Solution:

$4^{n} > 17^{19}$

$\Rightarrow$ $16^{n/2} > 17^{19}$

Therefore, we can say that n/2 > 19

n > 38

Hence, option D is the correct answer.

Question 3: What is the largest positive integer n such that $\frac{n^2 + 7n + 12}{n^2 – n – 12}$ is also a positive integer?

a) 6

b) 16

c) 8

d) 12

Solution:

$\ \frac{\ n^2+3n+4n+12}{n^2-4n+3n-12}$

=$\ \frac{\ n^{ }\left(n+3\right)+4\left(n+3\right)}{n^{ }\left(n-4\right)+3\left(n-4\right)}$

=$\ \frac{\left(\ n+4\right)\left(n+3\right)}{\left(n-4\right)\left(n+3\right)}$

=$\ \frac{\left(\ n+4\right)}{\left(n-4\right)}$

=$\ \frac{\left(\ n-4\right)+8}{\left(n-4\right)}$

=$\ 1+\frac{8}{\left(n-4\right)}$ which will be maximum when n-4 =8

n=12

Question 4: How many pairs (m, n) of positive integers satisfy the equation $m^2 + 105 = n^2$?

Solution:

$n^2-m^2=105$

(n-m)(n+m) = 1*105, 3*35, 5*21, 7*15, 15*7, 21*5, 35*3, 105*1.

n-m=1, n+m=105  ==> n=53, m=52

n-m=3, n+m=35 ==> n=19, m=16

n-m=5, n+m=21  ==> n=13, m=8

n-m=7, n+m=15 ==> n=11, m=4

n-m=15, n+m=7 ==> n=11, m=-4

n-m=21, n+m=5 ==> n=13, m=-8

n-m=35, n+m=3 ==> n=19, m=-16

n-m=105, n+m=1 ==> n=53, m=-52

Since only positive integer values of m and n are required. There are 4 possible solutions.

Question 5: How many factors of $2^4 \times 3^5 \times 10^4$ are perfect squares which are greater than 1?

Solution:

$2^4 \times 3^5 \times 10^4$

=$2^4 \times 3^5 \times 2^4*5^4$

=$2^8 \times 3^5 \times 5^4$

For the factor to be a perfect square, the factor should be even power of the number.

In $2^8$, the factors which are perfect squares are $2^0, 2^2, 2^4, 2^6, 2^8$ = 5

Similarly, in $3^5$, the factors which are perfect squares are $3^0, 3^2, 3^4$ = 3

In $5^4$, the factors which are perfect squares are $5^0, 5^2, 5^4$ = 3

Number of perfect squares greater than 1 = 5*3*3-1

=44

Question 6: In a six-digit number, the sixth, that is, the rightmost, digit is the sum of the first three digits, the fifth digit is the sum of first two digits, the third digit is equal to the first digit, the second digit is twice the first digit and the fourth digit is the sum of fifth and sixth digits. Then, the largest possible value of the fourth digit is

Solution:

Let the six-digit number be ABCDEF

F = A+B+C, E= A+B, C=A, B= 2A, D= E+F.

Therefore D = 2A+2B+C = 2A + 4A + A= 7A.

A cannot be 0 as the number is a 6 digit number.

A cannot be 2 as D would become 2 digit number.

Therefore A is 1 and D is 7.

Question 7: The product of two positive numbers is 616. If the ratio of the difference of their cubes to the cube of their difference is 157:3, then the sum of the two numbers is

a) 58

b) 85

c) 50

d) 95

Solution:

Assume the numbers are a and b, then ab=616

We have, $\ \ \frac{\ a^3-b^3}{\left(a-b\right)^3}$ = $\ \frac{\ 157}{3}$

=> $\ 3\left(a^3-b^3\right)\ =\ 157\left(a^3-b^3+3ab\left(b-a\right)\right)$

=> $154\left(a^3-b^3\right)+3*157*ab\left(b-a\right)$ = 0

=> $154\left(a^3-b^3\right)+3*616*157\left(b-a\right)$ = 0        (ab=616)

=>$a^3-b^3+\left(3\times\ 4\times\ 157\left(b-a\right)\right)$    (154*4=616)

=> $\left(a-b\right)\left(a^2+b^2+ab\right)\ =\ 3\times\ 4\times\ 157\left(a-b\right)$

=> $a^2+b^2+ab\ =\ 3\times\ 4\times\ 157$

Adding ab=616 on both sides, we get

$a^2+b^2+ab\ +ab=\ 3\times\ 4\times\ 157+616$

=> $\left(a+b\right)^2=\ 3\times\ 4\times\ 157+616$ = 2500

=> a+b=50

Question 8: How many 3-digit numbers are there, for which the product of their digits is more than 2 but less than 7?

Solution:

Let the number be ‘abc’. Then, $2<a\times\ b\times\ c<7$. The product can be 3,4,5,6.

We can obtain each of these as products with the combination 1,1, x where x = 3,4,5,6. Each number can be arranged in 3 ways, and we have 4 such numbers: hence, a total of 12 numbers fulfilling the criteria.

We can factories 4 as 2*2 and the combination 2,2,1 can be used to form 3 more distinct numbers.

We can factorize 6 as 2*3 and the combination 1,2,3 can be used to form 6 additional distinct numbers.

Thus a total of 12 + 3 + 6 = 21 such numbers can be formed.

Question 9: The mean of all 4-digit even natural numbers of the form ‘aabb’,where $a>0$, is

a) 4466

b) 5050

c) 4864

d) 5544

Solution:

The four digit even numbers will be of form:

1100, 1122, 1144 … 1188, 2200, 2222, 2244 … 9900, 9922, 9944, 9966, 9988

Their sum ‘S’ will be (1100+1100+22+1100+44+1100+66+1100+88)+(2200+2200+22+2200+44+…)….+(9900+9900+22+9900+44+9900+66+9900+88)

=> S=1100*5+(22+44+66+88)+2200*5+(22+44+66+88)….+9900*5+(22+44+66+88)

=> S=5*1100(1+2+3+…9)+9(22+44+66+88)

=>S=5*1100*9*10/2 + 9*11*20

Total number of numbers are 9*5=45

.’. Mean will be S/45 = 5*1100+44=5544.

Option D

Question 10: If a, b and c are positive integers such that ab = 432, bc = 96 and c < 9, then the smallest possible value of a + b + c is

a) 49

b) 56

c) 59

d) 46

Solution:

Since $c<9$, we can have the following viable combinations for $b\times\ c\ =96$ (given our objective is to minimize the sum):

$48\times\ 2$ ; $32\times3$ ; $24\times\ 4$ ; $16\times6$ ; $12\times8$

Similarly, we can factorize $a\times\ b\ = 432$ into its factors. On close observation, we notice that $18\times24\ and\ 24\ \times\ 4\$ corresponding to $a\times b\ and\ b\times\ c\$ respectively together render us with the least value of the sum of $a+b\ +\ c\ \ =\ 18+24+4\ =46$

Hence, Option D is the correct answer.

Question 11: 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

Solution:

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

=> 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

Question 12: How many integers in the set {100, 101, 102, …, 999} have at least one digit repeated?

Solution:

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

Question 13: 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?

Solution:

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

Question 14: 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

Solution:

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

Question 15: 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

Solution:

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

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.

Question 16: How many 4-digit numbers, each greater than 1000 and each having all four digits distinct, are there with 7 coming before 3?

Solution:

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

Question 17: 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:

Solution:

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

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

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

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

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

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

$-4\le n\le5.xx$

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.

Question 18: 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

Solution:

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 the maximum value 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

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