# CAT Number System Questions [Most Important]

**Number System**Â is one of the most important topics in the **CAT Quant Section**. You can check out theseÂ ** CAT Number System Questions from **the **CAT previous year papers.** This article will look into some very important Number System questions PDF(**with solutions**) for CAT. If you want to practice these questions, you can download this CAT Number System Questions PDF (most important) along with the solutions below, which is completely Free.

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**Question 1:Â **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

**1)Â Answer:Â 4195**

**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 placeÂ 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.Â (4)

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

Since c cannot be greater than 9 considering c to be the maximum valueÂ 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 conditionÂ is 4195

**Question 2:Â **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:

**2)Â Answer:Â 7**

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

**3)Â Answer:Â 315**

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

a)Â 2018

b)Â 2019

c)Â 2017

d)Â 2020

**4)Â AnswerÂ (A)**

**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.

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

**Question 5:Â **How many of the integers 1, 2, â€¦ , 120, are divisible by none of 2, 5 and 7?

a)Â 42

b)Â 41

c)Â 40

d)Â 43

**5)Â AnswerÂ (B)**

**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,Â numberÂ of the integers 1, 2, â€¦ , 120, are divisible by none of 2, 5 and 7 = 120 – 60 – 12 – 7 = 41

Checkout: **CAT Free Practice Questions and Videos**

**Question 6:Â **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 forÂ N?

**6)Â Answer:Â 6**

**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 7:Â **How many integers in the set {100, 101, 102, …, 999} have at least one digitÂ repeated?

**7)Â Answer:Â 252**

**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 8:Â **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)Â 3

b)Â 1

c)Â 2

d)Â 4

**8)Â AnswerÂ (B)**

**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

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

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

Hence m-2n = 9-8 = 1

**Question 9:Â **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)Â 49

b)Â 56

c)Â 59

d)Â 46

**9)Â AnswerÂ (D)**

**Solution:**

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

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

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

Hence, Option D is the correct answer.

**Question 10:Â **The mean of all 4-digit even natural numbers of the form ‘aabb’,where $a>0$, is

a)Â 4466

b)Â 5050

c)Â 4864

d)Â 5544

**10)Â AnswerÂ (D)**

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

**11)Â Answer:Â 21**

**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 12:Â **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)Â 58

b)Â 85

c)Â 50

d)Â 95

**12)Â AnswerÂ (C)**

**Solution:**

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

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

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

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

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

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

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

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

Adding ab=616 on both sides, we get

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

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

=> a+b=50

**Question 13:Â **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 ofÂ fifth and sixth digits. Then, the largest possible value of the fourth digit is

**13)Â Answer:Â 7**

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

**14)Â Answer:Â 44**

**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 areÂ $3^0, 3^2, 3^4$ = 3

In $5^4$,Â 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 15:Â **How many pairs (m, n) of positive integers satisfy the equation $m^2 + 105 = n^2$?

**15)Â Answer:Â 4**

**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=105Â ==> n=53, m=52

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

n-m=5, n+m=21Â ==> 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.

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