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

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Number System for CAT is one of the key topics in the Quant section. Over the past few years, Number Systems have made a recurrent appearance in the CAT Quants Section. You can expect around 1-2 questions in the 22-question format of the CAT Quant sections. You can check out these Â  In this article, we will look into some important Number System Questions for CAT. These are a good source for practice; If you want to practice these questions, you can download this CAT Number System Questions PDF, which is completely Free.

• CAT Number SystemsÂ  – Tip 1: A few Number Systems CAT concepts appear in the CAT and other MBA entrance exams every year. If you’re starting the prep, firstly understand the CAT Number System Syllabus; Based on our analysis of the previous year CAT questions on Number System, this was number system weightage in CAT: 1 question was asked on this topic (in CAT 2021).
• CAT Number Systems – Tip 2: If you’re not very strong in this topic, learn at least the basics, and number systems cat tricks, so that if an easy question comes you will be able to answer it. Number System questions for CAT can be tackled with a strong foundation in this topic. Practice these CAT Number System questions PDF with solutions. Getting yourself acquainted with the basics of these concepts will help you solve the problems. Learn all the major formulae from these concepts. You can check out the Number System questions and answers PDF for CAT and learn all the Important Number System for CAT tricks Formulas PDF here.

Question 1:Â What are the last two digits of $7^{2008}$?

a)Â 21

b)Â 61

c)Â 01

d)Â 41

e)Â 81

Solution:

$7^4$ = 2401 = 2400+1
So, any multiple of $7^4$ will always end in 01
Since 2008 is a multiple of 4, $7^{2008}$ will also end in 01

Question 2:Â A shop stores x kg of rice. The first customer buys half this amount plus half a kg of rice. The second customer buys half the remaining amount plus half a kg of rice. Then the third customer also buys half the remaining amount plus half a kg of rice. Thereafter, no rice is left in the shop. Which of the following best describes the value of x?

a)Â $2 \leq x \leq 6$

b)Â $5 \leq x \leq 8$

c)Â $9 \leq x \leq 12$

d)Â $11 \leq x \leq 14$

e)Â $13 \leq x \leq 18$

Solution:

After the first sale, the remaining quantity would be (x/2)-0.5 and after the second sale, the remaining quantity is 0.25x-0.75

After the last sale, the remaining quantity is 0.125x-(7/8) which will be equal to 0

SoÂ 0.125x-(7/8) = 0 => x = 7

Question 3:Â How many even integers n, where $100 \leq n \leq 200$ , are divisible neither by seven nor by nine?

a)Â 40

b)Â 37

c)Â 39

d)Â 38

Solution:

Between 100 and 200 both included there are 51 even nos. There are 7 even nos which are divisible by 7 and 6 nos which are divisible by 9 and 1 no divisible by both. hence in total 51 – (7+6-1) = 39

There is one more method through which we can find the answer. Since we have to find even numbers, consider the numbers which are divisible by 14, 18 and 126 between 100 and 200. These are 7, 6 and 1 respectively.

Question 4:Â The number of positive integers n in the range $12 \leq n \leq 40$ such that the product (n -1)*(n – 2)*â€¦*3*2*1 is not divisible by n is

a)Â 5

b)Â 7

c)Â 13

d)Â 14

Solution:

positive integers n in the range $12 \leq n \leq 40$ such that the product (n -1)*(n – 2)*â€¦*3*2*1 is not divisible by n, implies that n should be a prime no. So there are 7 prime nos. in given range. Hence option B.

Question 5:Â Let T be the set of integers {3,11,19,27,â€¦451,459,467} and S be a subset of T such that the sum of no two elements of S is 470. The maximum possible number of elements in S is

a)Â 32

b)Â 28

c)Â 29

d)Â 30

Solution:

No. of terms in series T , 3+(n-1)*8 = 467 i.e. n=59.

Now S will have atleast have of 59 terms i.e 29 .

Also the sum of 29th term and 30th term is less than 470.

Hence, maximum possible elements in S is 30.

Question 6:Â The sum of four consecutive two-digit odd numbers, when divided by 10, becomes a perfect square. Which of the following can possibly be one of these four numbers?

a)Â 21

b)Â 25

c)Â 41

d)Â 67

e)Â 73

Solution:

Maximum sum of the four numbers <= 384=99+97+95+93
384/10 = 38.4
So, the perfect square is a number less than 38.4
The possibilities are 36, 25, 16 and 9
For the sum to be 360, the numbers can be 87, 89, 91 and 93
The sum of four consecutive odd numbers cannot be 250
For the sum to be 160, the numbers can be 37,39,41 and 43
The sum of 4 consecutive odd numbers cannot be 90
So, from the options, the answer is 41.

Question 7:Â For a positive integer n, let $P_n$ denote the product of the digits of n, and $S_n$ denote the sum of the digits of n. The number of integers between 10 and 1000 for which $P_n$ + $S_n$ = n is

a)Â 81

b)Â 16

c)Â 18

d)Â 9

Solution:

Let n can be a 2 digit or a 3 digit number.

First letÂ n be a 2 digit number.
So n = 10x + y and Pn = xy and Sn = x + y
Now, Pn + Sn = n
Therefore, xy + x + y = 10x + y , we have y = 9 .
Hence there are 9 numbers 19, 29,..Â ,99, so 9 cases .

Now ifÂ n isÂ a 3 digit number.
Let n = 100x + 10y + z
So Pn = xyz and Sn = x + y + z
Now, for Pn + Sn = n ; Â xyz + x + y + z = 100x + 10y + z ; so.Â xyz = 99x + 9y .
For above equation there is no value for which the above equation have an integer (singleÂ digit) value.

Hence option D.

Question 8:Â Let S be a set of positive integers such that every element n of S satisfies the conditions
A. 1000 <= n <= 1200
B. every digit in n is odd
Then how many elements of S are divisible by 3?

a)Â 9

b)Â 10

c)Â 11

d)Â 12

Solution:

The no. has all the digits as odd no. and is divisible by 3. So the possibilities are

1113
1119
1131
1137
1155
1173
1179
1191
1197
Hence 9 possibilities .

Question 9:Â Of 128 boxes of oranges, each box contains at least 120 and at most 144 oranges. X is the maximum number of boxes containing the same number of oranges. What is the minimum value of X?

a)Â 5

b)Â 103

c)Â 6

d)Â Cannot be determined

Solution:

Each box contains at least 120 and at most 144 oranges.

So boxes may contain 25 different numbers of oranges among 120, 121, 122, …. 144.

Lets start counting.

1st 25 boxes contain different numbers of oranges and this is repeated till 5 sets as 25*5=125.

Now we have accounted for 125 boxes. Still 3 boxes are remaining. These 3 boxes can have any number of oranges from 120 to 144.

Already every number is in 5 boxes. Even if these 3 boxes have different number of oranges, some number of oranges will be in 6 boxes.

Hence the number of boxes containing the same number of oranges is at least 6.

Question 10:Â Let n be the number of different five-digit numbers, divisible by 4 with the digits 1, 2, 3, 4, 5 and 6, no digit being repeated in the numbers. What is the value of n?

a)Â 144

b)Â 168

c)Â 192

d)Â None of these

Solution:

To be divisible by 4 , last 2 digits of the 5 digit no. should be divisible by 4 . So possibilities are 12,16,32,64,24,36,52,56 which are 8 in number. Remaining 3 digits out of 4 can be selected in $^4C_3$ ways and further can be arranged in 3! ways . So in total = 8*4*6 = 192

Question 11:Â Let D be recurring decimal of the form, $D = 0.a_1a_2a_1a_2a_1a_2…$, where digits $a_1$ and $a_2$ lie between 0 and 9. Further, at most one of them is zero. Then which of the following numbers necessarily produces an integer, when multiplied by D?

a)Â 18

b)Â 108

c)Â 198

d)Â 288

Solution:

Case 1: $a_1=0$
So, D equals $0.0a_20a_20a_2…$
So, 100D equals $a_2.0a_20a_2…$
So, 99D equals $a_2$

Case 2: $a_2=0$
So, DÂ equals $0.a_10a_10a_1…$
So, 100D equals $a_10.a_10a_1….$
So, 99D equals $a_10$

So, in both the cases, 99D is an integer. From the given options, only option C satisfies this condition (198=2*99) and hence the correct answer is C.

Question 12:Â If $x^2 + y^2 = 0.1$ and |x-y|=0.2, then |x|+|y| is equal to:

a)Â 0.3

b)Â 0.4

c)Â 0.2

d)Â 0.6

Solution:

$(x – y)^2 = x^2 + y^2 – 2xy$

$0.04 = 0.1 – 2xy => xy = 0.03$

So, |xy| = 0.03

$(|x| + |y|)^2 = x^2 + y^2 + 2|xy| = 0.1 + 0.06 = 0.16$

So, |x|+|y| = 0.4

Question 13:Â What is the greatest power of 5 which can divide 80! exactly?

a)Â 16

b)Â 20

c)Â 19

d)Â None of these

Solution:

The highest power of 5 in 80! = [80/5] + [$80/5^2$] = 16 + 3 = 19

So, the highest power of 5 which divides 80! exactly = 19

Question 14:Â If x is a positive integer such that 2x +12 is perfectly divisible by x, then the number of possible values of x is

a)Â 2

b)Â 5

c)Â 6

d)Â 12

Solution:

If 2x+12 is perfectly divisible by x, then 12 must be divisible by x.
Hence, there are six possible values of x : (1,2,3,4,6,12)

Question 15:Â If a number 774958A96B is to be divisible by 8 and 9, the respective values of A and B will be

a)Â 7 and 8

b)Â 8 and 0

c)Â 5 and 8

d)Â None of these

Solution:

According to the divisible rule of 9, theÂ sum of all digits should be divisible by 9.
i.e. 55+A+B = 9k
So sum can be either 63 or 72.
For 63, A+B should be 8.
In given options, option B has values of A and B whose sum is 8 and by putting them we are having a number which divisible by both 9 and 8.

Question 16:Â If n is an integer, how many values of n will give an integral value of $\frac{(16n^2+ 7n+6)}{n}$ ?

a)Â 2

b)Â 3

c)Â 4

d)Â None of these

Solution:

Expression can be reduced to 16n + 7 + $\frac{6}{n}$
Now to make above value Â an integer n can be 1,2,3,6,-1,-2,-3,-6

Question 17:Â $n^3$ is odd. Which of the following statement(s) is/are true?
I. $n$ is odd.
II.$n^2$ is odd.
III.$n^2$ is even.

a)Â I only

b)Â II only

c)Â I and II

d)Â I and III

Solution:

if $n^3$ is odd then $n$ will be odd. let’s say it is $2k+1$
then $n^2$ will be = $(4k^2 + 4k + 1)$ which will be odd

Question 18:Â How many five digit numbers can be formed from 1, 2, 3, 4, 5, without repetition, when the digit at the unitâ€™s place must be greater than that in the tenâ€™s place?

a)Â 54

b)Â 60

c)Â 17

d)Â 2 Ã— 4!

Solution:

Possible numbers with unit’s place as 5 = $4 \times 3 \times 2 \times 1 = 24$

Possible numbers with unit’s place as 4 and ten’s place 3,2,1 = $3 \times 3 \times 2 \times 1 = 18$

Possible numbers with unit’s place as 3 and ten’s place 2,1 = $2 \times 3 \times 2 \times 1 = 12$

Possible numbers with unit’s place as 3 and ten’s place 1 = $1 \times 3 \times 2 \times 1 = 6$

Total possible values = 24+18+12+6 = 60

Question 19:Â 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)Â 0

b)Â 1

c)Â 2

d)Â None of these

Solution:

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 4Â is 3
The remainder when n is divided by 5Â is 4 and
The remainder when n is divided by 6Â 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

Question 20:Â How many five-digit numbers can be formed using the digits 2, 3, 8, 7, 5 exactly once such that the number is divisible by 125?

a)Â 0

b)Â 1

c)Â 4

d)Â 3

Solution:

As we know for a number to be divisible by 125, its last three digits should be divisible by 125
So for a five digit number, with digits 2,3,8,7,5 its last three digits should be 875 and 375
Hence only 4 numbers are possible with its three digits as 875 and 375
I.e. 23875, 32875, 28375, 82375
Question 21:Â What is the digit in the unitâ€™s place of $2^{51}$?

a)Â 2

b)Â 8

c)Â 1

d)Â 4

Solution:

The last digit of powers of 2 follow a pattern as given below.

The last digit of $2^1$ is 2
The last digit of $2^2$ is 4
The last digit of $2^3$ is 8
The last digit of $2^4$ is 6

The last digit of $2^5$ is 2
The last digit of $2^6$ is 4
The last digit of $2^7$ is 8
The last digit of $2^8$ is 6

Hence, the last digit of $2^{51}$ is 8

Question 22:Â If $a, b, c,$ and $d$ are integers such that $a+b+c+d=30$ then the minimum possible value of $(a – b)^{2} + (a – c)^{2} + (a – d)^{2}$Â  is

Solution:

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

Question 23:Â If the product of three consecutive positive integers is 15600 then the sum of the squares of these integers is

a)Â 1777

b)Â 1785

c)Â 1875

d)Â 1877

Solution:

$(x -1)x(x+1) = 15600$
=> $x^3 – x= 15600$
The nearest cube to 15600 is 15625 = $25^3$
We can verify that x = 25 satisfies the equation above.
Hence the three numbers are 24, 25, 26. Sum of their squares = 1877

Question 24:Â While multiplying three real numbers, Ashok took one of the numbers as 73 instead of 37. As a result, the product went up by 720. Then the minimum possible value of the sum of squares of the other two numbers is

Solution:

We know that one of the 3 numbers is 37.
Let the product of the other 2 numbers be x.
It has been given that 73x-37x = 720
36x = 720
x = 20

Product of 2 real numbers is 20.
We have to find the minimum possible value of the sum of the squares of the 2 numbers.
Let x=a*b
It has been given that a*b=20

The least possible sum for a given product is obtained when the numbers are as close to each other as possible.
Therefore, when a=b, the value of a and b will be $\sqrt{20}$.

Sum of the squares of the 2 numbers = 20 + 20 = 40.

Therefore, 40 is the correct answer.

Question 25:Â The number of integers x such that $0.25 \leq 2^x \leq 200$ and $2^x + 2$ is perfectly divisible by either 3 or 4, is

Solution:

At $x = 0, 2^x = 1$ which is in the given range [0.25, 200]

$2^x + 2$ = 1 + 2 = 3 Which is divisible by 3. Hence, x = 0 is one possible solution.

At $x = 1, 2^x = 2$ which is in the given range [0.25, 200]

$2^x + 2$ = 2 + 2 = 3 Which is divisible by 4. Hence, x = 1 is one possible solution.

At $x = 2, 2^x = 4$ which is in the given range [0.25, 200]

$2^x + 2$ = 4 + 2 = 6 Which is divisible by 3. Hence, x = 2 is one possible solution.

At $x = 3, 2^x = 8$ which is in the given range [0.25, 200]

$2^x + 2$ = 8 + 2 = 3 Which is not divisible by 3 or 4. Hence, x = 3 can’t be a solution.

At $x = 4, 2^x = 16$ which is in the given range [0.25, 200]

$2^x + 2$ = 16 + 2 = 18 Which is divisible by 3. Hence, x = 4 is one possible solution.

At $x = 5, 2^x = 32$ which is in the given range [0.25, 200]

$2^x + 2$ = 32 + 2 = 34 Which is not divisible by 3 or 4. Hence, x = 5 can’t be a solution.

At $x = 6, 2^x = 64$ which is in the given range [0.25, 200]

$2^x + 2$ = 64 + 2 = 66 Which is divisible by 3. Hence, x = 6 is one possible solution.

At $x = 7, 2^x = 128$ which is in the given range [0.25, 200]

$2^x + 2$ = 128 + 2 = 130 Which is not divisible by 3 or 4. Hence, x = 7 can’t be a solution.

At $x = 8, 2^x = 256$ which is not in the given range [0.25, 200]. Hence, x can’t take any value greater than 7.

Therefore, all possible values of x = {0,1,2,4,6}. Hence, we can say that ‘x’ can take 5 different integer values.

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