# CAT Number System Questions PDF [Important]

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Number System is one of the most important topics in the Quant Section of the CAT Exam. You can check out these Number SystemÂ  in CAT Previous year papers. In this Section, you can watch these Number System basics. This article will look into some important Number System Questions for CAT. These are good sources for practice; If you want to practice these questions, you can download this CAT Number System Questions PDF below, which is completely Free.

Question 1:Â How many pairs of positive integers m, n satisfy 1/m + 4/n = 1/12 , where n is an odd integer less than 60?

a)Â 6

b)Â 4

c)Â 7

d)Â 5

e)Â 3

Solution:

1/m + 4/n = 1/12
So, 1/m = 1/12 – 4/n
So, m = 12n/(n-48)
Since m is positive, n should be greater than 48
Also, since n is an odd number, it can take only 49, 51, 53, 55, 57 and 59
If n = 49, 51, 57 then m is an integer, else it is not an integer
So, there are 3 pairs of values for which the equation is satisfied

Question 2:Â In a tournament, there are n teams $T_1 , T_2 ….., T_n$ with $n > 5$. Each team consists of k players, $k > 3$. The following pairs of teams have one player in common: $T_1$ & $T_2$ , $T_2$ & $T_3$ ,……, $T_{n-1}$ & $T_n$ , and $T_n$ & $T_1$ . No other pair of teams has any player in common. How many players are participating in the tournament, considering all the n teams together?

a)Â n (k – 1)

b)Â k (n – 1)

c)Â n (k – 2)

d)Â k (k – 2)

e)Â (n – 1)(k – 1)

Solution:

The number of players in all the teams put together = k * n

The number of players that are common is 1*n = n

So, the number of players in the tournament = kn – n = n(k-1)

Question 3:Â Consider four digit numbers for which the first two digits are equal and the last two digits are also equal. How many such numbers are perfect squares?

a)Â 3

b)Â 2

c)Â 4

d)Â 0

e)Â 1

Solution:

Let the number be xxyy
xxyy = 1000x + 100x + 10y +y = 1100x+11y = 11(100x+y)
Since xxyy is a perfect square, and 11 is one of the factors, it should be a multiple of 121
So, xxyy = 121k, where k is also a perfect square.
For k = 4, xxyy is a 3 digit number and for k > 82, xxyy is a five digit number
Between 4 and 82, only for k = 64, the number is of the form xxyy
121*64 = 7744
So, there is only 1 number 7744 which is of the form xxyy and a perfect square.

Alternatively:

The number should be definitely more than 32 and less than 100 as the square is a two digit number.

A number of such form can be written as $(50 \pm a)$ and $100 – a$ where $0 \leq a \leq 100$
So, the square would be of form $(50 \pm a)^2 = 2500 + a^2 \pm 100a$ or $(100 – a)^2$ i.e. $10000 + a^2 + 200 a$

In both cases, only $a^2$ contributes to the tens and ones digit. Among squares from 0 to 25, only 12 square i.e. 144 has repeating tens and ones digit. So, the number can be 38, 62, or 88. Checking these squares only 88 square is in the form of xxyy i.e. 7744.

Question 4:Â The integers 1, 2, â€¦, 40 are written on a blackboard. The following operation is then repeated 39 times: In each repetition, any two numbers, say a and b, currently on the blackboard are erased and a new number a + b – 1 is written. What will be the number left on the board at the end?

a)Â 820

b)Â 821

c)Â 781

d)Â 819

e)Â 780

Solution:

Let the first operation be (1+40-1) = 40, the second operation be (2+39-1) = 40 and so on

So, after 20 operations, all the numbers are 40. After 10 more operations, all the numbers are 79

Proceeding this way, the last remaining number will be 781

Question 5: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 6: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 7:Â Three consecutive positive integers are raised to the first, second and third powers respectively and then added. The sum so obtained is perfect square whose square root equals the total of the three original integers. Which of the following best describes the minimum, say m, of these three integers?

a)Â $1 \leq m \leq 3$

b)Â $4 \leq m \leq 6$

c)Â $7 \leq m \leq 9$

d)Â $10 \leq m \leq 12$

e)Â $13 \leq m \leq 15$

Solution:

Let us assume that three positive consecutive integers are x, x+1, x+2. They are raised to first, second and third powers respectively.

$x^{1} + (x+1)^{2} + (x+2)^{3} = (x + (x+1) +(x+2))^{2}$

$x^{1} + (x+1)^{2} + (x+2)^{3}$ = $(3x + 3)^{2}$

$x^{3} + 7x^{2} + 15x + 9$ = $9x^{2} + 9 + 18x$

After simplifying you get,

$x^{3} – 2x^{2} – 3x = 0$

=> x=0,3,-1

Since x is a positive integer, it can only be 3.

So, the minimum of the three integers is 3. Option a) is the correct answer.

Question 8:Â The number of non-negative real roots of $2^x – x – 1 = 0$ equals

a)Â 0

b)Â 1

c)Â 2

d)Â 3

Solution:

$2^x – x – 1 = 0$ for this equation only 0 and 1 i.e 2 non-negative solutions are possible. Or we can plot the graph of $2^x$ and x+1 and determine the number of points of intersection and hence the solutin.

Question 9:Â Twenty-seven persons attend a party. Which one of the following statements can never be true?

a)Â There is a person in the party who is acquainted with all the twenty-six others.

b)Â Each person in the party has a different number of acquaintances.

c)Â There is a person in the party who has an odd number of acquaintances.

d)Â In the party, there is no set of three mutual acquaintances.

Solution:

From the options a, c and d all can possibly occur. Hence option b. Besides, if all people have different number of acquaintances, then first person will have 26 acquaintance, second person will have 25 acquaintance, third person will have 24 and so on till 27 th person will have 0 acquaintance. 0 acquaintance is practically not possible.

Question 10: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 11:Â A positive whole number M less than 100 is represented in base 2 notation, base 3 notation, and base 5 notation. It is found that in all three cases the last digit is 1, while in exactly two out of the three cases the leading digit is 1. Then M equals

a)Â 31

b)Â 63

c)Â 75

d)Â 91

Solution:

SinceÂ Â in all three cases the last digit is 1, the number should give remainder 1 when divided individually by 2,3,5 . So the no. may be 31 or 91 . Now 31 in base 2,3 and 5 give first digit as 1 in all the 3 cases while 91 givesÂ exactly two out of the three cases the leading digit as 1. Hence option D.

Question 12:Â How many three digit positive integers, with digits x, y and z in the hundred’s, ten’s and unit’s place respectively, exist such that x < y, z < y and x $\neq$ 0 ?

a)Â 245

b)Â 285

c)Â 240

d)Â 320

Solution:

x, y and z in the hundred’s, ten’s and unit’s place. So y should start from 2

If y=2 , possible values of x=1 and z = 0,1 .So 2 cases 120,121.

Also if y=3 , possible values of x=1,2 and z=0,1,2.

Here 6 three digit nos. possible .

Similarly for next cases would be 3*4=12,4*5=20,5*6=30,…..,8*9=72 . Adding all we get 240 cases.

Question 13:Â There are 8436 steel balls, each with a radius of 1 centimeter, stacked in a pile, with 1 ball on top, 3 balls in the second layer, 6 in the third layer, 10 in the fourth, and so on. The number of horizontal layers in the pile is

a)Â 34

b)Â 38

c)Â 36

d)Â 32

Solution:

For the given problem ,

$\sum {n(n+1)/2} = 8436$ which is

$\sum {n^2/2} + \sum{n/2} = 8436$ which is equal to

n*(n+1)(2n+1)/12 + n*(n+1)/4 = 8436 , solving we get n=36.

Solving the equation might be lengthy. you can substitute the values in the optionsÂ to arrive at the answer.

Question 14:Â In a certain examination paper, there are n questions. For j = 1,2 â€¦n, there are $2^{n-j}$ students who answered j or more questions wrongly. If the total number of wrong answers is 4095, then the value of n is

a)Â 12

b)Â 11

c)Â 10

d)Â 9

Solution:

Let there only be 2 questions.

Thus there are $2^{2-1}$ = 2 students who have done 1 or more questions wrongly, 2$^{2-2}$ = 1 students who have done all 2 questions wrongly .

Thus total number of wrong answers = 2 + 1 = 3= $2^n – 1$.

Now let there be 3 questions. Then j = 1,2,3

Number of students answering 1 or more questions incorrectly = 4

Number of students answering 2 or more questions incorrectly = 2

Number of students answering 3 or more questions incorrectly = 1

Total number of incorrect answers = 1(3)+(2-1)*2+(4-2)*1 = 7 = $2^3-1$

According to the question , the total number of wrong answers = 4095 = $2^{12} – 1$.

Hence Option A.

Question 15: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 16: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 17:Â On January 1, 2004 two new societies S1 and S2 are formed, each n numbers. On the first day of each subsequent month, S1 adds b members while S2 multiples its current numbers by a constant factor r. Both the societies have the same number of members on July 2, 2004. If b = 10.5n, what is the value of r?

a)Â 2.0

b)Â 1.9

c)Â 1.8

d)Â 1.7

n+6b =$nr^6$ and b=10.5n ,
63n+n = $nr^6$
$r^6 = 64$