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Number Systems is one of the key topics in the CAT Quantitative Ability (QA) Section. It is essential that you know the basics of the CAT Number Systems well and practice the questions. Also, do check out all the Number System questions from the CAT Previous Papers with detailed video solutions. This article will look into some Number System Questions for the CAT Exam. If you want to practice these important Number System questions, you can download the PDF, 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 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 The remainder, when $(15^{23} + 23^{23})$ is divided by 19, is

a)Â 4

b)Â 15

c)Â 0

d)Â 18

Solution:

The remainder when $15^{23}$ is divided by 19 equals $(-4)^{23}$
The remainder when $23^{23}$ is divided by 19 equals $4^{23}$
So, the sum of the two equals$(-4)^{23}+(4)^{23}=0$

Question 6:Â The number of employees in Obelix Menhir Co. is a prime number and is less than 300. The ratio of the number of employees who are graduates and above, to that of employees who are not, can possibly be:

a)Â 101:88

b)Â 87:100

c)Â 110:111

d)Â 85:98

e)Â 97:84

Solution:

The addition of numerator and denominatpr should give a prime no. Only option E gives that.

3 is a factor of 189 and 183 => A and D eliminated

17 is a factor of 187 and 221 => B and C eliminated

181 is prime.

Question 7:Â Let $x = \sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4- \ to \ infinity}}}}$. Then x equals

a)Â 3

b)Â $(\sqrt{13} – 1)/2$

c)Â $(\sqrt{13} + 1)/2$

d)Â $\sqrt{13}$

Solution:

$x = \sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4- \ to \ infinity}}}}$

=> $x = \sqrt{4+\sqrt{4-x}}$

=> $x^2 = 4 + \sqrt{4-x}$

=>$x^4 + 16 – 8x^2 = 4 – x$

=> $x^4 – 8x^2 + x +12 = 0$

On substituting options, we can see that option C satisfies the equation.

Question 8:Â A set of consecutive positive integers beginning with 1 is written on the blackboard. A student came along and erased one number. The average of the remaining numbers is $\frac{602}{17}$. What was the number erased?

a)Â 7

b)Â 8

c)Â 9

d)Â None of these

Solution:

Since the number starts from 1 if there are n numbers then initial average = $\dfrac{n+1}{2}$.

Average of N natural number can be either an integer {ab} or {ab.50} type. For example average of first 10 number = 5.5 whereas the average of first 11 natural numbers is 6.
Even if we erased the largest number change in average will be always less than 0.5.
Here we are given the average is 602/17 or 35$\frac{7}{17}$ Hence we can say that average must have been 35.5 or 35 before.
Case 1: If the average was 35.5 before the erasing process.
We know that average of 1st N natural number = $\dfrac{N+1}{2}$
35.5 = $\dfrac{N+1}{2}$
N = 70.
Sum of these 70 numbers = 70*71/2 = 35*71 = 2485.
Sum of the 69 numbers which we are left with after removing a number = (602/17)*69 = 2443.41. Which is not possible as the sum of natural numbers will always be an integer. Hence, we can say that case is not possible.

Case 1: If the average was 35 before the erasing process.
We know that average of 1st N natural number = $\dfrac{N+1}{2}$
35 = $\dfrac{N+1}{2}$
N = 69.
Sum of these 69 numbers = 69*70/2 = 35*69 = 2415.
Sum of the 68 numbers which we are left with after removing a number = (602/17)*68 = 2408.
Hence, we can say that the erased number = 2415 – 2408 = 7.

Question 9:Â Convert the number 1982 from base 10 to base 12. The result is:

a)Â 1182

b)Â 1912

c)Â 1192

d)Â 1292

Solution:

Quotient of 1982/12 = 165, remainder = 2
Quotient of 165/12 = 13, remainder = 9
Quotient of 13/12 = 1, remainder = 1
Remainder of 1/12 = 1
So, the required number in base 12 = 1192

Question 10: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 11: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 12: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 13:Â Mr.X enters a positive integer Y(>1) in an electronic calculator and then goes on pressing the square . root key repeatedly. Then

a)Â The display does not stabilize

b)Â The display becomes closer to 0

c)Â The display becomes closer to 1

d)Â May not be true and the answer depends on the choice of Y

Solution:

Take 64 for example.

$\sqrt{64}$ = 8

$\sqrt{8}$ = 2.828

$\sqrt{2.828}$ is close to 1.7

So, we can see that the result is tending towards 1.

Question 14: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

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