# CAT Level Questions on Number System

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## CAT Level Questions on Number System

Download important Number System Questions for CAT PDF based on previously asked questions in CAT exam. Practice Number System Questions PDF for CAT exam.

Question 1: In 2011, Plasma – a pharmaceutical company – allocated $Rs.4.5 \times 10^{7}$ for Research and Development. In 2012, the company allocated Rs.60,000,000 for Research and Development. If each year the funds are evenly divided among $2 \times 10^{2}$ departments, how much more will each department receive this year than it did last year?

a) $Rs.2.0 \times 10^{5}$

b) $Rs.7.5 \times 10^{5}$

c) $Rs.7.5 \times 10^{4}$

d) $Rs.2.5 \times 10^{7}$

Question 2: In the following series, what numbers should replace the question marks?
-1, 0, 1, 0, 2, 4, 1, 6, 9, 2, 12, 16, ? ? ?

a) 11, 18, 27

b) -1, 0, 3

c) 3, 20, 25

d) Cannot be ascertained

Question 3: Arrange the following letters to form a meaningful word.

a) 1 3 5 2 4 6 8 7

b) 8 6 7 1 4 2 3 5

c) 4 2 3 5 8 6 7 1

d) 5 3 7 1 8 4 2 6

Question 4: The micromanometer in a certain factory can measure the pressure inside the gas chamber from 1 unit to 999999 units. Lately this instrument has not been working properly. The problem with the instrument is that it always skips the digit 5 and moves directly from 4 to 6. What is the actual pressure inside the gas chamber if the micromanometer displays 0030l6?

a) 2201

b) 2202

c) 2600

d) 2960

e) None of the above options

Question 5: If x and y are real numbers, then the minimum value of $x^{2}+ 4xy+ 6y^{2}-4y+ 4$ is

a) ‒4

b) 0

c) 2

d) 4

e) None of the above

Question 6: A, B and C are defined as follows.A=$( 2.000004) \div ((2.000004)^2+ 4.000008)$ ;B = $(3.000003) \div ((3.000003)^2+9.000009)$C= $(4.000002) \div ((4.000002)^2 + 8.000004)$ Which of the following is true about the values of the above three expressions?

a) All of them lie between 0.18 and 0.2

b) A is twice of C

c) C is the smallest

d) B is the smallest

Question 7: If 8 + 12 = 2, 7 + 14 = 3 then 10 + 18 = ?

a) 10

b) 4

c) 6

d) 18

Question 8: What is the greatest power of 5 which can divide 80! exactly?

a) 16

b) 20

c) 19

d) None of these

Question 9: Let N = $55^3 + 17^3 – 72^3$. N is divisible by:

a) both 7 and 13

b) both 3 and 13

c) both 17 and 7

d) both 3 and 17

Question 10: In some code, letters a, b, c, d and e represent numbers 2, 4, 5, 6 and 10. We just do not know which letter represents which number. Consider the following relationships:
I. a + c = e,
II. b – d = d and
III. e + a = b
Which of the following statements is true?

a) b = 4, d = 2

b) a = 4, e = 6

c) b = 6, e = 2

d) a = 4, c = 6

Total amount allotted in 2011 = 4.5*10$^5$
Total number of departments = 200
Thus, the amount received by each department = 225000
Total amount allotted in 2012 = 60000000
The amount received by each department = 300000
Thus the excess of amount each department is getting = 300000 – 225000 = 750000  = Rs. 7.5*10$^4$
Hence, option C is the correct answer.

The given series is a combination of 3 series:-
1st containing all 3n+1 terms i.e. 1st, 4th, 7th and so on terms:-
-1, 0, 1, 2 and thus, the next number will be 3.
2nd containing all 3n+2 terms i.e. 2nd, 5th, 8th and so on terms:-
0, 2, 6, 12 as we can see 2n is being added to each term to get the next term and thus the next term will be 20.
3rd containing all 3n terms i.e. 3rd, 6th, 9th and so on terms:-
1, 4, 9, 16 an we can see these are squares of the natural numbers and thus, the next term will be 25.
Hence, ?,?,? will be 3, 20, 25
Hence, option C is the correct answer.

OVERHEAD is the word that can be formed from the given letters.
Hence, option C is the correct answer.

The meter skips all the numbers in which there is a 5

From, (1 to 99) -> 5 occurs 10 times in tens place and 10 times in units place (which also includes 55)

=> Total occurrence = $10 + 10 – 1 = 19$

Similarly from (100 to 199), from (200 to 299)…, from (400 to 499) , from (600 to 699),…from (900 to 999)

It occurs = $8 \times 19 = 152$

Now, from (500 to 599), there are 100 numbers, => micromanometer reading can change from 499 to 600.

Thus, total numbers skipped from (1 to 999) = $19 + 152 + 100 = 271$

Similarly, from (1000 to 1999) = 271

and from (2000 to 2999) = 271

Also, 3005 and 3015 are also skipped

=> Total number of skips = $271 + 271 + 271 + 2 = 815$

$\therefore$ Actual pressure = $3016 – 815 = 2201$

Expression : $x^{2}+ 4xy+ 6y^{2}-4y+ 4$

= $(x^2 + 4xy + 4y^2) + (2y^2 – 4y + 4)$

= $(x + 2y)^2 + \frac{1}{2} (4y^2 – 8y + 8)$

= $(x + 2y)^2 + \frac{1}{2} [(2y)^2 – 2 (2y) (2) + (2)^2 + 4]$

= $(x + 2y)^2 + \frac{1}{2} [(2y – 2)^2 + 4]$

= $(x + 2y)^2 + \frac{1}{2} (2y – 2)^2 + 2$

Since, $x$ and $y$ are real, => Min value of $(x + 2y)^2 = 0$

Minimum value of $(2y – 2)^2 = 0$

$\therefore$ Minimum value of expression = $0 + 0 + 2 = 2$

Given expressions can be reduced as follows

A = $\frac{1}{4.000004}$

B = $\frac{1}{6.000003}$

C = $\frac{1}{6.000002}$

Among all of them B is smallest.

8+12 =( 20 = 2+0) = 2
7+14 =( 21 = 2+1) = 3
10+18= (28= 2+8) = 10

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

$55^3 + 17^3 – 72^3$ = $(55-72)k + 17^3$. This is divisible by 17

Remainder when $55^3$ is divided by 3 = 1

Remainder when $17^3$ is divided by 3 = -1

Remainder when $72^3$ is divided by 3 = 0

So, $55^3 + 17^3 – 72^3$ is divisible by 3

So, the answer is d) 3 and 17