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# Number System Questions for SSC CHSL and MTS 2022 – Download PDF

Here you can download SSC CHSL & MTS 2022 – important for SSC CHSL & MTS Number System Questions PDF by Cracku. Very Important SSC 2022 exams and These questions will help your SSC preparation. So kindly download the PDF for reference and do more practice.

Question 1:Â What is the value of
$\frac{3}{4}Â ofÂ \left(\frac{1}{3} \div \frac{1}{2}\right) + \left(2 – \frac{2}{5}\right) \times \frac{3}{2} + \frac{2}{3}$?

a)Â $\frac{107}{30}$

b)Â $\frac{103}{25}$

c)Â $\frac{109}{17}$

d)Â $\frac{101}{6}$

Question 2:Â What is the value of $\frac{2 \div 3 \times (1 + 3) + 5 – 6}{2Â ofÂ 3 \div 5 \times 4 + 3 – 2}$?

a)Â $\frac{36}{89}$

b)Â $\frac{31}{73}$

c)Â $\frac{25}{87}$

d)Â $\frac{27}{92}$

Question 3:Â What is the mode of given data?
4, 3, 7, 13, 16, 23, 3, 4, 7, 4, 3, 3, 9, 6, 9, 6

a)Â 9

b)Â 4

c)Â 3

d)Â 6

Question 4:Â What is the mode of the given data?
4, 3, 4, 3, 2, 2, 2, 5, 5, 3, 4, 6, 4, 3, 3

a)Â 3

b)Â 2

c)Â 5

d)Â 4

Question 5:Â What is the value of
$36 \div 8 \times 4 + 2 \div 4 – 1 + 5Â ofÂ 3 \div (4 \times 2 – 3) – 3$?

a)Â $\frac{31}{2}$

b)Â $18$

c)Â $\frac{35}{2}$

d)Â $16$

Question 6:Â What is the value of $\frac{39 \div 26 + 22 \div 11 \times 2 + 4 \times 3}{2Â ofÂ 5 – 3(7 + 10 \div 2 – 3 \times 3)}$?

a)Â $\frac{39}{2}$

b)Â $\frac{49}{2}$

c)Â $\frac{61}{2}$

d)Â $\frac{35}{2}$

Question 7:Â What is the value of $(24 + 16 \times 5 – 8Â ofÂ 4) \div 84 \times 48 \div 24 \times 6 + 4 + 3 ?$

a)Â $\frac{139}{3}$

b)Â $\frac{121}{7}$

c)Â $\frac{56}{3}$

d)Â $\frac{156}{5}$

Question 8:Â If $X : Y : Z = 1 : 2 : 3$ and, $X^2 + Y^2 + Z^2 = 224$, then what is the value of $X + Y + Z$?

a)Â 24

b)Â 48

c)Â 36

d)Â 32

Question 9:Â What is the value of $(3 \times 4Â ofÂ 12 \div 2) \div 9 \times 4 + 4 \div 8 + 3 \times 2 ?$

a)Â $\frac{37}{2}$

b)Â $\frac{77}{2}$

c)Â $\frac{89}{3}$

d)Â $\frac{94}{3}$

Question 10:Â If $A = 8 \div 4 \times (3 – 1) + 6 \times 3 \div 2Â ofÂ 3Â andÂ B = 4 \div 8 \times 2 + 7 \times 3$, then what is the value of $A + B$?

a)Â 33

b)Â 29

c)Â 31

d)Â 35

Question 11:Â What is the least number of four digits which is exactly divisible by 2, 4, 6 and 8?

a)Â 1016

b)Â 1024

c)Â 1008

d)Â 1096

Question 12:Â What is the value of $\frac{3}{4} \div \left(\frac{1}{2} + \frac{1}{16}\right) + \frac{2}{3} of \frac{4}{9} \div \left(\frac{1}{3} – \frac{11}{81}\right) + \frac{1}{4} \times \frac{2}{3}$?

a)Â 3

b)Â 1

c)Â 2

d)Â 4

Question 13:Â What is the difference of mean and median of the given data : 4, 13, 8, 15, 9, 21, 18, 23, 35, 1?

a)Â 0.7

b)Â 1.7

c)Â 1.2

d)Â 2.1

Question 14:Â 60% of a number is 168, then what is the number?

a)Â 280

b)Â 320

c)Â 240

d)Â 200

Question 15:Â What is the value of: $5Â Â ofÂ 5Â ofÂ 5 \div 5 + 5 – 6 \div 3 \times 4 + 2 + (3 \div 6 \times 2)?$

a)Â 21

b)Â 25

c)Â 28

d)Â 19

Question 16:Â The mode of 2, 2, 3, 3, 5, 5, 5, 7, 8, 8, 9, 10 is:

a)Â 5

b)Â 2

c)Â 3

d)Â 6

Question 17:Â The mode of the following data is 36. What is the value of x ?

a)Â 11

b)Â 15

c)Â 13

d)Â 12

Question 18:Â When 6892, 7105 and 7531 are divided by the greatest number x, then the remainder in each case is y. What is the value of (x – y)?

a)Â 123

b)Â 137

c)Â 147

d)Â 113

Question 19:Â The sum of the perfect square between 120 and 300 is:

a)Â 1400

b)Â 1024

c)Â 1296

d)Â 1204

Question 20:Â The difference between the greatest and the least four digit numbers that begins with 3 and ends with 5 is:

a)Â 990

b)Â 900

c)Â 909

d)Â 999

$\frac{3}{4}Â ofÂ \left(\frac{1}{3} \div \frac{1}{2}\right) + \left(2 – \frac{2}{5}\right) \times \frac{3}{2} + \frac{2}{3}$

$\frac{3}{4}\times(\frac{1}{3}\times\frac{2}{1}) + (\frac{10 – 2}{5})\times\frac{3}{2} + \frac{2}{3}$

$\frac{3}{4}\times\frac{2}{3} + \frac{8}{5}\times\frac{3}{2} + \frac{2}{3}$

$\frac{6}{12} + \frac{24}{10} + \frac{2}{3}$

$\frac{30 + 144 + 40}{60}$

$\frac{214}{60}$ = $\frac{107}{30}$

Therefore option A is the answer.

$\frac{2 \div 3 \times (1 + 3) + 5 – 6}{2Â ofÂ 3 \div 5 \times 4 + 3 – 2}$

$\frac{\frac{2}{3}\times4 + 5 – 6}{2\times\frac{3}{5}\times 4 + 3 – 2}$

$\frac{\frac{8}{3} + 5 – 6}{\frac{24}{5} + 3 – 2}$

$\frac{\frac{8 + 15 – 18}{3}}{\frac{24 + 15 – 10}{5}}$

$\frac{8 + 15 – 18}{3}\times\frac{5}{24 + 15 – 10}$

$\frac{5}{3}\times\frac{5}{29}$

$\frac{25}{87}$

Number 3 is repeated more number of times when compared to other numbers.Therefore 3 is the answer.

The mode of a data set is the number that occurs most frequent in the set

To find the mode :

Step 1: arrange numbers in ascending order

2, 2 , 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 5, 5, 6

Step 2 : count how many times each number occurs

2 three times

3 five times

4 four times

5 two times

6 one time

Step 3 : The number that occurs the most is the mode

3 is the mode

$36 \div 8 \times 4 + 2 \div 4 – 1 + 5 of 3 \div (4 \times 2 – 3) – 3$

$= 36 \div 8 \times 4 + 2 \div 4 – 1 + 5 of 3 \div 5 – 3$

$=36 \div 8 \times 4 + 2 \div 4 – 1 + 15 \div 5 – 3$

$=\frac{9}{2} \times 4 + \frac{1}{2} – 1 + 3Â – 3$

$=18 +Â \frac{1}{2} – 1 + 3 – 3$

$=18 + \frac{1}{2} + 3 – 4$

$= \frac{35}{2}$

$\frac{39 \div 26 + 22 \div 11 \times 2 + 4 \times 3}{2Â ofÂ 5 – 3(7 + 10 \div 2 – 3 \times 3)}$

$\frac{\frac{39}{26} + \frac{22}{11} \times 2 + 4 \times 3}{2 \times 5 – 3(7 + \frac{10}{2} – 3\times 3)}$

$\frac{\frac{3}{2} + 4 + 12}{10 – 3(7 + 5 -9)}$

$\frac{\frac{3 + 8 + 24}{2}}{10 – 3(3)}$

$\frac{\frac{35}{2}}{10 – 9}$

$\frac{35}{2}$

$(24 + 16 \times 5 – 8Â ofÂ 4) \div 84 \times 48 \div 24 \times 6 + 4 + 3$

$\frac{(24 + 16\times5 – 8\times4)}{84}\times\frac{48}{24}\times6 + 4 + 3$

$\frac{24 + 80 + 32}{7} + 4 + 3$

$\frac{72 + 28 + 21}{7}$

$\frac{121}{7}$

$X : Y : Z = 1 : 2 : 3$

let X = a, Y= 2a , Z= 3a

nowÂ $X^2 + Y^2 + Z^2 = 224$ =Â $\left(a\right)^{2} +\left(2a\right)^{2} +\left(3a\right)^{2} = 224$

$a^2 + 4a^2 + 9a^2 = 224$

$14a^2 = 224$Â , $a^2 = \frac{224}{14}$Â ,Â $a^2 = 16$

a=4

$X + Y + Z$ = $a+2a+3a= 6a$ = $6\times 4$ = 24

using the BODMAS rule { priority brackets > of > division > multiplication > addition > subtraction}

solving the bracket first (1st priority brackets)

$(3 \times 4Â ofÂ 12 \div 2)$ , now since ‘of’ is the priority hence it should be solved first

simplifyingÂ  it we get

$(3 \times 4\times12 \div 2)$ (here 4 of 12 is $4\times12$ ) =$(3 \times 4\times 6)$

substituting in originalÂ  question we get

$(3 \times 4\times 6) \div 9 \times 4 + 4 \div 8 + 3 \times 2$

simpliying it further we get

$\frac{(3 \times 4\times 6)}{9} \times 4 +\frac{4}{8}+ 3 \times 2$

= 32+$\frac{1}{2}$+ 6 = $\frac{77}{2}$

Applying the BODMAS { priority brackets > of > division > multiplication > addition > subtraction }

To solve A , first solve the subtraction in the brackets i.e (3-1) = 2

simplifying A, we get

$A = 8 \div 4 \times 2 + 6 \times 3 \div 2Â ofÂ 3Â$ = $\frac{8}{4}\times 2 + \frac{6 \times 3}{6}$ ( here 2 of 3 is $2\times 3$ = 6)

A= 7

similarly applying BODMAS we solve for B

$B = 4 \div 8 \times 2 + 7 \times 3$ =Â $B = \frac{4}{8} \times 2 + 7 \times 3$ = 22

B = 22

A+B = 7+22 = 29

For a number to be divisible 2,4,6,8 should be multiple of 2 and 3,as numbers 2,4,8 have common factor 2 and number 6 is a multiple of 2 and 3.

So, from the options given we get 1008 as a multiple of 2 and 3 both.

Hence option C is a correct choice

$\frac{3}{4} \div \left(\frac{1}{2} + \frac{1}{16}\right) + \frac{2}{3} of \frac{4}{9} \div \left(\frac{1}{3} – \frac{11}{81}\right) + \frac{1}{4} \times \frac{2}{3}$

$\Rightarrow$ $\frac{3}{4} \div \frac{9}{16} + \frac{8}{27} \div \frac{16}{81} + \frac{1}{6}$

$\Rightarrow$ $\frac{3}{4}\times\frac{16}{9}+\frac{8}{27}\times\frac{81}{16}+\frac{1}{6}$

$\Rightarrow$ $\frac{4}{3}+\frac{3}{2}+\frac{1}{6}$

$\Rightarrow$ 3.

Mean:

No. of samples (n) = 10

Mean = $\frac{\sum x}{n} = \frac{4+13+8+15+9+21+18+23+35+1}{10}= \frac{147}{10} = 14.7$

Median:

Arranging the data in ascending order, we get:

1, 4, 8, 9, 13, 15, 18, 21, 23, 35

n = 10 (even)

Therefore, median is the average of 5th and 6th term.

Median = $Â Â \frac{13+15}{2} Â = 14$

Mean – Median = 14.7 – 14 = 0.7

Therefore, Option A is correct.

60% of the number is 168.

Let’s assume the number is ‘y’.

60% of y =Â 168

0.6y = 168

y =Â 280

$5\times5\times\frac{5}{5}+5-\frac{6}{3}\times4+2+\frac{3}{6}\times2$

$25+5-8+2+1$

25

Mode : The value that appears most often in a set of given data values.

Given Data :Â 2, 2, 3, 3, 5, 5, 5, 7, 8, 8, 9, 10

• Most number repeated in above data is 5.

So, Mode of the given data is 5.

Hence,Â OptionÂ AÂ is correct.

As per given data,

Class interval of 30-40 has highest frequency, thatswhy it is modal class

As we know,

M =Â $l+\left\{\frac{\left(f_1-f_0\right)}{2f_1-f_0-f_2}\right\}\times\ h$

where, h= size of the class interval,

l = lower limit of the modal class,

$f_{1=}$ frequency of the modal class,

$f_{_0=}$ frequency of the class preceding the modal class

$f_{_2=}$ frequency of the class succeeding the modal class

putting the values from the given dataÂ :

$36=30+\frac{\left(16-10\right)}{2\times\ 16-10-x}\times\ 10$

$36-30=\frac{6}{22-x}\times\ \ 10$

$22-x=10$

$x=12$

Hence, Option D is correct.

We have to find HCF of given numbers : 6892, 7105, 7531

7105 – 6892 = 213

7531 – 7105 = 426

426 – 213 = 213

So, Either the difference or the factor of difference is theÂ HCF of those given number.

Here , 213 is the HCF.

When 6892, 7105, 7531 is divided by 213 we get 76 as an remainder

So, x = 213 and y = 76

According to Question :

x – y = 213 – 76 = 137

Hence,Â Option B is correct.Â Â

Sum of the squares of n consecutive numbers =

The sum of the perfect square between 120 and 300 =Â $11^2+12^2+13^2+14^2+15^2+16^2+17^2$

$=\frac{17\left(17+1\right)\left(2\left(17+1\right)\right)}{6}-\frac{10\left(10+1\right)\left(2\left(10\right)+1\right)}{6}$

$=\frac{17\left(18\right)\left(35\right)}{6}-\frac{10\left(11\right)\left(21\right)}{6}$

$=51\times35-11\times35$

$=35\left(51-11\right)$

$=35\left(40\right)$

$=1400$

Hence, the correct answer is Option A

$\therefore\$The difference between the greatest and the least four digit numbers that begins with 3 and ends with 5 = 3995 – 3005 = 990