SNAP Mathematical Reasoning Questions PDF

Mathematical Reasoning is an important topic in the Quant section of the SNAP Exam. You can also download this Free Mathematical Reasoning Questions for SNAP PDF (with answers) by Cracku. These questions will help you to practice and solve the Mathematical Reasoning questions in the SNAP exam. Utilize this PDF practice set, which is one of the best sources for practising.

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Question 1: In a question on division, the divisor is 6 times the quotient and 3 times the remainder. If the remainder is 40, then find the dividend.

a) 2445

b) 2440

c) 2455

d) 2450

1) Answer (B)

Solution:

Given, remainder = 40

Divisor is 3 times the remainder

$\Rightarrow$  divisor = 3 x 40 = 120

Divisor is 6 times the quotient

$\Rightarrow$  120 = 6 x quotient

$\Rightarrow$  quotient = 20

Dividend = divisor x quotient + remainder

$\Rightarrow$ Dividend = 120 x 20 + 40

$\Rightarrow$ Dividend = 2400 + 40

$\Rightarrow$ Dividend = 2440

Hence, the correct answer is Option B

Question 2: In an examination, Anita scored 31% marks and failed by 16 marks. Sunita scored 40% marks and obtained 56 marks more than those required to pass. Find the minimum marks required to pass.

a) 3116

b) 264

c) 3944

d) 7100

2) Answer (B)

Solution:

Anita scored 31% marks and failed by 16 marks and Sunita scored 40% marks and obtained 56 marks more than those required to pass.

Difference of percentage = difference of marks

40% – 31% = 16 + 56 (Anita is failed and Sunita is passed so marks will be added)

9% = 72 marks

56 marks = $\frac{9}{72} \times 56 = 7%$

Sunita scored 40% marks and obtained 56 marks more than those required to pass so,

Minimum passing marks = 40% – 7% = 33%

7% = 56 marks

33% = $\frac{56}{7} \times 33 = 264$

Question 3: The value of $ \frac{(0.67 \times 0.67 \times0.67)  \times (0.33 \times 0.33 \times 0.33)}{(0.67 \times 0.67) \div (0.67 \times 0.33) \div (0.33 \times 0.33)} $

a) 11

b) 0.34

c) 1.1

d) 3.4

3) Answer (B)

Solution:

Mathematical Operators provided in the question seems to be incorrect. Please review the question again and provide correct data.

Question 4: In a division sum, the divisor ‘d’ is 10 times the quotient ‘q’ and 5 times the remainder’r’. If r = 46, the dividend will be

a) 5042

b) 5348

c) 5336

d) 4276

4) Answer (C)

Solution:

We know that

Dividend = $ (Divisor \times Quotient) +Remainder $ —- (1)

Given that Divisor = 10 times the Quotient

=> Divisor= 10Q —- (2)

and Divisor = 5 times the remainder

=> Divisor = 5R = 5(46) = 230

Substituting divisor value in (2), we get, Q= 23

Substituting all values in equation (1), we get

Dividend = $(230 \times 23) + 46 $ = 5336

Question 5: The value of $\frac{18.43 \times 18.43 – 6.57 \times 6.57}{11.86}$ is:

a) 23.62

b) 25

c) 26

d) 24.12

5) Answer (B)

Solution:

$\frac{18.43 \times 18.43 – 6.57 \times 6.57}{11.86}$

$\frac{18.43^{2}-6.57^{2}}{11.86}$

$\frac{(18.43+6.57)(18.43 – 6.57)}{11.86}$

$\frac{25\times11.86}{11.86}=25$

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Question 6: The value of the expression $\frac{1}{4} \left\{\left(a + \frac{1}{a}\right)^2 – \left(a – \frac{1}{a}\right)^2\right\}$ is:

a) $\frac{1}{2}$

b) $\frac{1}{4}$

c) 1

d) 4

6) Answer (C)

Solution:

$\frac{1}{4} \left\{\left(a + \frac{1}{a}\right)^2 – \left(a – \frac{1}{a}\right)^2\right\}$

$\frac{1}{4}\left\{a^2+\frac{1}{a^2}+2-\left(a^2+\frac{1}{a^2}-2\right)\right\}$

$\frac{1}{4}\left\{a^2+\frac{1}{a^2}+2-a^2-\frac{1}{a^2}+2\right\}$

$\frac{1}{4}\times4=1$

Question 7: The simplified value of $\frac{\left(3\frac{1}{5} + \frac{3}{5}\right) \div \frac{8}{5}}{1\frac{1}{7} \div \left\{\frac{5}{7} + \left(\frac{1}{7} \div \frac{1}{3}\right)\right\}}$ is:

a) $\frac{19}{7}$

b) $\frac{19}{16}$

c) $\frac{19}{8}$

d) $\frac{19}{64}$

7) Answer (C)

Solution:

$\frac{\left(3\frac{1}{5} + \frac{3}{5}\right) \div \frac{8}{5}}{1\frac{1}{7} \div \left\{\frac{5}{7} + \left(\frac{1}{7} \div \frac{1}{3}\right)\right\}}= \frac{\left(\frac{16}{5} + \frac{3}{5}\right) \div \frac{8}{5}}{\frac{8}{7} \div \{\frac{5}{7}+\frac{3}{7}\}}$

$=\frac{\frac{19}{5}\div\frac{8}{5}}{\frac{8}{7}\div\frac{8}{7}}$

$=\frac{19}{5}\div\frac{8}{5}$

$=\frac{19}{8}$

Hence, the correct answer is Option C

Question 8: The value of $3 \times 2 \div 3  of  12 – 3 \div 2 \times (2 – 3) \times 2 + 3 \div 2   of  3$ is:

a) $2\frac{1}{3}$

b) $-3\frac{2}{3}$

c) $-2\frac{1}{3}$

d) $3\frac{2}{3}$

8) Answer (D)

Solution:

Apply bodmas rule to this problem to simplify

BODMAS RULE. BODMAS is an acronym and it stands for Bracket, Of, Division, Multiplication, Addition and Subtraction. In certain regions, PEDMAS (Parentheses, Exponents, Division, Multiplication, Addition and Subtraction) is the synonym of BODMAS. It explains the order of operations to solve an expression

First solve the bracket then orders then division then multiplication then addition then subtraction

3×2÷3of12-3÷2×(2-3)×2+3÷2of3

3×$\frac{2}{36}$-$\frac{3}{2}$×(-1)×2+3÷2×3

3×$\frac{1}{18}$-$\frac{3}{2}$×(-1)×2+$\frac{3}{6}$

$\frac{1}{6}$+$\frac{6}{2}$+$\frac{1}{2}$

Take LCM of 6,2,2 you will get 6

= $\frac{1+18+3}{6}$

=$\frac{22}{6}$

=$\frac{11}{3}$

Question 9: The value of $\frac{3 \div \left\{5 – 5 \div (6 – 7) \times 8 + 9\right\}}{4 + 4 \times 4 \div 4  of  4}$ is:

a) $\frac{1}{3}$

b) $\frac{1}{45}$

c) $\frac{1}{90}$

d) $\frac{1}{18}$

9) Answer (C)

Solution:

Apply BODMAS rule to this type of problems

According to Bodmas rule, if an expression contains brackets ((), {}, []) we have to first solve or simplify the bracket followed by of (powers and roots etc.), then division, multiplication, addition and subtraction from left to right.

Numerator = $3 \div \left\{5-5\div(6-7)\times8+9\right\} = 3\div \left\{5-5\div(-1)\times8+9\right\}$
$= 3\div\left\{5+5\times8+9\right\} = 3\div\left\{5+40+9\right\}$
$=\dfrac{3}{54} = \dfrac{1}{18}$

Denominator = $4+4\times4\div4 \text{of} 4 = 4+4\times4\div16 = 4+1 = 5$

$\dfrac{3 \div \left\{5 – 5 \div (6 – 7) \times 8 + 9\right\}}{4 + 4 \times 4 \div 4 of 4} = \dfrac{(\dfrac{1}{18})}{5} = \dfrac{1}{90}$

Question 10: The value of $ 6 – 6 \div 6 \times 6 + (6 \div 6  of  6) \times 6 – \left(3\frac{2}{3} \div \frac{11}{30}  of  \frac{2}{3}\right) \div 5$ is:

a) 0

b) -1

c) -2

d) 2

10) Answer (C)

Solution:

Apply bodmas rule to this problem to simplify

BODMAS RULE. BODMAS is an acronym and it stands for Bracket, Of, Division, Multiplication, Addition and Subtraction. In certain regions, PEDMAS (Parentheses, Exponents, Division, Multiplication, Addition and Subtraction) is the synonym of BODMAS. It explains the order of operations to solve an expression

$ 6 – 6 \div 6 \times 6 + (6 \div 6  of  6) \times 6 – \left(3\frac{2}{3} \div \frac{11}{30}  of  \frac{2}{3}\right) \div 5$

First solve the bracket then orders then division then multiplication then addition then subtraction

=6-1×6+(6÷6×6)×6-($\frac{11}{3}$÷$\frac{11}{30}$×$\frac{2}{3}$)÷5

=6-6+($\frac{6}{36}$)×6-($\frac{11}{3}$×$\frac{90}{22}$)÷5

=0+$\frac{1}{6}$×6-($\frac{30}{2}$)÷5

=0+1-$\frac{15}{5}$

=0+1-3

=-2

Question 11: The value of $3.8 – (4.2 \div 0.7 \times 3) + 5 \times 2 \div 0.5$ is:

a) 15.6

b) 5.8

c) 21.8

d) 18.4

11) Answer (C)

Solution:

4.2/0.7 =6
using BODMAS we get
3.8-6*3+5*4
=3.8+2
=5.8

Question 12: The value of $4.5 – (3.2 \div 0.8 \times 5) + 3 \times 4 \div 6$ is:

a) -13.5

b) 4.2

c) -8.5

d) 5.7

12) Answer (A)

Solution:

we have to solve this question according to the vbodmas rule

$4.5 – (3.2 \div 0.8 \times 5) + 3 \times 4 \div 6$

=$4.5-(4\times5)+2$

=4.5-20+2

=6.5-20

= -13.5

Question 13: The value of
$\frac{9}{15}  of  \left(\frac{2}{3} \div \frac{2}{3}  of  \frac{3}{2}\right) \div \left(\frac{3}{4} \times \frac{3}{4} \div \frac{3}{4}  of  \frac{4} {3}\right)  of  \left(\frac{5}{4} \div \frac{5}{2} \times \frac{2}{5}  of  \frac{4}{5}\right)$ is:

a) $\frac{20}{9}$

b) $\frac{4}{25}$

c) $\frac{18}{125}$

d) $\frac{40}{9}$

13) Answer (D)

Solution:

$\frac{9}{15} of \left(\frac{2}{3} \div \frac{2}{3} of \frac{3}{2}\right) \div \left(\frac{3}{4} \times \frac{3}{4} \div \frac{3}{4} of \frac{4} {3}\right) of \left(\frac{5}{4} \div \frac{5}{2} \times \frac{2}{5} of \frac{4}{5}\right)$

We know that “of” can be replace by multiply,

$\Rightarrow \frac{9}{15} of \left(\frac{2}{3} \div 1 \right) \div \left(\frac{3}{4} \times \frac{3}{4} \div 1 \right) of \left(\frac{5}{4} \div \frac{5}{2} \times \frac{8}{25} \right)$

Now, solving small brakets,

$\Rightarrow \frac{9}{15} of \left(\frac{2}{3} \right) \div \left(\frac{9}{16} \right) of \left(\frac{10}{20} \times \frac{8}{25} \right)$

$\Rightarrow \frac{9}{15} of \left(\frac{2}{3} \right) \div \left(\frac{9}{16} \right) of \left(\frac{4}{25} \right)$

$\Rightarrow \dfrac{\frac{18}{45}}{ \left(\frac{9}{100} \right) }$

$\Rightarrow \frac{18\times 100}{45\times 9}$

$\Rightarrow \frac{2\times 20}{9}$

$\Rightarrow \frac{40}{9}$

Question 14: The value of $(5 + 3 \div 5 \times 5) \div (3 \div 3  of  6)  of  (4 \times 4 \div 4  of  4 + 4 \div 4 \times 4)$ is:

a) $7\frac{1}{3}$

b) $8\frac{1}{5}$

c) $9\frac{3}{5}$

d) $6\frac{2}{3}$

14) Answer (C)

Solution:

As per the question,

$(5 + 3 \div 5 \times 5) \div (3 \div 3 of 6) of (4 \times 4 \div 4 of 4 + 4 \div 4 \times 4)$

Now,

$\Rightarrow (5 + 3 \div 5 \times 5) \div (3 \div 3 of 6) of (4 \times 4 \div (4 \times 4)+ \dfrac{4}{4} \times 4))$

$\Rightarrow (5 + 3 \div 5 \times 5) \div (3 \div 3 of 6) of (4 \times \dfrac{4}{16} + 4)$

$\Rightarrow (5 + 3 \div 5 \times 5) \div (3 \div 3 of 6) of (5)$

$\Rightarrow (5 + 3 \div 5 \times 5) \div (3\div 18) of (5)$

$\Rightarrow (5 + 3 \div 5 \times 5) \div (\dfrac{1}{6}) of (5)$

$\Rightarrow \dfrac{(5 + 3 \div 5 \times 5)}{(\dfrac{5}{6})}$

$\Rightarrow \dfrac{(5 + \dfrac{3}{5} \times 5)}{ (\dfrac{5}{6})}$

$\Rightarrow \dfrac{(5 + 3)\times 6}{5}$

$\Rightarrow \dfrac{48}{ 5}$

$\Rightarrow 9\dfrac{3}{5}$

Question 15: The value of $2\frac{7}{8} \div \left(3\frac{5}{6} \div \frac{2}{7}  of  2\frac{1}{3}\right) \times \left[\left(2\frac{6}{7}  of  4\frac{1}{5} \div \frac{2}{3}\right) \times \frac{5}{9}\right]$ is:

a) $\frac{1}{4}$

b) 4

c) $\frac{1}{23}$

d) 5

15) Answer (D)

Solution:

Given that,

$2\frac{7}{8} \div \left(3\frac{5}{6} \div \frac{2}{7} of 2\frac{1}{3}\right) \times \left[\left(2\frac{6}{7} of 4\frac{1}{5} \div \frac{2}{3}\right) \times \frac{5}{9}\right]$

$\Rightarrow 2\frac{7}{8} \div \left(3\frac{5}{6} \div \frac{2}{7} of 2\frac{1}{3}\right) \times \left[\left(\frac{20}{7} of \frac{21}{5} \div \frac{2}{3}\right) \times \frac{5}{9}\right]$

$\Rightarrow 2\frac{7}{8} \div \left(3\frac{5}{6} \div \frac{2}{7} of 2\frac{1}{3}\right) \times \left[\left(\frac{20\times 21}{7\times 5} \div \frac{2}{3}\right) \times \frac{5}{9}\right]$

$\Rightarrow 2\frac{7}{8} \div \left(3\frac{5}{6} \div \frac{2}{7} of 2\frac{1}{3}\right) \times \left[\left(\frac{20\times 21\times 3}{7\times 5\times 2}\right) \times \frac{5}{9}\right]$

$\Rightarrow 2\frac{7}{8} \div \left(3\frac{5}{6} \div \frac{2}{7} of 2\frac{1}{3}\right) \times 10$

$\Rightarrow 2\frac{7}{8} \div \left(\frac{23}{6} \div \frac{2}{7} of \frac{7}{3}\right) \times 10$

$\Rightarrow 2\frac{7}{8} \div \left(\frac{23}{6} \div \frac{2}{3}\right) \times 10$

$\Rightarrow \frac{23}{8} \div \left(\frac{23\times 3}{6\times 2}\right) \times 10$

$\Rightarrow \frac{23}{8} \div \left(\frac{23}{4}\right) \times 10$

$\Rightarrow \frac{23\times 4}{8\times 23} \times 10$

$\Rightarrow 5$

Question 16: The value of $5 \div 5  of  5 \times 2 + 2 \div 2  of  2 \times 5 – (5 – 2) \div 6 \times 2$ is:

a) $\frac{9}{5}$

b) $\frac{23}{2}$

c) 19

d) $\frac{19}{10}$

16) Answer (D)

Solution:

$5 \div 5 of 5 \times 2 + 2 \div 2 of 2 \times 5 – (5 – 2) \div 6 \times 2$

$\Rightarrow 5 \div 25 \times 2 + 2 \div 4 \times 5 – (5 – 2) \div 6 \times 2$

$\Rightarrow 5 \div 25 \times 2 + 2 \div 4 \times 5 – (3) \div 6 \times 2$

$\Rightarrow 5 \div 25 \times 2 +\dfrac{2}{4} \times 5 – (3) \div 6 \times 2$

$\Rightarrow \dfrac{5}{25} \times 2 +\dfrac{2}{4} \times 5 – \dfrac{3}{6} \times 2$

$\Rightarrow \dfrac{1}{5} \times 2 +\dfrac{1}{2} \times 5 – \dfrac{1}{2} \times 2$

$\Rightarrow \dfrac{1}{5} \times 2 +\dfrac{1}{2} \times 5 – \dfrac{1}{2} \times 2$

$\Rightarrow \dfrac{2}{5} +\dfrac{ 5 }{2} – \dfrac{2}{2}$

$\Rightarrow \dfrac{19}{10}$

Question 17: The value of $2 \times 3 \div 2  of  3 \times 2 \div (4 + 4 \times 4 \div 4  of  4 – 4 \div 4 \times 4)$ is:

a) 8

b) 1

c) 4

d) 2

17) Answer (D)

Solution:

Following BODMAS ,

$2 \times 3 \div 2 of 3 \times 2 \div (4 + 4 \times 4 \div 4 of 4 – 4 \div 4 \times 4) $

= $2 \times 3 \div 6 \times 2 \div (4 + 4 \times 4 \div 16 – 4 \div 4 \times 4) $

= $2 \times  \frac{1}{2} \times 2 \div (4 + 4 \times \frac{1}{4} – 4 \div 4 \times 4) $

= $2 \div (4 + 1 – 1 \times 4) $

= $2 \div (5 – 4) $

=2

So , the answer would be option d) 2.

Question 18: The value of $15.2 + 5.8 \div 2.9 \times 2 – 3.5 of 2 \div 0.5$ is equal to:

a) 4.8

b) 3.2

c) 5.2

d) 5.4

18) Answer (C)

Solution:

$15.2 + 5.8 \div 2.9 \times 2 – 3.5 of 2 \div 0.5$

= $15.2 + 2  \times 2 – 14$

= $15.2 + 4 – 14$

=5.2

So , the answer would be option c)5.2

Question 19: The value of $62 – 5$ of $(18 – 14) + 5 \times 7$ is equal to:

a) 18

b) 28

c) 15

d) 77

19) Answer (D)

Solution:

Expression : $62 – 5$ of $(18 – 14) + 5 \times 7$

= $62-(5\times4)+(5\times7)$

= $62-20+35=77$

=> Ans – (D)

Question 20: The value of $(6 \times 15) \div (2 \times 3) –  2^2 + 3$ is:

a) 12

b) 14

c) 4

d) 6

20) Answer (B)

Solution:

Expression : $(6 \times 15) \div (2 \times 3) –  2^2 + 3$

= $\frac{90}{6}-4+3$

= $15-1=14$

=> Ans – (B)

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