# Bodmas questions for RRB Group D Set-2 PDF

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## Bodmas questions for RRB Group D Set-2 PDF

Download Top-15 RRB Group-D Bodmas questions Set-2 PDF. RRB Group – D Bodmas questions based on asked questions in previous exam papers very important for the Railway Group-D exam.

Question 1: $4 + \frac{1}{6} \times$ [ { $– 12 \times ( 24 – 13 – 3)$ } $\div ( 20 – 4)$ ] = ?

a) 4

b) 6

c) 5

d) 3

Question 2: The value of 16 – [ 5 – 2 { 14 of 2 – $( 8 \div 4 \times 2 – 1 + 3$ } ] is :

a) 35

b) 45

c) 55

d) 65

Question 3: The value of $14 \div$ { ( 5 of 2 – 3 ) } $\times 4 ( 7 – 2 )$ is :

a) 44

b) 40

c) $\frac{1}{10}$

d) $\frac{14}{19}$

Question 4: 0.296 + 2.96 + 29.6 + 296 = ?

a) 327.856

b) 328.756

c) 328.856

d) 327.756

Question 5: $45-[38-(80\div4-(8-12\div3)\div4)]=?$

a) 25

b) 27

c) 26

d) 28

Question 6: $(8-(28-53))\div(-4\times5-(-9))=?$

a) 11

b) -3

c) 3

d) -11

Question 7: $\frac{8.73 \times 8.73 \times 8.73 + 4.27 \times 4.27 \times 4.27}{8.73 \times 8.73 -8.73 \times 4.27 + 4.27 \times 4.27}$ is equal to —-

a) 11

b) 13

c) $\frac {11}{7}$

d) None of these

Question 8: $100 \times 10 – 100 + 2000 \div 100 = ?$

a) 29

b) 920

c) 980

d) 1000

Question 9: $(1-\frac{1}{2}) (1-\frac{1}{3}) (1-\frac{1}{4}) …..(1-\frac{1}{40}) = ?$

a) $\frac{1}{40}$

b) $\frac{1}{20}$

c) Countless

d) Zero

Question 10: $5^{5} \div (5^{3} \times 2^{2})$ is

a) 125

b) 31.25

c) 6.25

d) 12.5

Question 11: $2^{4} \times 3^{4} \times 8^{4} \div (2 \times 3 \times 8)^{3}$ is

a) 576

b) 48

c) 96

d) 192

Instructions

What approximate value should come in place of question-mark (?) in the following questions ? (You are expected to calculate the exact value)

Question 12: $(9321+5406+1001)\div(498+929+660)$ = ?

a) 13.5

b) 4.5

c) 16.5

d) 7.5

Question 13: $561204\times58 = ? \times 55555$

a) 606

b) 646

c) 586

d) 716

Question 14: $(7857 + 3596 + 4123)\div96$=?

a) 155.06

b) 162.25

c) 151.83

d) 165.70

Question 15: Evaluate :- $(-216 \times 1728)^{\frac{1}{3}}$

a) -72

b) 27

c) 72

d) -27

$4 + \frac{1}{6} \times$ [ { $– 12 \times ( 24 – 13 – 3)$ } $\div ( 20 – 4)$ ]
=$4 + \frac{1}{6} \times$ [ { $– 12 \times 8$ } $\div 16$ ]
=$4 + \frac{1}{6} \times -6$
=4-1
=3

16 – [ 5 – 2 { 14 of 2 – $( 8 \div 4 \times 2 – 1 + 3$ } ]
=16 – [ 5 – 2 { 28 – $( 2 \times 2 – 1 + 3$ } ]
=16 – [ 5 – 2 { 28 – $( 3 + 3$ } ]
=16-[5-44]
=11+44
=55

$14 \div$ { ( 5 of 2 – 3 ) } $\times 4 ( 7 – 2 )$
=$14 \div$   7  $\times 4 ( 5)$
=2$\times 4 ( 5)$
=2$\times 20$
=40

0.296 + 2.96 + 29.6 + 296 = 328.856

$45-[38-(80\div4-(8-12\div3)\div4)] = 45 – (38 – (20 – \dfrac{8-4}{4}))$
$= 45 – (38 -(20-\dfrac{4}{4})) = 45-(38-(20-1))$
$= 45-(38-19) = 45-19 = 26$

$(8-(28-53)) = 8-28+53 = 33$
$(-4\times5-(-9)) = -20+9 = -11$
Therefore, $(8-(28-53))\div(-4\times5-(-9))=\dfrac{33}{-11} = -3$

($a ^ {3} + b ^ {3}$) / ($a ^ {2} – ab + b ^ {2}$) = a + b

Substituting 8.73 for a and 4.27 for b, we get the answer

By BODMAS, Division and Multiplication will be done before addition and subtraction.

Hence, 100 x 10 – 100 + 2000 / 100 = 1000 – 100 + 20 = 920

1 – 1/2 = 1/2

1 – 1/3 = 2/3

1 – 1/4 = 3/4

The denominator of the first term gets cancelled by the numerator of the second term and so on…

So, the final value = 1/40

$5^{5} \div (5^{3} \times 2^{2})$=$5^{2}/2^{2}$

=25/4=6.25

$2^{4} \times 3^{4} \times 8^{4} \div (2 \times 3 \times 8)^{3}$

=$\frac{2^{16} \times 3^{4}}{2^{12} \times 3^{3}}=2^{4} \times 3=48$

9321+5406+1001= 15728
498+929+660= 2087
15728$\div$2087= 7.536
So option D is the right answer.

$561204\times58 = ? \times 55555$

implies ? = $561204\times58 \div 55555$ = 586

7857 + 3596 + 4123 = 15576

15576 / 96 = 162.25

$(-216 \times 1728)^{\frac{1}{3}}$
=$(-6^{3} \times 12^{3})^{\frac{1}{3}}$