Decimals Questions for Railway Group-D PDF
Download Top-15 RRB Group-D Decimal Questions PDF. RRB GROUP-D Decimal questions based on asked questions in previous exam papers very important for the Railway Group-D exam.
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Question 1: The value of (0.7 × 0.7 × 0.7 – 0.6 × 0.6 × 0.6) ÷ (0.7 × 0.7 + 0.6 × 0.6 + 0.7 × 0.6) is:
a) 0.1
b) 1
c) 1.3
d) 1.1
Instructions
Question 2: What is the value of 2047.235+231+21.2323 ?
a) 2299.4673
b) 2199.4673
c) 2219.4673
d) 2200.4673
Question 3: If $(\frac{1}{5})^{3y}=0.008$, then the a value of $(0.25)^{y}$ is:
a) 0.25
b) 6.25
c) 2.5
d) 53
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Question 4: What is x, if $\ x^{3}=1.5^{3} – 0.9^{3}-2.43$
a) -0.5
b) 0.6
c) -0.7
d) -1.6
Instructions
Question 5: What is the value of 254.58+732.254+631.279 ?
a) 1418.113
b) 1718.113
c) 1518.113
d) 1618.113
Question 6: What is the value of 996.45+658.214+72.326 ?
a) 1719.99
b) 1726.99
c) 1626.99
d) 1826.99
RRB Group D previous year papers
Question 7: What is the value of 66.66% of 80% of 900 is
a) 400
b) 480
c) 450
d) 720
Question 8: The value of the following expression.
$(47 \times 588) \div (28 \times 120) = ?$
a) 6.284
b) 8.285
c) 7.625
d) 8.225
Question 9: Simplify : $5 – 6.5 \div 13 + 2.3 \times 0.8 + 0.4$
a) 6.83
b) 6.74
c) 7.38
d) 5.38
RRB Group-D Important Questions (download PDF)
Instructions
What approximate value will come in place of the question mark (?) in the following questions.? (You are not required to calculate the exact value).
Question 10: $16.02^{2} + 144 + 23.96 + ? = 783.867$
a) 316
b) 262
c) 258
d) 360
e) 344
Instructions
What will come in place of the question mark (?) in the given questions ?
Question 11: $(\frac{42.25}{0.5}+\frac{16.2}{0.2})\times?=1986$
a) 16
b) 8
c) 12
d) 10
e) 14
Question 12: $\frac{1.12 \times 2.24 – 0.78 \times 1.56}{4 \times (1.12 – 0.78)} = ?$
a) 1.75
b) 1.5
c) 0.115
d) 0.85
e) 0.95
Instructions
What will come in place of the question mark (?) in the following questions ?
Question 13: 7831.63 + 2593.59 + 194.67 – 5452.21 = ?
a) 5167.86
b) 5267.68
c) 5167.68
d) 5267.86
e) None of these
Question 14: $27.27$% of $2112$ $+$ $55.55$% of $3276$ $-$ $23.33$% of $1950$ $=$ ?
a) 1857
b) 1946
c) 1971
d) 1941
e) None of these
Instructions
Find the approximate value of ?
Question 15: $(15.012)^3 – (55.072)^2+ 88$% of $999.989 = ?$
a) 1140
b) 1230
c) 1170
d) 1260
e) 1200
General Science Notes for RRB Exams (PDF)
Answers & Solutions:
1) Answer (A)
Expression : (0.7 × 0.7 × 0.7 – 0.6 × 0.6 × 0.6) ÷ (0.7 × 0.7 + 0.6 × 0.6 + 0.7 × 0.6)
Substituting 0.7 with a and 0.6 with b, we get :
= $\frac{a^3 – b^3}{a^2 + b^2 + ab}$
Using, $(x^3 – y^3) = (x – y) (x^2 + y^2 + xy)$
= $\frac{(a – b) (a^2 + b^2 + ab)}{(a^2 + b^2 + ab)}$
= $a – b = 0.7 – 0.6$
= $0.1$
2) Answer (A)
By adding we get 2278.235+21.2323=2299.4673
Hence, option A is the correct answer.
3) Answer (A)
Given : $(\frac{1}{5})^{3y}=0.008$
=> $(\frac{1}{5})^{3y}=\frac{8}{1000}=\frac{1}{125}$
=> $(\frac{1}{5})^{3y}=(\frac{1}{5})^3$
=> $3y=3$
=> $y=\frac{3}{3}=1$
$\therefore$ $(0.25)^{y}=(0.25)^1=0.25$
=> Ans – (A)
4) Answer (B)
Given : $\ x^{3}=1.5^{3} – 0.9^{3}-2.43$
= $(1.5)^3-(0.9)^3-3\times1.5\times0.9\times0.6$
= $(1.5)^3-(0.9)^3-3(1.5)(0.9)(1.5-0.9)$
= $(1.5-0.9)^3=(0.6)^3$
=> $x=0.6$
=> Ans – (B)
5) Answer (D)
By adding we get 986.834+631.279=1618.113
Hence, option D is the correct answer.
6) Answer (B)
By adding we get 1654.664+72.326=1726.99
Hence, option B is the correct answer.
7) Answer (B)
By simplification we get
=(2/3)*(8/10)*900
=480
Hence, option B is the correct answer.
8) Answer (D)
By Applying BODMAS we have 47*21/120
=47*7/40
=329/40
=8.225
9) Answer (B)
=$5 – 6.5 \div 13 + 2.3 \times 0.8 + 0.4$
=$5 – 0.5 + 2.3 \times 0.8 + 0.4$
=$5 – 0.5 + 1.84 + 0.4$
=$4.5 1.84 + 0.4$
=6.74
10) Answer (D)
Expression : $16.02^{2} + 144 + 23.96 + ? = 783.867$
=> $16^2 + 144 + 24 + x = 784$
=> $256 + 168 + x = 784$
=> $x = 784 – 424 = 360$
11) Answer (C)
Expression : $(\frac{42.25}{0.5}+\frac{16.2}{0.2})\times?=1986$
=> $(84.5 + 81) \times x = 1986$
=> $165.5 \times x = 1986$
=> $x = \frac{1986}{165.5} = 12$
12) Answer (E)
Expression : $\frac{1.12 \times 2.24 – 0.78 \times 1.56}{4 \times (1.12 – 0.78)} = ?$
= $\frac{2 (1.12)^2 – 2 (0.78)^2}{4 (1.12 – 0.78)}$
= $\frac{2 (1.12 – 0.78) (1.12 + 0.78)}{4 (1.12 – 0.78)}$ [Using, $a^2 – b^2 = (a – b) (a + b)$]
= $\frac{1.12 + 0.78}{2}$
= $\frac{1.9}{2} = 0.95$
13) Answer (C)
Expression : 7831.63 + 2593.59 + 194.67 – 5452.21 = ?
= 10619.89 – 5452.21
= 5167.68
14) Answer (D)
$\large\frac{1}{11}$ $=$ $9.09$%
$\Rightarrow$ $\large\frac{3}{11}$ $=$ $27.27$%
$27.27$% of $2112$ $=$ $\large\frac{3}{11}$ $\times2112$ $=$ $576$
$\large\frac{1}{9}$ $=$ $11.11$%
$\large\frac{5}{9}$ $=$ $55.55$%
$55.55$% of $3276$ $=$ $\large\frac{5}{9}$ $\times3276$ $=$ $1820$
$\large\frac{1}{30}$ $=$ $3.33$%
$\large\frac{7}{30}$ $=$ $23.33$%
$23.33$% of $1950$ = $\large\frac{7}{30}$ $\times1950$ $=$ $455$
$27.27$% of $2112$ $+$ $55.55$% of $3276$ $-$ $23.33$% of $1950$ $=$ $576$ $+$ $1820$ $-$ $455$ $=$ $1941$
15) Answer (B)
The given expression is:-
$(15.012)^3 – (55.072)^2 + 88$% of $999.989 = ?$
It can be approximated as:-
$(15)^3 – (55)^2 + 88$% of $1000 = ?$
$3375 – 3025 + 880 = ?$
Thus, ? =1230.
Hence, option B is the correct answer.
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