Maths Quiz Questions for RRB Group-D PDF
Download Top-15 RRB Group-D Maths Quiz Questions PDF. RRB GROUP-D Maths questions based on asked questions in previous exam papers very important for the Railway Group-D exam.
Download Maths Quiz Questions for RRB Group-D PDF
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Question 1: Find the value of $tan 56^\circ – tan 11^\circ – tan 56^\circ.tan 11^\circ$
a) 0
b) 1
c) -1
d) 2
Question 2: If sec θ + tan θ = -2, then find sec θ – tan θ = ?
a) -1/2
b) 2
c) 1/2
d) -2
Question 3: Find the value of $1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{3}}}}$.
a) $\dfrac{18}{11}$
b) $\dfrac{16}{11}$
c) $\dfrac{27}{11}$
d) $\dfrac{12}{11}$
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Question 4: If $x+\frac{1}{x} = 2$ then find $x^8+\frac{1}{x^8}$?
a) 8
b) 4
c) 2
d) 16
Question 5: A bag contains Re.1 and Rs.2 coins in the ratio 5 : 8 amounting to Rs.420. Find the number of Re.2 coins.
a) 150
b) 160
c) 120
d) 100
Question 6: The value of $122 + 345 – 3 \div 1116 \times 372 $ is :
a) 466
b) 469
c) 446
d) 460
Question 7: Find the total number of factors of $18^{25} \times 24^{32}$
a) 9360
b) 9600
c) 10126
d) 8480
RRB Group D previous year papers
Question 8: The reciprocal of sum of reciprocals of $\dfrac{7}{11}$ and $\dfrac{5}{8}$ is
a) $\dfrac{111}{35}$
b) $\dfrac{35}{111}$
c) $\dfrac{27}{101}$
d) $\dfrac{35}{121}$
Question 9: Srujay made an error by multiplying a number with 5/6 inplace of 6/5. What is the percentage error ?
a) 31.55%
b) 30.55%
c) 32.55%
d) 33.33%
Question 10: Find the total surface area of cuboid 40mmx30mmx15mm ?
a) 4000 sq.mm
b) 4500 sq.mm
c) 5000 sq.mm
d) 5500 sq.mm
Question 11: The length and the breadth of the rectangle are 14m and 16m. If the cost of fencing is Rs 6 per m, then find the cost of fencing the rectangle.
a) Rs300
b) Rs180
c) Rs360
d) Rs240
RRB Group-D Important Questions (download PDF)
Question 12: In a rhombus, the length of the two diagonals are 8cm and 10cm respectively. Find the area of the rhombus?
a) 30
b) 20
c) 40
d) 50
Question 13: If 1 apple, 2 bananas and 7 oranges cost Rs 34, then find the cost of 4 apples 8 bananas and 28 oranges?
a) 102
b) 154
c) 136
d) 148
Question 14: If a = log 2 and b = log 3 and c = log 5, then find the value of log l800 in terms of a, b and c.
a) 2a + 2b + 3c
b) a + 2b + 3c
c) 3a + 2b + c
d) 3a + 2b + 2c
Question 15: If three coins are tossed, what is the probability that atleast 2 heads occur?
a) 0.5
b) 0.25
c) 0.2
d) 0.1
General Science Notes for RRB Exams (PDF)
Answers & Solutions:
1) Answer (B)
$56-11 = 45$
Applying ‘tan’ on both sides
$tan (56-11)^\circ = tan 45^\circ$ → (1)
We know that $tan (A-B)^\circ = \dfrac{tan A^\circ – tan B^\circ}{1+tan A^\circ.tan B^\circ}$
Then, (1) becomes
$\dfrac{tan 56^\circ – tan 11^\circ}{1+tan 56^\circ.tan 11^\circ} = 1$ (Since, $tan 45^\circ = 1$)
$tan 56^\circ – tan 11^\circ = 1+tan 56^\circ.tan 11^\circ$
⇒ $tan 56^\circ – tan 11^\circ – tan 56^\circ.tan 11^\circ = 1$
2) Answer (A)
$sec^2 θ – tan^2 θ = 1$
$(sec θ – tan θ)(sec θ + tan θ) = 1$
$(sec θ – tan θ) (-2) = 1$
$(sec θ – tan θ) = -1/2$
So the answer is option A.
3) Answer (A)
$1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{3}}}}$
$= 1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{\dfrac{4}{3}}}}$
$= 1+\dfrac{1}{1+\dfrac{1}{1+\frac{3}{4}}}$
$= 1+\dfrac{1}{1+\dfrac{1}{\dfrac{7}{4}}}$
$= 1+\dfrac{1}{1+\dfrac{4}{7}}$
$= 1+\dfrac{1}{\dfrac{11}{7}}$
$= 1+\dfrac{7}{11} = \dfrac{18}{11}$
4) Answer (C)
$x+\frac{1}{x} = 2$
Squaring on both sides
$(x+\frac{1}{x})^2 = 4$
$x^2+\frac{1}{x^2}+2 = 4$
$x^2+\frac{1}{x^2} = 2$
Squaring on both sides
$(x^2+\frac{1}{x^2})^2 = 4$
$x^4+\frac{1}{x^4}+2 = 4$
$x^4+\frac{1}{x^4} = 2$
Squaring on both sides
$(x^4+\frac{1}{x^4})^2 = 4$
$x^8+\frac{1}{x^8}+2 = 4$
$x^8+\frac{1}{x^8} = 2$
So the answer is option C.
5) Answer (B)
Let the number of Re.1 coins be 5x.
Total value of Re.1 coins = Rs.5x*1 = Rs.5x
Let the number of Rs.2 coins be 8x.
Total value of Rs.2 coins = Rs.8x*2 = Rs.16x
Total money in the bag = Rs.5x+Rs.16x = Rs.21x
Given, Rs.21x = Rs.420
⇒ x = 20
Hence, The number of Rs.2 coins = 8x = 8*20 = 160
6) Answer (A)
Using BODMAS rule we can write the equation as,
$467 – \frac{1}{372} \times 372 $
$467 – 1 = 466$
Hence, option A is the correct answer.
7) Answer (C)
$18^{25} \times 24^{32} = 2^{25} \times 3^{50} \times 2^{96} \times 3^{32}$
This equals $2^{121} \times 3^{82}$
Hence, the number of factors equals $122 \times 83 = 10126$
8) Answer (B)
Reciprocals of given numbers are $\dfrac{11}{7}$ and $\dfrac{8}{5}$
$\dfrac{11}{7}+\dfrac{8}{5} = \dfrac{55+56}{35} = \dfrac{111}{35}$
Required number = Reciprocal of $\dfrac{111}{35} = \dfrac{35}{111}$
9) Answer (B)
Let us assume the number to be 30 as it is lcm of 5 and 6.
Original value will be (6/5)*30=36
Wrong value obtained is (5/6)*30=25
Percentage error =((36-25)/36)*100
=30.55%
10) Answer (B)
Total surface area of cuboid = 2(lb+bh+hl) = 2[(40)(30)+(30)(15)+(15)(40)] = 2[1200+450+600] = 2[2250] = 4500 sq.mm
So the answer is option B.
11) Answer (C)
Perimeter of rectangle = 2(14+16)m = 60m
Total cost of fencing = Rs (6*60)= Rs 360
12) Answer (C)
Area of the rhombus = 1/2 *8 *10 = 40
13) Answer (C)
Let the cost of 1 apple, 1 banana and 1 orange be x, y, z respectively.
x + 2y + 7z = 34 —(a)
We have to infer the value of 4x + 8y + 28z.
Multiplying equation (a) by 4,
4x+8y+28z = 34*4 = 136
Hence, the required cost is 136.
14) Answer (D)
1800 = $2^3 * 3^2 * 5^2$
=> log 1800 = log ($2^3 * 3^2 * 5^2$)
=> log 1800 = log $2^3$ + log $3^2$ + log $5^2$
=> log 1800 = 3 log 2 + 2 log 3 + 2 log 5
=> log 1800 = 3a + 2b + 2c
15) Answer (A)
The total number of possibilities are TTT, TTH, THT, HTT, THH, HTH, HHT and HHH.
We need to select those cases in which there are at least 2 Hs.
There are 4 such cases out of 8.
=> Probability = 4/8 = 0.5
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