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# 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.

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) $\dfrac{12}{11}$

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

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

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

$56-11 = 45$
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, $tan 45^\circ = 1$)

$tan 56^\circ – tan 11^\circ = 1+tan 56^\circ.tan 11^\circ$
⇒ $tan 56^\circ – tan 11^\circ – tan 56^\circ.tan 11^\circ = 1$

$sec^2 θ – tan^2 θ = 1$

$(sec θ – tan θ)(sec θ + tan θ) = 1$

$(sec θ – tan θ) (-2) = 1$

$(sec θ – tan θ) = -1/2$

So the answer is option A.

$1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{3}}}}$

$= 1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{\dfrac{4}{3}}}}$

$= 1+\dfrac{1}{1+\dfrac{1}{1+\frac{3}{4}}}$

$= 1+\dfrac{1}{1+\dfrac{1}{\dfrac{7}{4}}}$

$= 1+\dfrac{1}{1+\dfrac{4}{7}}$

$= 1+\dfrac{1}{\dfrac{11}{7}}$

$= 1+\dfrac{7}{11} = \dfrac{18}{11}$

$x+\frac{1}{x} = 2$

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.

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

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.

$18^{25} \times 24^{32} = 2^{25} \times 3^{50} \times 2^{96} \times 3^{32}$
This equals $2^{121} \times 3^{82}$
Hence, the number of factors equals $122 \times 83 = 10126$

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 $\dfrac{111}{35} = \dfrac{35}{111}$

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%

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.

Perimeter of rectangle = 2(14+16)m = 60m
Total cost of fencing = Rs (6*60)= Rs 360

Area of the rhombus = 1/2 *8 *10 = 40

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.

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