**Maths Questions for SSC CGL Tier – 2 PDF**

Download SSC CGL Tier 2 Maths Questions PDF. Top 20 Maths questions based on previous exam papers very important for the SSC exam.

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**Question 1: **Three persons A, B and C have different amounts of rupees with them. If A takes ₹6 from C, A will have equal amount as B has. A and B together have total ₹74. How many rupees does B have ?

a) 44

b) 30

c) 40

d) 34

**Question 2: **The difference between the Mode and the Median of a data is 25. What is the difference between the Median and the Mean?

a) 10.5

b) 16

c) 12.5

d) 14

**Question 3: **A dice is rolled two times. Find the probability of getting a composite number on first roll and a prime number on second roll?

a) $\frac{1}{2}$

b) $\frac{1}{6}$

c) $\frac{1}{9}$

d) $\frac{1}{4}$

**Question 4: **AB is a tangent to a circle with centre O. If the radius at the circle is 7 cm and the length of AB is 24 cm, the what is the length (in cm.) of OA ?

a) 25

b) 26

c) 28

d) ~~31~~

**Question 5: **What Is the compound interest (in Rs.) on a principal sum of Rs. 2800 for 2 years at the rate of 12% per annum?

a) 687.18

b) 634.46

c) 712.32

d) 568.68

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**Question 6: **One third of a certain journey is covered at the speed of 80 km/hr, one fourth of the journey at the speed of 50 km/hr And the rest at the speed of 100 km/hr, what will be the average speed (in km/hr) for the whole journey ?

a) 75

b) 67

c) 66.66

d) 76.66

**Question 7: **The marked price of an article is Rs. 8480. If a discount of 12.5% is given, then what will be the selling price (in Rs.) of the article ?

a) 7420

b) 6890

c) 6360

d) 7380

**Question 8: **Pipe A can fill a tank in 12 hours and pipe B can fill the tank in 18 hours. If both the pipes are opened on alternate hours and if pipe B is opened first, then in how much time (in hours) the tank will be full?

a) $14\frac{1}{3}$

b) $14\frac{2}{3}$

c) $14\frac{1}{2}$

d) $14\frac{2}{5}$

**Question 9: **A certain amount grows at an annual interest rate of 12%, compounded monthly. Which of the following equations can be solved to find the number of years, y, that it would take for the investment to increase by a factor of 64 ?

a) $64=(1.01)^{12y}$

b) $\frac{1}{64}= (1.04)12y$

c) $64=(1.04)^{12y}$

d) $8=(1.01)^{6y}$

**Question 10: **A balance of a trader weighs 20% less than it should be. Still the trader mark-up his goods to get the overall profit of 35%. What is mark-up on the cost price ?

a) 7%

b) 8%

c) 9%

d) 8.5%

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**Question 11: **Chord PQ is the perpendicular bisector of radius OA of circle with center O (A is a point on the edge of the circle). If the length of Arc PAQ = $\frac{2\pi}{3}$. What is the length of chord PQ ?

a) 2

b) $\sqrt{3}$

c) $2\sqrt{3}$

d) 1

**Question 12: **A vertical pole AB is standing at the centre B of a square PQRS. If PR subtends an angle of $90^{0}$at the top A of the pole, then the angle subtended by a side of the square at A is:

a) $30^{0}$

b) $45^{0}$

c) $60^{0}$

d) None of these

**Question 13: **Anil started a business with an investment of Rs. 25,000. After 3 months, Vishal joined his business with a capital of Rs. 30,000. At the end of the year, they have made a profit of Rs. 19,000. What will be Anil’s share in the profit ?

a) Rs. 10,000

b) Rs. 12,500

c) Rs. 10,250

d) Rs. 14,000

**Question 14: **If $\frac{\sin \theta}{1 + \cos \theta} + \frac{1 + \cos \theta}{\sin \theta} = \frac{4}{\sqrt{3}}, 0^\circ < \theta < 90^\circ$, then the value of $(\tan \theta + \sec \theta)^{-1}$ is:

a) $2 – \sqrt{3}$

b) $3 – \sqrt{2}$

c) $2 + \sqrt{3}$

d) $3 + \sqrt{2}$

**Question 15: **The value of $\left(\frac{sinA}{1-cosA} + \frac{1-cosA}{sinA}\right) \div \left(\frac{cot^2A}{1+cosecA} + 1\right)$ is:

a) $\frac{3}{2}$

b) $\frac{1}{2}$

c) 1

d) 2

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**Question 16: **The greatest number that will divide 146, 248 and 611 leaving remainders 2,8 and 11 respectively is:

a) 47

b) 144

c) 24

d) 612

**Question 17: **What is the ratio between the fourth proportional to 1.2, 2.8 and 3.3 and the mean proportional between 1.6 and 48.4?

a) 7:8

b) 4:5

c) 11:12

d) 6:7

**Question 18: **The 3rd and 6th term of an arithmetic progression are 13 and -5 respectively. What is the 11th term?

a) -29

b) -41

c) -47

d) -35

**Question 19: **If $\sin \theta = \sqrt{3} \cos \theta, 0^\circ < \theta < 90^\circ$, then the value of $2 \sin^2 \theta + \sec^2 \theta + \sin \theta \sec \theta + cosec \theta$ is:

a) $\frac{33 + 10\sqrt{3}}{6}$

b) $\frac{19 + 10\sqrt{3}}{6}$

c) $\frac{33 + 10\sqrt{3}}{3}$

d) $\frac{19 + 10\sqrt{3}}{3}$

**Question 20: **An amount is distributed among Sona, Mona and Raj such as the amount obtained by Mona is 92% of the amount obtained by Sona. The ratio between the amount obtained by Mona and Raj is 23:39 respectively. If the difference between the amount obtained by Sona and Raj is Rs. 434 . Then find out the amount obtained by Mona.

a) Rs. 771

b) Rs. 789

c) Rs. 747

d) Rs. 713

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**Answers & Solutions:**

**1) Answer (C)**

Let initially, A has $₹ x$, B has $₹ y$ and C has $₹ z$

According to question,

$x+6 = y$ and $x+y = 74$,

So by solving equation, We get

$\therefore y=40$

**2) Answer (C)**

We now that 3 median -2 mean=mode

Given mode-median=25

mode=median+25

therefore 3 median-2 mean=median+25

2 median-2 mean=25

median-mean=12.5

**3) Answer (B)**

Number of composite numbers in a dice = 2 (4 and 6)

Hence, Probability of getting a composite number = $\dfrac{2}{6} = \dfrac{1}{3}$

Number of prime numbers in a dice = 3 (2, 3 and 5)

Hence, Probability of getting a prime number = $\dfrac{3}{6} = \dfrac{1}{2}$

Therefore, probability of getting a composite number on first roll and a prime number on second roll = $\dfrac{1}{3}\times\dfrac{1}{2} = \dfrac{1}{6}$

**4) Answer (A)**

Given : OB is radius of circle = 7 cm and tangent AB = 24 cm

To find : OA = ?

Solution : The radius of a circle intersects the tangent at right angle, => $\angle OBA = 90^\circ$

Thus in $\triangle$ OAB,

=> $(OA)^2=(OB)^2+(AB)^2$

=> $(OA)^2=(7)^2+(24)^2$

=> $(OA)^2=49+576=625$

=> $OA=\sqrt{625}=25$ cm

=> Ans – (A)

**5) Answer (C)**

Principal sum = Rs. 2800

Rate of interest = 12% and time = 2years

Compound interest = $P[(1+\frac{R}{100})^T-1]$

= $2800[(1+\frac{12}{100})^2-1]$

= $2800[(\frac{28}{25})^2-1]$

= $2800\times(\frac{784-625}{625})$

= $2800\times\frac{159}{625}=Rs.$ $712.32$

=> Ans – (C)

**6) Answer (A)**

Let the total distance = $12x$ km

Distance covered at 80 km/hr = $\frac{12x}{3}=4x$ km

=> Time taken = $\frac{4x}{80}=\frac{x}{20}$ hours

Distance covered at 50 km/hr = $\frac{12x}{4}=3x$ km

=> Time taken = $\frac{3x}{50}$ hours

Distance covered at 100 km/hr = $12x-4x-3x=5x$ km

=> Time taken = $\frac{5x}{100}=\frac{x}{20}$ hours

Thus, total time = $\frac{x}{20}+\frac{3x}{50}+\frac{x}{20}$

= $\frac{x}{10}+\frac{3x}{50}=\frac{8x}{50}$

$\therefore$ Average speed = total distance/total time

= $\frac{12x}{\frac{8x}{50}}$

= $12\times\frac{50}{8}=75$ km/hr

=> Ans – (A)

**7) Answer (A)**

Marked price = Rs. 8480 and discount % = 12.5%

=> Selling price = $8480-\frac{12.5}{100}\times8480$

= $8480-\frac{8480}{8}=8480-1060$

= $Rs.$ $7420$

=> Ans – (A)

**8) Answer (C)**

Let capacity of tank = L.C.M. (12,18) = 36 litres

Pipe A can fill a tank in 12 hours, => Pipe A’s efficiency = $\frac{36}{12}=3$ litres/hr

Similarly, pipe B’s efficiency = $\frac{36}{18}=2$ litres/hr

Now, in **2** hours tank filled is (B opened first) = $2+3=5$ litres

$\because$ $5\times7=35$, hence 35 litres of tank is filled in 14 hours.

Now, B is opened and it will fill the remaining 1 litre in $\frac{1}{2}$ hour.

$\therefore$ Total time taken = $14\frac{1}{2}$ hours

=> Ans – (C)

**9) Answer (A)**

Rate of interest = 12% p.a. = 1% per month

Time = $12y$ months

Let principal = Re 1 and thus amount = Rs. 64

$\therefore$ $A=P(1+\frac{R}{100})^T$

=> $64=1(1+\frac{1}{100})^{12y}$

=> $64=(1.01)^{12y}$

=> Ans – (A)

**10) Answer (B)**

Profit % = 35%

Thus, mark up on cost price = $[35-20-(\frac{35\times20}{100})]\%$

= $15-7=8\%$

=> Ans – (B)

**11) Answer (B)**

.

PQ is perpendicular bisector of OA. Also, OP = OQ (radii)

Hence, OPAQ is a rhombus. ————–(i)

Also, $2\angle PAQ=$ reflex $\angle POQ$ [The angle subtended at the centre by an arc is twice to that at the circumference]

=> $2\angle PAQ=360^\circ-\angle POQ$

=> $2\angle PAQ+\angle POQ=360^\circ$

From (i), we have $\angle PAQ=\angle POQ$

=> $3\angle POQ=360^\circ$

=> $\angle POQ=120^\circ=\frac{2\pi}{3}$

We know that, $r=\frac{l}{\theta}$

=> $r=\frac{\frac{2\pi}{3}}{\frac{2\pi}{3}}=1$ unit

In $\triangle$ POB,

=> $sin(\angle POB)=\frac{PB}{OP}$

=> $sin(60^\circ)=\frac{PB}{1}$

=> $PB=\frac{\sqrt3}{2}$

$\therefore$ Chord PQ = $2\times(PB)=2\times\frac{\sqrt3}{2}=\sqrt3$

=> Ans – (B)

**12) Answer (C)**

The pole is standing at the centre of the square, => PA = PR

=> $\angle$ APB = $\angle$ ARB = $45^\circ$

Let the side of the square = $x$ units

=> PR (diagonal) = $\sqrt2x$ units

Hence, PB = $\frac{x}{\sqrt2}$ units

Now, in $\triangle$ APB,

=> $tan(\angle APB)=\frac{AB}{PB}$

=> $tan(45^\circ)=1=\frac{AB}{PB}$

=> $AB=PB=\frac{x}{\sqrt2}$

Thus, $PA=\sqrt{(\frac{x}{\sqrt2})^2+(\frac{x}{\sqrt2})^2}$

=> $PA=\sqrt{\frac{x^2}{2}+\frac{x^2}{2}}=\sqrt{x^2}=x$

Similarly, $QA=x$ units

Hence, PA = PQ = QA = $x$

$\therefore$ $\angle$ PAQ = $60^\circ$

=> Ans – (C)

**13) Answer (A)**

Anil invested Rs. 25,000 for 12 months and Vishal invested Rs. 30,000 for 9 months

=> Ratio of profits = $(25,000\times12):(30,000\times9)$

= $300:270=10:9$

Total profit = Rs. 19,000

$\therefore$ Anil’s share in the profit = $\frac{10}{(10+9)}\times19,000$

= $10\times1000=Rs.$ $10,000$

=> Ans – (A)

**14) Answer (A)**

$\frac{\sin \theta}{1 + \cos \theta} + \frac{1 + \cos \theta}{\sin \theta} = \frac{4}{\sqrt{3}}$

Put the $\theta = 60\degree$,

$\frac{\sin 60\degree}{1 + \cos 60\degree} + \frac{1 + \cos 60\degree}{\sin 60\degree} = \frac{4}{\sqrt{3}}$

$\frac{\frac{\sqrt{3}}{2}}{1 +\frac{1}{2}} + \frac{1 + \frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{4}{\sqrt{3}}$

$\frac{\frac{3}{4} + \frac{9}{4}}{\frac{3}{2} \times \frac{3}{4}} = \frac{4}{\sqrt{3}}$

$\frac{4}{\sqrt{3}} = \frac{4}{\sqrt{3}}$

Now,

$(\tan \theta + \sec \theta)^{-1}$

Put the $\theta = 60\degree$,

= $(\tan 60\degree + \sec 60\degree)^{-1}$

= $(\sqrt{3} + 2)^{-1}$

= $2 – \sqrt{3}$

**15) Answer (D)**

$\left(\frac{sinA}{1-cosA} + \frac{1-cosA}{sinA}\right) \div \left(\frac{cot^2A}{1+cosecA} + 1\right)$

Let the value of $\theta = 45\degree$,

$\left(\frac{\frac{1}{\sqrt{2}}}{1- \frac{1}{\sqrt{2}}} + \frac{1- \frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}}\right) \div \left(\frac{1}{1+\sqrt{2}} + 1\right)$

=$\left(\frac{\frac{1}{2} + (1- \frac{1}{\sqrt{2}})^2}{(\frac{1}{\sqrt{2}})(1- \frac{1}{\sqrt{2}})}\right) \div \left(\frac{1 + 1+\sqrt{2}}{1+\sqrt{2}}\right)$

= $\left(\frac{\frac{1}{2} + 1 + \frac{1}{2} – \sqrt{2}}{(\frac{1}{\sqrt{2}})(1- \frac{1}{\sqrt{2}})}\right) \div \left(\frac{2+\sqrt{2}}{1+\sqrt{2}}\right)$

= $\left(\frac{2 – \sqrt{2}}{(\frac{1}{\sqrt{2}}- \frac{1}{2})}\right) \div \left(\frac{2+\sqrt{2}}{1+\sqrt{2}}\right)$

= $\left(\frac{2 – \sqrt{2}}{\frac{2 – \sqrt{2}}{2\sqrt{2}}}\right) \div \left(\frac{2+\sqrt{2}}{1+\sqrt{2}}\right)$

= $2\sqrt{2} \times \frac{1+\sqrt{2}}{2+\sqrt{2}}$ = 2

**16) Answer (C)**

The greatest number that will divide 146, 248 and 611 leaving remainders 2,8 and 11 respectively

= H.C.F. of (146-2), (248-8), (611-11)

= H.C.F. (144,240,600) = **24**

=> Ans – (C)

**17) Answer (A)**

The fourth proportional to 1.2, 2.8 and 3.3 is $\dfrac{2.8\times3.3}{1.2} = 7.7$

The mean proportion between 1.6 and 48.4 is $\sqrt{1.6\times48.4} = 8.8$

Therefore, The required ratio = 7.7:8.8 = 7:8.

**18) Answer (D)**

$T_{3}$ = a + 2d = 13——-(1)

$T_{6}$ = a + 5d = -5——-(2)

on solving (1) AND (2)

d = -6 & a = 25

$T_{11}$ = a + 10d = 25 + 10(-6) = -35

so the answer is option D.

**19) Answer (A)**

$\sin \theta = \sqrt{3} \cos \theta, 0^\circ < \theta < 90^\circ$

$\Rightarrow \frac{\sin \theta}{\cos \theta} = \sqrt{3}$

$\Rightarrow \tan \theta = \sqrt{3}$

$\Rightarrow \theta = 60\degree$

Now,

$2 \sin^2 \theta + \sec^2 \theta + \sin \theta \sec \theta + \cosec \theta$

= $2 \sin^2 60\degree + \sec^2 60\degree + \tan 60\degree + \cosec60\degree$

= $2 \times (\frac{\sqrt{3}}{2})^2 + 2^2 + \sqrt{3} + \frac{2}{\sqrt{3}}$

= $\frac{3}{2} + 4 + \sqrt{3} + \frac{2}{\sqrt{3}}$

= $\frac{3\sqrt{3} + 8\sqrt{3} + 6 + 4}{2\sqrt{3}}$

= $\frac{11\sqrt{3} + 10}{2\sqrt{3}}$

= $\frac{33 + 10\sqrt{3}}{6}$

$\therefore$ The correct answer is option A.

**20) Answer (D)**

Let’s assume the amount obtained by Sona is 25y.

The amount obtained by Mona is 92% of the amount obtained by Sona.

Then the amount obtained by Mona = 25y of 92%

= $25y \times \frac{92}{100}$

= $y \times \frac{92}{4}$

= 23y

The ratio between the amount obtained by Mona and Raj is 23:39 respectively.

So we can say that the amount obtained by Raj will be 39y.

If the difference between the amount obtained by Sona and Raj is Rs. 434 .

39y – 25y = 434

14y = 434

y = 31

The amount obtained by Mona = 23y = $23\times31$

= Rs. 713

Hence, option d is the correct answer.

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