{"id":42986,"date":"2020-10-01T18:41:59","date_gmt":"2020-10-01T13:11:59","guid":{"rendered":"https:\/\/cracku.in\/blog\/?p=42986"},"modified":"2020-10-01T18:41:59","modified_gmt":"2020-10-01T13:11:59","slug":"maths-questions-for-ssc-cgl-tier-2-pdf","status":"publish","type":"post","link":"https:\/\/cracku.in\/blog\/maths-questions-for-ssc-cgl-tier-2-pdf\/","title":{"rendered":"Maths Questions for SSC CGL Tier – 2 PDF"},"content":{"rendered":"

Maths Questions for SSC CGL Tier – 2 PDF<\/strong><\/span><\/h1>\n

Download SSC CGL Tier 2 Maths Questions PDF. Top 20 Maths questions based on previous exam papers very important for the SSC exam.<\/p>\n

Download Maths Questions for SSC CGL Tier – 2 PDF<\/a><\/p>\n

Get 10 SSC CGL Tier-2 Mocks for Rs. 149<\/a><\/p>\n

Take SSC CGL Tier-2 Mock Tests<\/a><\/p>\n

Download SSC CGL Tier-2 Previous Papers PDF<\/a><\/p>\n

Question 1:\u00a0<\/b>Three persons A, B and C have different amounts of rupees with them. If A takes \u20b96 from C, A will have equal amount as B has. A and B together have total \u20b974. How many rupees does B have ?<\/p>\n

a)\u00a044<\/p>\n

b)\u00a030<\/p>\n

c)\u00a040<\/p>\n

d)\u00a034<\/p>\n

Question 2:\u00a0<\/b>The difference between the Mode and the Median of a data is 25. What is the difference between the Median and the Mean?<\/p>\n

a)\u00a010.5<\/p>\n

b)\u00a016<\/p>\n

c)\u00a012.5<\/p>\n

d)\u00a014<\/p>\n

Question 3:\u00a0<\/b>A dice is rolled two times. Find the probability of getting a composite number on first roll and a prime number on second roll?<\/p>\n

a)\u00a0$\\frac{1}{2}$<\/p>\n

b)\u00a0$\\frac{1}{6}$<\/p>\n

c)\u00a0$\\frac{1}{9}$<\/p>\n

d)\u00a0$\\frac{1}{4}$<\/p>\n

Question 4:\u00a0<\/b>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 ?<\/p>\n

a)\u00a025<\/p>\n

b)\u00a026<\/p>\n

c)\u00a028<\/p>\n

d)\u00a0<\/del>31<\/p>\n

Question 5:\u00a0<\/b>What Is the compound interest (in Rs.) on a principal sum of Rs. 2800 for 2 years at the rate of 12% per annum?<\/p>\n

a)\u00a0687.18<\/p>\n

b)\u00a0634.46<\/p>\n

c)\u00a0712.32<\/p>\n

d)\u00a0568.68<\/p>\n

Take SSC CGL Tier-2 Mock Tests<\/a><\/p>\n

Question 6:\u00a0<\/b>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 ?<\/p>\n

a)\u00a075<\/p>\n

b)\u00a067<\/p>\n

c)\u00a066.66<\/p>\n

d)\u00a076.66<\/p>\n

Question 7:\u00a0<\/b>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 ?<\/p>\n

a)\u00a07420<\/p>\n

b)\u00a06890<\/p>\n

c)\u00a06360<\/p>\n

d)\u00a07380<\/p>\n

Question 8:\u00a0<\/b>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?<\/p>\n

a)\u00a0$14\\frac{1}{3}$<\/p>\n

b)\u00a0$14\\frac{2}{3}$<\/p>\n

c)\u00a0$14\\frac{1}{2}$<\/p>\n

d)\u00a0$14\\frac{2}{5}$<\/p>\n

Question 9:\u00a0<\/b>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 ?<\/p>\n

a)\u00a0$64=(1.01)^{12y}$<\/p>\n

b)\u00a0$\\frac{1}{64}= (1.04)12y$<\/p>\n

c)\u00a0$64=(1.04)^{12y}$<\/p>\n

d)\u00a0$8=(1.01)^{6y}$<\/p>\n

Question 10:\u00a0<\/b>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 ?<\/p>\n

a)\u00a07%<\/p>\n

b)\u00a08%<\/p>\n

c)\u00a09%<\/p>\n

d)\u00a08.5%<\/p>\n

Download SSC CGL Tier-2 Previous Papers PDF<\/a><\/p>\n

Take SSC CGL Tier-2 Mock Tests<\/a><\/p>\n

Question 11:\u00a0<\/b>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 ?<\/p>\n

a)\u00a02<\/p>\n

b)\u00a0$\\sqrt{3}$<\/p>\n

c)\u00a0$2\\sqrt{3}$<\/p>\n

d)\u00a01<\/p>\n

Question 12:\u00a0<\/b>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:<\/p>\n

a)\u00a0$30^{0}$<\/p>\n

b)\u00a0$45^{0}$<\/p>\n

c)\u00a0$60^{0}$<\/p>\n

d)\u00a0None of these<\/p>\n

Question 13:\u00a0<\/b>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\u2019s share in the profit ?<\/p>\n

a)\u00a0Rs. 10,000<\/p>\n

b)\u00a0Rs. 12,500<\/p>\n

c)\u00a0Rs. 10,250<\/p>\n

d)\u00a0Rs. 14,000<\/p>\n

Question 14:\u00a0<\/b>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:<\/p>\n

a)\u00a0$2 – \\sqrt{3}$<\/p>\n

b)\u00a0$3 – \\sqrt{2}$<\/p>\n

c)\u00a0$2 + \\sqrt{3}$<\/p>\n

d)\u00a0$3 + \\sqrt{2}$<\/p>\n

Question 15:\u00a0<\/b>The value of $\\left(\\frac{sinA}{1-cosA} + \\frac{1-cosA}{sinA}\\right) \\div \\left(\\frac{cot^2A}{1+cosecA} + 1\\right)$ is:<\/p>\n

a)\u00a0$\\frac{3}{2}$<\/p>\n

b)\u00a0$\\frac{1}{2}$<\/p>\n

c)\u00a01<\/p>\n

d)\u00a02<\/p>\n

Download SSC CGL General Science Notes PDF<\/a><\/p>\n

Question 16:\u00a0<\/b>The greatest number that will divide 146, 248 and 611 leaving remainders 2,8 and 11 respectively is:<\/p>\n

a)\u00a047<\/p>\n

b)\u00a0144<\/p>\n

c)\u00a024<\/p>\n

d)\u00a0612<\/p>\n

Question 17:\u00a0<\/b>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?<\/p>\n

a)\u00a07:8<\/p>\n

b)\u00a04:5<\/p>\n

c)\u00a011:12<\/p>\n

d)\u00a06:7<\/p>\n

Question 18:\u00a0<\/b>The 3rd and 6th term of an arithmetic progression are 13 and -5 respectively. What is the 11th term?<\/p>\n

a)\u00a0-29<\/p>\n

b)\u00a0-41<\/p>\n

c)\u00a0-47<\/p>\n

d)\u00a0-35<\/p>\n

Question 19:\u00a0<\/b>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:<\/p>\n

a)\u00a0$\\frac{33 + 10\\sqrt{3}}{6}$<\/p>\n

b)\u00a0$\\frac{19 + 10\\sqrt{3}}{6}$<\/p>\n

c)\u00a0$\\frac{33 + 10\\sqrt{3}}{3}$<\/p>\n

d)\u00a0$\\frac{19 + 10\\sqrt{3}}{3}$<\/p>\n

Question 20:\u00a0<\/b>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.<\/p>\n

a)\u00a0Rs. 771<\/p>\n

b)\u00a0Rs. 789<\/p>\n

c)\u00a0Rs. 747<\/p>\n

d)\u00a0Rs. 713<\/p>\n

18000+ Questions – Free SSC Study Material<\/a><\/p>\n

Free SSC Preparation (Videos Youtube)<\/a><\/p>\n

Answers & Solutions:<\/strong><\/span><\/p>\n

1)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

Let initially,\u00a0 A has $\u20b9 x$, B has $\u20b9 y$ and C has $\u20b9 z$<\/p>\n

According to question,<\/p>\n

$x+6 = y$\u00a0 and $x+y = 74$,<\/p>\n

So by solving equation, We get<\/p>\n

$\\therefore y=40$<\/p>\n

2)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

We now that 3 median -2 mean=mode
\nGiven mode-median=25
\nmode=median+25
\ntherefore 3 median-2 mean=median+25
\n2 median-2 mean=25
\nmedian-mean=12.5<\/p>\n

3)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Number of composite numbers in a dice = 2 (4 and 6)
\nHence, Probability of getting a composite number = $\\dfrac{2}{6} = \\dfrac{1}{3}$
\nNumber of prime numbers in a dice = 3 (2, 3 and 5)
\nHence, Probability of getting a prime number = $\\dfrac{3}{6} = \\dfrac{1}{2}$
\nTherefore,\u00a0probability of getting a composite number on first roll and a prime number on second roll = $\\dfrac{1}{3}\\times\\dfrac{1}{2} = \\dfrac{1}{6}$<\/p>\n

4)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

<\/figure>\n

Given\u00a0: OB is radius of circle = 7 cm and tangent AB = 24 cm<\/p>\n

To find\u00a0: OA = ?<\/p>\n

Solution\u00a0: The radius of a circle intersects the tangent at right angle, => $\\angle OBA = 90^\\circ$<\/p>\n

Thus in $\\triangle$ OAB,<\/p>\n

=> $(OA)^2=(OB)^2+(AB)^2$<\/p>\n

=> $(OA)^2=(7)^2+(24)^2$<\/p>\n

=> $(OA)^2=49+576=625$<\/p>\n

=> $OA=\\sqrt{625}=25$ cm<\/p>\n

=> Ans – (A)<\/p>\n

<\/figure>\n

5)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

Principal sum = Rs. 2800<\/p>\n

Rate of interest = 12% and time = 2years<\/p>\n

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

=\u00a0$2800[(1+\\frac{12}{100})^2-1]$<\/p>\n

= $2800[(\\frac{28}{25})^2-1]$<\/p>\n

= $2800\\times(\\frac{784-625}{625})$<\/p>\n

= $2800\\times\\frac{159}{625}=Rs.$ $712.32$<\/p>\n

=> Ans – (C)<\/p>\n

6)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

Let the total distance = $12x$ km<\/p>\n

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

=> Time taken = $\\frac{4x}{80}=\\frac{x}{20}$ hours<\/p>\n

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

=> Time taken = $\\frac{3x}{50}$ hours<\/p>\n

Distance covered at 100 km\/hr = $12x-4x-3x=5x$ km<\/p>\n

=> Time taken = $\\frac{5x}{100}=\\frac{x}{20}$ hours<\/p>\n

Thus, total time = $\\frac{x}{20}+\\frac{3x}{50}+\\frac{x}{20}$<\/p>\n

= $\\frac{x}{10}+\\frac{3x}{50}=\\frac{8x}{50}$<\/p>\n

$\\therefore$ Average speed = total distance\/total time<\/p>\n

= $\\frac{12x}{\\frac{8x}{50}}$<\/p>\n

= $12\\times\\frac{50}{8}=75$ km\/hr<\/p>\n

=> Ans – (A)<\/p>\n

7)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

Marked price = Rs. 8480 and discount % = 12.5%<\/p>\n

=> Selling price = $8480-\\frac{12.5}{100}\\times8480$<\/p>\n

= $8480-\\frac{8480}{8}=8480-1060$<\/p>\n

= $Rs.$ $7420$<\/p>\n

=> Ans – (A)<\/p>\n

8)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

Let capacity of tank = L.C.M. (12,18) = 36 litres<\/p>\n

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

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

Now, in 2<\/strong> hours tank filled is (B opened first) = $2+3=5$ litres<\/p>\n

$\\because$ $5\\times7=35$, hence 35 litres of tank is filled in 14 hours.<\/p>\n

Now,\u00a0B is opened and it will fill the remaining 1 litre in $\\frac{1}{2}$ hour.<\/p>\n

$\\therefore$ Total time taken =\u00a0$14\\frac{1}{2}$ hours<\/p>\n

=> Ans – (C)<\/p>\n

9)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

Rate of interest = 12% p.a. = 1% per month<\/p>\n

Time = $12y$ months<\/p>\n

Let principal = Re 1 and thus amount = Rs. 64<\/p>\n

$\\therefore$ $A=P(1+\\frac{R}{100})^T$<\/p>\n

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

=> $64=(1.01)^{12y}$<\/p>\n

=> Ans – (A)<\/p>\n

10)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

Profit % = 35%<\/p>\n

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

=\u00a0$15-7=8\\%$<\/p>\n

=> Ans – (B)<\/p>\n

11)\u00a0Answer\u00a0(B)<\/strong><\/p>\n

.<\/p>\n

PQ is perpendicular bisector of OA. Also, OP = OQ (radii)<\/p>\n

Hence, OPAQ is a rhombus. ————–(i)<\/p>\n

Also, $2\\angle PAQ=$ reflex $\\angle POQ$ \u00a0 \u00a0 [The angle subtended at the centre by an arc is twice to that at the circumference]<\/p>\n

=> $2\\angle\u00a0PAQ=360^\\circ-\\angle POQ$<\/p>\n

=>\u00a0$2\\angle\u00a0PAQ+\\angle POQ=360^\\circ$<\/p>\n

From (i), we have $\\angle\u00a0PAQ=\\angle\u00a0POQ$<\/p>\n

=> $3\\angle\u00a0POQ=360^\\circ$<\/p>\n

=> $\\angle POQ=120^\\circ=\\frac{2\\pi}{3}$<\/p>\n

We know that, $r=\\frac{l}{\\theta}$<\/p>\n

=> $r=\\frac{\\frac{2\\pi}{3}}{\\frac{2\\pi}{3}}=1$ unit<\/p>\n

In $\\triangle$ POB,<\/p>\n

=> $sin(\\angle POB)=\\frac{PB}{OP}$<\/p>\n

=> $sin(60^\\circ)=\\frac{PB}{1}$<\/p>\n

=> $PB=\\frac{\\sqrt3}{2}$<\/p>\n

$\\therefore$ Chord PQ = $2\\times(PB)=2\\times\\frac{\\sqrt3}{2}=\\sqrt3$<\/p>\n

=> Ans – (B)<\/p>\n

12)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

<\/figure>\n

The pole is standing at the centre of the square, => PA = PR<\/p>\n

=> $\\angle$ APB =\u00a0$\\angle$\u00a0ARB = $45^\\circ$<\/p>\n

Let the side of the square = $x$ units<\/p>\n

=> PR (diagonal) = $\\sqrt2x$ units<\/p>\n

Hence, PB = $\\frac{x}{\\sqrt2}$ units<\/p>\n

Now, in\u00a0$\\triangle$\u00a0APB,<\/p>\n

=> $tan(\\angle APB)=\\frac{AB}{PB}$<\/p>\n

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

=> $AB=PB=\\frac{x}{\\sqrt2}$<\/p>\n

Thus, $PA=\\sqrt{(\\frac{x}{\\sqrt2})^2+(\\frac{x}{\\sqrt2})^2}$<\/p>\n

=> $PA=\\sqrt{\\frac{x^2}{2}+\\frac{x^2}{2}}=\\sqrt{x^2}=x$<\/p>\n

Similarly, $QA=x$ units<\/p>\n

Hence, PA = PQ = QA = $x$<\/p>\n

$\\therefore$\u00a0$\\angle$\u00a0PAQ = $60^\\circ$<\/p>\n

=> Ans – (C)<\/p>\n

13)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

Anil invested Rs. 25,000 for 12 months and Vishal invested Rs. 30,000 for 9 months<\/p>\n

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

= $300:270=10:9$<\/p>\n

Total profit = Rs. 19,000<\/p>\n

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

= $10\\times1000=Rs.$ $10,000$<\/p>\n

=> Ans – (A)<\/p>\n

14)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

$\\frac{\\sin \\theta}{1 + \\cos \\theta} + \\frac{1 + \\cos \\theta}{\\sin \\theta} = \\frac{4}{\\sqrt{3}}$
\nPut the $\\theta = 60\\degree$,
\n$\\frac{\\sin 60\\degree}{1 + \\cos 60\\degree} + \\frac{1 + \\cos 60\\degree}{\\sin 60\\degree} = \\frac{4}{\\sqrt{3}}$
\n$\\frac{\\frac{\\sqrt{3}}{2}}{1 +\\frac{1}{2}} + \\frac{1 + \\frac{1}{2}}{\\frac{\\sqrt{3}}{2}} = \\frac{4}{\\sqrt{3}}$
\n$\\frac{\\frac{3}{4} +\u00a0\\frac{9}{4}}{\\frac{3}{2} \\times\u00a0\\frac{3}{4}} =\u00a0\\frac{4}{\\sqrt{3}}$
\n$\\frac{4}{\\sqrt{3}}\u00a0= \\frac{4}{\\sqrt{3}}$
\nNow,
\n$(\\tan \\theta + \\sec \\theta)^{-1}$
\nPut the $\\theta = 60\\degree$,
\n= $(\\tan 60\\degree + \\sec 60\\degree)^{-1}$
\n= $(\\sqrt{3} + 2)^{-1}$
\n= $2 –\u00a0\\sqrt{3}$<\/p>\n

15)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

$\\left(\\frac{sinA}{1-cosA} + \\frac{1-cosA}{sinA}\\right) \\div \\left(\\frac{cot^2A}{1+cosecA} + 1\\right)$
\nLet the value of $\\theta = 45\\degree$,
\n$\\left(\\frac{\\frac{1}{\\sqrt{2}}}{1- \\frac{1}{\\sqrt{2}}} + \\frac{1- \\frac{1}{\\sqrt{2}}}{\\frac{1}{\\sqrt{2}}}\\right) \\div \\left(\\frac{1}{1+\\sqrt{2}} + 1\\right)$<\/p>\n

=$\\left(\\frac{\\frac{1}{2} +\u00a0(1- \\frac{1}{\\sqrt{2}})^2}{(\\frac{1}{\\sqrt{2}})(1- \\frac{1}{\\sqrt{2}})}\\right)\u00a0\\div \\left(\\frac{1 +\u00a01+\\sqrt{2}}{1+\\sqrt{2}}\\right)$<\/p>\n

= $\\left(\\frac{\\frac{1}{2} + 1 +\u00a0\\frac{1}{2} – \\sqrt{2}}{(\\frac{1}{\\sqrt{2}})(1- \\frac{1}{\\sqrt{2}})}\\right) \\div \\left(\\frac{2+\\sqrt{2}}{1+\\sqrt{2}}\\right)$<\/p>\n

= $\\left(\\frac{2 – \\sqrt{2}}{(\\frac{1}{\\sqrt{2}}- \\frac{1}{2})}\\right) \\div \\left(\\frac{2+\\sqrt{2}}{1+\\sqrt{2}}\\right)$<\/p>\n

= $\\left(\\frac{2 – \\sqrt{2}}{\\frac{2 – \\sqrt{2}}{2\\sqrt{2}}}\\right) \\div \\left(\\frac{2+\\sqrt{2}}{1+\\sqrt{2}}\\right)$<\/p>\n

= $2\\sqrt{2} \\times \\frac{1+\\sqrt{2}}{2+\\sqrt{2}}$ = 2<\/p>\n

16)\u00a0Answer\u00a0(C)<\/strong><\/p>\n

The greatest number that will divide 146, 248 and 611 leaving remainders 2,8 and 11 respectively<\/p>\n

= H.C.F. of (146-2), (248-8), (611-11)<\/p>\n

= H.C.F. (144,240,600) =\u00a024<\/strong><\/p>\n

=> Ans – (C)<\/p>\n

17)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

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

The mean proportion between 1.6 and 48.4 is $\\sqrt{1.6\\times48.4} = 8.8$<\/p>\n

Therefore, The required ratio = 7.7:8.8 = 7:8.<\/p>\n

18)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

$T_{3}$ = a + 2d = 13——-(1)<\/p>\n

$T_{6}$ = a + 5d = -5——-(2)<\/p>\n

on solving (1) AND (2)<\/p>\n

d = -6 & a = 25<\/p>\n

$T_{11}$ = a + 10d = 25 + 10(-6) = -35<\/p>\n

so the answer is option D.<\/p>\n

19)\u00a0Answer\u00a0(A)<\/strong><\/p>\n

$\\sin \\theta = \\sqrt{3} \\cos \\theta, 0^\\circ < \\theta < 90^\\circ$
\n$\\Rightarrow\u00a0\u00a0\\frac{\\sin \\theta}{\\cos \\theta} = \\sqrt{3}$
\n$\\Rightarrow \\tan \\theta = \\sqrt{3}$
\n$\\Rightarrow\u00a0\\theta = 60\\degree$
\nNow,
\n$2 \\sin^2 \\theta + \\sec^2 \\theta + \\sin \\theta \\sec \\theta + \\cosec \\theta$
\n=\u00a0$2 \\sin^2 60\\degree + \\sec^2 60\\degree + \\tan\u00a060\\degree + \\cosec60\\degree$
\n= $2 \\times (\\frac{\\sqrt{3}}{2})^2 + 2^2 +\u00a0\\sqrt{3} + \\frac{2}{\\sqrt{3}}$
\n= $\\frac{3}{2} + 4 + \\sqrt{3} + \\frac{2}{\\sqrt{3}}$
\n= $\\frac{3\\sqrt{3} + 8\\sqrt{3} + 6 +\u00a04}{2\\sqrt{3}}$
\n= $\\frac{11\\sqrt{3} + 10}{2\\sqrt{3}}$
\n= $\\frac{33 + 10\\sqrt{3}}{6}$
\n$\\therefore$ The correct answer is option A.<\/p>\n

20)\u00a0Answer\u00a0(D)<\/strong><\/p>\n

Let\u2019s assume the amount obtained by Sona is 25y.
\nThe amount obtained by Mona is 92% of the amount obtained by Sona.
\nThen the amount obtained by Mona = 25y of 92%
\n= $25y \\times \\frac{92}{100}$<\/p>\n

= $y \\times \\frac{92}{4}$<\/p>\n

= 23y
\nThe ratio between the amount obtained by Mona and Raj is 23:39 respectively.
\nSo we can say that the amount obtained by Raj will be 39y.
\nIf the difference between the amount obtained by Sona and Raj is Rs. 434 .
\n39y – 25y = 434
\n14y = 434
\ny = 31<\/p>\n

The amount obtained by Mona = 23y = $23\\times31$
\n= Rs. 713
\nHence, option d is the correct answer.<\/p>\n

Free SSC Online Coaching<\/a><\/p>\n

DOWNLOAD APP FOR SSC FREE MOCKS<\/a><\/p>\n

We hope this Maths Questions pdf for SSC CGL Tier 2 exam will be highly useful for your Preparation.<\/p>\n","protected":false},"excerpt":{"rendered":"

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. Take SSC CGL Tier-2 Mock Tests Download SSC CGL Tier-2 Previous Papers PDF Question 1:\u00a0Three persons A, B and C have different amounts […]<\/p>\n","protected":false},"author":41,"featured_media":42988,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_mi_skip_tracking":false,"footnotes":""},"categories":[169,125,9,504,378,1493,1459,1611,1741,1268],"tags":[3529,462,4217,358,461,2000],"class_list":{"0":"post-42986","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-downloads","8":"category-featured","9":"category-ssc","10":"category-ssc-cgl","11":"category-ssc-chsl","12":"category-ssc-cpo","13":"category-ssc-gd","14":"category-ssc-je","15":"category-ssc-mts","16":"category-ssc-stenographer","17":"tag-maths-questions","18":"tag-ssc-cgl","19":"tag-ssc-chl-tier-2","20":"tag-ssc-chsl","21":"tag-ssc-exams","22":"tag-ssc-mocks"},"better_featured_image":{"id":42988,"alt_text":"Maths Questions for SSC CGL Tier - 2 PDF","caption":"Maths Questions for SSC CGL Tier - 2 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