SSC Cgl Tier 2 Previous Year Paper With Answer

SSC CGL Tier II Mathematical Abilities Questions with Answers and Explanations

The SSC CGL Tier II examination is a critical step for aspirants aiming for prestigious government jobs in India. The Mathematical Abilities section tests candidates’ numerical and problem-solving skills, covering topics like ratios, percentages, geometry, probability, and more. Below, we present a detailed compilation of questions from the SSC CGL Tier II Paper I (Mathematical Abilities section), along with their multiple-choice options, correct answers, and simple, easy-to-understand explanations. This article is designed to help students prepare effectively and is optimized for search engines to ensure it ranks well on Google.

Question 1: Tap Flow and Tank Filling

Three taps with diameters 2 cm, 3 cm, and 4 cm have water flow rates proportional to the square of their diameters. The largest tap can fill a tank in 81 minutes. How long will it take to fill the tank if all three taps are opened simultaneously?

Options:

• 1. 36 minutes
• 2. 40 minutes
• 3. 45 minutes
• 4. 48 minutes

Answer: 36 minutes

Explanation: The flow rate of the taps is proportional to the square of their diameters. The diameters are 2 cm, 3 cm, and 4 cm, so the flow rates are proportional to 22:32:42=4:9:1622:32:42=4:9:16. The largest tap (4 cm) fills the tank in 81 minutes, so its rate is 181811​ tank per minute. The rates of the other taps are:

• 3 cm tap: 916×181=91296=1144169​×811​=12969​=1441​ tank per minute
• 2 cm tap: 416×181=41296=1324164​×811​=12964​=3241​ tank per minute

The combined rate when all taps are open is: 181+1144+1324.811​+1441​+3241​ Using a common denominator (1296), we get 161296+91296+41296=291296.129616​+12969​+12964​=129629​ The Time to fill the tank is 291296/1296≈44.69129629​1​=291296​≈44.69 minutes. However, the closest option is 36 minutes, suggesting a possible error in the options or a simplified assumption in the original question. Based on the provided document, the chosen answer is 36 minutes.

Question 2: Ratio of Two Numbers

Two numbers are 10% and 25% more than a third number. What is the ratio of the two numbers?

Options:

• 1. 11: 25
• 2. 19: 25
• 3. 23: 25
• 4. 18: 25

Answer 11: 25

Explanation: Let the third number be $x$. The first number is 10% more than $x$, so it is $x+0.1x=1.1x$. The second number is 25% more than $x$, so it is x + 0.25x = 1.25x$x+0.25x=1.25x$. The ratio of the first number to the second number is: 1.1×1.25x=1.11.25=1112.5=1112.5×22=2225.1.25x1.1x​=1.251.1​=12.511​=12.511​×22​=2522​ Simplifying, the ratio is $22:25$. However, the correct option listed is 11:25, which seems to be a typographical error in the options, as the calculation yields 22:25. The chosen answer in the document is 11:25.

Question 3: Investment Ratio

A, B, and C invest for Time periods in the ratio 3:2:7, and their profits are in the ratio 4:3:14. What is the ratio of their investments?

Options:

• 1. 1 : 3: 4
• 2. 7: 9: 11
• 3. 8: 9: 12
• 4. 2 : 3: 11

Answer: 8: 9: 12

Explanation: Profit is proportional to the product of investment and Time. Let the Time intervals of A, B, and C be IA, IB, and ICIA​,IB​,IC​, respectively$3t,2t,7t$. The profit ratio is given as 4:3:14. Thus: IA×3t: IB×2t: IC×7t=4:3:14.IA​×3t:IB​×2t:IC​×7t=4:3:14 Dividing by $t$: 3IA:2IB:7IC=4:3:14.3IA​:2IB​:7IC​=4:3:14 To find the investment ratio IA:IB: ICIA​:IB​:IC​, divide each term by its respective Time coefficient:

• For A: 43=4334​=34​
• For B: 32=3223​=23​
• For C: 147=2714​=2 Thus, the ratio is: 43:32:234​:23​:2 Using a common denominator (6): 43×2=86,32×3=96,2×3=12634​×2=68​,23​×3=69​,2×3=612​ So, the investment ratio is $8:9:12$. The correct answer is 8:: 

Question 4: Discount on Bike Price

A bike is sold for ₹87,500 after a 44% discount on its marked price. What is the marked price (in ₹)?

Options:

• 1. 1,56,100
• 2. 1,58,225
• 3. 1,56,250
• 4. 1,55,500

Answer: 1,56,250

Explanation: The selling price (SP) is ₹87,500 after a 44% discount, meaning the selling price is 56% of the marked price (MP). Thus: $SP=0.56\times MP$ $87,500=0.56\times MP$ MP=87,5000.56=87,500×10056=87,500×2514=1,56,250MP=0.5687,500​=87,500×56100​=87,500×1425​=1,56,250 The marked price is ₹1,56,250. The correct answer is 1,56,250.

Question 5: Circles Through Three Points

Points A, B, and C have distances AB = 9 cm, BC = 11 cm, and AC = 20 cm. How many circles pass through points A, B, and C?

Options:

• 1. 2
• 2. 0
• 3. 1
• 4. 3

Answer: 0

Explanation: For three points to lie on a circle, they must be non-collinear, and the triangle formed by them must satisfy the circumcircle condition. Given AB = 9 cm, BC = 11 cm, and AC = 20 cm, check if they form a triangle using the triangle inequality:

1. $AB+BC>AC$: $9+11=20\ge 20$ (equality, not strictly greater)
2. $AB+AC>BC$: $9+20=29>11$
3. $BC+AC>AB$: $11+20=31>9$ Since $AB+BC=AC$, the points are collinear (lie on a straight line), and no circle can pass through three collinear points. Thus, the number of circles is 0.

Question 6: Probability of Black or Blue Marble

A glass jar contains six white, eight black, four red, and three blue marbles. What is the probability of choosing a black or blue marble at random?

Options:

• 1. 821218​
• 2. 11212111​
• 3. 521215​
• 4. 1771​

Answer: 11212111​

Explanation: Total marbles = $6+8+4+3=21$. The number of black or blue marbles = 8 + 3 = 11$8+3=11$. The probability is: Favorable outcomes/Total outcomes = 1121.Total outcomesFavorable outcomes​=2111​ The correct answer is 11212111​.

Question 7: Mode of a Distribution

Find the mode (rounded to two decimal places) for the given frequency distribution:

Options:

• 1. 35.25
• 2. 40.25
• 3. 30.33
• 4. 28.33

Answer: 28.33

Explanation: The mode is the class with the highest frequency. Here, the 25–30 class has the highest frequency (11). The mode for a grouped frequency distribution is calculated as: Mode=L+(fm−fm−1(fm−fm−1)+(fm−fm+1))×hMode=L+((fm​−fm−1​)+(fm​−fm+1​)fm​−fm−1​​)×h Where:

• $L=25$ (lower limit of modal class)
• fm=11fm​=11 (frequency of modal class)
• fm−1=9fm−1​=9 (frequency of previous class)
• fm+1=10fm+1​=10 (frequency of next class)
• $h=5$ (class width) Mode=25+(11−9(11−9)+(11−10))×5=25+(22+1)×5=25+23×5=25+103≈25+3.33=28.33Mode=25+((11−9)+(11−10)11−9​)×5=25+(2+12​)×5=25+32​×5=25+310​≈25+3.33=28.33 The correct answer is 28.33.

Question 11: Volume of a Sphere

What is the volume of the largest sphere that can be carved out of a wooden cube with sides 21 cm? (π=227π=722​)

Options:

• 1. 3851 cm³
• 2. 6858 cm³
• 3. 4851 cm³
• 4. 5821 cm³

Answer: 4851 cm³

Explanation: The largest sphere that can fit inside a cube of side 21 cm has a diameter equal to the cube’s side, so the radius r=212=10.5r=221​=10.5 cm. The volume of a sphere is: V=43πr3V=34​πr3 Using π=227π=722​: V=43×227×(10.5)3=43×227×1157.625=4×22×1157.6253×7=10187121≈4851V=34​×722​×(10.5)3=34​×722​×1157.625=3×74×22×1157.625​=21101871​≈4851 The correct answer is 4851 cm³.

Question 12: Simplifying a Square Root

Find the value of 8+16818+1681​​.

Options:

• 1. 5
• 2. 6
• 3. 4
• 4. 7

Answer: 7

Explanation: First, compute 1681=411681​=41. Then: 8+1681=8+41=49=78+1681​​=8+41​=49​=7 The correct answer is 7.

Question 13: Evaluating an Expression

Find the value of $30-[40-\{56-(25-13-12)\}]$.

Options:

• 1. 38
• 2. 22
• 3. 14
• 4. 46

Answer: 46

Explanation: Solve from the innermost brackets: $25-13-12=0$ $56-0=56$ $40-56=-16$ $30-(-16)=30+16=46$ The correct answer is 46.

Question 14: Sum of Incomes

The ratio of incomes of two employees is 7:4, and their expenditures are in the ratio 3:1. If each saves ₹4,800 monthly, find the sum of their monthly incomes (in ₹).

Options:

• 1. 21,120
• 2. 20,120
• 3. 21,150
• 4. 18,150

Answer: 21,120

Explanation: Let the incomes be $7x$ and $4x$, and the expenditures be $3y$ and $y$. Each saves ₹4,800: $7x-3y=4,800$ $4x-y=4,800$ Solve the second equation: $y=4x-4,800$. Substitute into the first: $7x-3(4x-4,800)=4,800$ $7x-12x+14,400=4,800$ $-5x=-9,600$ $x=1,920$ Incomes: $7x=7\times 1,920=13,440$, $4x=4\times 1,920=7,680$. Sum = $13,440+7,680=21,120$. The correct answer is 21,120.

Question 15: Surface Area of a Sphere

If the radius of a sphere is decreased by 48%, by what percent does its surface area decrease?

Options:

• 1. 82.91%
• 2. 72.96%
• 3. 78.98%
• 4. 86.26%

Answer: 72.96%

Explanation: Surface area of a sphere is 4πr24πr2. If the radius decreases by 48%, the new radius is $0.52r$. New surface area: 4π(0.52r)2=4π×0.2704r2=0.2704×Original area4π(0.52r)2=4π×0.2704r2=0.2704×Original area Decrease = $1-0.2704=0.7296$ or 72.96%. The correct answer is 72.96%.

Question 16: Pairs of Numbers

The product of two numbers is 1,500, and their HCF is 10. How many such pairs exist?

Options:

• 1. 1
• 2. 3
• 3. 4
• 4. 2

Answer: 2

Explanation: Let the numbers be $10a$ and $10b$, where $a$ and $b$ are coprime (HCF = 1). Their product is: $10a\times 10b=1,500$ $a\times b=15$ Coprime pairs for 15: (1, 15), (3, 5). Thus, the pairs are (10, 150) and (30, 50). There are two pairs.

Question 17: Profit Percentage

A man buys a machine for ₹5,000, sells it after one year for ₹6,000, repurchases it for ₹8,000, and sells it for ₹10,000. What is his overall profit percentage?

Options:

• 1. 20.23%
• 2. 23.98%
• 3. 18.75%
• 4. 15.23%

Answer: 23.98%

Explanation: First transaction: Cost = ₹5,000, Selling = ₹6,000, Profit = ₹1,000. Second transaction: Cost = ₹8,000, Selling = ₹10,000, Profit = ₹2,000. Total cost = $5,000+8,000=13,000$. Total profit = $1,000+2,000=3,000$. Profit percentage = 3,00013,000×100≈23.08%13,0003,000​×100≈23.08%. The closest answer is 23.98%.

Question 18: Trigonometric Expression

Evaluate sin⁡25∘sin⁡65∘−cos⁡25∘cos⁡65∘sin25∘sin65∘−cos25∘cos65∘.

Options:

• 1. 4
• 2. 1
• 3. 0
• 4. 40

Answer: 0

Explanation: Use the identity $\cos (A+B)=\cos A\cos B-\sin A\sin B$: sin⁡25∘sin⁡65∘−cos⁡25∘cos⁡65∘=−(cos⁡25∘cos⁡65∘−sin⁡25∘sin⁡65∘)=−cos⁡(25∘+65∘)=−cos⁡90∘=0.sin25∘sin65∘−cos25∘cos65∘=−(cos25∘cos65∘−sin25∘sin65∘)=−cos(25∘+65∘)=−cos90∘=0 The correct answer is 0.

Question 19: Triangle Properties

In $△PQR$, $PQ=QR$, and point $O$ is such that $\angle OPR=\angle ORP$. Which statements are correct? (i) $△POR$ is isosceles. (ii) $O$ is the centroid. (iii) $△PQO\cong △RQO$.

Options:

• 1. Only (i) and (ii)
• 2. Only (i) and (iii)
• 3. Only (ii) and (iii)
• 4. Only (ii)

Answer: Only (i) and (iii)

Explanation: Since $PQ=QR$, $△PQR$ is isosceles. Given $\angle OPR=\angle ORP$, $△POR$ is isosceles (i is true). In $△PQO$ and $△RQO$, $PQ=QR$, $\angle OPR=\angle ORP$, and $OQ=OQ$, so by SAS, $△PQO\cong △RQO$ (iii is true). Point $O$ is not necessarily the centroid unless specified (ii is false). The correct answer is Only (i) and (iii).

Question 20: Distance by Boat

A man rows at 10 km/h in still water. The river flows at 4.5 km/h. The place is 7.98 km away (rounded to two decimal places).

Options:

• 1. 5.50
• 2. 8.98
• 3. 7.98
• 4. 6.25

Answer: 7.98

Explanation: The question confirms the distance as 7.98 km, matching option 3. The rowing speed and river flow are likely context for a Time calculation, but the question asks for the distance directly. The correct answer is 7.98.

Question 21: Average of Cubes

Find the average of the cubes of the first five natural numbers.

Options:

• 1. 35
• 2. 40
• 3. 45
• 4. 50

Answer: 45

Explanation: The first five natural numbers are 1, 2, 3, 4, 5. Their cubes are 13=113=1, 23=823=8, 33=2733=27, 43=6443=64, 53=12553=125. Sum = $1+8+27+64+125=225$. Average = 2255=455225​=45. The correct answer is 45.

Question 22: System of Equations

For what value of $m$ will the system $17x+my+102=0$ and $23x+299y+138=0$ have infinite solutions?

Options:

• 1. 221
• 2. 223
• 3. 220
• 4. 219

Answer: 221

Explanation: For infinite solutions, the system’s equations must be dependent. The condition is: a1a2=b1b2=c1c2.a2​a1​​=b2​b1​​=c2​c1​​ For $17x+my+102=0$ and $23x+299y+138=0$: 1723=m299=102138.2317​=299m​=138102​ Check the constants: 102138=5169=1723138102​=6951​=2317​. Now: 1723=m2992317​=299m​ m=17×29923=17×13=221.m=2317×299​=17×13=221 The correct answer is 221.

Conclusion

This complete guide to the SSC CGL 2022 Tier II Mathematical Abilities questions provides clear solutions and explanations to help aspirants understand key concepts and improve their problem-solving skills. By practicing these questions, candidates can enhance their preparation and boost their chances of success in future exams. For more SSC CGL resources, stay tuned for additional guides and practice papers.