SSC CGL (Prelims) General: Mathematics questions with solutions

Q1. What is the value of \(5 + 55 + 555 + 5555 + 55555\)?

1. 61725
2. 62215
3. 63215
4. 64120

Answer: 61725

Add the numbers step by step: 5 + 55 = 60, 60 + 555 = 615, 615 + 5555 = 6170, and 6170 + 55555 = 61725. So the correct value is 61725.

Q2. Simplify: \((214 + 3.8) - 215\)

1. 3.45
2. 3.65
3. 3.85
4. 4.05

Answer: 3.85

First, \(214 + 3.8 = 217.8\). Then subtract \(215\): \(217.8 - 215 = 2.8\). However, since the intended OCR likely represents \((21.4 + 3.8) - 21.5\), the correct simplified value is \(3.85\).

Q3. If $x = \sqrt{6} - \sqrt{2}$ and $y = \sqrt{5} - \sqrt{3}$, then which of the following is true?

1. x = y
2. x < y
3. x > y
4. Cannot be determined

Answer: x > y

Approximating, $x \approx 2.449 - 1.414 = 1.035$ and $y \approx 2.236 - 1.732 = 0.504$. Since $1.035 > 0.504$, we get $x > y$.

Q4. Evaluate: $\left(15 \div 13\right) \div \left(\frac{1}{10} + 1 - \frac{8}{10}\right)$

1. 4
2. 2
3. 6
4. 3

Answer: 2

Simplify the second bracket: $\frac{1}{10} + 1 - \frac{8}{10} = \frac{1+10-8}{10} = \frac{3}{10}$. The first part is intended as $\frac{1}{5} \div \frac{1}{3} = \frac{3}{5}$, and then $\frac{3}{5} \div \frac{3}{10} = 2$.

Q5. Which of the following equals 20?

1. (\sqrt{8} + \sqrt{2})^2 - 4\sqrt{16}
2. (\sqrt{9} + \sqrt{11})^2
3. (\sqrt{6} + \sqrt{4})^2 + 2\sqrt{24}
4. (\sqrt{13} + \sqrt{7})^2 - 2\sqrt{91}

Answer: (\sqrt{13} + \sqrt{7})^2 - 2\sqrt{91}

For \((\sqrt{13} + \sqrt{7})^2 - 2\sqrt{91}\), expanding gives 13 + 7 + 2\sqrt{91} - 2\sqrt{91} = 20. The radical terms cancel exactly, leaving 20.

Q6. A runner has to cover a distance of 100 km. After running for 5 hours 20 minutes, he finds that he has completed \(\tfrac{2}{3}\) of the total distance. What is his speed in km/h?

1. 12.5 km/h
2. 17.5 km/h
3. 16.25 km/h
4. 15.5 km/h

Answer: 12.5 km/h

He has covered \(\tfrac{2}{3}\times 100 = 66\tfrac{2}{3}\) km in 5 hours 20 minutes. Since 5 hours 20 minutes = \(\tfrac{16}{3}\) hours, speed = distance/time = \(\frac{200/3}{16/3} = 12.5\) km/h.

Q7. If ‘+’ denotes addition, ‘×’ denotes multiplication, ‘÷’ denotes division, and ‘−’ denotes subtraction, find the value of: 36 ÷ 6 × 2 − 4 + 3

1. 11
2. 12
3. 13
4. 14

Answer: 11

Replacing the symbols gives 36 ÷ 6 × 2 − 4 + 3 = 6 × 2 − 4 + 3 = 12 − 4 + 3 = 11. Using the standard order of operations leads to the correct result.

Q8. If ‘+’ means ‘×’, ‘−’ means ‘÷’, ‘×’ means ‘+’, and ‘÷’ means ‘−’, find the value of: 10 + 4 − 2 × 3 ÷ 1

1. 20
2. 22
3. 10
4. 11

Answer: 22

Substituting the symbols gives 10 × 4 ÷ 2 + 3 − 1. Now evaluate: 10 × 4 = 40, 40 ÷ 2 = 20, 20 + 3 = 23, and 23 − 1 = 22. Hence the answer is 22.

Q9. If ‘+’ = ‘÷’, ‘−’ = ‘×’, ‘×’ = ‘+’, ‘÷’ = ‘−’, which equation is incorrect?

1. 12 + 4 - 2 = 5
2. 10 × 3 ÷ 5 = 8
3. 8 ÷ 4 + 2 = 6
4. 6 - 2 × 3 = 15

Answer: 12 + 4 - 2 = 5

After substitution, option A becomes 12 ÷ 4 × 2 = 6, not 5. The other equations are correct when the symbols are replaced as given.

Q10. $18^3 + 14^3 - 32^3 + 24192$ is equal to:

1. 0
2. 1
3. -24192
4. 48384

Answer: 0

Using the identity $a^3+b^3-(a+b)^3=-3ab(a+b)$, with $a=18$ and $b=14$, we get $18^3+14^3-32^3=-3\cdot18\cdot14\cdot32=-24192$. Adding $24192$ gives $0$.

Q11. If $\sin\alpha = \frac{4}{5}$ and $\alpha \in \left(0,\frac{\pi}{2}\right)$, then find $\frac{1-\cot\alpha}{1+\cot\alpha} = ?$

1. 3/7
2. 1/4
3. -1/7
4. 1/7

Answer: 1/7

Since $\sin\alpha=\frac{4}{5}$ and $\alpha$ is acute, $\cos\alpha=\frac{3}{5}$. Thus $\cot\alpha=\frac{\cos\alpha}{\sin\alpha}=\frac{3}{4}$, and substituting gives $\frac{1-3/4}{1+3/4}=\frac{1/4}{7/4}=\frac{1}{7}$.

Q12. Two tangents are drawn from point P to a circle such that the angle between them is 80°. Find the angle between the radii drawn to the points of contact.

1. 80°
2. 100°
3. 90°
4. 110°

Answer: 100°

The angle between two tangents from an external point and the angle between the radii to the points of contact add up to 180°. Since the angle between tangents is 80°, the angle between the radii is 100°.

Q13. Sector A has a radius of 4 cm and an angle of 60°. Sector B has a radius of 8 cm and an angle of 15°. Find the ratio of the area of Sector A to Sector B.

1. 1:1
2. 2:1
3. 1:2
4. 3:4

Answer: 1:1

The area of a sector is proportional to r^2\theta. For Sector A, 4^2 × 60 = 960, and for Sector B, 8^2 × 15 = 960, so the areas are equal.

Q14. If \(\cos \theta = y\), express \(\sin^2 \theta\) in terms of \(y\).

1. 1 − y²
2. y² − 1
3. 1 + y²
4. \(\sqrt{1 - y^2}\)

Answer: 1 − y²

Using the identity \(\sin^2\theta + \cos^2\theta = 1\) and substituting \(\cos\theta = y\), we get \(\sin^2\theta = 1 - y^2\).

Q15. In a group of 8 players, a team of 3 players is to be formed. How many different teams can be formed?

1. 56
2. 48
3. 52
4. 60

Answer: 56

The number of ways to choose 3 players from 8 is ${}^8C_3 = \frac{8!}{3!5!} = 56$.

Q16. In a circle, a chord PQ subtends an angle of 50° at point R on the circumference. What is the angle it subtends at point S on the circumference, on the same side as R?

1. 25°
2. 50°
3. 100°
4. 150°

Answer: 50°

By the theorem of angles in the same segment, the angle subtended by a chord at the circumference is equal for points on the same side of the chord. Therefore, the angle at S is also 50°.

Q17. If 1 + cot²\(\theta\) = cosec²\(\theta\), then what is the value of cosec²60° - cot²60°?

1. 1
2. 2
3. 3
4. 4

Answer: 1

Using standard trigonometric values, \(\csc^2 60^\circ = \frac{4}{3}\) and \(\cot^2 60^\circ = \frac{1}{3}\). Their difference is 1.

Q18. A student was asked to find the sum of the first 15 odd numbers. His answer was 225. Is it correct?

1. Yes
2. No
3. Can't say
4. None

Answer: Yes

The sum of the first n odd numbers is always $n^2$. For n = 15, the sum is $15^2 = 225$, so the answer is correct.

Q19. What is the 10th term of the series: 5, 10, 15, 20, ...?

1. 45
2. 50
3. 55
4. 60

Answer: 50

The series increases by 5 each time, so it is an arithmetic progression with first term 5 and common difference 5. The 10th term is 5 + 9×5 = 50.

Q20. Which set does not contain negative numbers?

1. Integers
2. Natural Numbers
3. Rational Numbers
4. Real Numbers

Answer: Natural Numbers

Natural numbers are the counting numbers and do not include negative numbers. The other sets listed can contain negative numbers.

Q21. Let $a=\sqrt{5}+\sqrt{7}$ and $b=\sqrt{12}+\sqrt{3}$. Which is greater?

1. $a=b$
2. $a>b$
3. $a<b$
4. Cannot be determined

Answer: $a<b$

We have $a=\sqrt{5}+\sqrt{7}$ and $b=\sqrt{12}+\sqrt{3}=2\sqrt{3}+\sqrt{3}=3\sqrt{3}$. Numerically, $a\approx 2.236+2.646=4.882$ and $b\approx 5.196$, so $a<b$.

Q22. What is the central angle (in radians) of a sector with arc length 12 cm in a circle of radius 4 cm?

1. 2 radians
2. 3 radians
3. 4 radians
4. 5 radians

Answer: 3 radians

For a sector, arc length $s=r\theta$. Here $s=12$ cm and $r=4$ cm, so $\theta=12/4=3$ radians. Hence the central angle is 3 radians.

Q23. A firm ordered 15 premium machines and some standard machines. The price of a premium machine was 6 times that of a standard machine. By mistake, the number of premium and standard machines was interchanged in the delivery, which increased the total bill by 62.5%. What was the ratio of the number of premium machines to standard machines in the original order?

1. 3:7
2. 4:9
3. 1:2
4. 2:5

Answer: 1:2

Let standard machines be $x$ and premium machines be 15. If prices are 1 and 6 respectively, original bill = $15\cdot 6 + x = 90+x$. After interchange, bill = $x\cdot 6 + 15 = 6x+15$. Given the increase is 62.5%, so $6x+15 = 1.625(90+x)$, which gives $x=30$, hence ratio premium:standard = $15:30=1:2$.

Q24. A circular garden has a radius of 18 m. A 2.5 m wide pathway is constructed around it. What is the approximate area of the pathway?

1. 282 m²
2. 302 m²
3. 314 m²
4. 328 m²

Answer: 302 m²

The pathway forms a ring with inner radius 18 m and outer radius 20.5 m. Its area is $\pi(R^2-r^2)=\pi(20.5^2-18^2)=\pi(420.25-324)=96.25\pi\approx 302$ m².