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SSC CHSL Tier 1 2025 13 Nov Shift 3rd

Verified by Ajay Kumar

25 questions · Choose an option, then open the answer and solution when you are ready.

Question 51

A bicycle is sold for ₹1800 at a loss of 10%. At what price should it be sold to gain 15%?

₹2000
₹2100
₹2250
₹2300

Answer: D

Solution

Cost Price = \(\frac{1800\times100}{90}\) = ₹2000. Required Selling Price = \(\frac{2000\times115}{100}\) = ₹2300. Therefore, the correct answer is ₹2300.

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Question 52

A, B, and C started a business. A contributed Rs. 12,000, B contributed Rs. 15,000, and C contributed Rs. 18,000. After 6 months, B withdrew Rs. 5,000, while C added Rs. 10,000. If the total profit at the end of the year is Rs. 54,000, how much did C receive?

₹26,147
₹24,901
₹14,000
₹17,890

Answer: A

Solution

Profit is divided according to capital × time. A = 12000 × 12 = 144000. B = 15000 × 6 + 10000 × 6 = 150000. C = 18000 × 6 + 28000 × 6 = 276000. C's share = \(54000\times\frac{276}{570}\) = approximately ₹26,147.37.

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Question 53

Simplify the expression: \(\sin x\cot x+\cos x\tan x\).

-2
\(\sin x\cos x\)
\(2\cos x\)
\(\sin x+\cos x\)

Answer: D

Solution

Using \(\cot x=\frac{\cos x}{\sin x}\) and \(\tan x=\frac{\sin x}{\cos x}\), we get \(\sin x\cot x+\cos x\tan x=\cos x+\sin x=\sin x+\cos x\).

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Question 54

If the ratio of chairs to tables is 3:2 and there are 30 chairs, how many tables are there?

Answer: A

Solution

3 parts = 30 chairs. One part = 30 ÷ 3 = 10. Number of tables = 2 × 10 = 20.

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Question 55

The compound interest on ₹8,000 for 1 year 6 months at 20% p.a., compounded half-yearly, is:

₹2,880
₹2,400
₹2,648
₹3,000

Answer: C

Solution

Half-yearly rate = 20 ÷ 2 = 10%. Number of periods = 3. Amount = \(8000(1+\frac{10}{100})^3\) = ₹10,648. Compound Interest = ₹10,648 − ₹8,000 = ₹2,648.

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Question 56

A man deposits ₹1,50,000 at a CI rate of 10% p.a. After the first year's interest is credited, he withdraws 50% of the interest earned. What is his account balance at the end of the second year?

₹1,65,000
₹1,72,000
₹1,73,250
₹1,75,000

Answer: C

Solution

First-year interest = \(150000\times\frac{10}{100}\) = ₹15,000. Withdrawal = ₹7,500. Remaining balance = ₹1,65,000 − ₹7,500 = ₹1,57,500. Second-year interest = \(157500\times\frac{10}{100}\) = ₹15,750. Final balance = ₹1,57,500 + ₹15,750 = ₹1,73,250.

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Question 57

The ratio of two numbers is 3:4 and their LCM is 120. What is the sum of the two numbers?

Answer: A

Solution

Let the numbers be 3x and 4x. Their LCM = 12x = 120, so x = 10. Sum = 3x + 4x = 7 × 10 = 70.

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Question 58

The average age of 20 students is 18 years. A group of 5 new students joins, increasing the average to 20 years. Find the average age of the 5 new students.

Answer: C

Solution

Original total age = 20 × 18 = 360 years. New total age = 25 × 20 = 500 years. Total age of the 5 new students = 500 − 360 = 140 years. Average = 140 ÷ 5 = 28 years.

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Question 59

Which of the following is the least?

18% of 450
20% of 380
25% of 300
30% of 240

Answer: D

Solution

A = 18% of 450 = 81. B = 20% of 380 = 76. C = 25% of 300 = 75. D = 30% of 240 = 72. The least value is 72.

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Question 60

Ravi goes to school at a speed of 5 km/h and returns home at 3 km/h. If the total time taken is 2 hours, what is the distance between his home and school?

2.75 km
3.25 km
3.75 km
4.25 km

Answer: C

Solution

Let the one-way distance be d km. Then \(\frac{d}{5}+\frac{d}{3}=2\). Therefore, \(\frac{8d}{15}=2\), so d = 30 ÷ 8 = 3.75 km.

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Question 61

A, B, C together finish a job in 8 days. A and B together take 12 days, while B and C together take 16 days. In how many days can C alone finish the job?

18 days
20 days
24 days
30 days

Answer: C

Solution

A + B + C work rate = 1/8 and A + B work rate = 1/12. C's work rate = \(\frac18-\frac1{12}=\frac1{24}\). Therefore, C alone takes 24 days.

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Question 62

In a sale, a shopkeeper offers a 15% discount on clothes if you buy more than two items. Priya buys four shirts for ₹2040. If the shopkeeper still makes a 20% profit, find the profit the shopkeeper would make if he had sold the clothes at full price.

₹600
₹700
₹800
₹900

Answer: B

Solution

Marked Price = \(\frac{2040\times100}{85}\) = ₹2400. Cost Price = \(\frac{2040\times100}{120}\) = ₹1700. Profit at full price = ₹2400 − ₹1700 = ₹700.

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Question 63

At simple interest, a certain amount of money doubles in five years. What is the interest rate?

Answer: A

Solution

When money doubles, Simple Interest = Principal. Using \(R=\frac{SI\times100}{P\times T}\), we get \(R=\frac{100}{5}=20\%\) per annum.

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Question 64

A solid right circular cylinder is melted down and recast into a right circular cone with the same base radius. If the cylinder's height is h, what is the height of the cone?

Answer: A

Solution

Volumes are equal: \(\pi r^2h=\frac13\pi r^2H\). Cancelling the common terms gives h = H/3, so H = 3h.

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Question 65

What is the value of the expression \(\sqrt{(12345)^2+12345+12346}\)?

12345
12346
12347
12344

Answer: B

Solution

Let n = 12345. Then \(\sqrt{n^2+n+(n+1)}=\sqrt{n^2+2n+1}=\sqrt{(n+1)^2}=n+1=12346\).

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Question 66

A container can hold 50 liters of milk. First, 5 liters of milk are removed and the same amount of water is poured in to refill the container. The mixture is then thoroughly stirred. This entire process of removing 5 liters of liquid and replacing it with water is carried out once again. After these two successive operations, how many liters of pure milk remain in the container?

40 liters
45 liters
40.5 liters
38.5 liters

Answer: C

Solution

Milk remaining = \(V(1-\frac{x}{V})^n\). Here V = 50, x = 5 and n = 2. Remaining milk = \(50(1-\frac5{50})^2=50(0.9)^2=40.5\) liters.

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Question 67

The numerical value of the volume of a hemisphere is equal to the numerical value of its total surface area. What is the radius of this hemisphere in units?

8 units
4.5 units
5.5 units
6 units

Answer: B

Solution

Volume = \(\frac23\pi r^3\) and total surface area = \(3\pi r^2\). Equating them gives \(\frac23\pi r^3=3\pi r^2\). Therefore, r = 4.5 units.

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Question 68

A car travels at a constant speed. It covers \(\frac25\) of a journey in 2 hours. How long will it take to cover the remaining journey?

2 hours
5 hours
4 hours
3 hours

Answer: D

Solution

Remaining journey = \(1-\frac25=\frac35\). Time for the remaining journey = \(2\times\frac{3/5}{2/5}=3\) hours.

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Question 69

In \(\triangle ABC\), the coordinates of the vertices are A(1,1), B(5,5), and C(1,9). Find the coordinates of the centroid.

(2.33, 5.00)
(2.00, 5.53)
(3.32, 4.00)
(1.98, 5.23)

Answer: A

Solution

Centroid = \((\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})\) = \((\frac{1+5+1}{3},\frac{1+5+9}{3})\) = approximately (2.33, 5.00).

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Question 70

Two similar triangles have areas in the ratio of 49:81. What is the ratio of their corresponding altitudes?

2401:6561
49:81
7:81
7:9

Answer: D

Solution

For similar triangles, the ratio of corresponding altitudes is the square root of the area ratio. Required ratio = \(\sqrt{49:81}=7:9\).

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Question 71

What is the slope of the line that passes through the points (2, 4) and (6, 10)?

2
\(\frac32\)
\(\frac25\)
-5

Answer: B

Solution

Slope = \(\frac{y_2-y_1}{x_2-x_1}=\frac{10-4}{6-2}=\frac64=\frac32\).

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Question 72

Solve: \(\frac{x}{(x+1)}+\frac1x=3\).

\(x=\frac{-5+\sqrt2}{2}\)
\(x=\frac{-1+\sqrt3}{2}\)
\(x=\frac{-3+\sqrt7}{2}\)
\(x=\frac{-9+\sqrt5}{2}\)

Answer: B

Solution

Multiplying both sides by x(x + 1) gives \(x^2+x+1=3x^2+3x\), so \(2x^2+2x-1=0\). Using the quadratic formula, \(x=\frac{-1\pm\sqrt3}{2}\). The matching listed option is \(x=\frac{-1+\sqrt3}{2}\).

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Question 73

Two triangles are similar. The ratio of their corresponding side lengths is 3:5. If the length of an altitude of the smaller triangle is 6 cm, what is the length of the corresponding altitude of the larger triangle?

4 cm
7 cm
10 cm
15 cm

Answer: C

Solution

Corresponding altitudes have the same ratio as corresponding sides. \(\frac35=\frac6h\). Therefore, \(3h=30\), so h = 10 cm.

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Question 74

Solve: \(16^{3/2}+16^{-3/2}=?\)

\(\frac{4097}{64}\)
\(\frac{97}{64}\)
\(\frac{497}{14}\)
\(\frac{237}{64}\)

Answer: A

Solution

\(16^{3/2}=(\sqrt{16})^3=64\). Also, \(16^{-3/2}=\frac1{64}\). Therefore, \(64+\frac1{64}=\frac{4097}{64}\).

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Question 75

Simplify: \(\frac{\cos(x)}{1+\sin(x)}+\frac{\sin(x)}{1+\cos(x)}\).

\(\frac{1}{(1+\sin(x))(1+\cos(x))}\)
1
\(\tan(x)\)
\(\frac{\cos(x)+\sin(x)+1}{(1+\sin(x))(1+\cos(x))}\)

Answer: D

Solution

Taking the LCM gives \(\frac{\cos x(1+\cos x)+\sin x(1+\sin x)}{(1+\sin x)(1+\cos x)}\). Using \(\sin^2x+\cos^2x=1\), the numerator becomes \(\cos x+\sin x+1\). Therefore, the answer is \(\frac{\cos x+\sin x+1}{(1+\sin x)(1+\cos x)}\).

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