SSC CHSL2023Quantitative AptitudePage 8 of 16

SSC CHSL Tier 1 2023 Quantitative Aptitude Questions with Solutions

Questions from every shift of the year, grouped by section. Questions 351–400 of 781, each with the correct answer and a step-by-step explanation.

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351Quantitative Aptitude
A blended tea in a barrel contains Assam, Darjeeling and Nilgiri Hills in the ratio 2:3:1. Another barrel contains Darjeeling, Nilgiri Hills and Kangra Valley in the ratio 5:4:3. If equal weights of both blended tea mix together to make a new brand, then the weight of Kangra Valley per kilogram in the new brand is:
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Answer: (D) 125 gm

Explanation

To find the weight of Kangra Valley per kilogram in the new brand, we mix equal parts of both barrels, say 0.5 kg of each to make 1 kg of the new brand. Since Barrel 1 contains no Kangra Valley, and Barrel 2 contains Kangra Valley in the ratio of 3 out of 12 parts (or 1/4th of its weight), the weight of Kangra Valley in 0.5 kg of Barrel 2 is 0.5 kg * (1/4) = 1/8 kg, which equals 125 gm. Thus, 1 kg of the new brand contains 125 gm of Kangra Valley.

352Quantitative Aptitude
The following table shows the heights of six students. Study the table carefully and answer the following question. Name | Height (in cm) Vanaja | 117 Sailaja | 125 Uma | 133 Sudha | 140 Amala | 132 Leela | 121 How many students are there whose height is between that of Vanaja and Uma?
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Answer: (C) Three

Explanation

The height of Vanaja is 117 cm and the height of Uma is 133 cm. To find the number of students whose height is between that of Vanaja and Uma, we identify the students with heights strictly greater than 117 cm and less than 133 cm.

From the given table, the students falling within this range (117 cm < height < 133 cm) are Sailaja with a height of 125 cm, Amala with a height of 132 cm, and Leela with a height of 121 cm. This gives a total of three students.

353Quantitative Aptitude
At what annual rate (rounded off to the nearest integer) of simple interest will a sum of money become five times its initial value in 18 years?
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Answer: (C) 22%

Explanation

Using the simple interest formula: Amount = Principal + Simple Interest

Since the sum of money becomes five times its initial value, the final amount is 5P, where P is the principal. This gives the Simple Interest: SI = 5P - P = 4P

The formula for simple interest is: SI = (P × R × T) / 100

Substituting the values SI = 4P and T = 18 years into the formula: 4P = (P × R × 18) / 100 4 = 18R / 100 R = 400 / 18 R = 22.22%

Rounding 22.22% to the nearest integer yields 22%.

354Quantitative Aptitude
If one diagonal of a rhombus is equal to its side, then the diagonals of the rhombus are in the ratio of:
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Answer: (D) 1 : √3

Explanation

If one diagonal of a rhombus is equal to its side, the rhombus is made of two equilateral triangles joined base-to-base, making the half-diagonals a/2 and (a√3)/2. Thus, the ratio of the diagonals is d1 : d2 = a : a√3, which simplifies to 1 : √3.

355Quantitative Aptitude
If cos θ + sec θ = √3 , then the value of cos²θ + sec³θ is:
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Answer: (A) 0

Explanation

Let x = cos θ. Since sec θ = 1/cos θ, the given relation is: x + 1/x = √3

Using the algebraic identity for the sum of cubes: (x + 1/x)³ = x³ + 1/x³ + 3(x + 1/x)

Substituting the given value of x + 1/x: (√3)³ = x³ + 1/x³ + 3(√3) 3√3 = x³ + 1/x³ + 3√3

Subtracting 3√3 from both sides gives: x³ + 1/x³ = 0

Thus, cos³θ + sec³θ = 0.

356Quantitative Aptitude
The following table gives the percentage of marks obtained by seven students in six different subjects in an examination. The number in the brackets gives the maximum marks in each subject. Students | Maths (150) | Chemistry (130) | Physics (120) | Geography (100) | History (60) | Computer Science (40) --- | --- | --- | --- | --- | --- | --- Ayush | 90 | 50 | 90 | 60 | 70 | 80 Aman | 100 | 80 | 80 | 40 | 80 | 70 Sajal | 90 | 60 | 70 | 70 | 90 | 70 Rohit | 80 | 65 | 80 | 80 | 60 | 60 Muskan | 80 | 65 | 85 | 95 | 50 | 90 Tanvi | 70 | 75 | 65 | 85 | 40 | 60 Tarun | 65 | 35 | 50 | 77 | 80 | 80 If a student secures 80% or more in a subject, the student is considered to have received a distinction in the subject. Who among the students below received a distinction in exactly one of the six subjects?
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Answer: (C) Tanvi

Explanation

Tanvi secured the following percentage of marks in each of the six subjects: Maths: 70% Chemistry: 75% Physics: 65% Geography: 85% History: 40% Computer Science: 60%

Since a student is considered to have received a distinction if they secure 80% or more in a subject, Tanvi achieved a distinction only in Geography, which counts as exactly one subject.

357Quantitative Aptitude
Which of the following is the nearest number to 13051 and is divisible by 9?
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Answer: (C) 13059

Explanation

To find the number nearest to 13051 that is divisible by 9, we apply the divisibility rule for 9, which states that a number is divisible by 9 if the sum of its digits is divisible by 9.

Checking the sum of the digits for 13059: 1 + 3 + 0 + 5 + 9 = 18

Since 18 is divisible by 9, the number 13059 is divisible by 9. It is the only number among the options that satisfies this divisibility condition.

358Quantitative Aptitude
If the height and slant height of a cone are 21 cm and 29 cm, respectively. Find its volume. (Use π = ²²⁄₇ )
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Answer: (A) 8800 cm3

Explanation

The slant height l, height h, and radius r of a cone are related by the formula l² = r² + h².

Substituting the given values: 29² = r² + 21² 841 = r² + 441 r² = 400 r = 20 cm

The volume V of a cone is given by the formula V = (1/3) × π × r² × h.

Substituting the values into the formula: V = (1/3) × (22/7) × 400 × 21 V = 22 × 400 V = 8800 cm³

359Quantitative Aptitude
Ashima borrowed 75,000 from a nationalised bank on 26th March 2022 at the rate of 9% per annum simple interest. If she cleared the account on 7th June of the same year, then what amount did she pay?
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Answer: (B) 76,350

Explanation

To find the total amount paid, we first calculate the number of days for which the interest is charged from 26th March 2022 to 7th June 2022.

Number of days in March = 31 − 26 = 5 days Number of days in April = 30 days Number of days in May = 31 days Number of days in June = 7 days Total duration = 5 + 30 + 31 + 7 = 73 days

Since 2022 is a non-leap year, the time period in years is T = 73/365 = 1/5 year.

Using the simple interest formula: Simple Interest (SI) = (P × R × T) / 100

Substituting the given values: SI = (75000 × 9 × 1/5) / 100 SI = 150 × 9 SI = 1350

The total amount paid to clear the account is: Amount = Principal + Simple Interest Amount = 75000 + 1350 = 76350

360Quantitative Aptitude
If cos (3α) = sin (α - 22°) , where 3α < 90° , then what is the value of α ?
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Answer: (D) 28°

Explanation

Using the trigonometric identity cos(θ) = sin(90° - θ), the equation cos(3α) = sin(α - 22°) can be rewritten as:

sin(90° - 3α) = sin(α - 22°)

Since both angles are in the first quadrant, we can equate the arguments directly:

90° - 3α = α - 22°

Rearranging the terms to solve for α:

90° + 22° = α + 3α

112° = 4α

α = 112° ÷ 4

α = 28°

This value satisfies the condition 3α < 90° since 3 × 28° = 84° < 90°.

361Quantitative Aptitude
Two candidates A and B contested in an election. 90% of the voters cast their votes, out of which 2% of the votes were invalid. A got 9,408 votes which was 60% of the valid votes. Find the total number of votes.
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Answer: (C) 20,000

Explanation

Let the total number of voters be V.

With 80% of the voters casting their votes: Votes cast = 0.80 × V

Since 2% of the cast votes were invalid, the valid votes constitute 98% of the cast votes: Valid votes = 0.98 × 0.80 × V = 0.784 × V

Candidate A received 60% of the valid votes: Votes secured by A = 0.60 × 0.784 × V = 0.4704 × V

Given that A got 9,408 votes: 0.4704 × V = 9,408 V = 9,408 ÷ 0.4704 = 20,000

362Quantitative Aptitude
If (5a + ⁴⁄ₐ - 2) = 13 and a > 0 , what is the value of (25a² + (16)/(a²)) ?
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Answer: (C) 185

Explanation

Using the algebraic identity (x + y)² = x² + y² + 2xy, the given equation can be simplified and squared to find the required value.

First, simplify the given equation by adding 2 to both sides: 5a + 4/a = 15

Next, square both sides of this equation: (5a + 4/a)² = 15²

Expand the left side using the identity (x + y)² = x² + y² + 2xy: 25a² + 16/a² + 2 × 5a × (4/a) = 225

Simplify the middle term by canceling a: 25a² + 16/a² + 40 = 225

Subtract 40 from both sides to find the target value: 25a² + 16/a² = 225 - 40 25a² + 16/a² = 185

363Quantitative Aptitude
What is the value of x in the number 3426x if the number is divisible by 6 but NOT divisible by 5?
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Answer: (C) 6

Explanation

A number is divisible by 6 if it is divisible by both 2 and 3.

For the number 3426x to be divisible by 2, its last digit x must be an even number, meaning x must be 0, 2, 4, 6, or 8.

For the number 3426x to be divisible by 3, the sum of its digits must be divisible by 3. The sum of the digits is 3 + 4 + 2 + 6 + x = 15 + x. Since 15 is already divisible by 3, the value of x must also be a multiple of 3, meaning x must be 0, 3, 6, or 9.

To satisfy both divisibility conditions, x must be an even multiple of 3, which limits the possible values of x to 0 or 6.

Additionally, the number is not divisible by 5, which means its units digit x cannot be 0 or 5. Excluding 0 from the possible values leaves x = 6 as the only valid solution.

364Quantitative Aptitude
Rice worth 504 per kg and 540 per kg are mixed with the third variety in the ratio 1:1:2. If the mixture is worth 612 per kg, find the price (per kg in ½ ) of the third variety.
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Answer: (B) 702

Explanation

The weighted average formula for a mixture of three components is used to find the price of the third variety:

Average Price = (w₁p₁ + w₂p₂ + w₃p₃) / (w₁ + w₂ + w₃)

Here, the ratios of the three varieties are w₁ = 1, w₂ = 1, and w₃ = 2. The prices of the first two varieties are p₁ = 504 and p₂ = 540. The average price of the mixture is 612. Let the price of the third variety be p₃.

Substituting the values into the formula:

612 = (1 × 504 + 1 × 540 + 2 × p₃) / (1 + 1 + 2)

612 = (1044 + 2p₃) / 4

Multiplying both sides by 4:

2448 = 1044 + 2p₃

2p₃ = 2448 - 1044

2p₃ = 1404

p₃ = 702

365Quantitative Aptitude
The average weight of the group of 12 players increases by 1.5kg when a player who weighs 63kg is replaced by a new player. What is the weight of the new player?
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Answer: (B) 81kg

Explanation

The weight of the new player can be calculated directly using the replacement formula:

Weight of new player = Weight of replaced player + (Total number of players × Increase in average weight)

Substituting the given values into the formula:

Weight of new player = 63 + 12 × 1.5 Weight of new player = 63 + 18 Weight of new player = 81 kg

366Quantitative Aptitude
Simplify the given expression. ((x³ - y³)(x + y))/(x² + xy + y²)
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Answer: (B) x² - y²

Explanation

Using the algebraic identity for the difference of cubes: x³ - y³ = (x - y)(x² + xy + y²)

Substitute this identity into the given expression: ((x³ - y³)(x + y)) / (x² + xy + y²) = ((x - y)(x² + xy + y²)(x + y)) / (x² + xy + y²)

Cancel the common term (x² + xy + y²) from the numerator and the denominator: (x - y)(x + y)

Applying the difference of squares formula: (x - y)(x + y) = x² - y²

367Quantitative Aptitude
If two circles of different radii touch externally, then what is the maximum number of common tangents that can be drawn to the two circles?
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Answer: (A) 3

Explanation

When two circles of different radii r₁ and r₂ touch each other externally, the distance d between their centers is exactly equal to the sum of their radii, so d = r₁ + r₂.

For this configuration, we can draw two types of common tangents: Two direct common tangents, which touch both circles on the same side and do not cross the line segment connecting the centers of the two circles. One transverse common tangent, which passes through the point of contact where the two circles touch externally and crosses the line segment connecting the centers.

Adding these together, the maximum number of common tangents that can be drawn to the two externally touching circles is 2 + 1 = 3.

368Quantitative Aptitude
Find the value of the following expression. 2(tan² A cos² A + cos² A)
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Answer: (D) 2

Explanation

Using the basic trigonometric identity, tan A is equal to sin A / cos A. Substituting this into the given expression simplifies the term tan² A cos² A:

tan² A cos² A = (sin² A / cos² A) × cos² A = sin² A

Now, substituting this back into the original expression:

2(tan² A cos² A + cos² A) = 2(sin² A + cos² A)

Since sin² A + cos² A = 1 for any angle A, the expression simplifies directly to:

2(1) = 2

369Quantitative Aptitude
If A, B, C are the acute angles, such that tan(A + B - C) = (1)/(√3), cos(B + C - A) = ½, and sin(C + A - B) = (1)/(√2). The value of A + B - C will be equal to:
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Answer: (D) 135°

Explanation

The question contains a typographical error and intends to ask for the value of A + B + C instead of A + B - C. From the given trigonometric equations, we have A + B - C = 30 degrees, B + C - A = 60 degrees, and C + A - B = 45 degrees. Adding these three equations yields A + B + C = 30 + 60 + 45 = 135 degrees, which corresponds to Option D.

370Quantitative Aptitude
Study the table given below and answer the question. The table shows the number of Athletes who participated in sports event from 4 different countries over the years. Country ⇒ India | England | France | Australia Male Female | Male Female | Male Female | Male Female Years ↓ 2016 460 180 | 740 480 | 480 280 | 320 100 2017 430 330 | 630 420 | 840 120 | 220 80 2018 200 150 | 400 130 | 350 160 | 100 90 2019 360 120 | 550 250 | 900 600 | 600 300 What is the ratio of the average number of female athletes who participated from England over all the years together to the average number of female athletes who participated from France over all the years together?
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Answer: (D) 32 : 29

Explanation

To find the ratio of the average number of female athletes from England to France over all the years, we first calculate the total number of female athletes for each country because the ratio of the averages is equal to the ratio of their sums over the same number of years.

Total number of female athletes from England over the years 2016 to 2019 is: 480 + 420 + 130 + 250 = 1280

Average number of female athletes from England is: 1280 ÷ 4 = 320

Total number of female athletes from France over the years 2016 to 2019 is: 280 + 120 + 160 + 600 = 1160

Average number of female athletes from France is: 1160 ÷ 4 = 290

The ratio of these averages is: 320 : 290 = 32 : 29

371Quantitative Aptitude
If a+5b=13 and ab=9 , find the value of a² +25b² .
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Answer: (B) 79

Explanation

Using the algebraic identity for the square of a binomial: (x + y)² = x² + 2xy + y²

By setting x = a and y = 5b, the relation becomes: (a + 5b)² = a² + 2(a)(5b) + (5b)² (a + 5b)² = a² + 25b² + 10ab

Substituting the given values, a + 5b = 13 and ab = 9, into this identity: 13² = a² + 25b² + 10(9) 169 = a² + 25b² + 90

Subtracting 90 from both sides gives the required value: a² + 25b² = 169 - 90 a² + 25b² = 79

372Quantitative Aptitude
In a triangle ABC; 8 ∠ A = 6 ∠ B = 3 ∠ C. What are the degree measures of ∠ A, ∠ B and ∠ C ?
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Answer: (C) 36°, 48°, 96°

Explanation

The sum of angles in a triangle is 180°, which gives the relation: ∠A + ∠B + ∠C = 180°

We are given the relation: 8 ∠A = 6 ∠B = 3 ∠C

Let this relation be equal to a constant k: 8 ∠A = 6 ∠B = 3 ∠C = k

This allows us to express each angle in terms of k: ∠A = k/8 ∠B = k/6 ∠C = k/3

Substituting these expressions into the angle sum equation: k/8 + k/6 + k/3 = 180°

Using the least common multiple of 8, 6, and 3, which is 24, we combine the fractions: (3k + 4k + 8k) / 24 = 180° 15k / 24 = 180° 5k / 8 = 180° k = (180° × 8) / 5 k = 288°

Now, we calculate the individual angle measures: ∠A = 288° / 8 = 36° ∠B = 288° / 6 = 48° ∠C = 288° / 3 = 96°

373Quantitative Aptitude
When P is subtracted from each of the numbers 8, 6, 2 and 9, the numbers so obtained in this order are in proportion. What is the mean proportional between (3P - 6) and (9P - 4)?
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Answer: (B) 28

Explanation

When a constant P is subtracted from four numbers a, b, c, and d such that the resulting numbers are in proportion, the value of P is given by the formula:

P = (ad - bc) / ((a + d) - (b + c))

Substituting the values a = 8, b = 6, c = 2, and d = 9 into the formula:

P = (8 × 9 - 6 × 2) / ((8 + 9) - (6 + 2)) P = (72 - 12) / (17 - 8) P = 60 / 9 = 20/3

Now, substitute the value of P into the two expressions:

3P - 6 = 3 × (20/3) - 6 = 20 - 6 = 14 9P - 4 = 9 × (20/3) - 4 = 60 - 4 = 56

The mean proportional between two numbers x and y is given by √(x × y).

Mean proportional = √(14 × 56) Mean proportional = √(14 × 14 × 4) Mean proportional = 14 × 2 = 28

374Quantitative Aptitude
In a 140 m race, Pankaj runs at a speed of ⁵⁄₂ m/s. He gives a start of 20 m to Arun and still defeats him by 4 seconds. The speed of Arun is:
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Answer: (C) 2 m/s

Explanation

The time taken by Pankaj to complete the 140 m race at a speed of 5/2 m/s is calculated using the formula:

Time = Distance / Speed

Time taken by Pankaj = 140 / (5/2) = 140 × 2/5 = 56 seconds.

Arun has a start of 20 m, which means he only needs to cover a distance of:

Distance covered by Arun = 140 - 20 = 120 m.

Since Pankaj defeats Arun by 4 seconds, Arun takes 4 seconds more than Pankaj to cover his distance.

Time taken by Arun = 56 + 4 = 60 seconds.

The speed of Arun is:

Speed = Distance / Time = 120 / 60 = 2 m/s.

375Quantitative Aptitude
A bookseller gives 7 books free on buying 13 books. Find the percentage of discount given to the customer's.
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Answer: (C) 35%

Explanation

Discount Percentage = (Free Items / Total Items) × 100

Total items received by the customer = 13 + 7 = 20 Number of free items = 7

Discount Percentage = (7 / 20) × 100 = 35%

376Quantitative Aptitude
In what time will ₹3,20,000 amount to ₹4,05,000 if compounded at 12.5% per annum?
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Answer: (B) 2 years

Explanation

The formula for compound interest is: A = P(1 + R/100)ⁿ

Here, Principal (P) = ₹3,20,000, Amount (A) = ₹4,05,000, and Rate of interest (R) = 12.5% per annum.

Substituting these values into the formula: 4,05,000 = 3,20,000 × (1 + 12.5/100)ⁿ

Since 12.5/100 = 1/8, we can simplify the equation: 4,05,000 / 3,20,000 = (1 + 1/8)ⁿ 405 / 320 = (9/8)ⁿ

Dividing the numerator and the denominator on the left side by 5: 81 / 64 = (9/8)ⁿ

Expressing 81/64 as a power of 9/8: (9/8)² = (9/8)ⁿ

Comparing the exponents on both sides gives n = 2 years.

377Quantitative Aptitude
A can finish a work in 24 days and B in 30 days, individually. They work together for 11 days. The percentage of the work still to be finished will be:
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Answer: (C) 17.5%

Explanation

Let the total work be 120 units, which is the least common multiple of 24 and 30.

The daily work efficiency of A is 120 ÷ 24 = 5 units per day. The daily work efficiency of B is 120 ÷ 30 = 4 units per day.

Their combined daily efficiency is 5 + 4 = 9 units per day.

In 11 days, the total work completed by A and B together is 11 × 9 = 99 units.

The remaining work is 120 - 99 = 21 units.

The percentage of the work still to be finished is (21 ÷ 120) × 100 = 17.5%.

378Quantitative Aptitude
Two similar triangles are given i.e. Δ LMN ∼ Δ PQR, with measurement of angle and side as angle I = 40°, angle N = 80°, LM = 6 cm, LN = 8 cm and PQ = 7.5 cm. Find the value of angle Q and side PR, respectively.
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Answer: (D) 60°, 10 cm

Explanation

In similar triangles, corresponding angles are equal and corresponding sides are in proportion. Given ΔLMN ~ ΔPQR:

First, find the third angle of ΔLMN. The sum of angles in a triangle is 180°: Angle L + Angle M + Angle N = 180° 40° + Angle M + 80° = 180° Angle M = 180° - 120° = 60°

Since ΔLMN ~ ΔPQR, corresponding angles are equal: Angle Q = Angle M = 60°

Next, use the ratio of corresponding sides to find PR: LM/PQ = LN/PR

Substitute the given values into the ratio: 6/7.5 = 8/PR PR = (8 × 7.5)/6 PR = 60/6 PR = 10 cm

379Quantitative Aptitude
Two circles of radius 2.4 cm and 4 cm, respectively, have a common tangent. The distance between the centres of the two circles is 6.5 cm. If the common tangent does not intersect the line joining the centres, then find the length of a common tangent to the circles.
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Answer: (D) 6.3 cm

Explanation

The length of a direct common tangent to two circles with radii r₁ and r₂ and distance d between their centres is given by the formula:

L = √(d² - (r₂ - r₁)²)

Given: r₁ = 2.4 cm r₂ = 4.0 cm d = 6.5 cm

Substitute these values into the formula:

L = √(6.5² - (4.0 - 2.4)²) L = √(6.5² - 1.6²)

Using the algebraic identity a² - b² = (a - b)(a + b):

L = √((6.5 - 1.6) × (6.5 + 1.6)) L = √(4.9 × 8.1) L = √(39.69) L = 6.3 cm

380Quantitative Aptitude
Find the value of the following expression. ((1 + sec θ))/(sec θ) (1 - cos θ)
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Answer: (D) sin²θ

Explanation

We can simplify the given expression by writing sec θ in terms of cos θ using the relation sec θ = 1/cos θ.

The expression is: ((1 + sec θ) / sec θ) (1 - cos θ)

We can rewrite the first term as: (1/sec θ + 1) (1 - cos θ)

Since 1/sec θ = cos θ, this becomes: (cos θ + 1) (1 - cos θ)

Rearranging the terms: (1 + cos θ) (1 - cos θ)

Applying the algebraic identity (a + b)(a - b) = a² - b²: 1 - cos² θ

Using the fundamental Pythagorean identity sin² θ + cos² θ = 1, we can write: 1 - cos² θ = sin² θ

381Quantitative Aptitude
Study the given table and answer the question that follows. The table shows the total score scored by 8 batsmen in the year 2022. Batsman A B C D E F G H Total Score 1356 1301 1278 1156 1380 1281 1129 1365 Who is the second-highest scorer in the year 2022?
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Answer: (D) H

Explanation

To find the second-highest scorer, we compare the total scores of all the batsmen given in the table:

E: 1380 H: 1365 A: 1356 B: 1301 F: 1281 C: 1278 D: 1156 G: 1129

Arranging these scores in descending order: 1380 > 1365 > 1356 > 1301 > 1281 > 1278 > 1156 > 1129

The highest score is 1380, belonging to batsman E. The second-highest score is 1365, which belongs to batsman H.

382Quantitative Aptitude
An amount of ₹2 was put at simple interest at a certain rate for 4 years. If it had been put at a 6% higher rate for the same period, it would have fetched ₹600 more interest. What is the value of 2.5?
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Answer: (B) ₹6,250

Explanation

Let P be the principal amount. The increase in simple interest due to an increase in the interest rate is given by the formula:

Increase in Interest = P × Increase in Rate × Time / 100

Substituting the given values:

600 = P × 6 × 4 / 100 600 = 24P / 100 P = 600 × 100 / 24 P = 2500

The question asks for the value of 2.5 times this principal amount:

2.5 × P = 2.5 × 2500 = 6250

383Quantitative Aptitude
A shopkeeper marks his goods at 90% more than their cost price and allows a discount of 40% on the marked price. His gain percentage is:
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Answer: (C) 14%

Explanation

The net gain percentage when an article is marked up by m% and then sold at a discount of d% can be calculated using the successive percentage formula:

Net Gain % = m - d - (m × d)/100

Substituting the given values, where the markup percentage m = 90% and the discount percentage d = 40%:

Net Gain % = 90 - 40 - (90 × 40)/100 Net Gain % = 50 - 3600/100 Net Gain % = 50 - 36 Net Gain % = 14%

Alternatively, assuming the cost price is ₹100, a 90% markup makes the marked price ₹190. Offering a 40% discount on this marked price results in a selling price of:

190 × (1 - 0.40) = 190 × 0.60 = ₹114

The gain is the difference between the selling price and the cost price, which is:

114 - 100 = 14%

384Quantitative Aptitude
A sells a television to B for ₹4,860, thereby losing 19%. B sells it to C at a price which would have given A a 17% profit. Find the gain of B.
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Answer: (B) ₹2,160

Explanation

The cost price for A (CP) is calculated using the relation CP = SP / (1 - Loss Rate). Since A sells the television for ₹4,860 at a 19% loss: CP of A = 4860 / (1 - 0.19) = 4860 / 0.81 = ₹6,000

B buys the television at this selling price, so the cost price for B is ₹4,860.

B then sells the television to C at a price that would have given A a 17% profit on A's cost price: Selling price of B = CP of A × (1 + 0.17) = 6000 × 1.17 = ₹7,020

The gain of B is the difference between B's selling price and B's cost price: Gain of B = 7020 - 4860 = ₹2,160

385Quantitative Aptitude
A circle with centre O has a chord AB that is 20 cm in length. If the radius of the circle is 12 cm, then the area of triangle AOB is:
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Answer: (C) 20√(11)cm²

Explanation

In triangle AOB, OA and OB are radii of the circle, so OA = OB = 12 cm. The chord AB is the base of the triangle with a length of 20 cm.

Let OD be the perpendicular drawn from the centre O to the chord AB. Since the perpendicular from the centre of a circle to a chord bisects the chord, D is the midpoint of AB. Therefore, AD = DB = 10 cm.

In the right-angled triangle OAD, we can apply the Pythagorean theorem: OA² = OD² + AD² 12² = OD² + 10² 144 = OD² + 100 OD² = 44 OD = √44 = 2√11 cm

The area of triangle AOB is given by the formula: Area = ½ × base × height Area = ½ × AB × OD Area = ½ × 20 × 2√11 Area = 20√11 cm²

386Quantitative Aptitude
A hemispherical dome of a building needs to be painted. If the circumference of the base of the dome is 154 cm, then find the cost of painting it if the cost of painting is ₹4 per 100 cm² (use π = ²²⁄₇ ).
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Answer: (A) ₹150.92

Explanation

The circumference of the base of the hemispherical dome is given by the formula C = 2πr, where r is the radius of the base.

2 × (22/7) × r = 154 r = (154 × 7) / 44 r = 24.5 cm

The curved surface area (CSA) of a hemispherical dome is given by the formula 2πr².

CSA = 2 × (22/7) × 24.5 × 24.5 CSA = 2 × (22/7) × (49/2) × (49/2) CSA = 11 × 7 × 49 CSA = 3773 cm²

The cost of painting is ₹4 per 100 cm².

Total Cost = CSA × (4 / 100) Total Cost = 3773 × 0.04 Total Cost = ₹150.92

387Quantitative Aptitude
The cost of 3 breads and 1 packet of juice at a certain store is 323.50. At the same store, the cost of 5 breads and 1 packet of juice is 433.50. What will be the cost of 2 breads and 2 packets of juice at this store?
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Answer: (A) ₹427

Explanation

Let b be the cost of one bread and j be the cost of one packet of juice. Based on the given conditions, we can set up the following system of linear equations:

3b + j = 323.50 5b + j = 433.50

Subtracting the first equation from the second equation to eliminate j:

(5b + j) - (3b + j) = 433.50 - 323.50 2b = 110.00 b = 55.00

Now, substituting b = 55.00 back into the first equation:

3(55.00) + j = 323.50 165.00 + j = 323.50 j = 323.50 - 165.00 j = 158.50

To find the cost of 2 breads and 2 packets of juice, we calculate:

2b + 2j = 2(55.00) + 2(158.50) 2b + 2j = 110.00 + 317.00 2b + 2j = 427.00

388Quantitative Aptitude
In how many years will 40000 amounts to 46656 at the rate of 8% compound interest per annum?
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Answer: (B) 2

Explanation

Correct Answer: B

The compound interest formula is given by A = P(1 + r/100)ⁿ, where A is the amount, P is the principal, r is the annual interest rate, and n is the time in years.

Substituting the given values into the formula: 46656 = 40000(1 + 8/100)ⁿ 46656 / 40000 = (1.08)ⁿ 1.1664 = (1.08)ⁿ

To find the value of n, we can calculate the powers of 1.08: (1.08)² = 1.08 × 1.08 = 1.1664

Since (1.08)² matches 1.1664, the value of n is 2.

389Quantitative Aptitude
Two triangles MNO and XYZ are given similar with the ratio of their side as 9 : 4. If the area of the larger triangle is 243 sq.cm, then the area of smaller triangle will be:
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Answer: (C) 48sq.cm

Explanation

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Area of larger triangle / Area of smaller triangle = (Side of larger triangle / Side of smaller triangle)²

Substituting the given values into the formula:

243 / Area of smaller triangle = (9 / 4)²

243 / Area of smaller triangle = 81 / 16

Area of smaller triangle = 243 × 16 / 81

Area of smaller triangle = 3 × 16 = 48 sq.cm

390Quantitative Aptitude
If a = (1)/(a - √6) (a > 0) , then the value of (a + ¹⁄ₐ) is:
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Answer: (B) √(10)

Explanation

From the given equation, we have a(a - √6) = 1, which can be rewritten as a - 1/a = √6. Using the algebraic identity (a + 1/a)² = (a - 1/a)² + 4, we get (a + 1/a)² = 6 + 4 = 10. Since a > 0, we take the positive square root to get a + 1/a = √(10).

391Quantitative Aptitude
Manoj marked his goods 30% above the Cost Price and gives a 12% discount to the customers. The profit or loss percentage is:
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Answer: (C) 14.4% profit

Explanation

The net profit or loss percentage when an item is marked up by x% and then sold at a discount of y% can be directly calculated using the successive percentage formula:

Net Percentage = x - y - (x × y) / 100

Here, the markup percentage x is 30% and the discount percentage y is 12%. Substituting these values into the formula:

Net Percentage = 30 - 12 - (30 × 12) / 100 Net Percentage = 18 - 3.6 Net Percentage = 14.4%

Since the net percentage change is positive, it represents a profit of 14.4%.

392Quantitative Aptitude
If x = (1)/(x - 5) (x > 0) , then the value of x + ¹⁄ₓ is:
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Answer: (B) √(29)

Explanation

Given the equation: x = 1 / (x - 5)

Multiplying both sides by x - 5 gives: x(x - 5) = 1 x² - 5x = 1

Dividing both sides by x (since x > 0): x - 1/x = 5

To find the value of x + 1/x, we use the standard algebraic identity: (x + 1/x)² = (x - 1/x)² + 4

Substituting the value of x - 1/x: (x + 1/x)² = 5² + 4 (x + 1/x)² = 25 + 4 = 29

Taking the positive square root on both sides because x > 0: x + 1/x = √29

393Quantitative Aptitude
A shopkeeper marked a product, 40% above the cost price and sold the product for ₹3,790 by giving a discount of 10% . The profit percentage of the shopkeeper is:
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Answer: (B) 26%

Explanation

The profit percentage can be calculated directly using the successive percentage change formula:

Net percentage change = A + B + (A × B)/100

Here, the markup percentage A is +40% and the discount percentage B is -10%.

Substituting these values into the formula:

Net percentage change = 40 - 10 + (40 × (-10))/100 Net percentage change = 30 - 400/100 Net percentage change = 30 - 4 = 26%

Since the net percentage change is positive, the shopkeeper makes a profit of 26%. The selling price of ₹3,790 is redundant information.

394Quantitative Aptitude
A man invested ₹4,400 at 10% and ₹7,500 at 8% simple interest. What amount (in ₹) will he get from his investments after 12 years?
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Answer: (B) 24,380

Explanation

The total amount obtained after a period of time under simple interest is given by the formula:

Amount = Principal + Simple Interest Simple Interest = Principal × Rate × Time / 100

For the first investment: Principal = ₹4,400 Rate = 10% Time = 12 years Simple Interest = 4400 × 10 × 12 / 100 = ₹5,280

For the second investment: Principal = ₹7,500 Rate = 8% Time = 12 years Simple Interest = 7500 × 8 × 12 / 100 = ₹7,200

Total Simple Interest = 5280 + 7200 = ₹12,480 Total Principal = 4400 + 7500 = ₹11,900

Total Amount = Total Principal + Total Simple Interest = 11900 + 12480 = ₹24,380

395Quantitative Aptitude
Two cities P and Q are 181 km apart on a straight road. One man starts from P at 8:30 A.M. and travels toward Q at 30 km/h. If another man starts from Q at 8:54 A.M. and travels towards P at a speed of 35 km/h, then at what time will they meet?
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Answer: (A) 11:30 A.M.

Explanation

The time difference between the starts of the two men is from 8:30 A.M. to 8:54 A.M., which is 24 minutes or 24/60 = 0.4 hours.

During this time, the first man travels a distance of: Distance = Speed × Time = 30 km/h × 0.4 h = 12 km

The remaining distance between them at 8:54 A.M. is: 181 km - 12 km = 169 km

Since they are moving towards each other, their relative speed is the sum of their individual speeds: Relative Speed = 30 km/h + 35 km/h = 65 km/h

The time taken to cover the remaining distance after 8:54 A.M. is: Time = Distance / Relative Speed = 169 km / 65 km/h = 2.6 hours

Converting 2.6 hours to hours and minutes: 2.6 hours = 2 hours + 0.6 × 60 minutes = 2 hours and 36 minutes

Adding 2 hours and 36 minutes to 8:54 A.M.: 8:54 A.M. + 2 hours = 10:54 A.M. 10:54 A.M. + 36 minutes = 11:30 A.M.

396Quantitative Aptitude
The angle subtended by an arc at the circumference of the circle is 55° . Find the angle subtended by it at the centre.
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Answer: (B) 110°

Explanation

According to the circle theorem, the angle subtended by an arc at the center of a circle is twice the angle subtended by the same arc at any point on the remaining part of the circumference.

Given that the angle subtended at the circumference is 55°, the angle subtended at the center is calculated as:

Angle at the center = 2 × Angle at the circumference Angle at the center = 2 × 55° = 110°

397Quantitative Aptitude
The following table represents the number of five different types of cars A, B, C, D, and E manufactured by a company over five years. Read the table carefully and answer the question given below: Years/Cars | A | B | C | D | E --- | --- | --- | --- | --- | --- 2018 | 2550 | 2675 | 2400 | 2000 | 450 2019 | 2600 | 2800 | 2550 | 1900 | 375 2020 | 2775 | 2600 | 2600 | 2000 | 400 2021 | 2800 | 2750 | 2500 | 1975 | 475 2022 | 3000 | 2800 | 2650 | 2100 | 500 What is the percentage increment in the production of D- type of cars in the year 2022 in comparison to the previous year? (correct up to two decimal places)
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Answer: (D) 6.33%

Explanation

The percentage increment in the production of D-type cars is calculated using the formula:

Percentage increment = ((Production in 2022 - Production in 2021) / Production in 2021) × 100

From the given table, the production of D-type cars is: Production in 2021 = 1975 Production in 2022 = 2100

Substituting these values into the formula: Percentage increment = ((2100 - 1975) / 1975) × 100 Percentage increment = (125 / 1975) × 100 Percentage increment = (5 / 79) × 100 Percentage increment ≈ 6.33%

398Quantitative Aptitude
The ratio of the father's age to his son's age is 5:3. The product of the numbers representing their ages is 960. The ratio of their ages after 6 years will be:
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Answer: (A) 23:15

Explanation

Let the present age of the father and the son be 5x and 3x, respectively.

According to the given condition, the product of their ages is: 5x × 3x = 960 15x² = 960 x² = 64 x = 8

Thus, the present age of the father is 5 × 8 = 40 years, and the present age of the son is 3 × 8 = 24 years.

After 6 years: Father's age = 40 + 6 = 46 years Son's age = 24 + 6 = 30 years

The ratio of their ages after 6 years is: 46:30 = 23:15

399Quantitative Aptitude
The cost price of an article is 72% of the marked price. The profit percentage after allowing a discount of 6% on the marked price is (rounded off to 2 decimal places):
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Answer: (C) 30.56%

Explanation

Let the marked price (MP) of the article be 100.

The cost price (CP) is 72% of the marked price, so CP = 72.

The selling price (SP) after a discount of 6% on the marked price is SP = 100 - 6 = 94.

Profit = SP - CP = 94 - 72 = 22

Profit Percentage = (Profit / CP) × 100 Profit Percentage = (22 / 72) × 100 = 30.555... %

Rounding off to two decimal places, the profit percentage is 30.56%.

400Quantitative Aptitude
Simplify (36a² - 49b²)/(6a + 7b) .
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Answer: (C) 6a - 7b

Explanation

Using the difference of squares formula, x² - y² = (x - y)(x + y), the numerator can be rewritten as: 36a² - 49b² = (6a)² - (7b)² = (6a - 7b)(6a + 7b)

Substituting this back into the original expression: ((6a - 7b)(6a + 7b)) / (6a + 7b) = 6a - 7b

Answer key

Tap a number to jump to that question.

  1. 351D
  2. 352C
  3. 353C
  4. 354D
  5. 355A
  6. 356C
  7. 357C
  8. 358A
  9. 359B
  10. 360D
  11. 361C
  12. 362C
  13. 363C
  14. 364B
  15. 365B
  16. 366B
  17. 367A
  18. 368D
  19. 369D
  20. 370D
  21. 371B
  22. 372C
  23. 373B
  24. 374C
  25. 375C
  26. 376B
  27. 377C
  28. 378D
  29. 379D
  30. 380D
  31. 381D
  32. 382B
  33. 383C
  34. 384B
  35. 385C
  36. 386A
  37. 387A
  38. 388B
  39. 389C
  40. 390B
  41. 391C
  42. 392B
  43. 393B
  44. 394B
  45. 395A
  46. 396B
  47. 397D
  48. 398A
  49. 399C
  50. 400C

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