SSC CHSL2024Quantitative AptitudePage 6 of 13

SSC CHSL Tier 1 2024 Quantitative Aptitude Questions with Solutions

Questions from every shift of the year, grouped by section. Questions 251–300 of 648, each with the correct answer and a step-by-step explanation.

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251Quantitative Aptitude
The given table represents the percentage marks of three students in three subjects. Study the table and answer the question. | Subject | Geography | History | Economics | | :--- | :--- | :--- | :--- | | Rohan | 78 | 77 | 74 | | Sohan | 84 | 80 | 82 | | Mohan | 87 | 83 | 86 | The average marks obtain by Sohan in all three subject is equal to:
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Answer: (A) 82

Explanation

The average of a set of values is calculated using the formula:

Average = (Sum of all values) / (Total number of values)

For Sohan, the marks in the three subjects (Geography, History, and Economics) are 84, 80, and 82 respectively. Substituting these values into the formula:

Average = (84 + 80 + 82) / 3

Average = 246 / 3 = 82

252Quantitative Aptitude
When the integer n is divided by 5, the remainder is 3. What is the remainder if 6n is divided by 5?
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Answer: (C) 3

Explanation

By the division algorithm, the integer n can be written as n = 5q + 3, where q is an integer.

Multiplying this equation by 6 gives: 6n = 6(5q + 3) 6n = 30q + 18

Since 18 can be written as 15 + 3, we can rewrite the expression as: 6n = 30q + 15 + 3 6n = 5(6q + 3) + 3

Because 5(6q + 3) is a multiple of 5, the remainder when 6n is divided by 5 is 3.

253Quantitative Aptitude
ABC is a right-angled triangle, right angled at A in such a way that AB = AC and the bisector of angle B intersects the side AC at D. DM is perpendicular to BC, intersecting BC at M. If AB = 12 cm and BC = 12√2 cm, then find the length of side AD.
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Answer: (B) 12(√2 - 1)cm

Explanation

By the Angle Bisector Theorem, the internal bisector of an angle of a triangle divides the opposite side into segments that are proportional to the adjacent sides. Since BD is the bisector of angle B intersecting AC at D, we have:

AD / DC = AB / BC

Given that AB = AC = 12 cm and BC = 12√2 cm, we can express DC as AC - AD, which is 12 - AD. Substituting these values into the ratio:

AD / (12 - AD) = 12 / (12√2)

AD / (12 - AD) = 1 / √2

Cross-multiplying the terms gives:

AD × √2 = 12 - AD

AD × √2 + AD = 12

AD(√2 + 1) = 12

AD = 12 / (√2 + 1)

To rationalize the denominator, multiply the numerator and the denominator by (√2 - 1):

AD = 12(√2 - 1) / ((√2 + 1)(√2 - 1))

AD = 12(√2 - 1) / (2 - 1)

AD = 12(√2 - 1) cm

254Quantitative Aptitude
A metallic solid cuboid of dimensions 36cm× 22cm× 12cm is melted and recast in the form of cubes of side 6 cm. Find the number of the cubes so formed.
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Answer: (D) 44

Explanation

The number of cubes formed by melting a solid cuboid is obtained by dividing the volume of the cuboid by the volume of a single cube.

Volume of the cuboid = length × width × height Volume of a cube = side³

Substituting the given values: Number of cubes = (36 × 22 × 12) ÷ (6 × 6 × 6)

Simplifying the division: Number of cubes = (36 ÷ 6) × 22 × (12 ÷ 6) ÷ 6 Number of cubes = 6 × 22 × 2 ÷ 6 Number of cubes = 22 × 2 = 44

255Quantitative Aptitude
Identify a single-digit number other than 1, which may exactly divide the number 17³ + 18³ - 16³ - 15³
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Answer: (D) 2

Explanation

We can rewrite the expression by grouping the terms as follows: (17³ - 15³) + (18³ - 16³)

Using the algebraic identity a³ - b³ = (a - b)(a² + ab + b²):

For the first group: 17³ - 15³ = (17 - 15)(17² + 17 × 15 + 15²) = 2 × (17² + 17 × 15 + 15²)

For the second group: 18³ - 16³ = (18 - 16)(18² + 18 × 16 + 16²) = 2 × (18² + 18 × 16 + 16²)

Since both terms contain 2 as a common factor, their sum is also a multiple of 2, which means the entire expression is exactly divisible by 2.

256Quantitative Aptitude
The marked price of a battery- operated rickshaw is ₹90,000. It is available for sale at two successive discounts of 10% and 5% . What is the selling price (in ₹) of the battery- operated rickshaw?
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Answer: (A) 76,950

Explanation

The selling price SP after two successive discounts of d₁% and d₂% on the marked price MP is given by the formula:

SP = MP × (1 - d₁/100) × (1 - d₂/100)

Substituting the given values MP = 90,000, d₁ = 10, and d₂ = 5:

SP = 90,000 × (1 - 10/100) × (1 - 5/100) SP = 90,000 × 0.90 × 0.95 SP = 81,000 × 0.95 SP = 76,950

257Quantitative Aptitude
Simplify 4¼× 4¼ - 2× 4¼× ¼ +¼× ¼
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Answer: (A) 16

Explanation

The given expression is in the algebraic form a² - 2ab + b², where a = 4¼ and b = ¼.

Using the standard algebraic identity: a² - 2ab + b² = (a - b)²

Substitute the values of a and b: a - b = 4¼ - ¼ = 4

Squaring this result gives: (a - b)² = 4² = 16

258Quantitative Aptitude
8 apples and 10 oranges weigh 5 kg. 12 apples and 20 oranges weigh 9 kg. What is the weight (in kg, rounded off to the nearest integer) of 15 apples and 24 oranges?
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Answer: (B) 11

Explanation

Let a represent the weight of an apple and o represent the weight of an orange. This gives the following system of linear equations:

8a + 10o = 5 12a + 20o = 9

To eliminate o, multiply the first equation by 2:

16a + 20o = 10

Subtracting the second equation from this equation gives:

4a = 1 a = 0.25 kg

Substituting a = 0.25 into the first equation:

8(0.25) + 10o = 5 2 + 10o = 5 10o = 3 o = 0.3 kg

Now, calculate the weight of 15 apples and 24 oranges:

15a + 24o = 15(0.25) + 24(0.3) = 3.75 + 7.2 = 10.95 kg

Rounding 10.95 kg to the nearest integer gives 11.

259Quantitative Aptitude
Find the altitude (in cm) of side MT of triangle MNT with side MN = 36 cm, MT = 36 cm and NT = 48 cm.
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Answer: (B) 16√5

Explanation

The semi-perimeter s of the triangle MNT is: s = (MN + MT + NT) / 2 = (36 + 36 + 48) / 2 = 60 cm

Using Heron's formula, the area of the triangle is: Area = √(s(s - MN)(s - MT)(s - NT)) Area = √(60 × (60 - 36) × (60 - 36) × (60 - 48)) Area = √(60 × 24 × 24 × 12) Area = √(12 × 5 × 24² × 12) Area = 12 × 24 × √5 = 288√5 cm²

The area of a triangle is also given by the formula: Area = ½ × base × altitude Area = ½ × MT × h 288√5 = ½ × 36 × h 288√5 = 18 × h h = 288√5 / 18 = 16√5 cm

260Quantitative Aptitude
Pradeep sells two fans at ₹480 each and by doing so he gains 20% on one fan and loses 20% on the other. His loss on the whole (in ₹) is:
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Answer: (B) 40

Explanation

When two items are sold at the same selling price (SP) with one yielding a gain of x% and the other a loss of x%, there is always an overall loss.

Net loss percentage = x² / 100 % Net loss percentage = 20² / 100 % = 400 / 100 % = 4%

Total selling price of both fans = 2 × 480 = ₹960

Since there is a net loss of 4%, the total selling price represents 96% of the total cost price (CP).

Total CP = (Total SP / (100 - Net loss %)) × 100 Total CP = (960 / 96) × 100 = ₹1000

Total Loss in rupees = Total CP - Total SP Total Loss = 1000 - 960 = ₹40

261Quantitative Aptitude
In a triangle ABC, P and Q are two points on AB and AC, respectively, such that PQ parallel to BC. If AC = 5QC, then the ratio PQ : BC is equal to:
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Answer: (B) 4 : 5

Explanation

In triangle ABC, since PQ is parallel to BC, triangle APQ is similar to triangle ABC. By the property of similar triangles, the ratio of corresponding sides is equal:

PQ/BC = AQ/AC

We are given that AC = 5QC. Since Q is a point on AC:

AC = AQ + QC 5QC = AQ + QC AQ = 4QC

Using this, the ratio of AQ to AC is:

AQ/AC = 4QC/5QC = 4/5

Since PQ/BC = AQ/AC, the ratio PQ : BC is 4 : 5.

262Quantitative Aptitude
The centres of two circles are 84 cm apart. If the radii of these two circles are 38 cm and 26 cm, respectively, then which of the following options gives the length (in cm) of a direct common tangent of these two circles?
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Answer: (A) 48√3

Explanation

The length of a direct common tangent of two circles is given by the formula:

Length = √(d² - (r₁ - r₂)²)

where d is the distance between the centres of the two circles, and r₁ and r₂ are their radii.

Substituting the given values, d = 84 cm, r₁ = 38 cm, and r₂ = 26 cm:

Length = √(84² - (38 - 26)²) Length = √(84² - 12²)

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

Length = √((84 - 12) × (84 + 12)) Length = √(72 × 96) Length = √(24 × 3 × 24 × 4) Length = √(24² × 12) Length = √(24² × 4 × 3) Length = √(24² × 2² × 3) Length = √(48² × 3) Length = 48√3 cm

263Quantitative Aptitude
What is the value of the following expression? 420÷ [⁴⁄₉× (1 + ⅘)× 5] + [⁵⁄₇of²¹⁄₂₅÷ ¾of⅖
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Answer: (C) 107

Explanation

The expression can be evaluated step-by-step using the BODMAS rule, which dictates the order of operations as Brackets, Of, Division, Multiplication, Addition, and Subtraction.

First, simplify the expression inside the brackets for the first term: 1 + 4/5 = 9/5

Substitute this value back into the bracket: 4/9 × 9/5 × 5 = 4

The first part of the expression becomes: 420 ÷ 4 = 105

Next, evaluate the second part of the expression, where the "of" operation has higher precedence than division and is calculated first: 5/7 of 21/25 = (5/7) × (21/25) = 3/5 3/4 of 2/5 = (3/4) × (2/5) = 3/10

Now, divide the two resulting values: (3/5) ÷ (3/10) = (3/5) × (10/3) = 2

Finally, add the values of both parts together: 105 + 2 = 107

264Quantitative Aptitude
Find the mean proportional between 25 and 81.
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Answer: (D) 45

Explanation

The mean proportional between two positive numbers a and b is given by the formula √(a × b).

By substituting the given values, a = 25 and b = 81:

Mean proportional = √(25 × 81) = √25 × √81 = 5 × 9 = 45.

265Quantitative Aptitude
Ramu mixes 2 litres of water at 100°C and 18 litres of water at 32°C . What temperature will the water have after mixing?
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Answer: (C) 38.8°C

Explanation

According to the principle of calorimetry, when two masses of the same liquid at different temperatures are mixed, the final equilibrium temperature T_f is given by the weighted average formula:

T_f = (m₁T₁ + m₂T₂) / (m₁ + m₂)

Here, the density and specific heat capacity of water are constant, so we can directly use the volumes in litres as representative of their masses:

m₁ = 2 L, T₁ = 100 °C m₂ = 18 L, T₂ = 32 °C

Substituting these values into the formula:

T_f = (2 × 100 + 18 × 32) / (2 + 18) T_f = (200 + 576) / 20 T_f = 776 / 20 T_f = 38.8 °C

266Quantitative Aptitude
At an election involving two candidates, 68 votes are declared invalid. The winning candidate gets 52% of the valid votes and wins by 98 votes. The total number of votes polled is:
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Answer: (A) 2518

Explanation

Let V be the total number of valid votes.

The winning candidate gets 52% of the valid votes, which means the losing candidate gets: 100% - 52% = 48% of the valid votes

The difference between their vote shares is: 52% - 48% = 4% of the valid votes

Since the winner wins by 98 votes, we can set up the relation: 4% of V = 98 V × 4/100 = 98 V = (98 × 100) / 4 V = 98 × 25 V = 2450

The total number of votes polled is the sum of the valid votes and the invalid votes: Total votes polled = Valid votes + Invalid votes Total votes polled = 2450 + 68 = 2518

267Quantitative Aptitude
A sum of money becomes five times its original value in 15 years when invested at a certain simple interest rate. If the sum was invested twice the time at the same rate of interest, what would be the final amount?
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Answer: (A) The money becomes 9 times its original value

Explanation

Let the principal sum be P. The final amount A under simple interest is given by the formula A = P + S.I., where S.I. is the simple interest.

When the sum becomes 5 times its original value, the amount is 5P. The interest earned is: S.I. = 5P - P = 4P

Simple interest is directly proportional to time. When the investment duration is doubled, the simple interest earned also doubles.

New simple interest = 2 × 4P = 8P

The final amount after twice the time is: Amount = P + 8P = 9P

268Quantitative Aptitude
Ramesh's pocket money was reduced by 25% and then the reduced pocket money was increased by 20% . Find the net increase or decrease percentage in his original pocket money.
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Answer: (D) Decrease 10%

Explanation

The net percentage change when a value is successively changed by a% and b% is calculated using the formula:

Net change = a + b + ab/100

In this case, the first change is a reduction of 25%, so a = -25. The second change is an increase of 20%, so b = 20.

Substituting these values into the formula:

Net change = -25 + 20 + (-25 × 20)/100 Net change = -5 - 500/100 Net change = -5 - 5 Net change = -10%

A negative result of 10% corresponds to a net decrease of 10%.

269Quantitative Aptitude
R and S can finish a work in 10 days, S and T can finish it in 12 days and T and R can finish in 8 days. In how many days will the work be finished if they all work simultaneously?
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Answer: (B) 240/37

Explanation

To find the combined rate of work, we use the formula for work rate: Rate = 1 / Time.

Let the daily work rates of R, S, and T be r, s, and t respectively.

From the given data, we have: r + s = 1/10 s + t = 1/12 t + r = 1/8

Adding these three equations together gives: 2(r + s + t) = 1/10 + 1/12 + 1/8

To sum the fractions, find the least common multiple of the denominators 10, 12, and 8, which is 120: 1/10 + 1/12 + 1/8 = 12/120 + 10/120 + 15/120 = 37/120

Substitute this back into the equation: 2(r + s + t) = 37/120 r + s + t = 37/240

Since the combined daily rate of R, S, and T working together is 37/240, the total time required to complete the work is the reciprocal of this rate, which is 240/37 days.

270Quantitative Aptitude
The area of an isosceles right angled triangle is 16 ~m² . Its hypotenuse is ______.
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Answer: (B) 8 m

Explanation

The area of an isosceles right-angled triangle with hypotenuse h is given by the formula: Area = h²/4

Substituting the given area of 16 m² into the formula: 16 = h²/4 h² = 16 × 4 h² = 64 h = √64 = 8 m

271Quantitative Aptitude
R can finish a work in 6 days if he works 10 hours a day. S can finish it in 5 days if he works 8 hours a day. If both start a work and work daily for 6 hours, the work will be finished in ___ days.
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Answer: (B) 4

Explanation

Correct Answer: B

Total hours required by R to complete the work is 6 days × 10 hours per day = 60 hours.

Total hours required by S to complete the work is 5 days × 8 hours per day = 40 hours.

Using the combined work formula, the total number of hours T required when both work together is: T = (R × S) / (R + S) T = (60 × 40) / (60 + 40) = 2400 / 100 = 24 hours

Since they work daily for 6 hours, the number of days required to finish the work is: Number of days = Total hours / Daily hours = 24 / 6 = 4 days

272Quantitative Aptitude
From a circular sheet of circumference 264 cm, two equal maximum-sized circular plates are cut off. What will be the circumference of each plate? (Use π = ²²⁄₇ )
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Answer: (D) 132 cm

Explanation

Correct Answer: D

The circumference of the large circular sheet is C = 2πR, where R is its radius.

To cut two equal maximum-sized circular plates from this circular sheet, the two plates must lie along the diameter of the large sheet, touching each other at the center and being tangent to the outer boundary of the sheet.

This means the diameter of the large circle is equal to the sum of the diameters of the two equal smaller circles: 2R = 2r + 2r 2R = 4r r = R / 2

Thus, the radius of each smaller circular plate, r, is exactly half the radius of the large circular sheet, R.

Since the circumference of a circle is directly proportional to its radius, the circumference of each smaller circular plate, c, is: c = 2πr = 2π(R / 2) = C / 2

Substituting the given circumference of the large sheet: c = 264 / 2 = 132 cm

273Quantitative Aptitude
The speed of a boat in still water is 12 km/h, and the speed of the stream is 3 km/h. A man goes to a place by boat, 45 km and comes back to the starting point. Find the total time taken by him.
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Answer: (C) 8 hours

Explanation

The total time taken for a round trip is given by the formula: T = d / (u + v) + d / (u - v)

where d is the one-way distance, u is the speed of the boat in still water, and v is the speed of the stream.

Substituting the given values into the formula: T = 45 / (12 + 3) + 45 / (12 - 3) T = 45 / 15 + 45 / 9 T = 3 + 5 T = 8 hours

274Quantitative Aptitude
In an election a candidate secured 47% of the votes and lost to his only opponent by 276 votes. Find the total number of votes polled.
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Answer: (B) 4600

Explanation

Let the total number of votes polled be 100%.

The candidate who lost secured 47% of the total votes. Since there are only two candidates in the election, the winning candidate must have secured the remaining votes: 100% - 47% = 53%

The difference in the percentage of votes between the winner and the loser is: 53% - 47% = 6%

This difference of 6% corresponds to the margin of 276 votes.

To find the total number of votes, we calculate: Total votes = 276 × 100 / 6 = 4600

275Quantitative Aptitude
Ram invested a certain sum of money at simple interest. It amounted to ⁵⁄₄ of itself in 3 years. What will be its rate per cent per annum?
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Answer: (D) 8.33%

Explanation

Let the principal sum be P. The amount A after 3 years is 5/4 of the principal, which is 5/4 P.

The simple interest earned is given by: SI = A - P = 5/4 P - P = 1/4 P

Using the simple interest formula, SI = (P × R × T) / 100, where R is the rate per cent per annum and T is the time in years: 1/4 P = (P × R × 3) / 100

Dividing both sides by P: 1/4 = 3R / 100

Solving for R: R = 100 / (4 × 3) = 25 / 3 = 8.33% per annum

276Quantitative Aptitude
Tanvi can do a work in 25 days, and Tai can complete the same work in 30 days. They started the work together for 5 days. How much work is left?
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Answer: (B) 19/30

Explanation

Tanvi's 1-day work rate is 1/25 and Tai's 1-day work rate is 1/30.

Their combined 1-day work rate is: 1/25 + 1/30 = (6 + 5)/150 = 11/150

The amount of work completed by both of them working together for 5 days is: 5 × (11/150) = 11/30

The remaining work is calculated by subtracting the completed work from the total work, which is represented as 1: 1 - 11/30 = 19/30

277Quantitative Aptitude
Ram is 20 years older than Rahul. 'n' years ago Ram was thrice as old as Rahul. '2n' years from now the ratio of the ages of Ram and Rahul will be 15 : 11. What is the value of 'n'?
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Answer: (D) 15

Explanation

Let the current age of Ram be R and the current age of Rahul be r. The difference between their ages is constant at all times: R - r = 20

Let their ages 'n' years ago be R - n and r - n. At this time, Ram was thrice as old as Rahul: R - n = 3(r - n)

Since the age difference is always 20 years: (R - n) - (r - n) = 20 3(r - n) - (r - n) = 20 2(r - n) = 20 r - n = 10

This gives Rahul's current age as r = 10 + n, and Ram's current age as R = 30 + n.

In '2n' years from now, their ages will be: Rahul's age = r + 2n = 10 + 3n Ram's age = R + 2n = 30 + 3n

The ratio of their ages '2n' years from now is 15 : 11: (30 + 3n) / (10 + 3n) = 15 / 11

Cross-multiplying the terms: 11 × (30 + 3n) = 15 × (10 + 3n) 330 + 33n = 150 + 45n 12n = 180 n = 15

278Quantitative Aptitude
A sells an object to B at 10% profit, B sells it to C at 60% profit and C sells it to D at 12½% profit. If D paid ₹59.40, then at what price did A buy it?
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Answer: (A) ₹30

Explanation

The sequential transaction can be represented using the compounding formula for successive profits. Let the cost price for A be x.

The multipliers for the sequential profit percentages are: A to B at 10% profit: 1 + 10/100 = 11/10 B to C at 60% profit: 1 + 60/100 = 8/5 C to D at 12.5% profit: 1 + 12.5/100 = 9/8

The final price paid by D is given by: x × (11/10) × (8/5) × (9/8) = 59.40

Simplifying the fraction by cancelling 8 in the numerator and denominator: x × 99/50 = 59.40

Solving for x: x = (59.40 × 50) / 99 x = 0.60 × 50 x = 30

Thus, A purchased the object for ₹30.

279Quantitative Aptitude
A man can row a boat at speed 8 km/h in still water. It takes thrice as long to row upstream as it does to row downstream. Find the speed of the stream (in km/h).
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Answer: (A) 4

Explanation

Let the speed of the boat in still water be u = 8 km/h and the speed of the stream be v km/h.

The downstream speed is u + v and the upstream speed is u - v. Since the distance covered is the same, time is inversely proportional to speed.

t_upstream = 3 × t_downstream

1 / (u - v) = 3 / (u + v)

u + v = 3(u - v)

u + v = 3u - 3v

4v = 2u

v = u / 2

Substituting u = 8 km/h into the equation:

v = 8 / 2 = 4 km/h

280Quantitative Aptitude
Simplify the expression 720 + 48 ÷ 8 × 5 - 6 × 3 .
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Answer: (A) 732

Explanation

Using the BODMAS rule, the operations in the expression 720 + 48 ÷ 8 × 5 - 6 × 3 are performed in the sequence of Division, Multiplication, Addition, and then Subtraction.

First, perform the division: 48 ÷ 8 = 6

The expression becomes: 720 + 6 × 5 - 6 × 3

Next, perform the multiplication operations: 6 × 5 = 30 6 × 3 = 18

The expression now simplifies to: 720 + 30 - 18

Finally, perform addition followed by subtraction: 720 + 30 = 750 750 - 18 = 732

281Quantitative Aptitude
In an election contested by two candidates, one candidate got 42% of the total votes and still lost by 800 votes. Find the total number of votes that were cast.
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Answer: (A) 5000

Explanation

Let the total number of votes cast in the election be 100%.

The losing candidate secured 42% of the total votes. Since there were only two candidates, the winning candidate received the remaining portion: 100% - 42% = 58% of the total votes.

The difference in vote percentage between the winning and losing candidates is: 58% - 42% = 16% of the total votes.

This 16% margin corresponds to the losing margin of 800 votes. Therefore: 16% of Total Votes = 800 Total Votes = (800 × 100) ÷ 16 = 5000.

282Quantitative Aptitude
On buying 24 bangles Meena gets 6 bangles free of cost. By what percentage does Meena get a discount?
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Answer: (A) 20%

Explanation

Discount percentage is calculated as the ratio of free items received to the total number of items obtained, multiplied by 100.

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

Here, Meena gets 6 bangles free on buying 24 bangles. Total items obtained = 24 + 6 = 30 bangles Free items = 6 bangles

Substituting these values into the formula: Discount % = (6 / 30) × 100 = 1/5 × 100 = 20%

283Quantitative Aptitude
5% of 2% of 3 is equal to:
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Answer: (A) 0.003

Explanation

To find the required value, convert the percentages into decimals or fractions and perform the multiplication:

5% of 2% of 3 = (5 / 100) × (2 / 100) × 3

Multiplying the terms:

(5 × 2 × 3) / (100 × 100) = 30 / 10000 = 3 / 1000 = 0.003

284Quantitative Aptitude
The following table shows admitted students in an institute during the academic years 2015 to 2020 in different courses. Academic year | Engineering | Science | Arts | Management 2015 | 1200 | 850 | 450 | 240 2016 | 1362 | 821 | 525 | 275 2017 | 1351 | 870 | 624 | 286 2018 | 1409 | 921 | 615 | 293 2019 | 1421 | 931 | 795 | 300 2020 | 1400 | 950 | 750 | 320 Which of the following course had either a consistent growth or a consistent decline in the admissions from 2015 to 2020?
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Answer: (A) Management

Explanation

To determine which course had consistent growth or decline, we analyze the year-over-year admission data. For the Management course, the admissions from 2015 to 2020 are 240, 275, 286, 293, 300, and 320. This sequence shows a continuous increase in student enrollment every single year, as represented by the relation 240 < 275 < 286 < 293 < 300 < 320. This continuous upward trend represents a consistent growth in admissions over the specified academic years.

285Quantitative Aptitude
Solve (2× 2 + 2)/(2 + 2× 2)÷ (½ + ½of½)/(½ + ½of½)
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Answer: (C) 1⅛

Explanation

To solve the expression, we apply the BODMAS rule (Brackets, Of, Division, Multiplication, Addition, Subtraction).

First, simplify the first term: (2 × 2 + 2) / (2 × 2 - 2) = (4 + 2) / (4 - 2) = 6 / 2 = 3

Next, simplify the second term: In the numerator: 1/2 ÷ 1/2 of 1/2 Since "of" takes precedence over division: 1/2 of 1/2 = 1/4 1/2 ÷ 1/4 = 1/2 × 4 = 2

In the denominator: 1/2 + 1/2 of 1/2 Since "of" takes precedence over addition: 1/2 of 1/2 = 1/4 1/2 + 1/4 = 3/4

Combining the numerator and denominator of the second term: 2 / (3/4) = 2 × 4/3 = 8/3

Now, divide the first term by the simplified second term: 3 ÷ 8/3 = 3 × 3/8 = 9/8 = 1 ⅛

286Quantitative Aptitude
Which of the following statements is INCORRECT about the circle?
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Answer: (C) The length of an arc = (central angle made by the arc)/(360°)× π r²

Explanation

The length of an arc of a circle with radius r and central angle θ is given by the formula:

Arc length = (θ / 360°) × 2πr

The formula given in option C, (θ / 360°) × πr², represents the area of a sector of a circle rather than its arc length. This makes the statement mathematically incorrect.

287Quantitative Aptitude
The distance between the centres of two circles is 81 cm, and their radii are 42 cm and 51 cm. What is the length (in cm) of the direct common tangent to the circles?
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Answer: (A) 80.5

Explanation

The formula for the length of the direct common tangent (L) to two circles is: L = √(d² - (r₁ - r₂)²)

In this formula, d is the distance between the centres of the two circles, r₁ is the radius of the larger circle, and r₂ is the radius of the smaller circle.

Substituting the given values: d = 81 cm r₁ = 51 cm r₂ = 42 cm

We get: L = √(81² - (51 - 42)²) L = √(81² - 9²)

Using the algebraic identity a² - b² = (a - b)(a + b): L = √((81 - 9) × (81 + 9)) L = √(72 × 90) L = √6480

Calculating the square root: 80.5² = 6480.25 √6480 ≈ 80.5 cm

288Quantitative Aptitude
The value of a machine depreciates every year by 5% . If the present value of the machine is £80,000, what will be its value after 2 years?
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Answer: (A) £72,200

Explanation

The depreciated value of an asset after n years, where the present value is P and the depreciation rate is r, is given by the formula:

V = P × (1 - r)ⁿ

Substituting the given values into the formula: P = 80000 r = 5% = 0.05 n = 2

V = 80000 × (1 - 0.05)² V = 80000 × 0.95² V = 80000 × 0.9025 V = 72200

289Quantitative Aptitude
The average weight of 35 students in a class is 40 kgs. Five of the students of this class whose average weight is 42 kgs left the class. Find the average weight of the remaining 30 students.
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Answer: (A) 39.67 kgs

Explanation

The total weight of the initial 35 students is calculated by multiplying the number of students by their average weight: 35 × 40 = 1400 kgs

The total weight of the 5 students who left the class is: 5 × 42 = 210 kgs

Subtracting the weight of the departed students gives the total weight of the remaining 30 students: 1400 - 210 = 1190 kgs

The average weight of the remaining 30 students is: 1190 / 30 = 39.67 kgs

290Quantitative Aptitude
Two triangles ABC and DEF are similar. The smallest side of ABC is equal to 15 units. If the sides of ABC are in the ratio 3:4:5 , and the area of DEF is half of the area of ABC, then what is the largest side of DEF (in units)?
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Answer: (A) 25/√2

Explanation

The sides of triangle ABC are in the ratio 3:4:5. Let the sides of triangle ABC be 3x, 4x, and 5x.

Given that the smallest side of triangle ABC is 15 units: 3x = 15 x = 5

The largest side of triangle ABC is: 5x = 5 × 5 = 25 units

For similar triangles, the ratio of their areas is equal to the square of the ratio of their corresponding sides. Let the largest side of triangle DEF be L. Area(DEF) / Area(ABC) = (L / Largest side of ABC)²

Since the area of DEF is half of the area of ABC: ½ = (L / 25)²

Taking the square root on both sides: 1 / √2 = L / 25

Solving for L: L = 25 / √2

291Quantitative Aptitude
Find the cost of painting a ball that is in the shape of a sphere with a radius of 14 cm. The painting cost of the ball is 5 per square centimetre (take π = ²²⁄₇ ).
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Answer: (A) Rs.12,320

Explanation

The surface area of a sphere is given by the formula A = 4πr², where r is the radius of the sphere.

Substituting r = 14 cm and π = 22/7: Surface Area = 4 × (22/7) × 14 × 14 Surface Area = 4 × 22 × 2 × 14 Surface Area = 2464 cm²

The total cost of painting at the rate of ₹5 per square centimetre is calculated as: Total Cost = Surface Area × Rate Total Cost = 2464 × 5 = 12320

292Quantitative Aptitude
Two cubes, each of edge 5 cm, are joined face to face. Find the surface area of the cuboid thus formed.
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Answer: (A) 250 sq cm

Explanation

When two cubes of edge a = 5 cm are joined face to face, they form a cuboid with the following dimensions: Length (l) = 5 + 5 = 10 cm Breadth (b) = 5 cm Height (h) = 5 cm

The total surface area of a cuboid is given by the formula: Total Surface Area = 2(lb + bh + hl)

Substituting the values of l, b, and h into the formula: Total Surface Area = 2(10 × 5 + 5 × 5 + 5 × 10) Total Surface Area = 2(50 + 25 + 50) Total Surface Area = 2(125) Total Surface Area = 250 sq cm

293Quantitative Aptitude
If A: B = 2: 3 and B: C = 7: 8 , find A: C .
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Answer: (A) 7:12

Explanation

To find the ratio A:C, multiply the ratio of A:B by the ratio of B:C:

A/C = (A/B) × (B/C)

Substituting the given values:

A/C = 2/3 × 7/8

A/C = 14/24

Simplifying the fraction by dividing both the numerator and denominator by 2 yields 7/12.

294Quantitative Aptitude
Rohan can alone finish a work in 12 days and Hari can alone in 16 days. They both work for 3 days and then Rohan left. After that only Hari works and finishes the work. The total work will be completed in:
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Answer: (C) 12 days

Explanation

To find the total time to complete the work, we can use the rate of work method. Let the total work be the Least Common Multiple of the individual days taken by Rohan and Hari.

LCM of 12 and 16 is 48 units.

Rohan's efficiency = 48 ÷ 12 = 4 units per day Hari's efficiency = 48 ÷ 16 = 3 units per day

Combined efficiency of Rohan and Hari = 4 + 3 = 7 units per day

They work together for 3 days, completing: 7 units/day × 3 days = 21 units

Remaining work to be done after Rohan leaves: 48 - 21 = 27 units

Hari completes this remaining work alone: Time taken by Hari = 27 units ÷ 3 units/day = 9 days

Total time to complete the entire work is the sum of the days they worked together and the days Hari worked alone: 3 days + 9 days = 12 days

295Quantitative Aptitude
In a 2 km linear race, if P completes the race in 200 second and Q in 220 seconds, then the distance by which P beats Q is:
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Answer: (A) 181⁹⁄₁₁ metres

Explanation

In a 2 km race, the total distance is 2000 metres. P completes the race in 200 seconds, which means P reaches the finish line at t = 200 seconds.

Q takes 220 seconds to complete the same 2000-metre race. The speed of Q is calculated as: Speed of Q = Distance / Time = 2000 / 220 = 100/11 metres per second

When P finishes the race at 200 seconds, Q has been running for 200 seconds and still has 20 seconds remaining to finish the race (220 - 200 = 20 seconds). The distance by which P beats Q is the distance Q travels in this remaining 20 seconds.

Distance = Speed of Q × Remaining time Distance = (100/11) × 20 = 2000/11 metres

Converting the improper fraction 2000/11 into a mixed fraction: 2000 ÷ 11 = 181 with a remainder of 9, which is 181 9/11 metres.

296Quantitative Aptitude
A natural number n divides 732 and leaves 12 as a remainder. How many values of n are possible?
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Answer: (A) 20

Explanation

By the division algorithm, if a natural number n divides 732 and leaves a remainder of 12, then: 732 = q × n + 12 where q is an integer and the divisor n must be strictly greater than the remainder, so n > 12.

This simplifies to: q × n = 720

Therefore, n must be a factor of 720 that is strictly greater than 12.

To find the total number of factors of 720, we first determine its prime factorization: 720 = 2⁴ × 3² × 5¹

The total number of factors of 720 is given by the formula: (4 + 1) × (2 + 1) × (1 + 1) = 5 × 3 × 2 = 30

Now, we count the factors of 720 that are less than or equal to 12: 1, 2, 3, 4, 5, 6, 8, 9, 10, and 12.

There are 10 such factors.

To find the number of possible values of n where n > 12, we subtract these 10 factors from the total number of factors: 30 - 10 = 20

297Quantitative Aptitude
If cotα = (P)/(Q^′) then find the value of (Pcosα - Qsinα)/(Pcosα + Qsinα) = (P² - Q²)/(P² + Q²) +3.
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Answer: (A) 0

Explanation

Given that cot α = P/Q, we can simplify the trigonometric expression by dividing both the numerator and the denominator by sin α:

(P cos α - Q sin α) / (P cos α + Q sin α) = (P cot α - Q) / (P cot α + Q)

Substituting cot α = P/Q into this equation:

(P(P/Q) - Q) / (P(P/Q) + Q) = (P²/Q - Q) / (P²/Q + Q)

Multiplying the numerator and the denominator by Q:

(P² - Q²) / (P² + Q²)

Therefore, the difference between the two terms is:

(P cos α - Q sin α) / (P cos α + Q sin α) - (P² - Q²) / (P² + Q²) = 0

298Quantitative Aptitude
A trader buys 500 kg cotton for ₹8,000. If 10% of this cotton is spoiled due to rain, at what rate (₹/kg) should he sell the rest to earn a 20% profit?
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Answer: (A) 21⅓

Explanation

Cost Price (CP) = ₹8,000 To earn a profit of 20%, the required total Selling Price (SP) is: SP = 8,000 × (1 + 0.20) = ₹9,600

The total quantity of cotton is 500 kg. Since 10% is spoiled, the quantity of usable cotton left is: 500 × (1 - 0.10) = 450 kg

The selling rate per kg required to achieve the total SP is: Rate = Total SP ÷ Usable cotton Rate = 9,600 ÷ 450 = 64/3 = 21⅓ ₹/kg

299Quantitative Aptitude
The hypotenuse of a right triangle is 6m more than twice the shortest side. If the third side is 2 m less than the hypotenuse, find the area (in m²) of the triangle.
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Answer: (C) 120

Explanation

Let s be the length of the shortest side of the right triangle.

The hypotenuse h is given as: h = 2s + 6

The third side t is 2 m less than the hypotenuse: t = h - 2 = 2s + 4

By the Pythagorean theorem: s² + t² = h²

Substitute the expressions for t and h: s² + (2s + 4)² = (2s + 6)² s² + 4s² + 16s + 16 = 4s² + 24s + 36 s² - 8s - 20 = 0

Factoring the quadratic equation: (s - 10)(s + 2) = 0

Since a length must be positive, s = 10 m.

The two perpendicular sides are: Shortest side = 10 m Third side = 2(10) + 4 = 24 m

The area of a right-angled triangle is: Area = ½ × base × height Area = ½ × 10 × 24 = 120 m²

300Quantitative Aptitude
The table shows the percentage distribution of the population (only male and female) according to Gender and Literacy. State | Percentage literate | Gender Literate | Illiterate --- | --- | --- | --- | | | M : F | M : F A | 55 | 5 : 6 | 4 : 5 B | 72 | 1 : 1 | 3 : 2 C | 78 | 4 : 3 | 3 : 2 The total population in state C, if the illiterate female count in state C is 4,40,000, is:
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Answer: (B) 5000000

Explanation

Correct Answer: B

In State C, the percentage of literate population is 78%. This means the percentage of illiterate population is 100% - 78% = 22%.

The ratio of illiterate males to illiterate females in State C is 3 : 2. Thus, the illiterate female population is 2 / (3 + 2) = 2/5 of the total illiterate population.

Let the total population of State C be P. The illiterate female population can be written as: (2/5) × 22% of P = 4,40,000

Substitute 22% as 0.22: (2/5) × 0.22 × P = 4,40,000 0.088 × P = 4,40,000 P = 4,40,000 / 0.088 P = 5,000,000

Answer key

Tap a number to jump to that question.

  1. 251A
  2. 252C
  3. 253B
  4. 254D
  5. 255D
  6. 256A
  7. 257A
  8. 258B
  9. 259B
  10. 260B
  11. 261B
  12. 262A
  13. 263C
  14. 264D
  15. 265C
  16. 266A
  17. 267A
  18. 268D
  19. 269B
  20. 270B
  21. 271B
  22. 272D
  23. 273C
  24. 274B
  25. 275D
  26. 276B
  27. 277D
  28. 278A
  29. 279A
  30. 280A
  31. 281A
  32. 282A
  33. 283A
  34. 284A
  35. 285C
  36. 286C
  37. 287A
  38. 288A
  39. 289A
  40. 290A
  41. 291A
  42. 292A
  43. 293A
  44. 294C
  45. 295A
  46. 296A
  47. 297A
  48. 298A
  49. 299C
  50. 300B

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