SSC CHSL2012Quantitative AptitudePage 3 of 4

SSC CHSL Tier 1 2012 Quantitative Aptitude Questions with Solutions

Questions from every shift of the year, grouped by section. Questions 101–150 of 197, each with the correct answer and a step-by-step explanation.

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101Quantitative Aptitude
[Stem almost entirely lost — only a fragment survives: '...It will pass a tunnel of 250 metres long in:' The train's speed/length setup that would normally precede this is missing, merged/overwritten by question 114's text.]
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Answer: (C) 32 seconds

Explanation

Using the formula Speed = Distance / Time, we find the speed of the train when it passes the tree. Since the tree has negligible length, the distance covered is equal to the length of the train (150 meters):

Speed = 150 / 12 = 12.5 meters per second

To pass a 250-meter-long tunnel, the train must cover a total distance equal to the sum of its own length and the tunnel's length:

Total Distance = 150 + 250 = 400 meters

Using the formula Time = Distance / Speed, the time taken to cross the tunnel is:

Time = 400 / 12.5 = 32 seconds

102Quantitative Aptitude
The compound interest on ₹1800 at 10% per annum, for a certain period of time is ₹378. Find the time in years.
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Answer: (C) 2 years

Explanation

The formula for the total amount in compound interest is: A = P × (1 + R/100)ᵗ

Given the Principal P = ₹1800, Rate R = 10%, and Compound Interest CI = ₹378. The total Amount A is: A = P + CI = 1800 + 378 = 2178

Substituting these values into the formula: 2178 = 1800 × (1 + 10/100)ᵗ 2178 / 1800 = (11/10)ᵗ 1.21 = (1.1)ᵗ

Since 1.21 is the square of 1.1: (1.1)² = (1.1)ᵗ

Comparing the exponents, we get t = 2.

103Quantitative Aptitude
In how many years shall ₹2500, invested at the rate of 8% simple interest per annum, amount to ₹3300?
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Answer: (B) 4

Explanation

The simple interest SI obtained from a principal P invested at an annual rate of interest R for T years is given by the formula SI = (P × R × T) / 100.

The simple interest earned is the difference between the final amount and the principal: SI = ₹3300 - ₹2500 = ₹800

Substituting the known values into the simple interest formula: 800 = (2500 × 8 × T) / 100 800 = 200 × T T = 800 / 200 T = 4 years

104Quantitative Aptitude
If A, B and C denote respectively, the number of vertices, edges and faces of a cube, then A + B + C is:
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Answer: (D) 26

Explanation

For a standard three-dimensional cube, the geometric properties are defined by its vertices, edges, and faces.

A cube has 8 vertices, which represents the value of A. It has 12 edges, which represents the value of B. It has 6 faces, which represents the value of C.

To find the sum of these properties, we add them together:

A + B + C = 8 + 12 + 6 = 26

105Quantitative Aptitude
A horse is tied with a rope of length 7 m at one corner of a square field having side equal to 10 m. The minimum possible area of the square field that is left ungrazed is: [given π = 22/7]
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Answer: (B) 61.5 m²

Explanation

The total area of the square field is calculated by squaring the length of its side: Area of square = 10² = 100 m²

Since the horse is tied at one corner of the square, the area it can graze forms a quadrant of a circle with a radius equal to the length of the rope (r = 7 m). The angle at the corner of a square is 90°.

Area grazed = (1/4) × π × r² Area grazed = (1/4) × (22/7) × 7 × 7 Area grazed = (1/4) × 22 × 7 Area grazed = 154 / 4 = 38.5 m²

The area of the square field left ungrazed is the difference between the total area of the field and the grazed area: Area left ungrazed = 100 - 38.5 = 61.5 m²

106Quantitative Aptitude
ABCD is a rectangle with AB = h² and AD = 3p. If h is doubled and p is halved, then the:
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Answer: (B) area is doubled

Explanation

Area of a rectangle is given by the formula: Area = length × width

Initially, for rectangle ABCD, the area (A₁) is: A₁ = AB × AD = h² × 3p = 3ph²

When h is doubled, the new value of h is 2h, making the new length: AB' = (2h)² = 4h²

When p is halved, the new value of p is p/2, making the new width: AD' = 3(p/2) = 1.5p

The new area (A₂) is: A₂ = AB' × AD' = 4h² × 3(p/2) = 6ph²

Comparing the new area with the initial area: A₂ = 2 × (3ph²) = 2A₁

This shows that the area is doubled.

107Quantitative Aptitude
The number of bricks required to build a wall 8 m long, 6 m high and 22.5 cm thick, if each brick measures 25cm×11.25cm×6cm, is:
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Answer: (D) 6400

Explanation

The volume of a rectangular cuboid is given by the formula: Volume = length × width × height

To find the number of bricks required, all dimensions must be converted to the same unit. Converting the dimensions of the wall from meters to centimeters: Length of the wall = 8 m = 800 cm Height of the wall = 6 m = 600 cm Thickness of the wall = 22.5 cm

The volume of the wall is: 800 cm × 600 cm × 22.5 cm = 10,800,000 cm³

The dimensions of one brick are given as 25 cm × 11.25 cm × 6 cm. The volume of one brick is: 25 cm × 11.25 cm × 6 cm = 1687.5 cm³

The number of bricks required is calculated by dividing the total volume of the wall by the volume of a single brick: Number of bricks = Volume of the wall ÷ Volume of one brick Number of bricks = 10,800,000 ÷ 1687.5 = 6400

Alternatively, this can be solved directly by simplifying the expression: (800 × 600 × 22.5) ÷ (25 × 11.25 × 6) Dividing 800 by 25 equals 32. Dividing 600 by 6 equals 100. Dividing 22.5 by 11.25 equals 2. Multiplying these quotients together gives: 32 × 100 × 2 = 6400

108Quantitative Aptitude
If x + 1/x = 5, then the value of 2x/(3x² - 5x + 3) is:
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Answer: (C) 1/5

Explanation

To find the value of the given expression, divide both the numerator and the denominator by x:

2x / (3x² - 5x + 3) = 2 / (3x - 5 + 3/x)

Rearranging the terms in the denominator gives:

2 / (3(x + 1/x) - 5)

Substitute the given value x + 1/x = 5 into this simplified expression:

2 / (3(5) - 5) = 2 / (15 - 5) = 2/10 = 1/5

109Quantitative Aptitude
If p = 99, then value of p(p² + 3p + 3) is:
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Answer: (D) 999999

Explanation

By expanding the given expression: p(p² + 3p + 3) = p³ + 3p² + 3p

Using the algebraic identity for the cube of a binomial: (p + 1)³ = p³ + 3p² + 3p + 1

Subtracting 1 from both sides gives: (p + 1)³ - 1 = p³ + 3p² + 3p

Substituting p = 99 into this equivalent expression: (99 + 1)³ - 1 = 100³ - 1

Since 100³ is 1,000,000, the final calculation is: 1,000,000 - 1 = 999,999

110Quantitative Aptitude
When a = 4/3, the value of 27a³ - 108a² + 144a - 31 is:
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Answer: (B) -253

Explanation

The algebraic identity for the cube of a binomial is: (3a - 4)³ = 27a³ - 108a² + 144a - 64

The expression can be rewritten by isolating this perfect cube: 27a³ - 108a² + 144a - 317 = (3a - 4)³ - 253

Substituting a = 4/3 into the equation: 3a - 4 = 3(4/3) - 4 = 4 - 4 = 0

Therefore, the value of the expression is: 0³ - 253 = -253

111Quantitative Aptitude
If a=1, b=2 and c=-3, then the value of (a³+b³+c³-3abc) / (ab+bc+ca-(a²+b²+c²)) is:
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Answer: (C) 0

Explanation

The algebraic identity for the sum of cubes is: a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca)

The denominator of the given expression can be rewritten by factoring out a negative sign: ab + bc + ca - (a² + b² + c²) = -(a² + b² + c² - ab - bc - ca)

Thus, the given expression simplifies to: (a + b + c)(a² + b² + c² - ab - bc - ca) / -(a² + b² + c² - ab - bc - ca) = -(a + b + c)

Substituting the values a = 1, b = 2, and c = -3: -(1 + 2 + (-3)) = 0

112Quantitative Aptitude
D and E be two points on the sides AB and AC of the ΔABC such that DE ∥ BC and AD/BD = 2/3. Find the area of trapezium DECB : area of ΔABC.
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Answer: (D) 21:25

Explanation

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. Since DE is parallel to BC, ΔADE is similar to ΔABC.

We are given: AD/BD = 2/3

Since AB = AD + BD, the ratio of the side AD to AB is: AD/AB = AD / (AD + BD) = 2 / (2 + 3) = 2/5

Using the property of similar triangles, the ratio of the area of ΔADE to the area of ΔABC is: Area(ΔADE) / Area(ΔABC) = (AD/AB)² = (2/5)² = 4/25

If Area(ΔABC) is represented as 25 units, then Area(ΔADE) is 4 units.

The area of the trapezium DECB is the difference between the areas of the two triangles: Area(trapezium DECB) = Area(ΔABC) - Area(ΔADE) = 25 - 4 = 21 units

The ratio of the area of trapezium DECB to the area of ΔABC is: Area(trapezium DECB) : Area(ΔABC) = 21 : 25

113Quantitative Aptitude
Three medians AD, BE and CF of ΔABC intersect at G. Area of ΔABC is 108 sq.cm. The area of ΔAGB is:
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Answer: (B) 36cm²

Explanation

The centroid of a triangle is the point of intersection of its three medians. A fundamental property of the centroid is that it divides the triangle into three smaller triangles of equal area: ΔAGB, ΔBGC, and ΔCGA.

Therefore, the area of ΔAGB is exactly one-third of the total area of ΔABC.

Area(ΔAGB) = Area(ΔABC) ÷ 3

Substituting the given value of Area(ΔABC) = 108 cm²:

Area(ΔAGB) = 108 ÷ 3 = 36 cm²

114Quantitative Aptitude
In a triangle, distances from centroid to vertices are respectively 4cm, 6cm and 8cm. Find medians.
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Answer: (C) 6cm,9cm,12cm

Explanation

The centroid of a triangle divides each median in the ratio 2:1, measuring from the vertex to the midpoint of the opposite side. This means that the distance from a vertex to the centroid is two-thirds of the total length of that median.

Therefore, the length of each median is 3/2 times the distance from the corresponding vertex to the centroid.

Calculating the lengths of the medians: First median = (3/2) × 4 = 6 cm Second median = (3/2) × 6 = 9 cm Third median = (3/2) × 8 = 12 cm

115Quantitative Aptitude
A spherical balloon whose radius is r, subtends an angle α at the eye of an observer on the ground, when the angle of elevation of its centre is β. The height of its centre is:
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Answer: (B) r cosec(α/2) sin β

Explanation

Picture the balloon's center O at height h above the observer, with the line to the center making angle β with the ground. The balloon of radius r subtends angle α at the observer's eye, so the half-angle α/2 satisfies sin(α/2) = r divided by the distance from the observer to the center, giving that distance as r cosec(α/2). The height of the center is then this distance times sin β, so h = r cosec(α/2) sin β.

116Quantitative Aptitude
If sec θ + cosec(90° - θ) = 4, (0 < θ < 90°), then the value of tan θ is:
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Answer: (A) √3

Explanation

By using the trigonometric identity cosec(90° - θ) = sec θ, the given equation can be simplified as:

sec θ + sec θ = 4 2 sec θ = 4 sec θ = 2

Since θ is an acute angle (0 < θ < 90°), sec θ = 2 corresponds to cos θ = ½, which gives θ = 60°.

Substituting θ = 60° into tan θ: tan 60° = √3

117Quantitative Aptitude
The numerical value of 9/cosec²θ + 4cos²θ + 5/(1+tan²θ) is:
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Answer: (C) 9

Explanation

Using standard trigonometric identities, we can simplify the terms of the expression.

First, we know that 1/cosec²θ = sin²θ. Therefore: 9/cosec²θ = 9sin²θ

Second, using the fundamental identity 1 + tan²θ = sec²θ, we have: 5/(1+tan²θ) = 5/sec²θ

Since 1/sec²θ = cos²θ, this simplifies to: 5cos²θ

Substituting these back into the original expression gives: 9sin²θ + 4cos²θ + 5cos²θ

Combining the cos²θ terms yields: 9sin²θ + 9cos²θ

Factoring out 9, we get: 9(sin²θ + cos²θ)

Using the Pythagorean identity sin²θ + cos²θ = 1: 9(1) = 9

118Quantitative Aptitude
The solution set to the equation |x + 2| = √4 will be
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Answer: (C) 0,-4

Explanation

The given equation is |x + 2| = √4.

First, simplify the right side of the equation. The principal square root of 4 is 2, so the equation becomes: |x + 2| = 2

By definition of the absolute value, this equation can be split into two linear equations: x + 2 = 2 or x + 2 = -2

Solving the first equation: x = 2 - 2 x = 0

Solving the second equation: x = -2 - 2 x = -4

The solution set consists of 0 and -4.

119Quantitative Aptitude
A number is added with 5/7 and the sum is 1. The number is
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Answer: (B) 40% of 5/7

Explanation

Let the required number be x.

x + 5/7 = 1 x = 1 - 5/7 x = 2/7

To find what percentage of 5/7 this number is, we divide 2/7 by 5/7 and multiply by 100: Percentage = (2/7) ÷ (5/7) × 100% Percentage = 2/5 × 100% = 40%

120Quantitative Aptitude
A 250 m long train is running at a speed of 55 km/hr. It crossed a platform of length 520 m in
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Answer: (C) 50.4 sec

Explanation

To find the time taken by the train to cross the platform, the total distance to be covered is the sum of the length of the train and the length of the platform.

Total distance = Length of train + Length of platform Total distance = 250 + 520 = 770 m

Next, convert the speed of the train from km/hr to m/s by multiplying with 5/18: Speed = 55 × 5/18 = 275/18 m/s

Using the formula: Time = Distance ÷ Speed Time = 770 ÷ (275/18) Time = (770 × 18) ÷ 275 Time = (70 × 18) ÷ 25 Time = 1260 ÷ 25 = 50.4 seconds

121Quantitative Aptitude
If 12 men or 18 women can reap a field in 7 days, then in what time can 4 men and 8 women reap the same field?
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Answer: (D) 9 days

Explanation

The work done by 12 men is equal to the work done by 18 women. This gives the efficiency ratio: 12 men = 18 women 1 man = 1.5 women

We need to find the time taken by 4 men and 8 women to complete the same work. Converting men to women: 4 men + 8 women = (4 × 1.5) women + 8 women = 6 women + 8 women = 14 women

Using the work-time formula, M₁ × D₁ = M₂ × D₂, where M is the number of workers and D is the number of days: 18 women × 7 days = 14 women × D days 126 = 14 × D D = 126 / 14 = 9 days

122Quantitative Aptitude
A can build a wall in 30 days, while B alone can build it in 40 days. If they build it together and get a payment of ₹7,000, then the share of B is
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Answer: (A) ₹3,000

Explanation

The share of payment for work done together is distributed in the ratio of the individual efficiencies (or rates of work) of the workers.

Since efficiency is inversely proportional to the time taken to complete the work individually:

Efficiency of A : Efficiency of B = 1/30 : 1/40 = 4 : 3

The total payment of ₹7,000 is divided in the ratio 4 : 3.

Share of B = (3 / (4 + 3)) × ₹7,000 = 3/7 × ₹7,000 = ₹3,000

123Quantitative Aptitude
An edge of a cube is increased by 50%. Then percentage increase in total surface area is
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Answer: (D) 125%

Explanation

The total surface area of a cube is proportional to the square of its edge length, as represented by the formula A = 6a² where a is the edge length.

Since the surface area is a two-dimensional quantity, an increase of 50% in the edge length results in a successive percentage increase in the surface area. Using the successive percentage change formula:

Net percentage increase = x + y + (x × y) / 100

Substituting x = 50 and y = 50:

Net percentage increase = 50 + 50 + (50 × 50) / 100 = 100 + 25 = 125%

124Quantitative Aptitude
The area of the base of a rectangular tank is 6500 sq. cm and the volume of water contained in it is 2.6 cubic meters. The depth of water is
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Answer: (B) 4 m

Explanation

The volume of a rectangular tank is given by the formula:

Volume = Base Area × Depth

First, convert the area of the base from square centimeters to square meters. Since 1 square meter is equal to 10000 square centimeters, we have:

Base Area = 6500 ÷ 10000 = 0.65 m²

Now, substitute the volume and the base area into the formula to find the depth:

2.6 = 0.65 × Depth

Depth = 2.6 ÷ 0.65 = 4 m

125Quantitative Aptitude
The true discount on ₹1,500 for 3 years is same as the simple interest on ₹1,500 for 3 years at the same rate of interest. The rate of interest is
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Answer: (A) 8%

Explanation

The true discount on ₹1,860 for 3 years is equal to the simple interest on ₹1,500 for 3 years at the same rate. Let the rate of interest be R% per annum.

The formula for Simple Interest (SI) is: SI = (Principal × Rate × Time) / 100 SI = (1500 × R × 3) / 100 = 45R

The formula for True Discount (TD) is: TD = (Amount × Rate × Time) / (100 + (Rate × Time)) TD = (1860 × R × 3) / (100 + 3R) = 5580R / (100 + 3R)

Since the true discount is equal to the simple interest: 45R = 5580R / (100 + 3R)

Dividing both sides by R (since R ≠ 0): 45 = 5580 / (100 + 3R) 45 × (100 + 3R) = 5580 4500 + 135R = 5580 135R = 1080 R = 1080 / 135 R = 8%

126Quantitative Aptitude
The cash down value of an article is ₹12,500. It can be bought at credit for ₹14,300 payable 1 year hence at 10% per annum. The trader gets a profit of ___ in this transaction.
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Answer: (C) ₹500

Explanation

The present value of the credit payment due 1 year hence is calculated by discounting the amount at the given interest rate of 10% per annum.

Present Value = Amount ÷ (1 + R/100)

Substituting the given values: Present Value = 14300 ÷ (1 + 10/100) = 14300 ÷ 1.1 = 13000

The cash equivalent of the credit price is ₹13,000. Since the cash down value of the article is ₹12,500, the profit gained by the trader in terms of present value is:

Profit = Present Value - Cash Down Value Profit = 13000 - 12500 = 500

127Quantitative Aptitude
The marked price of a book is ₹165 and it is sold for ₹132. The percentage discount allowed is
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Answer: (C) 20%

Explanation

The marked price of the book is ₹165 and the selling price is ₹132.

The discount is calculated as: Discount = Marked Price - Selling Price Discount = 165 - 132 = ₹33

The percentage discount allowed is given by the formula: Discount % = (Discount / Marked Price) × 100 Discount % = (33 / 165) × 100 Discount % = 1/5 × 100 = 20%

128Quantitative Aptitude
In a mixture of 60 litres, the ratio of milk and water is 2:1. If this ratio is to be 1:2, then the quantity of water to be further added is
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Answer: (D) 60 litres

Explanation

Let the initial ratio of milk and water be M : W = 2 : 1. The total mixture is 60 litres, which represents 2 + 1 = 3 units. This means 1 unit is equal to 20 litres.

The initial quantity of milk is 2 × 20 = 40 litres, and the initial quantity of water is 1 × 20 = 20 litres.

Since only water is added to the mixture, the quantity of milk remains unchanged at 40 litres.

The new ratio of milk to water is 1 : 2.

If 1 unit of milk in the new mixture corresponds to 40 litres, then 2 units of water in the new mixture must correspond to 2 × 40 = 80 litres.

The quantity of water to be further added is the difference between the new and initial quantities of water: 80 litres − 20 litres = 60 litres.

129Quantitative Aptitude
Simplify: 2/3 : 3/4 : 1/6
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Answer: (A) 28:39:14

Explanation

The ratio consists of the mixed fractions: 2 ⅓ : 3 ¼ : 1 ⅙

Converting these mixed fractions into improper fractions: 2 ⅓ = 7/3 3 ¼ = 13/4 1 ⅙ = 7/6

The ratio becomes: 7/3 : 13/4 : 7/6

To simplify and eliminate the fractional denominators, multiply each term by the Least Common Multiple (LCM) of the denominators 3, 4, and 6, which is 12: (7/3) × 12 = 28 (13/4) × 12 = 39 (7/6) × 12 = 14

This yields the simplified ratio 28:39:14.

130Quantitative Aptitude
The number of men, women and boys employed in a factory are in the ratio 5:7:10. If the total number of workers is 660, find the number of men, women and boys.
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Answer: (B) 150,210,300

Explanation

Let the number of men, women, and boys employed in the factory be 5x, 7x, and 10x respectively.

The total number of workers is given as 660: 5x + 7x + 10x = 660 22x = 660 x = 30

Substituting the value of x: Number of men = 5 × 30 = 150 Number of women = 7 × 30 = 210 Number of boys = 10 × 30 = 300

131Quantitative Aptitude
The average expenditure of a man for the first 5 months is ₹12,000 and for the next seven months is ₹13,000. Find his average monthly income if he saves ₹29,000 during the year.
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Answer: (B) ₹15,000

Explanation

Average Monthly Income is calculated using the relation between total income, total expenditure, and total savings over the twelve months of the year.

Total Annual Expenditure = (5 × 12,000) + (7 × 13,000) = 60,000 + 91,000 = ₹1,51,000

Total Annual Income = Total Annual Expenditure + Total Savings = 1,51,000 + 29,000 = ₹1,80,000

Average Monthly Income = Total Annual Income / 12 = 1,80,000 / 12 = ₹15,000

132Quantitative Aptitude
If a train maintains an average speed of 40 km/hr, it arrives at its destination punctually. If however the average speed is 35 km/hr, it arrives 15 minutes late. The length of the journey is
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Answer: (D) 70 km

Explanation

Let the length of the journey be d km. Using the relation time = distance / speed, the time difference between the two speeds is:

d/35 - d/40 = 15/60

Simplifying the time difference of 15 minutes to ¼ hours:

d/35 - d/40 = ¼

Multiplying the entire equation by 280 (the least common multiple of 35 and 40) to clear the denominators:

8d - 7d = 70

d = 70

133Quantitative Aptitude
Three years ago, the average age of a family of 5 members was 17 years. A baby having been born, the average age of the family is the same today. What is the age of the baby?
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Answer: (B) 2 years

Explanation

Three years ago, the sum of the ages of the 5 family members was 5 × 17 = 85 years.

In the 3 years that have passed, each of the 5 members has grown 3 years older. Therefore, the sum of their ages today is 85 + (5 × 3) = 100 years.

With the addition of the baby, the family now has 6 members. The current average age of the family is the same as before, which is 17 years. The sum of the ages of all 6 members today is 6 × 17 = 102 years.

The age of the baby is the difference between the current sum of all 6 members and the current sum of the original 5 members, which is 102 - 100 = 2 years.

134Quantitative Aptitude
The cost of manufacture of an article is made up of materials, labour and overheads in the ratio 4:3:2. If the cost of labour is ₹45, find the total cost of the article. Find the profit percent, if the article is sold for ₹180.
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Answer: (D) 33 1/3 %

Explanation

The ratio of materials, labour, and overheads is 4:3:2. Let the costs of these three components be 4x, 3x, and 2x respectively.

Since the cost of labour is ₹45: 3x = 45 x = 15

The total cost of manufacturing the article is: 4x + 3x + 2x = 9x 9 × 15 = ₹135

The selling price of the article is ₹180. The profit earned is: Profit = Selling Price - Total Cost = 180 - 135 = ₹45

The profit percentage is: Profit % = (Profit / Total Cost) × 100 = (45 / 135) × 100 = (1/3) × 100 = 33 ⅓%

135Quantitative Aptitude
The difference between selling a shirt at profits 4% and 5% is ₹3. Find the cost price and the two selling prices of the shirt.
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Answer: (D) ₹300, ₹312, ₹315

Explanation

Let the Cost Price of the shirt be x.

The Selling Price at 4% profit is 1.04x. The Selling Price at 5% profit is 1.05x.

The difference between these two selling prices is: 1.05x - 1.04x = 3 0.01x = 3 x = 3 / 0.01 = 300

Thus, the Cost Price of the shirt is ₹300.

The first Selling Price at 4% profit is: 300 × 1.04 = ₹312

The second Selling Price at 5% profit is: 300 × 1.05 = ₹315

136Quantitative Aptitude
A toy is sold at a profit of 16%. Had it been sold for ₹10 more, then the profit would have been 20%. Then its cost price is
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Answer: (B) ₹250

Explanation

Let CP be the cost price of the toy.

The difference in profit percentages is: 20% - 16% = 4%

This 4% increase in the profit margin corresponds directly to the ₹10 increase in the selling price.

Using the relationship between the percentage change and the actual price change: 4% of CP = 10 0.04 × CP = 10 CP = 10 / 0.04 = 250

137Quantitative Aptitude
2/3 rd of 30% of 4 times of a is
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Answer: (B) 4a/5

Explanation

We translate the statement mathematically as: (2/3) × (30/100) × 4a

First, simplify 2/3 of 30%: (2/3) × (30/100) = (2/3) × (3/10) = 2/10 = 1/5

Next, multiply this result by 4a: (1/5) × 4a = 4a/5

138Quantitative Aptitude
65% of a number is 21 less than four-fifth of that number. What is the number?
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Answer: (C) 140

Explanation

Let the required number be x.

According to the given condition, 65% of the number is 21 less than four-fifth of the number.

Four-fifth of the number is written as (4/5)x, which is equivalent to 80% of x or 0.8x.

We can set up the equation: 0.8x - 0.65x = 21

Simplifying the left-hand side: 0.15x = 21

To find x, divide 21 by 0.15: x = 21 ÷ 0.15 x = 2100 ÷ 15 x = 140

139Quantitative Aptitude
A man walking at the rate of 6 km/hr crosses a bridge in 20 minutes. What is the length of the bridge in kilometers?
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Answer: (B) 2

Explanation

The length of the bridge can be calculated directly using the distance formula:

Distance = Speed × Time

First, convert the time from minutes to hours:

Time = 20 minutes = 20/60 hour = 1/3 hour

Now, substitute the speed of 6 km/hr and the time of 1/3 hour into the formula:

Distance = 6 km/hr × 1/3 hour = 2 km

140Quantitative Aptitude
A car is running at a speed of 36 km/hr. The distance covered by it in 32 seconds is
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Answer: (B) 320 meters

Explanation

To find the distance covered in meters, the speed must first be converted from kilometers per hour (km/hr) to meters per second (m/s) using the conversion factor 5/18:

Speed = 36 × (5/18) m/s = 10 m/s

Using the standard distance formula:

Distance = Speed × Time

Substituting the converted speed and time into the formula:

Distance = 10 m/s × 32 s = 320 meters

141Quantitative Aptitude
A sum of money amounts to ₹500 in 2 years and ₹540 in 3 years at compound interest per annum; then the rate percent, per annum, is
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Answer: (C) 8

Explanation

Between year 2 and year 3 the amount grows only by the interest earned in that single year, so the rate equals the increase divided by the amount at the start of that year. Here the amount rises from 500 to 540, an increase of 40, which is 40 out of 500, or 8 percent per annum.

142Quantitative Aptitude
A man had ₹2,000. He lent a part of this at 5% simple interest and the rest at 4%. The total interest he received in one year was ₹92. The amount of money (in rupees) he lent at 5% was
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Answer: (A) 1,200

Explanation

Using the method of allegation:

If the entire amount of ₹2,000 was lent at 5%, the interest earned in one year would be 5% of 2,000 = ₹100.

If the entire amount was lent at 4%, the interest earned would be 4% of 2,000 = ₹80.

The actual interest received is ₹92.

Applying the allegation rule, the ratio of the amount lent at 5% to the amount lent at 4% is: (92 - 80) : (100 - 92) = 12 : 8 = 3 : 2

The amount of money lent at 5% is: (3 / (3 + 2)) × 2,000 = (3/5) × 2,000 = 1,200

143Quantitative Aptitude
A rectangular circuit board is designed to have width w cm, perimeter p cm and area k sq. cm. Of the following, the equation which must be true is
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Answer: (D) 2w² - pw + 2k = 0

Explanation

Let the length of the rectangular circuit board be L cm. The perimeter p of a rectangle is given by the formula: p = 2(w + L)

Rearranging this formula to express L in terms of p and w: w + L = p/2 L = p/2 - w

The area k of a rectangle is given by the formula: k = w × L

Substituting the expression for L into the area formula: k = w × (p/2 - w) k = pw/2 - w²

Multiplying the entire equation by 2 to clear the fraction: 2k = pw - 2w²

Rearranging the terms into standard quadratic form: 2w² - pw + 2k = 0

144Quantitative Aptitude
If the radius of a sphere is increased by 50%, then the percentage increase in its volume is
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Answer: (B) 237.5

Explanation

The volume of a sphere depends on the cube of its radius. If the radius grows by 50 percent, the new radius is 1.5 times the original, so the new volume is 1.5 cubed, or 3.375 times the original volume. That is an increase of 2.375 times, which as a percentage is 237.5 percent.

145Quantitative Aptitude
A cylindrical container of radius 6 cm and height 15 cm is filled with ice-cream. The whole ice-cream has to be distributed to 10 children in equal cones with hemispherical tops. If the height of the conical portion is four times the radius of its base, then the radius of the base of the ice-cream cone is
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Answer: (A) 3 cm

Explanation

Let R and H be the radius and height of the cylindrical container, and r and h be the radius and height of the conical portion of the ice-cream cone.

The volume of the cylindrical container is: V_cyl = π × R² × H

Substituting R = 6 cm and H = 15 cm: V_cyl = π × 6² × 15 = 540π cm³

The total volume of one ice-cream cone, consisting of a conical portion of height h = 4r and a hemispherical top of radius r, is: V_cone = (1/3) × π × r² × h + (2/3) × π × r³

Substituting h = 4r into the equation: V_cone = (1/3) × π × r² × (4r) + (2/3) × π × r³ V_cone = (4/3) × π × r³ + (2/3) × π × r³ V_cone = 2π × r³

The volume of 10 such ice-cream cones is equal to the volume of the cylindrical container: 10 × 2π × r³ = 540π 20π × r³ = 540π r³ = 27 r = 3 cm

146Quantitative Aptitude
A hemispherical bowl of internal radius 15 cm contains a liquid. The liquid is to be filled into cylindrical-shaped bottles of diameter 5 cm and height 6 cm. The number of bottles necessary to empty the bowl is
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Answer: (D) 60

Explanation

The volume of a hemispherical bowl of radius R is given by the formula V_bowl = (2/3) × π × R³. Given the internal radius R = 15 cm: V_bowl = (2/3) × π × 15³ = (2/3) × π × 3375 = 2250π cm³

The volume of a cylindrical bottle of radius r and height h is given by the formula V_bottle = π × r² × h. Given the diameter of the bottle is 5 cm, its radius r = 2.5 cm. The height h = 6 cm: V_bottle = π × 2.5² × 6 = π × 6.25 × 6 = 37.5π cm³

The number of bottles necessary to empty the bowl is obtained by dividing the volume of the hemispherical bowl by the volume of a single cylindrical bottle: Number of bottles = V_bowl / V_bottle = 2250π / 37.5π = 60

147Quantitative Aptitude
The value of (1.7)³ + (3.9)³ - (5.6)³ is
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Answer: (B) -111.384

Explanation

If a + b + c = 0, then the algebraic identity states that a³ + b³ + c³ = 3abc.

Letting a = 1.7, b = 3.9, and c = -5.6, we find: a + b + c = 1.7 + 3.9 + (-5.6) = 5.6 - 5.6 = 0

Therefore, the given expression can be written as: (1.7)³ + (3.9)³ - (5.6)³ = (1.7)³ + (3.9)³ + (-5.6)³ = 3 × (1.7) × (3.9) × (-5.6) = 19.89 × (-5.6) = -111.384

148Quantitative Aptitude
Evaluate √(6 + √(6 + √(6 + …)))
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Answer: (B) 3

Explanation

Let the given infinite nested radical expression be x:

x = √(6 + √(6 + √(6 + …)))

Since the pattern repeats infinitely, the nested term inside the outermost square root is also equal to x:

x = √(6 + x)

Squaring both sides of the equation to eliminate the radical gives:

x² = 6 + x

Rearranging this into a standard quadratic equation:

x² - x - 6 = 0

Factoring the quadratic equation:

(x - 3)(x + 2) = 0

This yields two possible values: x = 3 or x = -2. Since the principal square root is always non-negative, the value of the expression must be positive, which leaves x = 3.

149Quantitative Aptitude
If x = (√3 + √2)/(√3 - √2) then the value of (x + 1/x)
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Answer: (B) 10

Explanation

Given that x = (√3 + √2)/(√3 - √2)

To rationalize the denominator, we multiply the numerator and the denominator by (√3 + √2): x = (√3 + √2)² / ((√3)² - (√2)²) x = (3 + 2 + 2√6) / (3 - 2) x = 5 + 2√6

The reciprocal of x is: 1/x = 1 / (5 + 2√6)

Rationalizing the denominator of 1/x by multiplying the numerator and denominator by (5 - 2√6): 1/x = (5 - 2√6) / (5² - (2√6)²) 1/x = (5 - 2√6) / (25 - 24) 1/x = 5 - 2√6

To find the value of x + 1/x: x + 1/x = (5 + 2√6) + (5 - 2√6) x + 1/x = 10

150Quantitative Aptitude
If (c - 3)³ = c³ + kc + 9 then the value of k is
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Answer: (B) -6

Explanation

Using the algebraic identity (a - b)² = a² - 2ab + b², we expand the expression (c - 3)²:

(c - 3)² = c² - 2(c)(3) + 3² (c - 3)² = c² - 6c + 9

By comparing this expanded form with the given equation:

c² + kc + 9 = c² - 6c + 9

Comparing the coefficients of c on both sides:

k = -6

Answer key

Tap a number to jump to that question.

  1. 101C
  2. 102C
  3. 103B
  4. 104D
  5. 105B
  6. 106B
  7. 107D
  8. 108C
  9. 109D
  10. 110B
  11. 111C
  12. 112D
  13. 113B
  14. 114C
  15. 115B
  16. 116A
  17. 117C
  18. 118C
  19. 119B
  20. 120C
  21. 121D
  22. 122A
  23. 123D
  24. 124B
  25. 125A
  26. 126C
  27. 127C
  28. 128D
  29. 129A
  30. 130B
  31. 131B
  32. 132D
  33. 133B
  34. 134D
  35. 135D
  36. 136B
  37. 137B
  38. 138C
  39. 139B
  40. 140B
  41. 141C
  42. 142A
  43. 143D
  44. 144B
  45. 145A
  46. 146D
  47. 147B
  48. 148B
  49. 149B
  50. 150B

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