SSC CHSL2016Quantitative AptitudePage 6 of 34

SSC CHSL Tier 1 2016 Quantitative Aptitude Questions with Solutions

Questions from every shift of the year, grouped by section. Questions 251–300 of 1,696, each with the correct answer and a step-by-step explanation.

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251Quantitative Aptitude
At what rate of compound interest per annum will a sum of Rs 50000 become Rs 73205 in 2 years?
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Answer: (A) 21 percent

Explanation

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

Substituting the given values, where Amount A = 73205, Principal P = 50000, and Time n = 2 years: 73205 = 50000 × (1 + R/100)²

Dividing both sides by 50000: (1 + R/100)² = 73205 / 50000 (1 + R/100)² = 14641 / 10000

Taking the square root on both sides: 1 + R/100 = √(14641 / 10000) 1 + R/100 = 121 / 100 1 + R/100 = 1.21 R/100 = 0.21 R = 21%

252Quantitative Aptitude
A bag has Rs 30.8 in the form of 1 rupee, 50 paise and 10 paise coins in the ratio of 6:3:2. What is the number of 50 paise coins?
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Answer: (C) 12

Explanation

Let the number of 1 rupee, 50 paise, and 10 paise coins be 6x, 3x, and 2x respectively.

The value of these coins in rupees is: Value of 1 rupee coins = 6x × 1 = 6x rupees Value of 50 paise coins = 3x × 0.5 = 1.5x rupees Value of 10 paise coins = 2x × 0.1 = 0.2x rupees

The total value of all the coins in the bag is: 6x + 1.5x + 0.2x = 30.8 7.7x = 30.8 x = 30.8 ÷ 7.7 = 4

The number of 50 paise coins is 3x: 3 × 4 = 12

253Quantitative Aptitude
Reflection of the point (-3,6) in the x-axis is
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Answer: (C) (-3,-6)

Explanation

The reflection of a point (x, y) in the x-axis is given by the formula (x, -y). Under this reflection, the horizontal distance from the y-axis remains unchanged, while the vertical position is mirrored across the x-axis.

For the point (-3, 6), substituting x = -3 and y = 6 into the formula yields the reflected coordinates (-3, -6).

254Quantitative Aptitude
The Simplified form of (b³x²a⁴z³)*(b⁴x³a³z²)/(a²b⁴z³) is
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Answer: (C) b³x⁵a⁵z²

Explanation

Using the laws of exponents where yᵐ × yⁿ = y⁽m+n) and yᵐ / yⁿ = y⁽m-n), the expression can be simplified by grouping the terms for each variable.

For a: (a⁴ × a³) / a² = a⁽4 + 3 - 2) = a⁵

For b: (b³ × b⁴) / b⁴ = b⁽3 + 4 - 4) = b³

For x: x² × x³ = x⁽2 + 3) = x⁵

For z: (z³ × z²) / z³ = z⁽3 + 2 - 3) = z²

Combining all the simplified terms results in b³x⁵a⁵z².

255Quantitative Aptitude
Two students appeared for an examination. One of them secured 13 marks more than the other and his marks were 76% of the sum of their marks. The marks obtained by them are
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Answer: (A) 19 and 6

Explanation

Let the marks obtained by the two students be x and x - 13.

The sum of their marks is: x + (x - 13) = 2x - 13

According to the given condition, the higher mark is 76% of the sum of their marks: x = 0.76 × (2x - 13) x = 1.52x - 9.88 1.52x - x = 9.88 0.52x = 9.88 x = 9.88 ÷ 0.52 x = 19

The marks of the other student are: 19 - 13 = 6

Thus, the marks obtained by the two students are 19 and 6.

256Quantitative Aptitude
What least number must be added to 1039, so that the sum obtained is completely divisible by 29?
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Answer: (B) 5

Explanation

To find the least number to be added to 1039 so that it is completely divisible by 29, divide 1039 by 29 to find the remainder.

1039 = 29 × 35 + 24

Here, the remainder is 24.

The least number that must be added to the dividend is given by the formula: Required number = Divisor - Remainder

Substituting the values: Required number = 29 - 24 = 5

Alternatively, the next multiple of 29 immediately greater than 1039 is: 29 × 36 = 1044

Subtracting the original number from this multiple gives the required value to be added: 1044 - 1039 = 5

257Quantitative Aptitude
If a merchant offers a discount of 4% on the list price, then she makes a loss of 10%. What % profit or % loss will she make if she sells at a discount of 20% of the list price?
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Answer: (A) 25 percent loss

Explanation

Let the list price of the article be MP and the cost price be CP.

The relationship between the marked price, cost price, selling price, discount, and loss is: SP = MP × (1 - discount percentage) SP = CP × (1 - loss percentage)

When a discount of 4% is offered, the merchant makes a loss of 10%: SP₁ = 0.96 × MP = 0.90 × CP CP = (0.96 / 0.90) × MP = (16/15) × MP

When a discount of 20% is offered, the new selling price SP₂ is: SP₂ = 0.80 × MP = (4/5) × MP

The net loss is the difference between CP and SP₂: CP - SP₂ = (16/15) × MP - (4/5) × MP CP - SP₂ = (16/15 - 12/15) × MP = (4/15) × MP

The loss percentage is given by the formula: Loss% = (Loss / CP) × 100 Loss% = ((4/15) × MP / ((16/15) × MP)) × 100 Loss% = (4 / 16) × 100 = 25%

258Quantitative Aptitude
An angle is smaller than its supplementary by 30°. What is the measure of the angle?
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Answer: (A) 75°

Explanation

Two angles are supplementary if their sum is 180°. Let the measure of the required angle be x. Its supplementary angle is therefore 180° - x.

The given condition states that the angle is smaller than its supplementary angle by 30°: (180° - x) - x = 30°

Simplifying this equation: 180° - 2x = 30° 2x = 150° x = 75°

This gives the measure of the angle as 75°.

259Quantitative Aptitude
A square is inscribed in a circle. If the side of the square is 14 cm, what is the area (in sq.cm) of the circle?
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Answer: (D) 98π

Explanation

Let the side of the square be s = 14 cm.

When a square is inscribed in a circle, the diagonal of the square is equal to the diameter of the circle.

The diagonal of a square is given by the formula: d = s√2

Substituting s = 14: d = 14√2 cm

Thus, the diameter of the circle is 14√2 cm. The radius r of the circle is half of the diameter: r = 7√2 cm

The area of the circle is given by: Area = πr² Area = π × (7√2)² Area = π × 98 Area = 98π sq.cm

260Quantitative Aptitude
A wooden bowl is in shape of a hollow hemisphere of internal radius 7 cm and thickness 1 cm. What is the total surface area (in sq.cm) of the bowl? (Take: π = 22/7)
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Answer: (B) 757.43

Explanation

The internal radius of the hollow hemisphere bowl is r = 7 cm, and the thickness is t = 1 cm. Thus, the external radius is R = r + t = 7 + 1 = 8 cm.

The total surface area of a hollow hemisphere bowl consists of the outer curved surface area, the inner curved surface area, and the area of the circular rim at the top.

Outer curved surface area = 2πR² Inner curved surface area = 2πr² Area of the circular rim = π(R² - r²)

Summing these components gives the formula for the total surface area: Total Surface Area = 2πR² + 2πr² + π(R² - r²) Total Surface Area = π(3R² + r²)

Substituting the given values into the formula: Total Surface Area = (22/7) × (3 × 8² + 7²) Total Surface Area = (22/7) × (3 × 64 + 49) Total Surface Area = (22/7) × (192 + 49) Total Surface Area = (22/7) × 241 Total Surface Area = 5302/7 Total Surface Area ≈ 757.43 sq.cm

261Quantitative Aptitude
Refer the below data table and answer the following Question. Weight (kg) Height (m) Ananya 59 1.63 Anchal 67 1.74 Anchita 66 1.66 Anisha 72 1.68 Who has the least weight to height ratio?
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Answer: (A) Ananya

Explanation

The ratio of weight to height is calculated by dividing weight in kilograms by height in meters. For Ananya, this ratio is 59 divided by 1.63, which equals approximately 36.20. This represents the lowest value among all the individuals listed in the table, as the other calculated ratios are 38.51, 39.76, and 42.86.

262Quantitative Aptitude
The following table shows the number of children in each house of a society. Number of Children Number of Houses 0 5 1 14 2 10 3 3 What is the average number of children per house?
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Answer: (A) 1.34

Explanation

The average number of children per house is calculated using the weighted mean formula:

Average = (Sum of (Number of Children × Number of Houses)) / (Total Number of Houses)

First, calculate the total number of houses: Total Houses = 5 + 14 + 10 + 3 = 32

Next, calculate the total number of children across all houses: Total Children = (0 × 5) + (1 × 14) + (2 × 10) + (3 × 3) = 0 + 14 + 20 + 9 = 43

Now, divide the total number of children by the total number of houses: Average = 43 / 32 = 1.34375

Rounding to two decimal places yields 1.34.

263Quantitative Aptitude
Refer the below data table and answer the following Question. Partners Present % share Anand 15 Basu 15 Chinmay 10 Dhiraj 15 Ejaz 45 If the company has issued six lakh shares between its five partners and if Chinmay offers to sell 15,000 of his shares to Ejaz, then Ejaz will have how many shares?
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Answer: (C) 285000 shares

Explanation

The total number of shares issued by the company is 6,00,000.

Ejaz initially holds a 45% share in the company. The number of shares initially held by Ejaz is calculated as: Initial shares = 45% of 6,00,000 = 0.45 × 6,00,000 = 2,70,000 shares

When Chinmay sells 15,000 of his shares to Ejaz, the new total number of shares held by Ejaz becomes: Total shares = 2,70,000 + 15,000 = 2,85,000 shares

264Quantitative Aptitude
Raj sells a machine for Rs 51 lakhs at a loss. Had he sold it for Rs 60 lakh, his gain would have been 8 times the earlier loss. What is the cost price of the machine?
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Answer: (B) Rs 52 lakhs

Explanation

Let the cost price of the machine be CP.

The loss incurred when the machine is sold for Rs 51 lakhs is given by: Loss = CP - 51

The gain obtained when the machine is sold for Rs 60 lakhs is given by: Gain = 60 - CP

According to the given condition, the gain is 8 times the earlier loss: Gain = 8 × Loss 60 - CP = 8 × (CP - 51) 60 - CP = 8CP - 408 9CP = 468 CP = 52 lakhs

265Quantitative Aptitude
A does 75% of a work in 50 days. He then calls in B and they together finish the remaining work in 10 days. How long B alone would take to complete the whole work?
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Answer: (C) 100 days

Explanation

A completes 75% (or 3/4) of the work in 50 days. A's 1 day's work = (3/4) / 50 = 3/200 of the total work.

The remaining work is 25% (or 1/4) of the total work, which A and B together finish in 10 days. Their combined 1 day's work = (1/4) / 10 = 1/40 of the total work.

To find B's rate of work: B's 1 day's work = Combined 1 day's work - A's 1 day's work B's 1 day's work = 1/40 - 3/200 B's 1 day's work = (5 - 3) / 200 = 2/200 = 1/100 of the total work.

Since B completes 1/100 of the work in one day, B alone would take 100 days to complete the entire work.

266Quantitative Aptitude
Factors of 48x3 - 8x2 - 93x - 45 are
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Answer: (C) (4x + 3)(4x + 3)(3x - 5)

Explanation

We can find the correct factors of the polynomial 48x³ - 8x² - 93x - 45 by expanding the expression (4x + 3)(4x + 3)(3x - 5).

First, square the term (4x + 3): (4x + 3)² = 16x² + 24x + 9

Next, multiply this result by (3x - 5): (16x² + 24x + 9)(3x - 5) = 16x²(3x - 5) + 24x(3x - 5) + 9(3x - 5) = 48x³ - 80x² + 72x² - 120x + 27x - 45 = 48x³ - 8x² - 93x - 45

This is identical to the given polynomial, confirming that (4x + 3)(4x + 3)(3x - 5) are its factors.

267Quantitative Aptitude
x and y are two numbers such that their mean proportion is 8 and third proportion is 512. What is the value of x and y?
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Answer: (B) 2 and 32

Explanation

The mean proportion of two numbers x and y is given by the formula: √(xy) = 8

Squaring both sides: xy = 64 x = 64/y

The third proportion of x and y is given by the formula: y²/x = 512 y² = 512x

Substituting the value of x into the third proportion equation: y² = 512 × (64/y) y³ = 32768 y = 32

Substituting the value of y back to find x: x = 64/32 x = 2

This gives the values of x and y as 2 and 32.

268Quantitative Aptitude
The difference between simple and compound interests compounded annually on a certain sum of money for 2 years at 12% per annum is Rs 72. What is the value of given sum (in Rs)?
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Answer: (C) 5000

Explanation

The difference between compound interest and simple interest for a period of 2 years is given by the formula:

Difference = P × (R / 100)²

where P is the principal sum and R is the rate of interest per annum.

Substituting the given values into the formula:

72 = P × (12 / 100)²

72 = P × 144 / 10000

P = (72 × 10000) / 144

P = 720000 / 144

P = 5000

269Quantitative Aptitude
The average weight of Shivshankar, Gopesh and Reena is 97 kg. If the average weight of Shivshankar and Gopesh is 93 kg and that of Gopesh and Reena is 82 kg, then the weight of Gopesh is
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Answer: (C) 59

Explanation

Total weight of Shivshankar, Gopesh and Reena = 97 × 3 = 291 kg

Total weight of Shivshankar and Gopesh = 93 × 2 = 186 kg

Total weight of Gopesh and Reena = 82 × 2 = 164 kg

By adding the weights of the two pairs: Shivshankar + 2 × Gopesh + Reena = 186 + 164 = 350 kg

Subtracting the total weight of all three individuals from this sum yields the weight of Gopesh: Weight of Gopesh = 350 - 291 = 59 kg

270Quantitative Aptitude
If cosec⁴A - cosec²A = x, then value of x is?
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Answer: (B) cot⁴A + cot²A

Explanation

The fundamental trigonometric identity relates cosecant and cotangent: cosec²A - cot²A = 1

This can be rewritten as: cosec²A = 1 + cot²A cosec²A - 1 = cot²A

The given expression is: x = cosec⁴A - cosec²A

By factoring out cosec²A: x = cosec²A (cosec²A - 1)

Substituting the trigonometric identities into this factored equation: x = (1 + cot²A) (cot²A) x = cot²A + cot⁴A x = cot⁴A + cot²A

271Quantitative Aptitude
What is the measure of an interior angle of a regular Octagon?
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Answer: (A) 135°

Explanation

The formula for the measure of each interior angle of a regular polygon with n sides is ((n - 2) × 180°) / n.

For a regular octagon, the number of sides is n = 8. Substituting this value into the formula:

((8 - 2) × 180°) / 8 = (6 × 180°) / 8 = 1080° / 8 = 135°

Alternatively, the sum of the exterior angles of any polygon is always 360°. Since a regular octagon has 8 equal exterior angles, the measure of each exterior angle is 360° / 8 = 45°. Because the interior and exterior angles at any vertex form a linear pair, the measure of each interior angle is 180° - 45° = 135°.

272Quantitative Aptitude
If 4(2x -4) - 2 > 3x - 1 ≥ 4x -7, then x can take which of the following values?
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Answer: (B) 6

Explanation

The compound inequality can be solved by splitting it into two parts.

First, solve the left inequality: 4(2x - 4) - 2 > 3x - 1 8x - 16 - 2 > 3x - 1 8x - 18 > 3x - 1 5x > 17 x > 17/5 x > 3.4

Next, solve the right inequality: 3x - 1 ≥ 4x - 7 -x ≥ -6 x ≤ 6

Combining both inequalities gives the range: 3.4 < x ≤ 6

The integer 6 lies within this range and satisfies both parts of the inequality.

273Quantitative Aptitude
Chords PQ and RS of a circle extended externally intersect at point T. Length of the segment PT, if lengths of QT, RT and ST are 9 cm, 18 cm and 6 cm respectively, is
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Answer: (D) 12 cm

Explanation

By the Secant-Secant Power Theorem, when two chords PQ and RS of a circle are extended externally to intersect at an external point T, the product of the lengths of the secant segments from the external point to the circle is equal for both secants. This gives the relation:

PT × QT = RT × ST

Substituting the given values, where QT = 9 cm, RT = 18 cm, and ST = 6 cm:

PT × 9 = 18 × 6

PT × 9 = 108

PT = 108 / 9

PT = 12 cm

274Quantitative Aptitude
What is the value of tan-240°?
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Answer: (A) -√3

Explanation

Using the trigonometric identity for negative angles, tan(-θ) = -tan(θ). Substituting θ = 240°:

tan(-240°) = -tan(240°)

The angle 240° lies in the third quadrant, where the tangent function is positive. Expressing 240° as 180° + 60° and using the identity tan(180° + θ) = tan(θ):

tan(240°) = tan(180° + 60°) = tan(60°)

The standard value of tan(60°) is √3, which gives:

tan(240°) = √3

Substituting this back into the first equation:

tan(-240°) = -√3

275Quantitative Aptitude
Rakesh goes on a trip on his motor-cycle and rides for 368 kms. If he rides for 5 hours at a speed of 49 km/hr, at what speed he travels for the remaining 3 hours of the journey?
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Answer: (B) 41 km/hr

Explanation

Using the relation Distance = Speed × Time, the distance covered in the first part of the journey is 49 km/hr × 5 hours = 245 km.

Subtracting this from the total distance of 368 km gives the remaining distance as 368 km - 245 km = 123 km.

The speed for the remaining 3 hours is calculated by dividing the remaining distance by the remaining time, which is 123 km / 3 hours = 41 km/hr.

276Quantitative Aptitude
What is the total surface area (in sq cm) of a hollow cylinder open at both ends, if its length is 22 cm, external radius is 7 cm and thickness is 1 cm? (Take n = 22/7)
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Answer: (A) 1879.43

Explanation

The total surface area of a hollow cylinder open at both ends consists of the outer curved surface area, the inner curved surface area, and the area of the two annular bases.

The formula for the total surface area (TSA) is: TSA = 2πRh + 2πrh + 2π(R² - r²)

This formula can be simplified as: TSA = 2π(R + r)(h + R - r)

The given values are: Length (h) = 22 cm External radius (R) = 7 cm Thickness = 1 cm Internal radius (r) = External radius - Thickness = 7 - 1 = 6 cm

Substituting these values into the simplified formula: TSA = 2 × (22/7) × (7 + 6) × (22 + 7 - 6) TSA = 2 × (22/7) × 13 × 23 TSA = 13156 / 7 TSA ≈ 1879.43 sq cm

277Quantitative Aptitude
What is the distance between the points (4,7) and (-1,-5)?
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Answer: (B) 13 units

Explanation

The distance d between two points (x₁, y₁) and (x₂, y₂) is given by the distance formula:

d = √((x₂ - x₁)² + (y₂ - y₁)²)

Substituting the coordinates of the given points (4, 7) and (-1, -5) into the formula:

d = √((-1 - 4)² + (-5 - 7)²)

d = √((-5)² + (-12)²)

d = √(25 + 144)

d = √169

d = 13 units

278Quantitative Aptitude
At 22% discount the selling price of a refrigerator is Rs 26000, what is the selling price if the discount is 40%?
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Answer: (D) Rs 20000

Explanation

The relationship between the selling prices and their corresponding discount percentages on the same marked price can be established using the ratio formula:

SP₂ / SP₁ = (100 - d₂) / (100 - d₁)

Here, SP₁ is the initial selling price of Rs 26000, d₁ is the initial discount of 22%, d₂ is the new discount of 40%, and SP₂ is the new selling price.

Substituting the values into the formula:

SP₂ / 26000 = (100 - 40) / (100 - 22)

SP₂ / 26000 = 60 / 78

Simplifying the fraction 60/78 by dividing the numerator and denominator by 6:

SP₂ / 26000 = 10 / 13

Solving for SP₂:

SP₂ = 26000 × (10 / 13)

SP₂ = 2000 × 10

SP₂ = 20000

279Quantitative Aptitude
26 is 0.4% of which number?
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Answer: (A) 6500

Explanation

Let the required number be x.

According to the percentage formula, we have: 0.4% of x = 26

Writing the percentage as a fraction: (0.4 / 100) × x = 26

Multiplying both sides by 100: 0.4 × x = 2600

Dividing both sides by 0.4: x = 2600 / 0.4 x = 26000 / 4 x = 6500

280Quantitative Aptitude
Divide 20 into two parts such that the sum of the square of the parts is 232. What is the value of the two parts?
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Answer: (A) 6, 14

Explanation

Let the two parts of 20 be x and 20 - x.

According to the given condition, the sum of the squares of these two parts is 232: x² + (20 - x)² = 232

Expanding the equation: x² + 400 - 40x + x² = 232 2x² - 40x + 168 = 0

Dividing the entire equation by 2: x² - 20x + 84 = 0

Factoring the quadratic equation: x² - 14x - 6x + 84 = 0 x(x - 14) - 6(x - 14) = 0 (x - 14)(x - 6) = 0

This yields x = 14 or x = 6.

281Quantitative Aptitude
What will be the equation of the perpendicular bisector of segment joining the points (5,-3) and (0,2)?
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Answer: (D) x - y = 3

Explanation

The midpoint of the segment joining (5, -3) and (0, 2) is calculated using the midpoint formula: M = ((5 + 0)/2, (-3 + 2)/2) = (5/2, -1/2)

The slope of the segment joining the two points is: m = (2 - (-3)) / (0 - 5) = 5 / -5 = -1

The slope of the perpendicular bisector is the negative reciprocal of this slope: m' = -1 / m = -1 / -1 = 1

Using the point-slope form with midpoint (5/2, -1/2) and slope 1: y - (-1/2) = 1 × (x - 5/2) y + ½ = x - 5/2

Rearranging the terms: x - y = 3

282Quantitative Aptitude
Refer the below data table and answer the following Question. Weight (kg) Height (m) Archita 67 1.63 Arpana 67 1.51 Arpita 57 1.56 Arundhati 73 1.59 Who has the least weight to height ratio?
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Answer: (C) Arpita

Explanation

The weight to height ratio is calculated by dividing the weight in kilograms by the height in meters. For Arpita, this ratio is 57 ÷ 1.56, which equals approximately 36.54. Comparing this to the other individuals, Archita has a ratio of 67 ÷ 1.63 ≈ 41.10, Arpana has a ratio of 67 ÷ 1.51 ≈ 44.37, and Arundhati has a ratio of 73 ÷ 1.59 ≈ 45.91. Arpita clearly has the lowest ratio.

283Quantitative Aptitude
The following table shows the number of children in each house of a society. Number of Children Number of Houses 0 7 1 11 2 15 3 5 What is the average number of children per house?
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Answer: (C) 1.47

Explanation

The average number of children per house is calculated using the formula: Average = Total number of children ÷ Total number of houses

The total number of houses is: 7 + 11 + 15 + 5 = 38

The total number of children across all houses is: (7 × 0) + (11 × 1) + (15 × 2) + (5 × 3) = 0 + 11 + 30 + 15 = 56

Applying the formula: Average = 56 ÷ 38 ≈ 1.47

284Quantitative Aptitude
Refer the below data table and answer the following Question. Partners present % share Anand 10 Basu 15 Chinmay 10 Dhiraj 15 Ejaz 50 If the company has issued six lakh shares between its five partners and if Chinmay offers to sell 15,000 of his shares to Dhiraj, then Dhiraj will have how many shares?
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Answer: (C) 105000 shares

Explanation

The total number of shares issued by the company is 6 lakh, which is equal to 600,000 shares.

Dhiraj initially holds a 15% share of these total shares.

Initial shares held by Dhiraj = 15% of 600,000 = (15/100) × 600,000 = 90,000 shares

When Chinmay sells 15,000 of his shares to Dhiraj, Dhiraj receives these shares, which increases his total shareholding.

New number of shares held by Dhiraj = 90,000 + 15,000 = 105,000 shares

285Quantitative Aptitude
If xy = 56 and x² + y² = 113, then what will be the value of (x + y)?
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Answer: (D) 15

Explanation

Using the standard algebraic identity: (x + y)² = x² + y² + 2xy

Substitute the given values of x² + y² = 113 and xy = 56 into the equation: (x + y)² = 113 + 2(56) (x + y)² = 113 + 112 (x + y)² = 225

Taking the square root on both sides yields: x + y = √225 x + y = 15

286Quantitative Aptitude
If 3A = 7B = 13C, then A : B : C is equal to
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Answer: (D) 91:39:21

Explanation

If mA = nB = pC, the ratio A : B : C is given by the inverse ratio 1/m : 1/n : 1/p.

For the relation 3A = 7B = 13C, the ratio A : B : C is: A : B : C = 1/3 : 1/7 : 1/13

Multiplying each term by the product of the denominators (3 × 7 × 13 = 273) simplifies the ratio to: A : B : C = (7 × 13) : (3 × 13) : (3 × 7) A : B : C = 91 : 39 : 21

287Quantitative Aptitude
Which of the following equations has 3 as a root?
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Answer: (A) x² - 7x + 12 = 0

Explanation

A number r is a root of a quadratic equation if substituting x = r into the equation satisfies it by yielding a value of zero.

Substituting x = 3 into the equation x² - 7x + 12 = 0 gives:

3² - 7(3) + 12 = 9 - 21 + 12 = 21 - 21 = 0

Since the equation is satisfied, 3 is a root of this equation.

288Quantitative Aptitude
Amar is 2 times as good a workman as Badal and therefore is able to finish a job in 15 days less than Badal. In how many days can they together complete the job?
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Answer: (C) 10 days

Explanation

The ratio of efficiency of Amar and Badal is 2 : 1. Since efficiency and time taken are inversely proportional, the ratio of time taken by Amar and Badal to complete the work is 1 : 2.

Let the time taken by Amar be x days and Badal be 2x days. The difference in their time taken is given as 15 days: 2x - x = 15 x = 15

Thus, Amar takes 15 days and Badal takes 30 days to complete the work.

Using the combined work formula, the time taken by both working together is: Time = (a × b) / (a + b) Time = (15 × 30) / (15 + 30) = 450 / 45 = 10 days

289Quantitative Aptitude
What is the equation of the line which intercepts x-axis and y-axis at 3/4 and -2/3 respectively?
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Answer: (C) 8x - 9y = 6

Explanation

The intercept form of the equation of a line is x/a + y/b = 1, where a is the x-intercept and b is the y-intercept.

Given: a = 3/4 b = -2/3

Substituting these values into the intercept form: x / (3/4) + y / (-2/3) = 1

Simplifying the fractions: 4x/3 - 3y/2 = 1

Multiplying the entire equation by the least common multiple of the denominators 3 and 2, which is 6, to clear the fractions: 6 × (4x/3) - 6 × (3y/2) = 6 × 1 8x - 9y = 6

290Quantitative Aptitude
What is the value of tan3A?
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Answer: (A) (3tanA - tan³A)/(1 - 3tan²A)

Explanation

The triple angle formula for tangent is derived using the compound angle identity:

tan(x + y) = (tan x + tan y) / (1 - tan x tan y)

By expressing 3A as 2A + A, we apply this identity:

tan 3A = (tan 2A + tan A) / (1 - tan 2A tan A)

Using the double angle identity, tan 2A = 2tan A / (1 - tan²A), and substituting this into the equation:

tan 3A = (2tan A / (1 - tan²A) + tan A) / (1 - (2tan A / (1 - tan²A)) × tan A)

To simplify, we multiply both the numerator and the denominator by (1 - tan²A):

tan 3A = (2tan A + tan A(1 - tan²A)) / (1 - tan²A - 2tan²A)

Expanding and combining like terms yields:

tan 3A = (2tan A + tan A - tan³A) / (1 - 3tan²A)

tan 3A = (3tan A - tan³A) / (1 - 3tan²A)

291Quantitative Aptitude
An angle is greater than its complementary by 20°. What will be the measure of the angle?
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Answer: (A) 55°

Explanation

Let the required angle be x.

Two angles are complementary if their sum is 90°. Therefore, the complement of the angle is 90° - x.

According to the given condition, the angle is 20° greater than its complement:

x - (90° - x) = 20°

2x - 90° = 20°

2x = 110°

x = 55°

292Quantitative Aptitude
The radius of a hemisphere is 21 cm, what is its volume? (Take n = 22/7)
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Answer: (C) 19404 cubic cms

Explanation

The volume of a hemisphere is given by the formula V = (2/3) × π × r³, where r is the radius of the hemisphere.

Substituting the given values r = 21 cm and π = 22/7 into the formula:

V = (2/3) × (22/7) × 21 × 21 × 21

V = 2 × 22 × 21 × 21

V = 44 × 441

V = 19404 cubic cm

293Quantitative Aptitude
If cot²A - cos²A = x, then value of x is
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Answer: (C) cot²Acos²A

Explanation

We can simplify the given trigonometric expression by writing cot²A in terms of sin²A and cos²A:

cot²A - cos²A = (cos²A / sin²A) - cos²A

Factoring out cos²A:

= cos²A (1 / sin²A - 1)

Using the reciprocal identity cosec²A = 1 / sin²A:

= cos²A (cosec²A - 1)

Using the standard trigonometric identity cosec²A - 1 = cot²A:

= cos²A cot²A

= cot²A cos²A

294Quantitative Aptitude
In a triangle the length of the side opposite the right angle is 6 cm, what is the length of the side opposite to the angle which measures 30°?
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Answer: (A) 3 cm

Explanation

In a right-angled triangle, the side opposite the right angle is the hypotenuse. The relationship between the side opposite to an angle θ and the hypotenuse is given by the trigonometric ratio:

sin θ = Opposite / Hypotenuse

Here, the angle θ is 30° and the length of the hypotenuse is 6 cm. Substituting these values into the formula:

sin 30° = Opposite / 6

Since sin 30° = ½, the equation becomes:

½ = Opposite / 6

Opposite = 6 × ½ = 3 cm

295Quantitative Aptitude
If 2x + 5 > 2 + 3x and 2x - 3 ≤ 4x - 5, then x can take which of the following values?
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Answer: (B) 2

Explanation

To find the possible values of x, we solve the two given inequalities individually.

For the first inequality: 2x + 5 > 2 + 3x Subtracting 2x from both sides gives: 5 > 2 + x Subtracting 2 from both sides gives: x < 3

For the second inequality: 2x - 3 ≤ 4x - 5 Subtracting 2x from both sides gives: -3 ≤ 2x - 5 Adding 5 to both sides gives: 2 ≤ 2x Dividing by 2 gives: x ≥ 1

Combining both inequalities, the range for x is: 1 ≤ x < 3

Since 2 lies within the interval [1, 3), it satisfies both inequalities.

296Quantitative Aptitude
If the radius of a circle is increased by 14% its area increases by
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Answer: (B) 29.96 percent

Explanation

The area of a circle is given by the formula A = πr², where r is the radius. When a quantity depending on the square of a variable increases by x%, the overall percentage increase in the quantity is given by the successive percentage formula:

Percentage increase = 2x + (x² / 100)

Given that the radius increases by x = 14%:

Percentage increase = 2(14) + (14² / 100) Percentage increase = 28 + (196 / 100) Percentage increase = 28 + 1.96 = 29.96%

297Quantitative Aptitude
What is the value of tan150°?
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Answer: (D) -1/√3

Explanation

Using the trigonometric identity for angles in the second quadrant, tan(180° - θ) = -tan θ.

Substituting θ = 30°: tan 150° = tan(180° - 30°) = -tan 30°

Since the standard trigonometric value of tan 30° is 1/√3, we get: tan 150° = -1/√3

298Quantitative Aptitude
When 0.151515....is converted into a fraction, then the result is
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Answer: (A) 5/33

Explanation

Let the recurring decimal be represented as x = 0.151515...

Since there are two repeating digits in the decimal pattern, multiply both sides by 100: 100x = 15.151515...

Now, subtract the first equation from the second: 100x - x = 15.151515... - 0.151515... 99x = 15 x = 15/99

Simplifying the fraction by dividing both the numerator and the denominator by their greatest common divisor, which is 3, yields: x = 5/33

299Quantitative Aptitude
Among three numbers, the first is twice the second and thrice the third. If the average of three numbers is 429, then what is the difference between the first and the third number?
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Answer: (B) 468

Explanation

Let the three numbers be x, y, and z.

According to the given conditions, the first number x is twice the second number y and thrice the third number z: x = 2y → y = x/2 x = 3z → z = x/3

The average of the three numbers is 429: (x + y + z) / 3 = 429 x + y + z = 1287

Substitute y and z in terms of x: x + x/2 + x/3 = 1287 (6x + 3x + 2x) / 6 = 1287 11x / 6 = 1287 x = (1287 × 6) / 11 x = 117 × 6 x = 702

The third number z is: z = x/3 = 702 / 3 = 234

The difference between the first and the third number is: x - z = 702 - 234 = 468

300Quantitative Aptitude
If (8x - 4) = (3x + 6), then the numerical value of (x + 1)³ is
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Answer: (C) 27

Explanation

To find the numerical value of (x + 1)³, we first solve the given linear equation for x:

8x - 4 = 3x + 6

Subtracting 3x from both sides of the equation:

5x - 4 = 6

Adding 4 to both sides:

5x = 10

Dividing by 5 gives:

x = 2

Substituting the value of x into the expression (x + 1)³:

(2 + 1)³ = 3³ = 27

Answer key

Tap a number to jump to that question.

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

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