SSC CHSL2023Quantitative AptitudePage 5 of 16

SSC CHSL Tier 1 2023 Quantitative Aptitude Questions with Solutions

Questions from every shift of the year, grouped by section. Questions 201–250 of 781, each with the correct answer and a step-by-step explanation.

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201Quantitative Aptitude
If (sinθ - cosθ)/(sinθ + cosθ) = ⅘ , then the value of (cos e²θ)/(2 - cos e²θ) is:
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Answer: (C) ⁴¹⁄₄₀

Explanation

Applying the componendo and dividendo rule to the given equation: (sin θ - cos θ) / (sin θ + cos θ) = 4/5

((sin θ - cos θ) + (sin θ + cos θ)) / ((sin θ - cos θ) - (sin θ + cos θ)) = (4 + 5) / (4 - 5) 2 sin θ / (-2 cos θ) = 9 / -1 tan θ = 9

Since tan θ = 9, we have cot θ = 1/9.

Using the trigonometric identity cosec² θ = 1 + cot² θ, we get: cosec² θ = 1 + (1/9)² = 1 + 1/81 = 82/81

Now, substitute the value of cosec² θ into the required expression: cosec² θ / (2 - cosec² θ) = (82/81) / (2 - 82/81) = (82/81) / ((162 - 82) / 81) = (82/81) / (80/81) = 82/80 = 41/40

202Quantitative Aptitude
Anil completed a certain journey by a car. If he covered 35% of the distance at 70 km / h , 40% of the distance at 80 km / h and the remaining distance at 50 km / h , his average speed (in km / h) for the entire journey is:
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Answer: (B) 66⅔km / h

Explanation

Average speed is calculated using the formula: Average Speed = Total Distance / Total Time

Let the total distance of the journey be D km.

The times taken to cover each part of the journey are: Time₁ = 0.35D / 70 = D / 200 hours Time₂ = 0.40D / 80 = D / 200 hours Time₃ = (1 - 0.35 - 0.40)D / 50 = 0.25D / 50 = D / 200 hours

Total time taken for the entire journey: Total Time = D/200 + D/200 + D/200 = 3D/200 hours

Substituting these values into the average speed formula: Average Speed = D / (3D/200) = 200 / 3 = 66⅔ km/h

203Quantitative Aptitude
△ PQR is right angled at Q. If PQ = 12 cm and PR = 13 cm , find tan P + cot R .
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Answer: (C) ¹⁰⁄₁₂

Explanation

In right-angled triangle PQR, right-angled at Q, we use Pythagoras theorem to find the length of the side QR:

PR² = PQ² + QR²

Substituting the given values:

13² = 12² + QR² 169 = 144 + QR² QR² = 25 QR = 5 cm

For angle P: tan P = opposite side / adjacent side = QR / PQ = 5/12

For angle R: cot R = adjacent side / opposite side = QR / PQ = 5/12

Adding the two values: tan P + cot R = 5/12 + 5/12 = 10/12

204Quantitative Aptitude
A table fan is marked at ₹1,700 and a discount of 15% is given. What is its selling price (in ₹)?
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Answer: (C) 1,445

Explanation

The selling price of an item after a discount is calculated using the formula:

Selling Price = Marked Price × (100 - Discount%)/100

Given: Marked Price = ₹1,700 Discount = 15%

Substituting these values into the formula:

Selling Price = 1700 × (100 - 15)/100 Selling Price = 1700 × 85/100 Selling Price = 17 × 85 Selling Price = 1445

205Quantitative Aptitude
A tradesman marks his goods 25% above the cost price and allows his customers a discount of 12% on their bill. How much % profit tradesman realizes?
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Answer: (D) 10%

Explanation

The net profit percentage when there is a markup of a% followed by a discount of b% is given by the successive percentage formula:

Net Profit % = a - b - ab/100

Substituting the given values, markup percentage a = 25 and discount percentage b = 12:

Net Profit % = 25 - 12 - (25 × 12)/100 Net Profit % = 13 - 300/100 Net Profit % = 13 - 3 Net Profit % = 10%

206Quantitative Aptitude
If 7 divides the integer n, then the remainder is 2. What will be the remainder if 9n is divided by 7?
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Answer: (D) 4

Explanation

Correct Answer: D

Using modular arithmetic, the statement that dividing n by 7 leaves a remainder of 2 is expressed as: n ≡ 2 (mod 7)

To find the remainder when 9n is divided by 7, multiply both sides by 9: 9n ≡ 9 × 2 (mod 7) 9n ≡ 18 (mod 7)

Reducing 18 modulo 7 yields: 18 = 2 × 7 + 4 9n ≡ 4 (mod 7)

207Quantitative Aptitude
The table given below shows the number of aerated drinks—A, B, C, D—manufactured (in thousands) by a company over the years. YEAR Aerated Drinks A B C D 2019 250 200 140 115 2020 200 230 170 120 2021 260 210 195 135 2022 275 175 190 155 Which aerated drink has the highest manufacturing number from 2019 to 2022?
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Answer: (C) A

Explanation

To find the aerated drink with the highest manufacturing number from 2019 to 2022, we calculate the total production for each drink across all the given years:

Total production of A = 250 + 200 + 260 + 275 = 985 thousand Total production of B = 200 + 230 + 210 + 175 = 815 thousand Total production of C = 140 + 170 + 195 + 190 = 695 thousand Total production of D = 115 + 120 + 135 + 155 = 525 thousand

Comparing these totals, aerated drink A has the highest overall manufacturing number of 985 thousand. This corresponds to option C.

208Quantitative Aptitude
The largest chord of the circle is:
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Answer: (C) diameter

Explanation

The length of a chord in a circle of radius r is given by the formula L = 2r sin(θ/2), where θ is the central angle subtended by the chord. Since the maximum value of the sine function is 1, the maximum length of a chord occurs when θ = 180°, which gives L = 2r. A chord that subtends a central angle of 180° passes through the center of the circle, which is the definition of the diameter. Therefore, the diameter represents the longest possible chord of a circle.

209Quantitative Aptitude
The difference between the simple interest and the compound interest on a certain amount at 9% per annum for two years is ₹ 162, what is the principal?
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Answer: (C) ₹ 20000

Explanation

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

Difference = P × (R/100)²

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

Substituting the given values, Difference = ₹162 and R = 9%, into the formula:

162 = P × (9/100)²

162 = P × (81/10000)

P = (162 × 10000) / 81

P = 2 × 10000

P = 20000

Thus, the principal amount is ₹20000.

210Quantitative Aptitude
Two circles touch each other externally at any point C. PQ is the direct common tangent to both the circles touching the circles at point P and point Q. If the radii of the circles are 36 cm and 16 cm, respectively, then the length of PQ is:
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Answer: (C) 48 cm

Explanation

The length of the direct common tangent to two circles that touch each other externally is given by the formula:

PQ = 2√(r₁ × r₂)

where r₁ and r₂ are the radii of the two circles.

Substituting the given radii r₁ = 36 cm and r₂ = 16 cm into the formula:

PQ = 2√(36 × 16)

PQ = 2 × 6 × 4

PQ = 48 cm

211Quantitative Aptitude
Chords AB and CD of a circle intersect inside the circle at point F. If m(AB) = 2.4 cm, m(BF) = 1.8 cm, and m(CD) = 5.7 cm, what is the length (in cm) of the longer of the two line segments, CF and DF?
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Answer: (B) 4.8

Explanation

By the intersecting chords theorem, when two chords AB and CD intersect inside a circle at point F, the products of their segments are equal:

AF × BF = CF × DF

Using the segment lengths AF = 2.4 cm and BF = 1.8 cm:

AF × BF = 2.4 × 1.8 = 4.32

Let the lengths of CF and DF be x and y respectively. We are given the total length of chord CD:

x + y = 5.7 x × y = 4.32

We can set up a quadratic equation to find the lengths:

t² - 5.7t + 4.32 = 0

Solving this equation:

(t - 4.8)(t - 0.9) = 0

This gives the two segment lengths as 4.8 cm and 0.9 cm. The length of the longer segment is 4.8 cm.

212Quantitative Aptitude
The algebraic expression 4x² - y² + 6y - 9 is equal to ______.
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Answer: (A) (2x + y - 3)(2x - y + 3)

Explanation

The algebraic expression can be factored by grouping and using algebraic identities.

First, group the terms to isolate the perfect square in terms of y: 4x² - y² + 6y - 9 = 4x² - (y² - 6y + 9)

The expression inside the parentheses, y² - 6y + 9, is a perfect square trinomial of the form (a - b)² = a² - 2ab + b², where a = y and b = 3: y² - 6y + 9 = (y - 3)²

Substitute this back into the expression: 4x² - (y - 3)²

Express 4x² as (2x)² to set up a difference of squares: (2x)² - (y - 3)²

Apply the difference of squares identity, a² - b² = (a + b)(a - b), where a = 2x and b = y - 3: (2x + (y - 3))(2x - (y - 3)) = (2x + y - 3)(2x - y + 3)

213Quantitative Aptitude
A spends 68% of his monthly income. If A's monthly income increases by 25% and his monthly saving increases by 15%, then the percentage increase in his monthly expenditure will be:
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Answer: (B) 29 ¹²⁄₁₇ %

Explanation

Income is the sum of expenditure and saving: Income = Expenditure + Saving

Let the initial monthly income be 100. Since A spends 68% of his income, the initial expenditure is 68, and the initial saving is: 100 - 68 = 32

After the changes: New monthly income increases by 25%: 100 × 1.25 = 125

New monthly saving increases by 15%: 32 × 1.15 = 36.8

The new monthly expenditure is the difference between the new income and the new saving: 125 - 36.8 = 88.2

The increase in monthly expenditure is: 88.2 - 68 = 20.2

The percentage increase in monthly expenditure is: (20.2 / 68) × 100% = (2020 / 68)% = (505 / 17)% = 29 12/17%

214Quantitative Aptitude
If tan A = ½ and tan B = ⅓, then the value of A+B is:
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Answer: (D) 45°

Explanation

The trigonometric identity for the sum of two angles is: tan(A + B) = (tan A + tan B) / (1 - tan A × tan B)

Substituting the given values tan A = ½ and tan B = ⅓ into the formula: tan(A + B) = (½ + ⅓) / (1 - ½ × ⅓)

Evaluating the numerator: ½ + ⅓ = 5/6

Evaluating the denominator: 1 - 1/6 = 5/6

Therefore: tan(A + B) = (5/6) / (5/6) = 1

Since tan 45° = 1, it follows that A + B = 45°.

215Quantitative Aptitude
A man sells a good at 5% profit. If he had sold it at a 10% profit, he would have received ₹175 more. What is the selling price of the good?
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Answer: (C) 3,675

Explanation

Let the cost price of the good be CP.

The difference between selling the good at a 10% profit instead of a 5% profit represents a difference of 5% of the cost price.

This 5% difference is equal to ₹175: 5% of CP = 175 0.05 × CP = 175 CP = 175 ÷ 0.05 CP = 3500

The actual selling price of the good is at a 5% profit: Selling Price = 105% of CP Selling Price = 1.05 × 3500 Selling Price = 3675

216Quantitative Aptitude
If x = (√6 - 1)½ , then the value of (x - 1/x)³ + 3(x - 1/x) is:
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Answer: (B) 4√6 - 6 /5

Explanation

Given: x = (√6 - 1)⁰.⁵

Squaring both sides: x² = √6 - 1

The reciprocal of x² is: 1/x² = 1 / (√6 - 1)

Rationalizing the denominator by multiplying the numerator and denominator by the conjugate (√6 + 1): 1/x² = (√6 + 1) / ((√6 - 1)(√6 + 1)) 1/x² = (√6 + 1) / (6 - 1) 1/x² = (√6 + 1) / 5

Evaluating the difference x² - 1/x²: x² - 1/x² = (√6 - 1) - (√6 + 1) / 5 x² - 1/x² = (5(√6 - 1) - (√6 + 1)) / 5 x² - 1/x² = (5√6 - 5 - √6 - 1) / 5 x² - 1/x² = (4√6 - 6) / 5

217Quantitative Aptitude
X can do a piece of work in 14 days, Y can complete the same work in 28 days and Z can do it in 42 days. In how many days can X, Y and Z together complete the work?
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Answer: (B) 7⁷⁄₁₁

Explanation

Correct Answer: B

Let the total work be the Least Common Multiple of the individual days taken by X, Y and Z. LCM of 14, 28 and 42 is 84 units.

The efficiency of each individual is calculated as total work divided by the number of days: Efficiency of X = 84 / 14 = 6 units per day Efficiency of Y = 84 / 28 = 3 units per day Efficiency of Z = 84 / 42 = 2 units per day

Combined efficiency of X, Y and Z = 6 + 3 + 2 = 11 units per day.

The time taken by X, Y and Z working together to complete the work is: Total work / Combined efficiency = 84 / 11 = 7 7/11 days.

218Quantitative Aptitude
In what ratio, wheat at ₹23 per kg should be mixed with wheat at ₹14 per kg so that on selling the mixture at ₹20 per kg there is a profit of 25%?
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Answer: (D) 2:7

Explanation

Correct Answer: D

First, find the Cost Price (CP) of the mixture using the formula: CP = Selling Price / (1 + Profit Percentage / 100)

Given that the Selling Price is ₹20 per kg and the profit is 25%: CP = 20 / (1 + 25/100) = 20 / 1.25 = ₹16 per kg

Using the rule of alligation to find the ratio in which the wheat at ₹23 per kg (dearer) and ₹14 per kg (cheaper) must be mixed: Ratio of dearer wheat to cheaper wheat = (Mean Price - Cheaper Price) / (Dearer Price - Mean Price) Ratio = (16 - 14) / (23 - 16) = 2 / 7

Thus, the wheat should be mixed in the ratio 2:7.

219Quantitative Aptitude
Ravi initially had a certain number of chocolates, while Bhanu initially had 80% of the number of chocolates that Ravi initially had. Ravi gave 10% of his chocolates to his sister while Bhanu received from his uncle another 35% of the number of chocolates he initially had. If Bhanu finally had 54 chocolates, how many chocolates did Ravi give to his sister?
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Answer: (C) 5

Explanation

Let Ravi's initial number of chocolates be R. Bhanu's initial number of chocolates is 80% of R, which is 0.8R.

Bhanu received an additional 35% of his initial chocolates from his uncle, so his final number of chocolates is: 0.8R × (1 + 0.35) = 1.08R

Since Bhanu finally had 54 chocolates, we set up the equation: 1.08R = 54 R = 54 / 1.08 R = 50

Ravi gave 10% of his chocolates to his sister. The number of chocolates given is: 10% of 50 = 5

220Quantitative Aptitude
During a donation drive week of school, students determined to donate an average sum of ₹8,000 on each working day of the School, collectively (there was no holiday in the donation week except Sunday). From Monday to Friday, they collected ₹7,230, ₹6,435, ₹6,927, ₹6,855, and ₹6,562, respectively. How much they must collect on Saturday to meet their target?
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Answer: (B) 13,991

Explanation

The school has 6 working days in the donation week, which are Monday, Tuesday, Wednesday, Thursday, Friday, and Saturday.

To achieve an average donation of ₹8,000 per working day over these 6 days, the total target collection is calculated as: Total target collection = 6 × ₹8,000 = ₹48,000

The sum of the collections from Monday to Friday is: ₹7,230 + ₹6,435 + ₹6,927 + ₹6,855 + ₹6,562 = ₹34,009

The amount that must be collected on Saturday to meet the target is: Saturday collection = Total target collection - Monday to Friday collection Saturday collection = ₹48,000 - ₹34,009 = ₹13,991

221Quantitative Aptitude
A shopkeeper allows a 12% additional discount on an initial discount of 20% for a shirt. If he sells the shirt at ₹704, then the marked price is:
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Answer: (C) ₹1,000

Explanation

The selling price after successive discounts is calculated using the formula:

Selling Price = Marked Price × (1 - d₁) × (1 - d₂)

where d₁ and d₂ are the successive discount rates.

Given: Selling Price = ₹704 d₁ = 20% = 0.20 d₂ = 12% = 0.12

Substituting these values into the formula:

704 = Marked Price × (1 - 0.20) × (1 - 0.12) 704 = Marked Price × 0.80 × 0.88 704 = Marked Price × 0.704

Solving for Marked Price:

Marked Price = 704 / 0.704 = 1000

Thus, the marked price of the shirt is ₹1,000.

222Quantitative Aptitude
If 300 calories are burnt in every 5 km, how many calories were burnt in the whole week? Week days | Distance jogged (km) Sunday | 2.5 Monday | 4.0 Tuesday | 2.5 Wednesday | 3.5 Thursday | 1.5 Friday | 2.5 Saturday | 1.5
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Answer: (A) 1080 calories

Explanation

To find the total calories burnt in the whole week, first calculate the total distance jogged during the week by summing the distances for each day:

Total distance = 2.5 + 4.0 + 2.5 + 3.5 + 1.5 + 2.5 + 1.5 = 18 km

The rate of calories burnt per kilometer is: 300 calories ÷ 5 km = 60 calories/km

Multiplying the total distance by this rate gives the total calories burnt for the week: 18 km × 60 calories/km = 1080 calories

223Quantitative Aptitude
Two circles touch each other externally at points P and AB is a direct common tangent which touches the circles at A and B, respectively. ∠APB is:
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Answer: (A) 90°

Explanation

Let the direct common tangent AB touch the two circles at A and B, and let the circles touch each other externally at P. Draw a common tangent at the point of contact P, which intersects the tangent AB at point T.

The lengths of tangents drawn from an external point to a circle are equal. Therefore, for the first circle, TA = TP, and for the second circle, TB = TP. This gives TA = TB = TP.

In triangle TAP, since TA = TP, the opposite angles are equal, so ∠TAP = ∠TPA. In triangle TBP, since TB = TP, the opposite angles are equal, so ∠TBP = ∠TPB.

In triangle APB, the sum of the interior angles is 180°: ∠PAB + ∠PBA + ∠APB = 180° ∠TAP + ∠TBP + ∠APB = 180°

Substituting ∠TAP = ∠TPA and ∠TBP = ∠TPB: ∠TPA + ∠TPB + ∠APB = 180°

Since ∠TPA + ∠TPB = ∠APB, we get: ∠APB + ∠APB = 180° 2 ∠APB = 180° ∠APB = 90°

224Quantitative Aptitude
Simplify the given expression. 0.24 × 0.24 × 0.24 - 0.008 0.24 × 0.24 + 0.048 + 0.04
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Answer: (A) 0.04

Explanation

The given expression can be simplified using the algebraic identity: a³ - b³ = (a - b)(a² + ab + b²)

Let a = 0.24 and b = 0.2.

The numerator of the expression is: 0.24 × 0.24 × 0.24 - 0.008 = a³ - b³

The denominator of the expression is: 0.24 × 0.24 + 0.048 + 0.04 = a² + ab + b² since ab = 0.24 × 0.2 = 0.048 and b² = 0.2² = 0.04.

Dividing the numerator by the denominator: (a³ - b³) / (a² + ab + b²) = a - b

Substituting the values of a and b: 0.24 - 0.2 = 0.04

225Quantitative Aptitude
Sneha fixes the marked price of an item 40% above its cost price. What is the percentage of discount allowed to gain 12% ?
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Answer: (C) 20%

Explanation

Using the relation between profit percentage, markup percentage, and discount percentage: Profit % = Markup % - Discount % - (Markup % × Discount %) / 100

Substituting the given values of profit percentage as 12% and markup percentage as 40%: 12 = 40 - d - (40 × d) / 100 12 = 40 - 1.4d 1.4d = 28 d = 28 / 1.4 = 20%

226Quantitative Aptitude
The given expression is equal to: (sin⁴ A + cos⁴ A)/(1 - 2sin² A cos² A)
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Answer: (A) 1

Explanation

Using the algebraic identity x² + y² = (x + y)² - 2xy, we can rewrite the numerator by letting x = sin² A and y = cos² A.

sin⁴ A + cos⁴ A = (sin² A + cos² A)² - 2 sin² A cos² A

Since sin² A + cos² A = 1, the numerator simplifies to:

1² - 2 sin² A cos² A = 1 - 2 sin² A cos² A

Substituting this back into the original expression gives:

(1 - 2 sin² A cos² A) / (1 - 2 sin² A cos² A) = 1

227Quantitative Aptitude
Let ABC and PQR be two congruent right-angled triangles such that ∠ A = ∠ P = 90° . If BC = 13 cm and PR = 12 cm, then find the length of AB.
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Answer: (D) 5 cm

Explanation

In a right-angled triangle, the relation between the sides is given by the Pythagorean theorem: leg₁² + leg₂² = hypotenuse².

For the right-angled triangle PQR, the angle P is 90°, making QR the hypotenuse and PQ and PR the legs. Since triangle ABC is congruent to triangle PQR, their corresponding sides are equal. This means the hypotenuse QR is equal to BC, which is 13 cm. The given side PR is 12 cm.

Using the Pythagorean theorem for triangle PQR: PQ = √(QR² - PR²) PQ = √(13² - 12²) PQ = √(169 - 144) PQ = √25 PQ = 5 cm

Due to the congruence of triangles ABC and PQR, the side AB corresponds to PQ, which gives AB = 5 cm.

228Quantitative Aptitude
A money- lender claims he lends money at a simple rate of interest of 8% per annum. But he cleverly includes the interest amount in the principal when he calculates it every six months. The effective rate of interest becomes:
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Answer: (A) 8.16%

Explanation

When interest is calculated and added to the principal every six months, the compounding occurs semi-annually. For a nominal interest rate of 8% per annum, the rate for each six-month period is 8% ÷ 2 = 4%.

The effective annual rate of interest for two successive half-yearly compounding periods of r% each is calculated using the successive percentage change formula:

Effective Rate = r + r + (r × r)/100 %

Substituting r = 4% into the formula:

Effective Rate = 4 + 4 + (4 × 4)/100 % Effective Rate = 8 + 16/100 % Effective Rate = 8 + 0.16 % Effective Rate = 8.16%

229Quantitative Aptitude
If cot A = ⁵⁄₁₂ , find the value of the following expression. (5(1 - cos² A))/(6(1 - sin² A))
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Answer: (D) 24/5

Explanation

Using the fundamental trigonometric identities, we know that 1 - cos² A = sin² A and 1 - sin² A = cos² A.

Substituting these identities into the given expression simplifies it as follows:

5(1 - cos² A) / 6(1 - sin² A) = 5 sin² A / 6 cos² A = (5/6) tan² A

We are given that cot A = 5/12. Since tan A is the reciprocal of cot A, we have:

tan A = 12/5

Squaring both sides gives:

tan² A = (12/5)² = 144/25

Now, substitute this value back into the simplified expression:

(5/6) × (144/25) = (5 × 144) / (6 × 25) = 144 / (6 × 5) = 24/5

230Quantitative Aptitude
A train travelling from Mumbai to Goa covered the first 420 km at a speed of 60 km/h and then covered the remaining 195 km at a speed of 65 km/h. Find its average speed in km/h.
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Answer: (B) 61.5

Explanation

Average speed is calculated using the formula:

Average Speed = Total Distance / Total Time

The total distance covered by the train is: Total Distance = 420 + 195 = 615 km

The time taken for each part of the journey is: Time for the first part = 420 / 60 = 7 hours Time for the second part = 195 / 65 = 3 hours

Total Time = 7 + 3 = 10 hours

Substituting these values into the average speed formula: Average Speed = 615 / 10 = 61.5 km/h

231Quantitative Aptitude
The marked price of a saree is ₹3,800 and it is available for ₹3,230. What is the rate of discount?
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Answer: (D) 15%

Explanation

The discount percentage is calculated using the formula: Discount % = ((Marked Price - Selling Price) / Marked Price) × 100

Given: Marked Price = ₹3,800 Selling Price = ₹3,230

Discount Amount = ₹3,800 - ₹3,230 = ₹570

Discount % = (570 / 3800) × 100 = 15%

232Quantitative Aptitude
The mean proportion of 6 and 54 is:
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Answer: (D) 18

Explanation

The mean proportion between two numbers a and b is defined by the formula √(a × b).

Substituting the values a = 6 and b = 54:

Mean proportion = √(6 × 54) = √324 = 18

233Quantitative Aptitude
In a right-angled triangle XYZ, if X = 60° and Y = 30°, then find the value of sin (X - Y).
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Answer: (B) ½

Explanation

The values of the angles are given as X = 60° and Y = 30°.

The difference between the two angles is calculated as: X - Y = 60° - 30° = 30°

Substituting this value into the expression gives: sin(X - Y) = sin(30°)

From the standard trigonometric ratio values, sin(30°) is equal to ½.

234Quantitative Aptitude
What will be the greatest number 32a78b, which is divisible by 3 but NOT divisible by 9? (Where a and b are single digit number).
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Answer: (D) 329784

Explanation

A number is divisible by 3 if the sum of its digits is divisible by 3, and it is divisible by 9 if the sum of its digits is divisible by 9. For the number 329784, the sum of the digits is 3 + 2 + 9 + 7 + 8 + 4 = 33. Because 33 is divisible by 3 but not divisible by 9, the number 329784 is divisible by 3 but not by 9. This number represents the greatest configuration because the thousands digit a is maximized at 9, and the unit digit b is 4, keeping the digit sum at 33 which avoids divisibility by 9 while maintaining divisibility by 3.

235Quantitative Aptitude
A solid cuboid is melted to form a number of cubes. If the length, breadth and height of a cuboid are 20 cm, 16 cm and 8 cm, and the edge of each cube is 4 cm, then the number of cubes is:
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Answer: (C) 40

Explanation

The total volume of material remains conserved when the solid cuboid is melted to form smaller cubes.

Volume of the cuboid = Length × Breadth × Height = 20 cm × 16 cm × 8 cm = 2560 cm³

Volume of each small cube = Side³ = 4 cm × 4 cm × 4 cm = 64 cm³

Number of cubes = Volume of cuboid ÷ Volume of each cube Number of cubes = 2560 ÷ 64 = 40

236Quantitative Aptitude
The cost price of 240 articles is the same as the selling price of x articles. If profit is 20% , find the value of x.
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Answer: (A) 200

Explanation

Let the cost price of one article be CP and the selling price of one article be SP.

We are given that the cost price of 240 articles is equal to the selling price of x articles: 240 × CP = x × SP SP / CP = 240 / x

The profit percentage is 20%, which means: SP = CP × (1 + Profit%/100) SP = CP × (1 + 20/100) SP = CP × 1.2 SP / CP = 6/5

Equating the two ratios for SP / CP: 240 / x = 6/5 x = (240 × 5) / 6 x = 40 × 5 x = 200

237Quantitative Aptitude
If (a + b + c) = 17 , and (a² + b² + c²) = 101 , find the value of (a - b)² + (b - c)² + (c - a)² .
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Answer: (B) 14

Explanation

Using the algebraic identity (a + b + c)² = a² + b² + c² + 2(ab + bc + ca), we substitute the given values:

17² = 101 + 2(ab + bc + ca) 289 = 101 + 2(ab + bc + ca) 2(ab + bc + ca) = 289 - 101 2(ab + bc + ca) = 188

The expression to be evaluated is: (a - b)² + (b - c)² + (c - a)² = a² - 2ab + b² + b² - 2bc + c² + c² - 2ca + a² = 2(a² + b² + c²) - 2(ab + bc + ca)

Substituting the values into this expression: 2(101) - 188 = 202 - 188 = 14

238Quantitative Aptitude
From a cask of milk containing 64 litres, 8 litres are drawn out and the cask is filled up with water. If the same process is repeated a second, then a third time, what will be the proportion of milk to water in the resulting mixture?
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Answer: (D) 343: 169

Explanation

The remaining quantity of milk after n operations of drawing out and replacing a liquid of initial quantity a with b quantity of another liquid is given by the formula:

Remaining Milk = a × (1 - b/a)ⁿ

Here, the initial quantity of milk (a) is 64 litres, the quantity drawn out and replaced (b) is 8 litres, and the process is performed 3 times in total (n = 3).

Remaining Milk = 64 × (1 - 8/64)³ = 64 × (1 - 1/8)³ = 64 × (7/8)³ = 64 × 343/512 = 343/8 litres

Since the total volume of the mixture remains 64 litres, the quantity of water in the mixture is:

Quantity of Water = Total Volume - Remaining Milk = 64 - 343/8 = (512 - 343)/8 = 169/8 litres

The proportion of milk to water in the resulting mixture is:

Milk : Water = (343/8) : (169/8) = 343 : 169

239Quantitative Aptitude
If (x² + 1)/(x) = 5 , then find the value of x⁴ + (1)/(x⁴) - 36 .
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Answer: (A) 491

Explanation

Given the equation (x² + 1)/x = 5, we can simplify it by dividing each term in the numerator by x to get x + 1/x = 5.

Squaring both sides of this equation using the algebraic identity (a + b)² = a² + 2ab + b² gives: (x + 1/x)² = 5² x² + 2 + 1/x² = 25 x² + 1/x² = 23

Squaring this result again to find the fourth power: (x² + 1/x²)² = 23² x⁴ + 2 + 1/x⁴ = 529 x⁴ + 1/x⁴ = 527

Substituting this value into the required expression x⁴ + 1/x⁴ - 36 yields: 527 - 36 = 491

240Quantitative Aptitude
Table given below shows the performance of students of class-10 according to NAS survey in Delhi. | | Below Basic | Basic | Proficient | Advance | |---|---|---|---|---| | MIL | 40 | 35 | 20 | 5 | | Maths | 19 | 45 | 25 | 11 | | Science | 41 | 29 | 25 | 5 | | Social Science | 34 | 21 | 33 | 12 | | English | 8 | 5 | 34 | 52 | The highest average performance is for ______.
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Answer: (B) Below Basic

Explanation

To find the performance category with the highest average, we calculate the sum and average percentage of students for each category across all five subjects.

For the Below Basic category, the sum of the percentages across the subjects MIL, Maths, Science, Social Science, and English is 40 + 19 + 41 + 34 + 8 = 142. Dividing this sum by the 5 subjects gives an average of 142 / 5 = 28.4%.

For the Basic category, the sum is 35 + 45 + 29 + 21 + 5 = 135, which gives an average of 135 / 5 = 27%.

For the Proficient category, the sum is 20 + 25 + 25 + 33 + 34 = 137, which gives an average of 137 / 5 = 27.4%.

For the Advance category, the sum is 5 + 11 + 5 + 12 + 52 = 85, which gives an average of 85 / 5 = 17%.

The Below Basic category has the highest average value of 28.4%.

241Quantitative Aptitude
The radius of a circle is 5 cm and the length of one of its chords is 8 cm. Find the distance of the chord from the centre.
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Answer: (A) 3 cm

Explanation

A perpendicular line drawn from the centre of a circle to a chord bisects the chord. For a chord of length 8 cm, the perpendicular from the centre divides it into two equal segments of 4 cm each.

The radius of the circle, the perpendicular distance from the centre to the chord, and the half-chord form a right-angled triangle. Applying Pythagoras' theorem:

d² + (L/2)² = R²

Here, R is the radius of 5 cm, and L/2 is the half-chord of 4 cm. Substituting these values into the formula:

d² + 4² = 5²

d² + 16 = 25

d² = 9

d = √9 = 3 cm

242Quantitative Aptitude
The length of the chord of a circle is 8 cm and the perpendicular distance between the centre and the chord is 3 cm then the diameter of the circle is equal to:
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Answer: (B) 10 cm

Explanation

A perpendicular drawn from the centre of a circle to a chord bisects the chord. Therefore, the length of half the chord is 8 / 2 = 4 cm.

This perpendicular, half of the chord, and the radius of the circle form a right-angled triangle. By applying the Pythagorean theorem, the radius r is calculated as:

r = √(3² + 4²) r = √(9 + 16) r = √25 r = 5 cm

The diameter of the circle is twice the radius:

Diameter = 2 × r = 2 × 5 = 10 cm

243Quantitative Aptitude
If a² + b² + c² = ab + bc + ca, then the value of (11a³ + 13b⁴ + 15c⁴)/(16a²b² + 19b²c² + 17c²a²) is:
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Answer: (C) ¾

Explanation

The algebraic identity states that: a² + b² + c² - ab - bc - ca = ½((a - b)² + (b - c)² + (c - a)²)

Since we are given a² + b² + c² = ab + bc + ca, it follows that: ½((a - b)² + (b - c)² + (c - a)²) = 0

For the sum of squares of real numbers to equal zero, each term must individually be zero: a - b = 0, b - c = 0, c - a = 0

This gives: a = b = c

Substituting a = b = c into the expression (where the numerator term 11a³ is 11a⁴ to maintain homogeneity of degree 4): (11a⁴ + 13b⁴ + 15c⁴) / (16a²b² + 19b²c² + 17c²a²) = (11a⁴ + 13a⁴ + 15a⁴) / (16a⁴ + 19a⁴ + 17a⁴) = 39a⁴ / 52a⁴ = 39/52 = 3/4

244Quantitative Aptitude
Find the length of a tangent drawn to a circle with radius 8 cm from a point 17 cm from the centre of the circle.
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Answer: (C) 15 cm

Explanation

Correct Answer: C

A tangent to a circle is perpendicular to the radius at the point of contact. This relationship forms a right-angled triangle where the hypotenuse is the distance from the external point to the centre of the circle (d = 17 cm), one leg is the radius of the circle (r = 8 cm), and the other leg is the length of the tangent (t).

Applying Pythagoras' theorem: d² = r² + t²

Substituting the given values into the formula: 17² = 8² + t² 289 = 64 + t² t² = 289 - 64 t² = 225 t = √225 = 15 cm

245Quantitative Aptitude
The average of 18 integers is found to be 41. It was detected that by mistake the integer 127 was copied as 341, while calculating the average. Then the correct average is:
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Answer: (C) 29⅑

Explanation

The correct average can be directly computed using the formula:

Correct Average = Incorrect Average + (Correct Value - Incorrect Value) / Number of Observations

Substituting the given values into the formula:

Correct Average = 41 + (127 - 341) / 18 Correct Average = 41 - 214 / 18 Correct Average = 41 - 107 / 9 Correct Average = 41 - 11 8/9 Correct Average = 29 1/9

246Quantitative Aptitude
Simplify: [⅕of(35 - 5) + 4]² -[9 - {7 + (2 - 3)} + 5]²
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Answer: (B) 36

Explanation

Applying the BODMAS rule, we simplify the expression by evaluating the innermost brackets first.

For the first term: ((1/5) of (35 - 5) + 4)² = ((1/5) of 30 + 4)² = (6 + 4)² = 10² = 100

For the second term: (9 - {7 + (2 - 3)} + 5)² = (9 - {7 - 1} + 5)² = (9 - 6 + 5)² = (3 + 5)² = 8² = 64

Subtracting the second term from the first term yields: 100 - 64 = 36

247Quantitative Aptitude
On an item having a marked price of 1,600, successive discounts of y% and z% were offered. As a result, the item was sold for 1,056. Which of the options below CANNOT be a set of values for y and z ?
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Answer: (A) y = 22, z = 15.5

Explanation

The total discount percentage required to reduce the marked price from 1,600 to the selling price of 1,056 is calculated as:

Total Discount = (1,600 - 1,056) ÷ 1,600 × 100 = 544 ÷ 1,600 × 100 = 34%

The formula for the net equivalent discount of two successive discounts of y% and z% is:

Net Discount = y + z - (y × z) ÷ 100

Substituting the values y = 22 and z = 15.5 into the formula:

Net Discount = 22 + 15.5 - (22 × 15.5) ÷ 100 Net Discount = 37.5 - 341 ÷ 100 Net Discount = 37.5 - 3.41 = 34.09%

Since the resulting net discount is 34.09% instead of the required 34%, these values cannot be the successive discounts offered on the item.

248Quantitative Aptitude
cos 2θ =
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Answer: (B) 2cos²θ - 1

Explanation

The double-angle formula for cosine is: cos 2θ = cos²θ - sin²θ

Using the fundamental trigonometric identity: sin²θ + cos²θ = 1 sin²θ = 1 - cos²θ

Substituting sin²θ into the double-angle formula gives: cos 2θ = cos²θ - (1 - cos²θ) cos 2θ = 2cos²θ - 1

This directly derives the identity.

249Quantitative Aptitude
Simplify the expression: (c + d)² - (c - d)²
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Answer: (A) 4cd

Explanation

Using the standard algebraic identity for the difference of squares, a² - b² = (a + b)(a - b), let a = c + d and b = c - d.

Substituting these values into the identity:

((c + d) + (c - d))((c + d) - (c - d))

Simplifying the terms inside the parentheses:

(c + d + c - d)(c + d - c + d) = (2c)(2d) = 4cd

Alternatively, expanding each squared term directly:

(c + d)² = c² + 2cd + d²

(c - d)² = c² - 2cd + d²

Subtracting the second expanded expression from the first yields:

(c² + 2cd + d²) - (c² - 2cd + d²) = c² + 2cd + d² - c² + 2cd - d² = 4cd

250Quantitative Aptitude
Aastha spends 15% of her total income on travelling and 20% of the rest income on food and then donates ₹380 and after that, she saves ₹2,000. Find her total income (in ₹).
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Answer: (A) 3,500

Explanation

Let the total income of Aastha be I.

She spends 15% of her total income on travelling. Remaining income = I - 0.15I = 0.85I

She spends 20% of this remaining income on food. Remaining income after food = 0.85I × (1 - 0.20) = 0.85I × 0.80 = 0.68I

From this remaining amount, she donates ₹380 and saves ₹2,000. 0.68I - 380 = 2000 0.68I = 2380 I = 2380 / 0.68 I = 3500

The total income of Aastha is ₹3,500.

Answer key

Tap a number to jump to that question.

  1. 201C
  2. 202B
  3. 203C
  4. 204C
  5. 205D
  6. 206D
  7. 207C
  8. 208C
  9. 209C
  10. 210C
  11. 211B
  12. 212A
  13. 213B
  14. 214D
  15. 215C
  16. 216B
  17. 217B
  18. 218D
  19. 219C
  20. 220B
  21. 221C
  22. 222A
  23. 223A
  24. 224A
  25. 225C
  26. 226A
  27. 227D
  28. 228A
  29. 229D
  30. 230B
  31. 231D
  32. 232D
  33. 233B
  34. 234D
  35. 235C
  36. 236A
  37. 237B
  38. 238D
  39. 239A
  40. 240B
  41. 241A
  42. 242B
  43. 243C
  44. 244C
  45. 245C
  46. 246B
  47. 247A
  48. 248B
  49. 249A
  50. 250A

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