SSC CHSL2015Quantitative AptitudePage 1 of 4

SSC CHSL Tier 1 2015 Quantitative Aptitude Questions with Solutions

Questions from every shift of the year, grouped by section. Questions 1–50 of 196, each with the correct answer and a step-by-step explanation.

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1Quantitative Aptitude
150 workers were engaged to finish a piece of work in a certain number of days. Four workers dropped on the second day, four more workers dropped on third day and so on. It takes 8 more days to finish the work no. Find the number of days in which the work was completed ?
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Answer: (D) 25

Explanation

Let d be the original number of days required to finish the work by 150 workers. Total work = 150d worker-days

When workers drop out daily, the number of workers working on consecutive days forms an arithmetic progression with first term a = 150 and common difference d_diff = -4: 150, 146, 142, ...

Let n be the total number of days taken to finish the work. Since it takes 8 more days, n = d + 8, which gives d = n - 8.

The sum of this arithmetic progression for n terms represents the total work done: Total work = (n/2) × [2a + (n - 1)d_diff] Total work = (n/2) × [2 × 150 + (n - 1)(-4)] Total work = (n/2) × [304 - 4n] Total work = n(152 - 2n)

Equating the two expressions for the total work: 150(n - 8) = n(152 - 2n) 150n - 1200 = 152n - 2n² 2n² - 2n - 1200 = 0 n² - n - 600 = 0 (n - 25)(n + 24) = 0

Solving for the positive value of n gives n = 25.

2Quantitative Aptitude
A mixture contains milk and water in the ratio 5 : 1. On adding 5 litres of water, the ratio of milk and water becomes 5 : 2. The quantity of milk in the mixture is
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Answer: (A) 25 litres

Explanation

Let the initial ratio of milk and water be 5 : 1, which means there are 5 parts of milk and 1 part of water.

When 5 litres of water is added to the mixture, the new ratio of milk and water becomes 5 : 2.

Since only water is added, the quantity of milk remains constant at 5 parts in both ratios. The water content increases from 1 part to 2 parts, which is an increase of 2 - 1 = 1 part.

This increase of 1 part corresponds directly to the 5 litres of water added. Therefore, 1 part is equal to 5 litres.

The quantity of milk in the mixture is represented by 5 parts. Quantity of milk = 5 × 5 = 25 litres.

3Quantitative Aptitude
The average of all the odd integers between 2 and 22 is
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Answer: (D) 12

Explanation

The odd integers between 2 and 22 are 3, 5, 7, 9, 11, 13, 15, 17, 19, and 21. These numbers form an arithmetic progression where the first term a is 3 and the last term l is 21.

For any arithmetic progression, the average of the terms is equal to the average of the first and the last term, given by the formula:

Average = (a + l) / 2

Substituting the values:

Average = (3 + 21) / 2 = 24 / 2 = 12

4Quantitative Aptitude
The least number that should be subtracted from the number 32146 to make it a perfect square is:
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Answer: (C) 105

Explanation

To find the least number to be subtracted, we calculate the square root of 32146 using the long division method.

Pairing the digits from right to left, we get 3, 21, and 46. First, the largest square number less than or equal to 3 is 1, with a remainder of 2. Bringing down 21 gives 221.

Doubling the quotient 1 gives 2. Since 27 × 7 = 189, we subtract 189 from 221 to get 32. Bringing down 46 gives 3246.

Doubling the quotient 17 to get 34, we find the next digit. Since 349 × 9 = 3141, we subtract 3141 from 3246, which leaves a remainder of 105.

The remainder 105 represents the excess value that prevents 32146 from being a perfect square. Subtracting 105 from 32146 results in 32041, which is the perfect square of 179.

5Quantitative Aptitude
The diameter of a 120 cm long roller is 84 cm. It takes 500 complete revolutions of the roller to level a ground. The cost of levelling the ground at ₹1.50 per sq.m. is:
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Answer: (A) 2376

Explanation

The roller is a cylinder where the length (h) is 1.2 m and the radius (r) is half of the 84 cm diameter, which is 42 cm or 0.42 m.

The area covered by the roller in one complete revolution is equal to its curved surface area: Area per revolution = 2πrh Area per revolution = 2 × (22/7) × 0.42 × 1.2 = 3.168 m²

The total area leveled in 500 revolutions is: Total area = 500 × 3.168 = 1584 m²

The total cost of leveling the ground at the rate of ₹1.50 per m² is: Total cost = 1584 × 1.50 = ₹2376

6Quantitative Aptitude
Mr. Dutta desired to deposit his retirement benefit of ₹3 lacs partly to a post office and partly to a bank at 10% and 6% interests respectively. If his monthly interest income was ₹2000, then the difference of his deposits in the post office and in the bank was:
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Answer: (A) Nil

Explanation

Let P be the amount deposited in the post office and B be the amount deposited in the bank. The total amount deposited is: P + B = 3,00,000

The monthly interest income is ₹2,000. Thus, the annual interest income is: 2,000 × 12 = ₹24,000

Since the annual interest rates for the post office and the bank are 10% and 6% respectively, we can express this as: 0.10 P + 0.06 B = 24,000

Multiplying this equation by 100 to clear decimals: 10 P + 6 B = 24,00,000 5 P + 3 B = 12,00,000

Now, multiplying the first equation (P + B = 3,00,000) by 3: 3 P + 3 B = 9,00,000

Subtracting this from the simplified interest equation: (5 P + 3 B) - (3 P + 3 B) = 12,00,000 - 9,00,000 2 P = 3,00,000 P = 1,50,000

Substituting P back into the total deposit equation gives: B = 1,50,000

The difference between the two deposits is: 1,50,000 - 1,50,000 = 0 (Nil)

7Quantitative Aptitude
Volume of a right circular cylinder of height 21 cm and base radios 5 cm is:
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Answer: (B) 1650cm³

Explanation

The volume V of a right circular cylinder with base radius r and height h is given by the formula V = πr²h.

Substituting the given values, where radius r = 5 cm, height h = 21 cm, and π = 22/7:

V = (22/7) × 5² × 21 V = (22/7) × 25 × 21 V = 22 × 25 × 3 V = 1650 cm³

8Quantitative Aptitude
A right triangle with sides 9 cm, 12 cm and 15 cm is rotated about the side of 9 cm to form a cone. The volume of the cone so formed is:
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Answer: (A) 324π cm³

Explanation

The volume V of a cone is given by the formula V = 1/3 × π × r² × h, where r is the radius of the base and h is the height.

Taking the side of 9 cm as the base radius (r) and the side of 12 cm as the height (h), we substitute these values into the formula:

V = 1/3 × π × 9² × 12 V = 1/3 × π × 81 × 12 V = 27 × 12 × π V = 324π cm³

9Quantitative Aptitude
If 12 men working 8 hours a day complete the work in 10 days, how long would 16 men working 7 1/2 hours a day take to complete the same work?
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Answer: (A) 8

Explanation

The total work is constant and can be expressed using the formula M₁ × D₁ × H₁ = M₂ × D₂ × H₂, where M is the number of men, D is the number of days, and H is the number of hours worked per day.

Substituting the given values: M₁ = 12, D₁ = 10, H₁ = 8 M₂ = 16, H₂ = 7 ½ = 7.5 D₂ = x

12 × 10 × 8 = 16 × x × 7.5 960 = 120 × x x = 960 ÷ 120 x = 8

10Quantitative Aptitude
The average age of mother and her six children is 12 years, which is reduced by 5 years if the age of the mother is excluded. The age of the mother (in yrs) is:
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Answer: (C) 42

Explanation

The total number of family members initially is 7 (1 mother and 6 children). Since their average age is 12 years, the sum of their ages is 7 × 12 = 84 years.

When the mother is excluded, the remaining 6 children have an average age that is reduced by 5 years, making the new average age 12 - 5 = 7 years. The sum of the children's ages is 6 × 7 = 42 years.

Subtracting the sum of the children's ages from the total sum gives the age of the mother: 84 - 42 = 42 years.

11Quantitative Aptitude
If x = a(sin θ + cos θ) and y = b(sin θ - cos θ) , then the value of (x²)/(a²) + (y²)/(b²) is:
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Answer: (C) 2

Explanation

Dividing the given equations by a and b respectively, we get: x/a = sin θ + cos θ y/b = sin θ - cos θ

Squaring both equations on both sides: x²/a² = (sin θ + cos θ)² = sin² θ + cos² θ + 2 sin θ cos θ y²/b² = (sin θ - cos θ)² = sin² θ + cos² θ - 2 sin θ cos θ

Adding these two squared equations: x²/a² + y²/b² = (sin² θ + cos² θ + 2 sin θ cos θ) + (sin² θ + cos² θ - 2 sin θ cos θ) x²/a² + y²/b² = 2(sin² θ + cos² θ)

Using the standard trigonometric identity sin² θ + cos² θ = 1: x²/a² + y²/b² = 2 × 1 = 2

12Quantitative Aptitude
If a² + b² + c² = 2(a - b - c) - 3 , then the value of a + b + c is;
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Answer: (B) - 1

Explanation

The given equation is: a² + b² + c² = 2(a - b - c) - 3

Rearranging all terms to the left side: a² + b² + c² - 2a + 2b + 2c + 3 = 0

By splitting the constant 3 into 1 + 1 + 1, we can group the terms to form perfect squares: (a² - 2a + 1) + (b² + 2b + 1) + (c² + 2c + 1) = 0 (a - 1)² + (b + 1)² + (c + 1)² = 0

Since the sum of squares of real numbers must be zero, each independent term must equal zero: a - 1 = 0 → a = 1 b + 1 = 0 → b = -1 c + 1 = 0 → c = -1

Calculating the required value: a + b + c = 1 + (-1) + (-1) = -1

13Quantitative Aptitude
If x + y = 4 , x² + y² = 14 and x > y , Then the correct value of x and y is:
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Answer: (C) 2 + √3,2 - √3

Explanation

Using the identity (x + y)² = x² + y² + 2xy, we substitute the given values: 4² = 14 + 2xy 16 = 14 + 2xy 2xy = 2 xy = 1

Next, we use the identity (x - y)² = (x + y)² - 4xy to find the difference between x and y: (x - y)² = 4² - 4(1) (x - y)² = 16 - 4 = 12 x - y = ±√12 = ±2√3

Since it is given that x > y, the value of x - y must be positive: x - y = 2√3

Now we solve the system of linear equations: x + y = 4 x - y = 2√3

Adding these two equations: 2x = 4 + 2√3 x = 2 + √3

Subtracting the second equation from the first: 2y = 4 - 2√3 y = 2 - √3

14Quantitative Aptitude
If C o s θ + S i n θ = m and S e c θ + C o s e c θ = n then the value of n(m² - 1) is equal to:
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Answer: (D) 2m

Explanation

Given: cos θ + sin θ = m sec θ + cosec θ = n

Expressing n in terms of sine and cosine: n = 1/cos θ + 1/sin θ n = (sin θ + cos θ) / (sin θ cos θ)

Substituting cos θ + sin θ = m: n = m / (sin θ cos θ) sin θ cos θ = m/n

Now, squaring both sides of the first equation: m² = (cos θ + sin θ)² m² = cos² θ + sin² θ + 2 sin θ cos θ m² = 1 + 2 sin θ cos θ m² - 1 = 2 sin θ cos θ

Substituting this expression into n(m² - 1): n(m² - 1) = n(2 sin θ cos θ) n(m² - 1) = n × 2(m/n) n(m² - 1) = 2m

15Quantitative Aptitude
The value of (1)/(1 + √2 + √3) + (1)/(1 - √2 + √3) is:
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Answer: (A) 1

Explanation

Let a = 1 + √3 and b = √2.

The given expression can be written in terms of a and b as: 1/(a + b) + 1/(a - b)

By taking the common denominator, we simplify this expression to: ((a - b) + (a + b)) / ((a + b)(a - b)) = 2a / (a² - b²)

Next, we calculate the terms in the denominator: a² = (1 + √3)² = 1 + 2√3 + 3 = 4 + 2√3 b² = (√2)² = 2 a² - b² = 4 + 2√3 - 2 = 2 + 2√3

Since 2a = 2(1 + √3) = 2 + 2√3, we have: a² - b² = 2a

Substituting this back into the simplified expression gives: 2a / 2a = 1

16Quantitative Aptitude
If the radius of a sphere is increased by 2 cm, then its surface area increases by 352 cm² . The radius of the sphere initially was: (use π = ²²⁄₇ )
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Answer: (C) 6 cm

Explanation

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

When the radius is increased by 2 cm, the new radius is r + 2. The change in the surface area is: 4π(r + 2)² - 4πr² = 352

Expanding and simplifying this equation: 4π(r² + 4r + 4 - r²) = 352 16π(r + 1) = 352

Using π = 22/7: 16 × (22/7) × (r + 1) = 352 (22/7) × (r + 1) = 22 r + 1 = 22 × (7/22) r + 1 = 7 r = 6 cm

17Quantitative Aptitude
The ratio of two numbers is 3: 4 and their HCF is 15. Then the sum of the two numbers is:
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Answer: (A) 105

Explanation

Let the two numbers be 3x and 4x, where x is their highest common factor. Since 3 and 4 are co-prime, their highest common factor is indeed x.

Given that the highest common factor is 15, we have x = 15.

The sum of the two numbers is 3x + 4x = 7x.

Substituting the value of x into the sum formula: Sum = 7 × 15 = 105

18Quantitative Aptitude
A shopkeeper fixes the price of an article at 30% higher than its actual cost. If he sells it at 10% discount on marked price then, the profit is:
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Answer: (B) 17%

Explanation

Let the cost price of the article be ₹100.

The shopkeeper marks the price 30% higher than the cost price, so the marked price is: 100 + 30% of 100 = ₹130

He then offers a 10% discount on this marked price: Discount = 10% of 130 = ₹13

The selling price of the article is: 130 - 13 = ₹117

The profit earned is: Selling Price - Cost Price = 117 - 100 = ₹17

Since the cost price is ₹100, this profit corresponds to a profit percentage of 17%.

Alternatively, using the successive percentage change formula: Net percentage change = x + y + (xy / 100) Here, x = +30% and y = -10%. Net percentage change = 30 - 10 + (30 × -10 / 100) = 20 - 3 = 17% profit.

19Quantitative Aptitude
The three successive angles of a cyclic quadrilateral are in the ratio 1:3:4, find the measure of the fourth angle ?
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Answer: (B) 72°

Explanation

Let the four successive angles of the cyclic quadrilateral be A, B, C, and D. In a cyclic quadrilateral, the sum of opposite angles is 180°, which gives A + C = 180° and B + D = 180°.

The first three successive angles A, B, and C are in the ratio 1:3:4. Let A = k, B = 3k, and C = 4k.

Using the relation for opposite angles A and C: A + C = 180° k + 4k = 180° 5k = 180° k = 36°

Angle B is calculated as: B = 3 × 36° = 108°

Using the relation for opposite angles B and D to find the fourth angle D: B + D = 180° 108° + D = 180° D = 180° - 108° = 72°

20Quantitative Aptitude
If the Cost Price of 25 chairs is equal to the Selling Price of 30 chairs, then the loss % is:
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Answer: (A) 16 2/3%

Explanation

Let the Cost Price of one chair be CP and the Selling Price of one chair be SP.

According to the question: 25 × CP = 30 × SP

This can be rewritten to find the ratio of Selling Price to Cost Price: SP / CP = 25 / 30 = 5 / 6

The formula for loss percentage is: Loss % = (CP - SP) / CP × 100 Loss % = (1 - SP / CP) × 100 Loss % = (1 - 5/6) × 100 = 1/6 × 100 = 16 ⅔%

21Quantitative Aptitude
The liquids, X and Y are mixed in the ratio of 3: 2 and the mixture is sold at ₹11 per litre at a profit of 10%. If the liquid X costs ₹2 more per litre then Y, the cost of X per litre is (in ₹):
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Answer: (B) 10.80

Explanation

The selling price of the mixture is ₹11 per litre at a profit of 10%. The cost price of the mixture is calculated using the relation:

Cost Price of mixture = Selling Price / (1 + Profit Rate) = 11 / (1 + 0.10) = 11 / 1.1 = ₹10 per litre

Let the cost price of liquid Y be ₹y per litre. Since liquid X costs ₹2 more per litre than Y, the cost price of liquid X is ₹(y + 2) per litre.

Since the liquids X and Y are mixed in the ratio 3:2, the weighted average of their individual costs yields the cost price of the mixture:

(3 × (y + 2) + 2 × y) / (3 + 2) = 10

(3y + 6 + 2y) / 5 = 10

5y + 6 = 50

5y = 44

y = 8.80

The cost price of liquid X is y + 2:

8.80 + 2 = ₹10.80 per litre

22Quantitative Aptitude
The value of the expression S i n²1° + S i n²11° + S i n²21° + S i n²31° + S i n²41° + S i n²45° +S i n²49° + S i n²59° + S i n²69° + S i n²79° + S i n²89°
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Answer: (B) 5 ½

Explanation

Using the complementary angle identity sin(90° - θ) = cos θ, we have sin² θ + sin² (90° - θ) = sin² θ + cos² θ = 1.

By pairing the terms whose angles sum to 90°, we can simplify the expression: (sin² 1° + sin² 89°) + (sin² 11° + sin² 79°) + (sin² 21° + sin² 69°) + (sin² 31° + sin² 59°) + (sin² 41° + sin² 49°) + sin² 45°

Substituting the identity for each of the five pairs and using the standard value sin 45° = 1/√2: 1 + 1 + 1 + 1 + 1 + (1/√2)² = 5 + ½ = 5 ½

23Quantitative Aptitude
The average of 8 numbers is 21. If each of the numbers is multiplied by 8, the average of the new set of numbers is
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Answer: (B) 168

Explanation

If a set of numbers has an average of A, and each number in the set is multiplied by a constant k, then the average of the new set of numbers is also multiplied by the same constant k.

New Average = Original Average × k

Given that the original average is 21 and each number is multiplied by 8, we substitute these values into the relation:

New Average = 21 × 8 = 168

This linear property of the arithmetic mean ensures that scaling every data point by a factor scales the entire mean by that same factor.

24Quantitative Aptitude
If x(x + y + z) = 20, y(x + y + z) = 30, | z(x + y + z) = 50, then the value of 2(x + y + z) is:
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Answer: (D) 20

Explanation

Let S = x + y + z. The given equations are:

xS = 20 yS = 30 zS = 50

Adding these three equations together:

xS + yS + zS = 20 + 30 + 50 S(x + y + z) = 100 S² = 100 S = ±10

Therefore, x + y + z = ±10.

Multiplying by 2 gives:

2(x + y + z) = 2(±10) = ±20

Taking the positive value present in the options, the result is 20.

25Quantitative Aptitude
If Cos A + Sin A = √2 Cos A then Cos A- Sin A is equal to: (Where 0°< A< 90° )
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Answer: (C) ; [√2 S i n A ; ]

Explanation

Using the standard trigonometric identity: (cos A + sin A)² + (cos A - sin A)² = 2(cos² A + sin² A)

Since cos² A + sin² A = 1, this simplifies to: (cos A + sin A)² + (cos A - sin A)² = 2

Substitute the given equation cos A + sin A = √2 cos A into the identity: (√2 cos A)² + (cos A - sin A)² = 2 2 cos² A + (cos A - sin A)² = 2 (cos A - sin A)² = 2 - 2 cos² A (cos A - sin A)² = 2(1 - cos² A)

Using the identity 1 - cos² A = sin² A: (cos A - sin A)² = 2 sin² A

Taking the positive square root on both sides for the acute angle A gives: cos A - sin A = √2 sin A

26Quantitative Aptitude
The straight line y=3x must pass through the point:
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Answer: (B) (0,0)

Explanation

A point lies on a given line if its coordinates satisfy the equation of the line. For the line y = 3x, substituting the coordinates of the origin, x = 0 and y = 0, we get 0 = 3 × 0, which is a true statement. This confirms that the line passes through the point (0,0).

27Quantitative Aptitude
Find out the wrong number in the sequence:
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Answer: (D) 200

Explanation

The given sequence is 40960, 10240, 2560, 640, 200, 40, 10.

Analyzing the relationship between consecutive terms shows that each number is obtained by dividing the preceding term by 4:

40960 ÷ 4 = 10240 10240 ÷ 4 = 2560 2560 ÷ 4 = 640

Following this consistent pattern, the next term should be:

640 ÷ 4 = 160

Instead of 160, the sequence lists 200. Applying the rule of division by 4 to the correct term 160 continues to yield the rest of the sequence correctly:

160 ÷ 4 = 40 40 ÷ 4 = 10

Thus, 200 is the incorrect number in the sequence.

28Quantitative Aptitude
AB is a diameter of a circle having centre at O, PQ is a chord which does not intersect AB. Join AP and BQ. If ∠ BAP = ∠ ABQ , then ABQP is a:
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Answer: (A) Cyclic trapezium

Explanation

Let O be the center of the circle, and let the diameter AB lie along the horizontal line. For any point P on the circle, the inscribed angle ∠BAP is equal to half of the central angle ∠BOP. Similarly, for any point Q on the circle, the inscribed angle ∠ABQ is equal to half of the central angle ∠AOQ.

Given that ∠BAP = ∠ABQ, it follows that their corresponding central angles are equal, meaning ∠BOP = ∠AOQ. Since O is the midpoint of the diameter AB, this equality of central angles means that the points P and Q are symmetric with respect to the perpendicular bisector of AB.

Due to this symmetry, the perpendicular distance from P to AB is equal to the perpendicular distance from Q to AB. Therefore, the chord PQ is parallel to the diameter AB. A quadrilateral with one pair of opposite sides parallel is a trapezium, and since all its vertices lie on the circle, ABQP is a cyclic trapezium.

29Quantitative Aptitude
In what proportions must water be added with milk to gain 20% by selling the mixture at cost price?
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Answer: (C) 1:5

Explanation

When water is added to milk and the mixture is sold at the cost price of milk, the percentage gain is given by the formula:

Gain% = (Quantity of Water / Quantity of Milk) × 100

Substituting the given gain of 20%:

20 = (Quantity of Water / Quantity of Milk) × 100

Quantity of Water : Quantity of Milk = 20 : 100 = 1 : 5

30Quantitative Aptitude
The centroid of an equilateral triangle ABC is G and AB=10 cm. The length of AG (in cm)
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Answer: (B) 10√3 3

Explanation

For an equilateral triangle of side a, the altitude h is given by the formula: h = (√3/2) × a

The centroid G divides the altitude/median in the ratio 2:1 from the vertex. Therefore, the distance from a vertex to the centroid AG is: AG = (2/3) × h = (2/3) × (√3/2) × a = a/√3

Given side a = 10 cm, we substitute the value: AG = 10/√3 cm

Rationalizing the denominator by multiplying the numerator and denominator by √3 gives: AG = (10√3)/3 cm

31Quantitative Aptitude
If (x - x tan² 30°)/(1 + tan² 30°) = sin² 30° + 4 cot² 45° - sec² 60° Then value of x is:
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Answer: (D) ½

Explanation

To find the value of x, substitute the standard trigonometric values into the given equation:

tan 30° = 1/√3 sin 30° = ½ cot 45° = 1 sec 60° = 2

Substituting these values into the left-hand side (LHS) of the equation: LHS = (x - x(1/√3)²) / (1 + (1/√3)²) LHS = (x - x/3) / (1 + 1/3) LHS = (2x/3) / (4/3) LHS = 2x/4 LHS = x/2

Substituting the values into the right-hand side (RHS) of the equation: RHS = (½)² + 4(1)² - (2)² RHS = ¼ + 4 - 4 RHS = ¼

Equating LHS and RHS: x/2 = ¼ x = 2/4 x = ½

32Quantitative Aptitude
A train is 250 m long. If the train takes 50 seconds to cross a tree by the railway line, then the speed of the train in km/hr is:
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Answer: (C) 18

Explanation

To find the speed of the train, we use the formula:

Speed = Distance / Time

When a train crosses a tree, the distance covered is equal to the length of the train.

Distance = 250 m Time = 50 s

Speed = 250 / 50 = 5 m/s

To convert the speed from meters per second (m/s) to kilometers per hour (km/hr), we multiply by the conversion factor 18/5:

Speed = 5 × 18/5 = 18 km/hr

33Quantitative Aptitude
The marked price of a CD is ₹250. It is sold for ₹225. The rate of discount is:
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Answer: (A) 10%

Explanation

The rate of discount is calculated using the formula:

Discount = Marked Price - Selling Price

Given that the Marked Price is ₹250 and the Selling Price is ₹225:

Discount = 250 - 225 = ₹25

The discount percentage is:

Discount % = (Discount / Marked Price) × 100

Discount % = (25 / 250) × 100 = 10%

34Quantitative Aptitude
Mohan purchased a bag with 20 percent discount on the labelled price. He sold it with 40 percent profit on the price he bought. The percentage of profit on the labelled price is:
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Answer: (D) 12%

Explanation

Let the labelled price of the bag be ₹100.

Since Mohan purchased the bag at a 20% discount on the labelled price, his cost price is: Cost Price = 100 - (20% of 100) = ₹80

He then sold the bag with a 40% profit on his cost price, so the selling price is: Selling Price = 80 + (40% of 80) = 80 + 32 = ₹112

The profit made relative to the labelled price is: Profit = Selling Price - Labelled Price = 112 - 100 = ₹12

The percentage of profit calculated on the labelled price is: Percentage Profit = (Profit / Labelled Price) × 100 = (12 / 100) × 100 = 12%

35Quantitative Aptitude
If 5416 x 6 is a perfect square, then the digit at 'x' is:
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Answer: (A) 9

Explanation

The given six-digit number is of the form 5416x6. Since it ends in 6, its square root must end in either 4 or 6.

To locate the range of the square root, we evaluate nearby perfect squares of multiples of 10. Since 700² = 490000 and 800² = 640000, the square root lies in the 700s. More precisely, because 730² = 532900 and 740² = 547600, the square root must lie between 730 and 740.

The only integers in this interval ending in 4 or 6 are 734 and 736. Computing their squares: 734² = 538756 736² = 541696

Matching 541696 with the template 5416x6 gives x = 9.

36Quantitative Aptitude
Ram babu donated 3% of his income to a charity and deposited 12% of the rest in bank. If now he has ₹12804, then his income was:
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Answer: (C) 15000

Explanation

Let the total income of Ram babu be I.

The remaining amount after donating 3% of his income to charity is: I × (1 - 0.03) = 0.97 × I

The remaining amount after depositing 12% of the rest in the bank is: 0.97 × I × (1 - 0.12) = 0.97 × 0.88 × I = 0.8536 × I

According to the question, this final remaining amount is equal to ₹12804: 0.8536 × I = 12804 I = 12804 ÷ 0.8536 I = 15000

37Quantitative Aptitude
In △ ABC , the internal bisectors of ∠ B and ∠ C meet at point D. If ∠ A = 80° , then ∠ BDC is of:
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Answer: (A) 130°

Explanation

For any triangle ABC, the angle formed at the intersection of the internal bisectors of ∠B and ∠C (point D, which is the incenter of the triangle) is given by the standard relation:

∠BDC = 90° + ½ ∠A

Substituting the given value of ∠A = 80° into the formula:

∠BDC = 90° + ½ × 80° ∠BDC = 90° + 40° ∠BDC = 130°

38Quantitative Aptitude
A motor boat covers a certain distance downstream in a river in 3 hours. It covers the same distance upstream in 3 hours and half. If the speed of the water is 1.5 km/h , then the speed of the boat in still water is:
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Answer: (C) 19.5 km/h

Explanation

The relationship between speed, distance, and time is given by: Distance = Speed × Time

Let the speed of the boat in still water be v km/h. The speed of the current is 1.5 km/h.

Downstream speed = v + 1.5 Upstream speed = v - 1.5

Since the downstream and upstream distances are equal: (v + 1.5) × 3 = (v - 1.5) × 3.5

Expanding the terms: 3v + 4.5 = 3.5v - 5.25

Rearranging the equation to solve for v: 3.5v - 3v = 4.5 + 5.25 0.5v = 9.75 v = 19.5

39Quantitative Aptitude
Let AX ½ BC of an equilateral triangle ABC. Then the sum of the perpendicular distances of the sides of ABC from any point inside the triangle is:
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Answer: (A) Equal to AX

Explanation

By Viviani's theorem, the sum of the perpendicular distances from any interior point to the sides of an equilateral triangle is always equal to the altitude of the triangle.

Let the side length of the equilateral triangle ABC be s, and let AX be the altitude to the side BC. The area of triangle ABC is given by ½ × s × AX.

Let P be any point inside the triangle, and let the perpendicular distances from P to the sides AB, BC, and CA be p₁, p₂, and p₃ respectively. The total area of the triangle ABC is the sum of the areas of the three smaller triangles PAB, PBC, and PCA:

Area(ABC) = Area(PAB) + Area(PBC) + Area(PCA)

½ × s × AX = ½ × s × p₁ + ½ × s × p₂ + ½ × s × p₃

Dividing both sides of the equation by ½ × s yields:

AX = p₁ + p₂ + p₃

This shows that the sum of the perpendicular distances from any point inside the triangle to its sides is equal to AX.

40Quantitative Aptitude
The sides of a triangle are in the ratio of 7:9:12 The difference between the lengths of largest and smallest sides is 15 cm. The length of the largest side would be:
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Answer: (A) 36 cm

Explanation

Let the lengths of the sides of the triangle be represented as 7x, 9x, and 12x.

The smallest side is 7x and the largest side is 12x. The difference between these two sides is: 12x - 7x = 5x

According to the problem, this difference is 15 cm: 5x = 15 x = 3

The length of the largest side is 12x. Substituting the value of x: 12 × 3 = 36 cm.

41Quantitative Aptitude
A boy aged 12 years is left with ₹100,000 which is under a trust. The trustees invest the money at 6% per annum and pay the minor boy a sum of ₹2500, for his pocket money at the end of each year. The expenses of trust come out to be ₹500 per annum. Find the amount that will be handed over to the minor boy after he attains the age of 18 years.
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Answer: (C) 118000

Explanation

The time duration for which the money is invested is 18 - 12 = 6 years.

The annual interest earned on the principal amount of ₹100,000 at 6% simple interest per annum is: Interest = 100,000 × 0.06 = ₹6000

The total annual deductions from this interest include the boy's pocket money and the trust expenses: Annual deductions = ₹2500 + ₹500 = ₹3000

The net annual amount added back to the trust is the difference between the annual interest earned and the annual deductions: Net annual addition = ₹6000 - ₹3000 = ₹3000

Over the period of 6 years, the total accumulated addition is: Total addition = 6 × 3000 = ₹18,000

The final amount handed over to the boy when he turns 18 is: Final amount = Initial principal + Total addition = 100,000 + 18,000 = ₹118,000

42Quantitative Aptitude
In △ ABC ∠ B = 60° ,and ∠ C = 40° ,AD and AE are respectively the bisector of ∠ A and perpendicular on BC. The measure of ∠ EAD is:
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Answer: (D) 10°

Explanation

In any triangle ABC, the angle between the altitude (perpendicular AE) and the angle bisector (AD) from the vertex A to the opposite side BC is given by the direct formula:

∠EAD = ½ |∠B - ∠C|

Substituting the given values:

∠EAD = ½ |60° - 40°| = ½ × 20° = 10°

43Quantitative Aptitude
The average of 13 results is 70. The average of first seven is 65 and that of the last seven is 75, the seventh result is:
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Answer: (A) 70

Explanation

The total sum of all 13 results is calculated by multiplying the number of results by their average: 13 × 70 = 910

The sum of the first seven results is: 7 × 65 = 455

The sum of the last seven results is: 7 × 75 = 525

Since the seventh result is included in both the first seven and the last seven results, adding these two sums together includes the seventh result twice. The sum of these 14 results is: 455 + 525 = 980

To find the seventh result, subtract the sum of the 13 results from this combined sum: 980 - 910 = 70

44Quantitative Aptitude
The contractor was engaged to construct a road in 16 days. After working for 12 days with 20 labours it was found that only ⅝ᵗʰ of the road had been constructed. To complete the work in stipulated the number of extra labours required is:
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Answer: (D) 16

Explanation

Using the work formula (M₁ × D₁) / W₁ = (M₂ × D₂) / W₂, we can determine the required number of labours to complete the remaining work in the stipulated time.

The initial parameters are: Number of initial labours (M₁) = 20 Days worked (D₁) = 12 Work completed (W₁) = 5/8

The remaining parameters are: Remaining work (W₂) = 1 - 5/8 = 3/8 Remaining days (D₂) = 16 - 12 = 4 Total labours required (M₂) = 20 + x, where x is the number of extra labours.

Substituting these values into the formula: (20 × 12) / (5/8) = (M₂ × 4) / (3/8)

Cancelling the denominator 8 from both sides: (20 × 12) / 5 = (M₂ × 4) / 3

Simplifying the equation: 4 × 12 = 4M₂ / 3 48 = 4M₂ / 3 M₂ = (48 × 3) / 4 M₂ = 36

Since the total number of labours needed to complete the work on time is 36 and 20 labours are already employed, the number of extra labours required is: 36 - 20 = 16

45Quantitative Aptitude
If p = - 0.12 , q = - 0.01 | r = - 0.015 , then the correct relation ship among the three is:
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Answer: (D) p<r<q

Explanation

To compare decimal values, it is easiest to represent them with the same number of decimal places: p = -0.120 q = -0.010 r = -0.015

Comparing their absolute values: 0.120 > 0.015 > 0.010

For negative numbers, a larger absolute value corresponds to a smaller algebraic value. Therefore, reversing the inequality gives: -0.120 < -0.015 < -0.010

Substituting the variables back into the inequality yields: p < r < q

46Quantitative Aptitude
The % money spent on food compared to house rent is by ?
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Answer: (C) 25%

Explanation

Based on the monthly family expenditure pie chart, the central angles representing the respective expenses are: Central angle for money spent on food = 150° Central angle for money spent on house rent = 120°

To find the percentage by which the money spent on food exceeds the money spent on house rent, use the percentage increase formula: Required percentage = ((Angle for Food - Angle for Rent) / Angle for Rent) × 100 Required percentage = ((150 - 120) / 120) × 100 = (30 / 120) × 100 = ¼ × 100 = 25%

47Quantitative Aptitude
The total money spent on clothes and miscellaneous items are
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Answer: (D) Cannot be determined

Explanation

The pie chart represents the distribution of a family's monthly expenditure across various categories in terms of percentages or degrees. To calculate the absolute monetary value spent on specific categories like clothes and miscellaneous expenses, the total monthly expenditure of the family must be known. Since the total value of the family's expenditure or income is not provided, it is impossible to determine the precise monetary amount spent on these items.

48Quantitative Aptitude
If the total amount spent is 7,200. Find the amount spent on food ?
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Answer: (A) 3000

Explanation

The family monthly expenditure pie chart indicates that the central angle corresponding to the expenditure on food is 150°. Since a complete circle represents 360°, the proportion of the total expenditure spent on food is the ratio of these two angles.

Amount spent on food = (Central angle of food ÷ 360°) × Total amount spent

Amount spent on food = (150° ÷ 360°) × 7,200 = 5/12 × 7,200 = 3,000

49Quantitative Aptitude
If for non- zero x, x² - 4x - 1 = 0 the value of x² + (1)/(x²) is:
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Answer: (D) 18

Explanation

Given the quadratic equation: x² - 4x - 1 = 0

Since x is non-zero, dividing both sides of the equation by x gives: x - 4 - 1/x = 0 x - 1/x = 4

Squaring both sides of this equation yields: (x - 1/x)² = 4² x² - 2 + 1/x² = 16 x² + 1/x² = 16 + 2 x² + 1/x² = 18

50Quantitative Aptitude
The length of two parallel sides of a trapezium are 15 cm and 20 cm If its area is 175 sq. cm , then its height is:
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Answer: (A) 10 cm

Explanation

The area of a trapezium is calculated using the formula:

Area = ½ × (a + b) × h

where a and b are the lengths of the parallel sides, and h is the height.

Substituting the given values into the formula:

175 = ½ × (15 + 20) × h

175 = ½ × 35 × h

350 = 35 × h

h = 350 / 35

h = 10 cm

Answer key

Tap a number to jump to that question.

  1. 1D
  2. 2A
  3. 3D
  4. 4C
  5. 5A
  6. 6A
  7. 7B
  8. 8A
  9. 9A
  10. 10C
  11. 11C
  12. 12B
  13. 13C
  14. 14D
  15. 15A
  16. 16C
  17. 17A
  18. 18B
  19. 19B
  20. 20A
  21. 21B
  22. 22B
  23. 23B
  24. 24D
  25. 25C
  26. 26B
  27. 27D
  28. 28A
  29. 29C
  30. 30B
  31. 31D
  32. 32C
  33. 33A
  34. 34D
  35. 35A
  36. 36C
  37. 37A
  38. 38C
  39. 39A
  40. 40A
  41. 41C
  42. 42D
  43. 43A
  44. 44D
  45. 45D
  46. 46C
  47. 47D
  48. 48A
  49. 49D
  50. 50A

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