Quantitative Aptitude PYQs

Given:

Ratio of sweets distributed among A, B, C, D, E = (1/2):(1/3):(1/4):(1/5):(1/7)

Formula Used:

To find the minimum total number of sweets, take the LCM of denominators and then find total parts.

Calculation:

Denominators = 2, 3, 4, 5, 7

⇒ LCM of 2, 3, 4, 5, 7 = 420

Now make all ratios with same denominator:

⇒ (1/2):(1/3):(1/4):(1/5):(1/7)

⇒ (210/420):(140/420):(105/420):(84/420):(60/420)

⇒ Ratio = 210 : 140 : 105 : 84 : 60

⇒ Total parts = 210 + 140 + 105 + 84 + 60

⇒ Total parts = 599

⇒ Minimum total sweets = 599

Minimum total number of sweets = 599

Given:

If speed is increased by 10 km/h, time reduced = 55 minutes

If speed is decreased by 8 km/h, time increased = 55 minutes

Formula Used:

Time = Distance / Speed

Calculation:

Let usual speed = x km/h

Distance = d km

Time taken at usual speed = d/x hours

If speed increased by 10 km/h ⇒ Time = d/(x + 10)

If speed decreased by 8 km/h ⇒ Time = d/(x - 8)

Given difference in both cases = 55 minutes = 55/60 = 11/12 hours

⇒ d/x - d/(x + 10) = 11/12 …(1)

⇒ d/(x - 8) - d/x = 11/12 …(2)

From (1): d = (11/12) × (x × (x + 10)) / 10

From (2): d = (11/12) × (x × (x - 8)) / 8

Equating both:

⇒ (x × (x + 10)) / 10 = (x × (x - 8)) / 8

⇒ 8(x² + 10x) = 10(x² - 8x)

⇒ 8x² + 80x = 10x² - 80x

⇒ 2x² - 160x = 0

⇒ x(x - 80) = 0

⇒ x = 80 (since speed ≠ 0)

Substitute in (1):

⇒ d = (11/12) × (80 × (80 + 10)) / 10

⇒ d = (11/12) × (80 × 90) / 10

⇒ d = (11/12) × 720

⇒ d = 11 × 60

⇒ d = 660 km

Distance = 660 km

Given:

Total score = 114 runs

Boundaries = 5

Sixes = 9

Formula Used:

Runs scored by running = Total score - (Runs from boundaries + Runs from sixes)

Percentage of runs scored by running = (Runs scored by running / Total score) × 100

Calculation:

⇒ Runs from boundaries = 5 × 4 = 20

⇒ Runs from sixes = 9 × 6 = 54

⇒ Runs scored by running = 114 - (20 + 54) = 114 - 74 = 40

⇒ Percentage of runs scored by running = (40 / 114) × 100

⇒ Percentage = (4000 / 114)

⇒ Percentage 35\(\frac{5}{57}\)

The percentage of his total score made by running between the wickets is 35\(\frac{5}{57}\).

Given:

One-third of one-fourth of one-fifth of a number = 18.

Formula Used:

Let the number be x. Then, (1/3) × (1/4) × (1/5) × x = 18.

We need to find three-tenth (3/10) of that number.

Calculation:

⇒ (1/3) × (1/4) × (1/5) × x = 18

⇒ x / (3 × 4 × 5) = 18

⇒ x / 60 = 18

⇒ x = 18 × 60

⇒ x = 1080

Now, three-tenth of that number:

⇒ (3/10) × x = (3/10) × 1080

⇒ 3 × 108 = 324

The value of three-tenth of the number is 324.

Given:

Time taken by A to complete the work = 10 days

Time taken by B to complete the work = 12 days

Time taken by C to complete the work = 15 days

B is assisted by A and C alternately.

Formula Used:

Work done per day = 1 / Time taken

Total work = LCM of individual times

Work done in alternating days = Work of B + Work of (A or C)

Calculation:

⇒ LCM of 10, 12, and 15 = 60 (Total work)

⇒ Work done by A in 1 day = 60 / 10 = 6 units

⇒ Work done by B in 1 day = 60 / 12 = 5 units

⇒ Work done by C in 1 day = 60 / 15 = 4 units

On 1^st day: B works with A.

⇒ Total work done in 1^st day = 5 + 6 = 11 units

On 2^nd day: B works with C.

⇒ Total work done in 2^nd day = 5 + 4 = 9 units

Work done in 2 days = 11 + 9 = 20 units

Total work = 60 units

⇒ Total number of such 2-day cycles = 60 / 20 = 3 cycles

⇒ Total time = 3 × 2 = 6 days

Thus, B can complete the work in 6 days when assisted by A and C alternately.

The correct answer is 6 days.

Given:

Profit = 20%

Formula used:

Profit% = [(S.P. - C.P.) ÷ C.P.] × 100

Calculations:

Let the cost price = x

So, Selling price = x + 20% of x = 1.2x

If both C.P. and S.P. are decreased by ₹50, then new profit = 25%

New C.P. = x - 50

New S.P. = 1.2x - 50

⇒ (1.2x - 50) - (x - 50) = 25% of (x - 50)

⇒ 0.2x = 0.25(x - 50)

⇒ 0.2x = 0.25x - 12.5

⇒ 0.25x - 0.2x = 12.5

⇒ 0.05x = 12.5

⇒ x = 12.5 ÷ 0.05

⇒ x = 250

∴ The original cost price of the fan is ₹250.

Given:

4/5 ÷ 3/10 of 4/9 − 4/5 × 1\(\frac{1}{9}\) ÷ 8/15 − 3/4 + 3/4 ÷ 1/2

Formula Used:

of = × , a/b ÷ c/d = a/b × d/c

Calculation:

⇒ 4/5 ÷ (3/10 × 4/9) = 4/5 ÷ 2/15 = 4/5 × 15/2 = 6

⇒ 4/5 × 10/9 = 8/9, ⇒ 8/9 ÷ 8/15 = 5/3

⇒ 3/4 ÷ 1/2 = 3/4 × 2/1 = 3/2

⇒ 6 − 5/3 − 3/4 + 3/2

⇒ 72/12 − 20/12 − 9/12 + 18/12

⇒ 61/12

∴ The correct answer is 61/12.

Given:

For a rectangular box:

Length = l

Breadth = b

Height = h

Formula Used:

Product of the areas of the 3 adjacent faces meeting at a point = (length × breadth) × (breadth × height) × (length × height)

= (l × b × h)²

Calculation:

⇒ Product of the areas of the 3 adjacent faces = (l × b × h) × (l × b × h)

⇒ Product of the areas of the 3 adjacent faces = (l × b × h)²

The correct answer is (length × breadth × height)².

Given:

Principal (P) = Rs. 8,000

Time (T) = 6 years

Rate of interest (R) = 12 paise per rupee per month

Formula Used:

Simple Interest (SI) = (P × R × T) / 100

Calculation:

Rate per annum = 12 × 12 = 144% (since 12 paise per rupee per month means 12% per year)

⇒ SI = (8000 × 144 × 6) / 100

⇒ SI = 8000 × 864 / 100

⇒ SI = Rs. 5760

The simple interest is Rs. 5760.

Given:

Radha scored 20% of marks and failed by 10 marks.

Seeta scored 30% of marks, which is 20 marks more than the pass mark.

Formula Used:

Total Marks = 100 × (Difference between percentage scores) / (Difference between marks)

Calculation:

Radha's marks = 20% of total marks = 0.2 × Total Marks.

Seeta's marks = 30% of total marks = 0.3 × Total Marks.

Pass marks = Radha's marks + 10 = Seeta's marks - 20.

⇒ 0.2 × Total Marks + 10 = 0.3 × Total Marks - 20.

⇒ 0.3 × Total Marks - 0.2 × Total Marks = 30.

⇒ 0.1 × Total Marks = 30.

⇒ Total Marks = 30 / 0.1.

⇒ Total Marks = 300.

The total marks of the examination are 300.

Given:

Principal (P) = 18000

Rate (R) = 10% per annum = 10/2 = 5% per half year

Time (T) = 1.5 years = 3 half years

Formula Used:

Amount = P × (1 + R/100)ⁿ

Calculation:

⇒ Amount = 18000 × (1 + 5/100)³

⇒ 18000 × (1.05)³

⇒ 18000 × 1.157625

⇒ 20837.25

Total Amount = Rs. 20,837.25

Given:

Final Sale Price = 19844

First discount = 12% ⇒ Remaining price = 88% = 0.88

Second discount = 18% ⇒ Remaining price = 82% = 0.82

Formula Used:

Sale Price = Marked Price × 0.88 × 0.82

Calculation:

⇒ 19844 = MP × 0.88 × 0.82

⇒ 19844 = MP × 0.7216

⇒ MP = 19844 ÷ 0.7216

⇒ MP = 27500

Marked Price = ₹ 27,500

Given:

Numbers: 397, 435, 5420

Remainders: 6, 10, 14

Formula Used:

Greatest number that divides the numbers leaving given remainders = HCF of (number - remainder)

Calculation:

⇒ Numbers after subtracting remainders:

397 - 6 = 391

435 - 10 = 425

5420 - 14 = 5406

⇒ Find HCF of 391, 425, and 5406:

Find the prime factorization:

391 = 17 × 23

425 = 5² × 17

5406 = 2 × 3 × 17 × 53

Common factor is 17.

⇒ HCF = 17

The greatest number that will divide 397, 435, and 5420 leaving remainders 6, 10, and 14 respectively is 17.

Given:

When folded into two equal halves, the perimeter of each part is:

1. 36 cm when folded along one set of sides.
2. 45 cm when folded along the other set of sides.

Formula Used:

Perimeter of a rectangle = 2 × (length + breadth).

Area of a rectangle = length × breadth.

Calculation:

Original sheet dimensions = a and b

Fold along side a: New dimensions = a/2 and b

Perimeter = 36 cm

⇒ 2(b + a/2) = 36

Fold along side b: New dimensions = a and b/2

Perimeter = 45 cm

⇒ 2(a + b/2) = 45

⇒ b + a/2 = 18

⇒ a + b/2 = 22.5

⇒ b = 18 − a/2

⇒ a + (18 − a/2)/2 = 22.5

⇒ a + 9 − a/4 = 22.5

⇒ 3a/4 = 13.5

⇒ a = 18

⇒ b = 18 − 9 = 9

Area = a × b = 18 × 9 = 162 cm²

∴ The correct answer is 162 cm²

Given:

Average age of grandparents = 69 years.

Average age of parents = 39 years.

Average age of grandchildren = 9 years.

Number of grandparents = 2.

Number of parents = 2.

Number of grandchildren = 2.

Formula Used:

Average age of the family = Total age of all family members / Total number of family members

Calculation:

⇒ Total age of grandparents = 69 × 2 = 138

⇒ Total age of parents = 39 × 2 = 78

⇒ Total age of grandchildren = 9 × 2 = 18

⇒ Total age of all family members = 138 + 78 + 18 = 234

⇒ Total number of family members = 2 + 2 + 2 = 6

⇒ Average age of the family = 234 / 6

⇒ Average age = 39 years

The average age of the family is 39 years.

Given:

The selling price is tripled, the profit becomes four times.

Formula Used:

Profit Percentage = (Profit / Cost Price) × 100

Calculation:

Let the Cost Price (CP) = 100 and the original Selling Price (SP) = 100 + Profit.

⇒ Assume original Profit = x.

⇒ SP = 100 + x.

When SP is tripled:

⇒ New SP = 3 × (100 + x).

⇒ New Profit = 3 × (100 + x) - 100.

Given, New Profit = 4 × Original Profit.

⇒ 3 × (100 + x) - 100 = 4x.

⇒ 300 + 3x - 100 = 4x.

⇒ 200 = x.

Original Profit = 200.

Profit Percentage = (Profit / Cost Price) × 100.

⇒ Profit Percentage = (200 / 100) × 100.

⇒ Profit Percentage = 200%.

The profit percentage is 200%.

Given:

Marked price of the article = Rs. 4,000

Three successive discounts: 20%, 25%, and x%

Final selling price = Rs. 1,968

Formula Used:

Final Price = Marked Price × (1 - Discount₁) × (1 - Discount₂) × (1 - Discount₃)

Calculation:

⇒ 1,968 = 4,000 × (1 - 20/100) × (1 - 25/100) × (1 - x/100)

⇒ 1,968 = 4,000 × 0.8 × 0.75 × (1 - x/100)

⇒ 1,968 = 4,000 × 0.6 × (1 - x/100)

⇒ 1,968 = 2,400 × (1 - x/100)

⇒ 1 - x/100 = 1,968 / 2,400

⇒ 1 - x/100 = 0.82

⇒ x/100 = 1 - 0.82

⇒ x/100 = 0.18

⇒ x = 18

The value of x is 18%.

Given:

Initial solution volume = 18 ml

Initial acid concentration = 50% = 0.5

Final acid concentration = 30% = 0.3

Water to be added = ? (in ml)

Formula Used:

Final acid content = Initial acid content

Let the amount of water added be x ml.

Calculation:

⇒ Initial acid content = 18 × 0.5 = 9 ml

⇒ Final solution volume = 18 + x

⇒ Final acid content = Final solution volume × Final acid concentration

⇒ 9 = (18 + x) × 0.3

⇒ 9 = 5.4 + 0.3x

⇒ 9 - 5.4 = 0.3x

⇒ 3.6 = 0.3x

⇒ x = 3.6 / 0.3

⇒ x = 12 ml

The correct amount of water to be added is 12 ml.

Given:

Ratio of passengers in Second AC : Sleeper Class = 3 : 5

Ratio of fares of Second AC : Sleeper Class = 30 : 17

Total amount collected = Rs. 8,750

Formula Used:

Total amount = (Number of passengers in Second AC × Fare of Second AC) + (Number of passengers in Sleeper Class × Fare of Sleeper Class)

Calculation:

Let the number of passengers in Second AC = 3x

Let the number of passengers in Sleeper Class = 5x

Let the fare of Second AC = 30y

Let the fare of Sleeper Class = 17y

⇒ Total amount = (3x × 30y) + (5x × 17y)

⇒ 8,750 = 90xy + 85xy

⇒ 8,750 = 175xy

⇒ xy = 8,750 / 175

⇒ xy = 50

Amount collected from Sleeper Class = Number of passengers in Sleeper Class × Fare of Sleeper Class

⇒ Amount collected from Sleeper Class = (5x × 17y)

⇒ Amount collected from Sleeper Class = 85xy

⇒ Amount collected from Sleeper Class = 85 × 50

⇒ Amount collected from Sleeper Class = Rs. 4,250

The amount collected from the Sleeper Class is Rs. 4,250.

Given:

Principal = 40000

First 2 years rate = 10% p.a.

Last 1 year rate = 12% p.a. (changed)

Formula Used:

Amount = P × (1 + R/100)^time

Calculation:

Original amount (all 3 years @10%):

⇒ 40000 × 1.1 × 1.1 × 1.1 = 53240

New amount (2 years @10% then 1 year @12%):

⇒ 40000 × 1.1 × 1.1 × 1.12 = 54208

Extra amount:

⇒ 54208 − 53240 = 968

Extra amount payable = ₹ 968