Quantitative Aptitude PYQs

Quantity I:

X’s share = Rs. [7000 × (2/5)] ⇒ Rs. 2800

Y’s share = Rs. [7000 × (3/5)] ⇒ Rs. 4200

Thrice the share of X = Rs. 8400

⇒ Difference between thrice the share of X and the share of Y = Rs. 8400 - Rs. 4200

= Rs. 4200

Quantity II:

Let the Q's share be Rs. x

P's share + Q's share + R's share + S's share = 1, 25, 000

⇒ (x/3) + x + 25, 000 + [x/3] = 1, 25, 000

⇒ (5x/3) + 25000 = 1, 25, 000

⇒ 5x/3 = 1, 25, 000 - 25000

⇒ 5x = 3, 00, 000

⇒ x = 60, 000

∴ Quantity I < Quantity II

Shortcut Trick

Shift in distance of construction = 20 km.

Reduction in delay = 3 hours − 2 hours = 1 hour.

Speed ratio at patch = 1 : 0.25 = 4 : 1 → Time ratio = 1 : 4 (Gap = 3 units).

For 20 km, 3 units of late time = 1 hour ∴ 1 unit (usual time for 20 km) = 1/3 hour.

Usual speed S = 20 ÷ (1/3) = 60 km/hr. Since 3 hours late for (100−d) km → (100−d)/15 − (100−d)/60 = 3 ⇒ d = 40 km.

Time for (2×40 + 40) = 120 km at 60 km/hr = 2 hours.

∴ The correct answer is 2 hr.

Alternate Method

Given: Total distance = 100 km, Reduced speed = 0.25S.

Formula: Time = Distance ÷ Speed | Delay = Time_slow − Time_usual

⇒ Delay for distance (100 − d) = (100 − d)/(0.25S) − (100 − d)/S = 3

⇒ (100 − d)/S × (4 − 1) = 3 ⇒ (100 − d)/S = 1 ... (Eq 1)

⇒ Delay for distance (100 − (d + 20)) = (80 − d)/(0.25S) − (80 − d)/S = 2

⇒ (80 − d)/S × 3 = 2 ⇒ (80 − d)/S = 2/3 ... (Eq 2)

⇒ Subtracting Eq 2 from Eq 1: (100 − d − 80 + d)/S = 1 − 2/3

⇒ 20/S = 1/3 ⇒ S = 60 km/hr

⇒ Put S in Eq 1: (100 − d)/60 = 1 ⇒ 100 − d = 60 ⇒ d = 40 km

⇒ Required Distance = 2d + 40 = 2(40) + 40 = 120 km

⇒ Time = 120 / 60 = 2 hours.

∴ The correct answer is 2 hr.

Additional Information

Speed-Time Relationship

When distance is constant, speed is inversely proportional to time (S ∝ 1/T). If speed becomes 1/n, time becomes n times the original.

Concept of Delay

Delay = Distance × [ (1/S_reduced) − (1/S_usual) ]. This formula helps isolate the specific patch causing the time difference.

Relative Delay Shift

A shift in the point of speed reduction directly correlates the change in delay to the distance between the two points.

Shortcut Trick

Ratio of Quantity (Mobiles : Laptops) = 60 : 100 = 3 : 5.

Ratio of Unit CP (Mobile : Laptop) = 100 : 180 = 5 : 9.

Ratio of Total CP (Mobile : Laptop) = (3 × 5) : (5 × 9) = 15 : 45 = 1 : 3.

Weighted Profit % = [(1 × 35) + (3 × 15)] ÷ (1 + 3) = (35 + 45) ÷ 4 = 20%.

Given 0.5X + 5 = 20 ⇒ 0.5X = 15 ⇒ X = 30.

∴ The correct answer is 30.

Alternate Method

Given: Total CP = Rs. 1,80,000 | Mobile quantity = Laptop quantity − 40% | Laptop CP = Mobile CP + 80% | Laptop Profit = 15% | Mobile Profit = 35%.

Formulas: Overall Profit % = (Total Profit ÷ Total CP) × 100 | Weighted Average = (n₁p₁ + n₂p₂) ÷ (n₁ + n₂).

⇒ Let Number of Laptops = 100n, then Number of Mobiles = 60n.

⇒ Let CP of one Mobile = 100c, then CP of one Laptop = 180c.

⇒ Total CP of Mobiles = 60n × 100c = 6000nc.

⇒ Total CP of Laptops = 100n × 180c = 18000nc.

⇒ Total Profit = (35% of 6000nc) + (15% of 18000nc) = 2100nc + 2700nc = 4800nc.

⇒ Overall Profit % = (4800nc ÷ 24000nc) × 100 = 20%.

⇒ 0.5X + 5 = 20 ⇒ 0.5X = 15 ⇒ X = 30.

∴ The correct answer is 30.

Additional Information

Weighted Average Profit

When different quantities are sold at different profit percentages, the net profit % is the weighted average of individual profit percentages based on their total cost values.

Percentage Change

Profit or loss is always calculated on the Cost Price (CP) unless stated otherwise. Profit = Selling Price (SP) − Cost Price (CP).

Shortcut Trick
Ratio of profit shares = Ratio of (Investment × Time).
Rohan's investment: [(x2 + 2x) × 4] + [(x2 + 4x) × 8] = 12x2 + 40x.
Karan's investment: [7x × 4] + [(8x + 10) × 8] = 92x + 80.
Ratio (12x2 + 40x) ÷ (92x + 80) = 8 ÷ 5.
On solving, 60x2 + 200x = 736x + 640 ⇒ 60x2 − 536x − 640 = 0.
∴ The correct answer is 10.

Alternate Method
Given: Rohan's initial investment = Rs. (x2 + 2x), Karan's initial = Rs. 7x.
Time = 12 months. Change after 4 months. Profit ratio Rohan : Karan = 8 : 5.
Formulas: Profit Share = Investment × Time.

⇒ Rohan total investment-months = (x2 + 2x) × 4 + (x2 + 2x + 2x) × 8
⇒ 4x2 + 8x + 8x2 + 32x = 12x2 + 40x
⇒ Karan total investment-months = 7x × 4 + (7x + x + 10) × 8
⇒ 28x + 64x + 80 = 92x + 80
⇒ (12x2 + 40x) ÷ (92x + 80) = 8 ÷ 5
⇒ 5(12x2 + 40x) = 8(92x + 80) ⇒ 60x2 + 200x = 736x + 640
⇒ 60x2 − 536x − 640 = 0 ⇒ 15x2 − 134x − 160 = 0
⇒ 15x2 − 150x + 16x − 160 = 0 ⇒ 15x(x − 10) + 16(x − 10) = 0
⇒ (x − 10)(15x + 16) = 0. Since x > 4, x = 10.
∴ The correct answer is 10.

Additional Information
Partnership Basics
Profit is divided in the ratio of the product of capital and time period: P1 : P2 = C1T1 : C2T2.
Change in Capital
When capital changes during the period, calculate equivalent capital for one month by summing (Amount × Duration) for each segment.

Shortcut Trick

Let the amount given to charity be C and remaining amount be R.

If charity is reduced by 20%, that 20% of C becomes the extra amount he has: 20% of C = ₹434.

⇒ C = 434 × 5 = ₹2170.

Given C ≈ 68.89% of R → 2170 ≈ 0.6889 × R ⇒ R ≈ ₹3150.

Total amount = C + R = 2170 + 3150 = ₹5320.

Required % = (1596 ÷ 5320) × 100 = 30%.

∴ The correct answer is 30%.

Alternate Method

Given: Charity (C) = 68.89% of Remaining (R); Change in savings = ₹434 when charity decreases by 20%.

⇒ 20% of C = 434

⇒ C = (434 ÷ 20) × 100 = ₹2170

⇒ 2170 = 0.6889 × R

⇒ R = 2170 ÷ 0.6889 ≈ ₹3150

⇒ Total Money = C + R = 2170 + 3150 = ₹5320

⇒ Percentage = (1596 ÷ 5320) × 100

⇒ Percentage = 0.3 × 100 = 30%

∴ The correct answer is 30%.

Additional Information

Percentage Conversion

To convert a fraction to a percentage, multiply by 100. To find a value from a percentage, multiply the total by (percentage/100).

Approximation in Percentages

When dealing with values like 68.89%, check if they represent standard fractions (e.g., 68.89% ≈ 31/45 or similar) to simplify calculations in banking exams.

Income and Expenditure Logic

Total Money = Amount Spent (Charity) + Amount Saved (Remaining). Any reduction in expenditure directly increases the savings by the same absolute amount.

Common explanation

For Kharagpur:

Percentage of students who play Tennis = (100 - 40 - 25) % = 35%

Similarly, 

 

Let the total number of students in Kanpur be 'x'.

So, total number of students in Kharagpur = x + 1200

ATQ,

(30% of x) - (20% of x) = 2480

Or, 10% of x = 2480

Or, 0.1 × x = 2480

So, x = (2480/0.1) = 24800

Total number of students in Kharagpur = x + 1200 = 24800 + 1200 = 26,000

Common explanation

For Kharagpur:

Percentage of students who play Tennis = (100 - 40 - 25) % = 35%

Similarly, 

 

Let total number of students in Bombay be 'x'.

Sum of number of students who play tennis from Madras and Bombay = 40% × 6000 + 30% × x = 5100

Or, 2400 + 0.3x = 5100

Or, 0.3x = 5100 - 2400

So, x = (2700/0.3) = 9000

Number of students who play football in Madras = 30% × 6000 = 1800

Number of students who play football in Bombay = 60% of x = 0.6 × 9000 = 5400

Required ratio = 1800:5400 = 1:3

Common explanation

For Kharagpur:

Percentage of students who play Tennis = (100 - 40 - 25) % = 35%

Similarly, 

 

Let the total number of students in Kanpur and Kharagpur be '3x' and '2x', respectively.

ATQ;

Sum of students who play tennis in the given two colleges

= (50% × 3x) + (35% × 2x) = 5280

Or, 1.5x  + 0.7x = 5280

Or, 2.2x = 5280

So, x = (5280/2.2) = 2400

Total number of students in Kanpur = 3x = 3 × 2400 = 7,200

Common explanation

For Kharagpur:

Percentage of students who play Tennis = (100 - 40 - 25) % = 35%

Similarly, 

 

Let the total number of students in Bombay be 'x'.

ATQ,

(60 + 30) % of x - (60 + 10) % of x = 1440

Or, 90% of x - 70% of x = 1440

Or, 20% of x = 1440

Or, 0.2 × x = 1440

So, x = (1440/0.2) = 7200

Number of students who play football from Bombay = 60% of x = 0.6 × 7200 = 4320

Common explanation

For Kharagpur:

Percentage of students who play Tennis = (100 - 40 - 25) % = 35%

Similarly, 

 

Let the total number of students in Madras and Kharagpur be 'x' and 'y', respectively.

Difference between the number of students who play tennis and golf from Madras:

(40% of x) - (30% of x) = 105 of 'x' = 500

Or, 0.1 × x = 500

So, x = (500/0.1) = 5000

Average of number of students who play football and tennis from Kharagpur

{(40% of y) + (35% of y)}/2 = 3600

Or, 75% of y = 3600 × 2

Or, 0.75 × y = 7200

So, y = (7200/0.75) = 9600

Required difference = 9600 – 5000 = 4600

Statement I:

Ratio of profit share of Aman, Akash and Anil at the end of 15 months:

= [(x - 1,200) × 15] : [(2x - 1,800) × 15] : [(x - 1,800) × 15]

= (x - 1,200) : (2x - 1,800) : (x - 1,800)

So, (x - 1,200) : (2x - 1,800) = 3 : 7

Or, 7 × (x - 1,200) = 3 × (2x - 1,800)

Or, 7x - 8,400 = 6x - 5,400

Or, 'x' = 3,000

Investment of Akash = 2 × 3,000 - 1,800 = Rs. 4,200

Investment of Anil = 3,000 - 1,800 = Rs. 1,200

Required difference = 4,200 - 1,200 = Rs. 3,000

So, data in statement I alone is sufficient to answer the question.

Statement II:

Ratio of profit share of Aman, Akash and Anil at the end of 17 months:

= [(x - 1,200) × 17] : [(2x - 1,800) × 17] : [(x - 1,800) × 17]

= (x - 1,200) : (2x - 1,800) : (x - 1,800)

So, (x - 1,200) = 0.25 × [(x - 1,200) + (2x - 1,800) + (x - 1,800)]

Or, x - 1,200 = (1/4) × (4x - 4,800)

Or, x - 1,200 = x - 1,200

We cannot solve this equation further.

So, data in statement II alone is not sufficient to answer the question.

Therefore, data in statement I alone is sufficient to answer the question, while the data in statement II alone is not sufficient to answer the question.

Hence, option a.

Statement I:

(20% of 60n² - 11n²) + 25% of (400 - 76) = 40n + 10n - 32n

Or, (12n² - 11n²) + 25% of 324 = 50n - 32n

Or, n² + 81 = 18n

Or, n² - 18n + 81 = 0

Or, n² - 9n - 9n + 81 = 0

Or, n(n - 9) - 9(n - 9) = 0

Or, (n - 9)(n - 9) = 0

So, n = 9

Therefore, required value = (9 + 5)² = 196

So, data in Statement I alone is sufficient to answer the question.

Statement II:

(17n + 34n - 50n - 70 + 64)² - 16 = (36 ÷ 18)n - (101 - 1) ÷ 4

Or, (n - 6)² - 16 = 2n - 25

Or, n² - 12n + 36 - 16 = 2n - 25

Or, n² - 12n + 20 = 2n - 25

Or, n² - 14n + 45 = 0

Or, n² - 9n - 5n + 45 = 0

Or, n(n - 9) - 5(n - 9) = 0

Or, (n - 9)(n - 5) = 0

So, n = 9 or n = 5

Since there are two possible values of n, Statement II alone is not sufficient.

Therefore, data in statement I alone is sufficient to answer the question, while the data in statement II alone is not sufficient to answer the question

Hence, option b.

Let the number of students in ‘A’, ‘B’ and ‘C’ be ‘p’, ‘q’ and ‘r’ respectively.

So, sum of the number of students in ‘A’, ‘B’ and ‘C’ = 180 × 3

So, (p + q + r) = 540

Statement I:

Given, (q − r) = k ........(I)

Or, (r − p) = k ........(II)

Or, q − r = r − p

So, p + q = 2r

So, (p + q) + r = 2r + r = 3r = 540

So, ‘r’ = 180

But we cannot determine the value of ‘k’

So, data in statement I alone is not sufficient to answer the question.

Statement II:

Case I:

We have, q = p + 120

Case II:

We have, q = p − 120

But we do not have any other data.

So, data in statement II alone is not sufficient to answer the question.

From Statement I and Statement II:

So, ‘r’ = 180

So, p + q = 540 − 180 = 360

And q − p = 120

So, q = (360 + 120) / 2 = 240

And ‘p’ = (360 − 120) / 2 = 120

So, total number of students in ‘A’ and ‘C’ = 120 + 180 = 300

Number of students in ‘B’ = 240

So, required percentage = {(300 − 240) / 300} × 100 = 20% less

The data in both statements I and II together is necessary to answer the question.

Hence, option e.

Statement I:

Let the marked price of article be Rs. m.

Selling price of article = Rs. (m - 540)

But we have no data regarding the cost price.

So, data in statement I alone is not sufficient to answer the question.

Statement II:

Let the cost price of article be Rs. C.

Marked price of article = Rs. (C + 900)

Selling price of article = 0.88 × (C + 900) = Rs. (0.88C + 792)

But we cannot solve further to obtain the cost price of the article.

So, data in statement II alone is not sufficient to answer the question.

Statement III:

Let the cost price of article be Rs. C.

Marked price of article = Rs. 1.25C

Selling price of article = Rs. 1.10C

But we cannot solve further to obtain the cost price of the article.

So, data in statement III alone is not sufficient to answer the question.

Statement I and II together:

Let the cost price of article be Rs. C.

Marked price of article = Rs. (C + 900)

Selling price of article = 0.88 × (C + 900) = Rs. (0.88C + 792)

So,

(C + 900) − (0.88C + 792) = 540

⇒ 0.12C + 108 = 540

⇒ 0.12C = 432

⇒ C = 3,600

Therefore, cost price of article is Rs. 3,600.

So, data in statement I and II together is sufficient to answer the question.

Statement II and III together:

Let the cost price of article be Rs. C.

From Statement II: Marked price = C + 900

From Statement III: Marked price = 1.25C

So,

1.25C = C + 900

⇒ 0.25C = 900

⇒ C = 3,600

Therefore, cost price of article is Rs. 3,600.

So, data in statement II and III together is sufficient to answer the question.

Statement I and III together:

Let the cost price of article be Rs. C.

Marked price of article = Rs. 1.25C

Selling price of article = Rs. 1.10C

So,

1.25C − 1.10C = 540

⇒ 0.15C = 540

⇒ C = 3,600

Therefore, cost price of article is Rs. 3,600.

So, data in statement I and III together is sufficient to answer the question.

Therefore, data given any of the two statements combined together is sufficient to answer the question.

Hence, option b.

Statement I:

Let the speed of train 'P' and 'Q' be '9v' and '7v' m/s respectively.

We cannot solve further.

So, data in statement I alone is not sufficient to answer the question.

Statement II:

Length of train 'Q' = 960 metres

We do not know whose speed is greater between 'P' and 'Q'.

We cannot solve further.

So, data in statement II alone is not sufficient to answer the question.

Statement III:

We do not know the speed of trains. So, we cannot solve further.

So, data in statement III alone is not sufficient to answer the question.

Statement I and II together:

Let the speed of train 'P' and 'Q' be '9v' and '7v' m/s respectively.

9v - 7v = 36 × (5/18) = 10

Or, 2v = 10

Or, 'v' = 5

So, speed of train 'P' and 'Q' is 45 m/s and 35 m/s respectively.

Length of train 'Q' = 960 metres

But we do not have any further data, so we cannot solve further.

So, data in statement I and II together is not sufficient to answer the question.

Statement II and III together:

Length of train 'Q' = 960 metres

Time taken by trains 'P' and 'Q' to cross each other = 24 seconds

But we do not know whose speed is greater between 'P' and 'Q'.

We cannot solve further.

So, data in statement II and III together is not sufficient to answer the question.

Statement I and III together:

Let the speed of train 'P' and 'Q' be '9v' and '7v' m/s respectively.

Sum of length of train 'P' and 'Q' = (9v + 7v) × 24 = 384v metres

We cannot solve further.

So, data in statement I and III together is not sufficient to answer the question.

Statement I, II and III together:

Let the speed of train 'P' and 'Q' be '9v' and '7v' m/s respectively.

9v - 7v = 36 × (5/18) = 10

Or, 2v = 10

Or, 'v' = 5

So, speed of train 'P' = 9 × 5 = 45 m/s

Speed of train 'Q' = 7 × 5 = 35 m/s

Sum of length of train 'P' and 'Q' = (45 + 35) × 24 = 80 × 24 = 1920 metres

Given that length of train 'Q' = 960 metres

So, length of train 'P' = 1920 - 960 = 960 metres

So, data in statement I, II and III together is sufficient to answer the question.

Therefore, data given in all three statements combined is sufficient to answer the question.

Hence, option e**.**

75.01 + 344.923 - 179.93 = ?³ of (√678 + 3.89)

75 + 345 - 180 = ?³ of (√676 + 4)

420 - 180 = ?³ × (26 + 4)

240 = ?³ × 30

?³ = 8

? = 2

74.89% of (16.03 of 49.91) = ?² + 28.98 of √17

75% of (16 of 50) = ?² + 29 of √16

0.75 × 800 = ?² + 29 × 4

600 = ?² + 116

?² = 600 - 116 = 484

? = ±22

√575.89 × 15.98² ÷ 47.78 = √? of (24.78% of 31.89)

√576 × 16² ÷ 48 = √? of 25% of 32

(24 × 16 × 16) ÷ 48 = √? × 8

16 = √?

? = 256

12.03² + 25.997² - 10.013² = ?² + 15.982² - √400.03

12² + 26² - 10² = ?² + 16² - √400

144 + 576 - 100 = ?² + 256 - 20

820 - 100 = ?² + 236

?² = 484

? = 22

127.76 ÷ 7.983 × 14.83 + 159.89 = 80.09 of ?

128 ÷ 8 × 15 + 160 = 80 of ?

16 × 15 + 160 = 80 of ?

240 + 160 = 80 of ?

400 = 80 of ?

? = 5

Hence, option b