Quantitative Aptitude PYQs

112 + 6 × 7 = 154

154 + 7 × 8 = 210

210 + 8 × 9 = 282

282 + 9 × 10 = 372

372 + 10 × 11 = 482

So, 282 should come in place of 280.

20 × 3 = 60

60 ÷ 2 = 30

30 × 3 = 90

90 ÷ 2 = 45

45 × 3 = 135

Therefore, 45 should come in place of 30.

1549 – 15² = 1324

1324 + 13² = 1493

1493 – 11² = 1372

1372 + 9² = 1453

1453 – 7² = 1404

Therefore, 1372 should come in place of 1374.

Shortcut Trick

Effective rate for P (2 years at 10%) = 10 + 10 + (10 × 10)/100 = 21%

Effective rate for Q (3 years at 10%) = 21 + 10 + (21 × 10)/100 = 33.1%

Difference in interest = 33.1% − 21% = 12.1% of X = ₹1452

⇒ X = (1452 ÷ 12.1) × 100 = ₹12000

SI on (12000 + 3000) = ₹15000 at 12% for 2 years = 15000 × 0.24 = ₹3600

∴ The correct answer is ₹3600.

Alternate Method

Given: Rate (R) = 10%, Time for P = 2 years, Time for Q = 3 years, Difference = ₹1452.

Formula: CI = P[(1 + R/100)ⁿ − 1] | SI = (P × R × T) / 100

⇒ Interest of Q = X[(1 + 10/100)³ − 1] = X[(1.1)³ − 1] = 0.331X

⇒ Interest of P = X[(1 + 10/100)² − 1] = X[(1.1)² − 1] = 0.21X

⇒ Difference = 0.331X − 0.21X = 0.121X

⇒ 0.121X = 1452

⇒ X = 1452 ÷ 0.121 = ₹12000

⇒ New Principal = X + 3000 = 12000 + 3000 = ₹15000

⇒ Simple Interest = (15000 × 12 × 2) / 100

⇒ SI = 15000 × 0.24 = ₹3600

∴ The correct answer is ₹3600.

Additional Information

Effective Rate of Interest

For CI, the effective rate for n years can be calculated using successive percentage formula: a + b + (ab/100).

Difference between CI and SI

For 2 years: Difference = PR²/100². For 3 years: Difference = [PR²(300 + R)]/100³.

Compound Interest Compounded Annually

Amount A = P(1 + R/100)ⁿ. Interest CI = A − P.

Shortcut Trick

Initial Tin is 40% → Copper is 60%.

If 60% = 120 kg, then 1% = 2 kg ∴ Tin = 40 × 2 = 80 kg.

Ratio (120 + 3x + 24) : (80 + 3x + 6) = 3 : 2.

(144 + 3x) : (86 + 3x) = 3 : 2 ⇒ 288 + 6x = 258 + 9x.

3x = 30 ⇒ x = 10.

∴ The correct answer is 10.

Alternate Method

Given:

Initial Copper = 120 kg, Tin % = 40%.

Added Tin = (3x + 6) kg, Added Copper = (3x + 24) kg.

Final ratio Copper : Tin = 3 : 2.

Calculations:

⇒ Total weight % = 100%. Since Tin is 40%, Copper = 100% − 40% = 60%.

⇒ 60% of total alloy = 120 kg ⇒ Total alloy = (120/60) × 100 = 200 kg.

⇒ Initial weight of Tin = 40% of 200 = 80 kg.

⇒ New quantity of Copper = 120 + 3x + 24 = (144 + 3x) kg.

⇒ New quantity of Tin = 80 + 3x + 6 = (86 + 3x) kg.

⇒ (144 + 3x) / (86 + 3x) = 3 / 2.

⇒ 2 × (144 + 3x) = 3 × (86 + 3x).

⇒ 288 + 6x = 258 + 9x.

⇒ 9x − 6x = 288 − 258.

⇒ 3x = 30 ⇒ x = 10.

∴ The correct answer is 10.

Additional Information

Mixture Ratio Rule

When components are added to a mixture, the new ratio is (Initial A + Added A) / (Initial B + Added B) = New Ratio A/B.

Percentage to Value

Value = (Given Component Value ÷ Given Component %) × Target %.

Common solution:

For A,

Total books = 240

Non-fiction books = ({5}/{11+5} x 240 = 75)

Fiction books = (240 – 75 = 165)

For B,

Let Fiction books and non-fiction books be y and x respectively.

[x + y = 690]

[y – x = 270]

So, [y = 480, x = 210]

For C,

Total books = 420, non-fiction books = 156

Fiction books = (420-156=264)

For D,

Total books = 540

Total non-fiction books = ({7/12} x 540 = 315)

Total Fiction books = (540-315=225)

For E,

Total books = 625

Total non-fiction books = 24% of 625 = 150

Total Fiction books = (625 – 150 = 475)

 

Calculation:

Total books sold by Z = (23/13) x 165 = 345

Fiction books sold by Z = 40% x 345 = 138

Non-fiction books sold by Z = 345 – 138 = 207

Common solution:

For A,

Total books = 240

Non-fiction books = ({5}/{11+5} x 240 = 75)

Fiction books = (240 – 75 = 165)

For B,

Let Fiction books and non-fiction books be y and x respectively.

[x + y = 690]

[y – x = 270]

So, [y = 480, x = 210]

For C,

Total books = 420, non-fiction books = 156

Fiction books = (420-156=264)

For D,

Total books = 540

Total non-fiction books = ({7/12} x 540 = 315)

Total Fiction books = (540-315=225)

For E,

Total books = 625

Total non-fiction books = 24% of 625 = 150

Total Fiction books = (625 – 150 = 475)

 

Calculation:

Defective books in E = 16% x 625 = 100

Defective non-fiction books = 20% x 100 = 20

Defective fiction books = 100 – 20 = 80

Difference with total books sold by C = 420 – 80 = 340

Common solution:

For A,

Total books = 240

Non-fiction books = ({5}/{11+5} x 240 = 75)

Fiction books = (240 – 75 = 165)

For B,

Let Fiction books and non-fiction books be y and x respectively.

[x + y = 690]

[y – x = 270]

So, [y = 480, x = 210]

For C,

Total books = 420, non-fiction books = 156

Fiction books = (420-156=264)

For D,

Total books = 540

Total non-fiction books = ({7/12} x 540 = 315)

Total Fiction books = (540-315=225)

For E,

Total books = 625

Total non-fiction books = 24% of 625 = 150

Total Fiction books = (625 – 150 = 475)

 

Calculation:

Average number of fiction books sold by A and E = (165 + 475)/2 = 320

Total fiction books sold by D and A = 225 + 165 = 390

Required percentage = (320/390) x 100 = 82.05% = 82% approx.

Common solution:

For A,

Total books = 240

Non-fiction books = ({5}/{11+5} x 240 = 75)

Fiction books = (240 – 75 = 165)

For B,

Let Fiction books and non-fiction books be y and x respectively.

[x + y = 690]

[y – x = 270]

So, [y = 480, x = 210]

For C,

Total books = 420, non-fiction books = 156

Fiction books = (420-156=264)

For D,

Total books = 540

Total non-fiction books = ({7/12} x 540 = 315)

Total Fiction books = (540-315=225)

For E,

Total books = 625

Total non-fiction books = 24% of 625 = 150

Total Fiction books = (625 – 150 = 475)

 

Calculation:

Non-fiction books sold by C and Fiction books sold by A together = 156 + 165 = 321

Total books sold by C and E together = 420 + 625 = 1045

Required ratio = 321:1045

Common solution:

For A,

Total books = 240

Non-fiction books = ({5}/{11+5} x 240 = 75)

Fiction books = (240 – 75 = 165)

For B,

Let Fiction books and non-fiction books be y and x respectively.

[x + y = 690]

[y – x = 270]

So, [y = 480, x = 210]

For C,

Total books = 420, non-fiction books = 156

Fiction books = (420-156=264)

For D,

Total books = 540

Total non-fiction books = ({7/12} x 540 = 315)

Total Fiction books = (540-315=225)

For E,

Total books = 625

Total non-fiction books = 24% of 625 = 150

Total Fiction books = (625 – 150 = 475)

 

Calculation:

Non-fiction books sold by C = 156

Books sold at Rs.25 = 25% of 156 = 39

Remaining books = 156 – 39 = 117

Revenue = 39 x 25 + 117 x 20 = Rs. 3315

Common solution:

In P,

Total students = 100%

60% = 180

1% = 3

Total students = 300

Total boys = 40% of 300 = 120

In Q,

Total students = 100%

30% = 120

1% = 4

Total students = 400

Total boys = 70% of 400 = 280

Total students attended the seminar = 40% of 400 = 160

In R,

Total students = 100%

45% = 90

1% = 2

Total students = 200

Total boys = 55% of 200 = 110

In S,

Total students = 100%

40% = 80

1% = 2

Total students = 200

Total boys = 60% of 200 = 120

Total students attended the seminar = 50% of 200 = 100

ATQ**,** Total students not attended the seminar in S = 200 − 100 = 100

 

Calculation:

Required percentage = 100/(180 + 280) x 100 = 21.74% = 22% approx.

Common solution:

In P,

Total students = 100%

60% = 180

1% = 3

Total students = 300

Total boys = 40% of 300 = 120

In Q,

Total students = 100%

30% = 120

1% = 4

Total students = 400

Total boys = 70% of 400 = 280

Total students attended the seminar = 40% of 400 = 160

In R,

Total students = 100%

45% = 90

1% = 2

Total students = 200

Total boys = 55% of 200 = 110

In S,

Total students = 100%

40% = 80

1% = 2

Total students = 200

Total boys = 60% of 200 = 120

Total students attended the seminar = 50% of 200 = 100

ATQ**,** Total students not attended the seminar in S = 200 − 100 = 100

 

Calculation:

ATQ,

In Q, ratio of girls: boys who attended the seminar = 3: 1

Total attended = 160

Boys attended = (1/4) x 160 = 40

In R, ratio of girls: boys who attended the seminar = 3: 2

Total attended = 80

Boys attended = 2/5 x 80 = 32

Required sum = 40 + 32 = 72

Common solution:

In P,

Total students = 100%

60% = 180

1% = 3

Total students = 300

Total boys = 40% of 300 = 120

In Q,

Total students = 100%

30% = 120

1% = 4

Total students = 400

Total boys = 70% of 400 = 280

Total students attended the seminar = 40% of 400 = 160

In R,

Total students = 100%

45% = 90

1% = 2

Total students = 200

Total boys = 55% of 200 = 110

In S,

Total students = 100%

40% = 80

1% = 2

Total students = 200

Total boys = 60% of 200 = 120

Total students attended the seminar = 50% of 200 = 100

ATQ**,** Total students not attended the seminar in S = 200 − 100 = 100

 

Calculation:

ATQ,

Difference between boys and girls in R = 110 – 90 = 20

According to the question,

Girls attended the seminar in Q = 2 × 20 = 40

Total students attended the seminar in Q = 160

Therefore, Boys attended the seminar in Q = 160 – 40 = 120

Total students in R and S together = 200 + 200 = 400

Required ratio = 120:400 = 3:10

Common solution:

In P,

Total students = 100%

60% = 180

1% = 3

Total students = 300

Total boys = 40% of 300 = 120

In Q,

Total students = 100%

30% = 120

1% = 4

Total students = 400

Total boys = 70% of 400 = 280

Total students attended the seminar = 40% of 400 = 160

In R,

Total students = 100%

45% = 90

1% = 2

Total students = 200

Total boys = 55% of 200 = 110

In S,

Total students = 100%

40% = 80

1% = 2

Total students = 200

Total boys = 60% of 200 = 120

Total students attended the seminar = 50% of 200 = 100

ATQ**,** Total students not attended the seminar in S = 200 − 100 = 100

 

Calculation:

ATQ,

Average number of boys in P and S = (120 + 120)/2 = 120

Difference between the students who attended the seminar in P and R = 105 – 80 = 25

Required percentage = (120/25) x 100 = 480%

Common solution:

In P,

Total students = 100%

60% = 180

1% = 3

Total students = 300

Total boys = 40% of 300 = 120

In Q,

Total students = 100%

30% = 120

1% = 4

Total students = 400

Total boys = 70% of 400 = 280

Total students attended the seminar = 40% of 400 = 160

In R,

Total students = 100%

45% = 90

1% = 2

Total students = 200

Total boys = 55% of 200 = 110

In S,

Total students = 100%

40% = 80

1% = 2

Total students = 200

Total boys = 60% of 200 = 120

Total students attended the seminar = 50% of 200 = 100

ATQ**,** Total students not attended the seminar in S = 200 − 100 = 100

 

Calculation:

ATQ,

Ratio of boys: girls in P = 120:180 = 2:3

Boys in E are 20% more than girls in R = 120% of 90 = 108

Thus, boys in E = 108

Since, boys: girls in E = 2: 3

Girls in E = 108 x (3/2) = 162

Shortcut Trick

Effective Compound Interest rate for 2 years (x + y + xy/100).

Scheme X (15%): 15 + 15 + (15 × 15)/100 = 32.25%

Scheme Y (25%): 25 + 25 + (25 × 25)/100 = 56.25%

Difference in interest = 56.25% − 32.25% = 24%

Given 24% of Q = ₹4800 ⇒ Q = 4800 ÷ 0.24 = ₹20000.

∴ The correct answer is ₹20000.

Alternate Method

Given: Principal = ₹Q, Time (n) = 2 years, R_X = 15%, R_Y = 25%

Formula: Amount (A) = P(1 + R/100)ⁿ | Interest (I) = A − P

⇒ Interest from Scheme X = Q[(1 + 15/100)² − 1]

⇒ I_X = Q[(1.15)² − 1] = Q[1.3225 − 1] = 0.3225Q

⇒ Interest from Scheme Y = Q[(1 + 25/100)² − 1]

⇒ I_Y = Q[(1.25)² − 1] = Q[1.5625 − 1] = 0.5625Q

⇒ Difference = 0.5625Q − 0.3225Q = 4800

⇒ 0.24Q = 4800

⇒ Q = 4800 / 0.24 = 20000

∴ The correct answer is ₹20000.

Additional Information

Effective Interest Rate

For 2 years, the total compound interest rate can be calculated as x + y + (xy/100) where x and y are the rates for consecutive years.

Difference between CI and SI

The difference between CI and SI for 2 years on a principal P at R% rate is given by P(R/100)².

Compounding Periods

If interest is compounded half-yearly, the rate becomes R/2 and the time becomes 2n.

Shortcut Trick

Ratio of Profit = Σ(Investment × Time)

A's equivalent capital = (12000 × 3) + (8000 × 9) = 36000 + 72000 = 108000

B's equivalent capital = (18000 × 3) + (24000 × 9) = 54000 + 216000 = 270000

Ratio A : B = 108000 : 270000 = 108 : 270 = 2 : 5

Given A's profit (2 units) = Rs 12000 ⇒ 1 unit = Rs 6000

Total profit (2 + 5 = 7 units) = 7 × 6000 = Rs 42000

∴ The correct answer is Rs 42000.

Alternate Method

Given → A's initial investment = 12000, B's initial investment = 18000. Time = 12 months.
Formulas: Profit Share ∝ Investment × Time Period

⇒ A's total investment-month = (12000 × 3) + (8000 × 9) = 1,08,000
⇒ B's total investment-month = (18000 × 3) + (24000 × 9) = 2,70,000
⇒ Profit Ratio (A:B) = 108000 : 270000 = 2 : 5
⇒ Since 2 units = 12000, Total units (2 + 5) = 7
⇒ Total Profit = (12000 ÷ 2) × 7 = 42000

∴ The correct answer is Rs 42000.

Additional Information

Partnership Basics

In a partnership, profit is distributed based on the product of capital and the time for which it was invested.

Change in Capital

When capital changes during the period, calculate equivalent capital for each duration: (C₁ × T₁) + (C₂ × T₂).

Quantity I

SI for first 2 years = (5000 x 8 x 2)/100 = 800

SI for last 4 years = (5000 x 10 x 4)/100 = 2000

Total SI = 800 + 2000 = 2800

Quantity II = Rs.2400

Hence, Quantity I > Quantity II

Shortcut Trick

Upstream speed (U) = 216 ÷ 6 = 36 kmph.

Stream speed (s) = 25% of Downstream (D) ⇒ s = D/4 ⇒ D = 4s.

As D − U = 2s ⇒ 4s − 36 = 2s ⇒ 2s = 36 ⇒ s = 18 kmph.

D = 4 × 18 = 72 kmph. U = 36 kmph.

Total distance (P + 6) + (P + 72) in 6 hours ⇒ (P + 6)/72 + (P + 72)/36 = 6.

Solving: (P + 6) + 2(P + 72) = 432 ⇒ 3P + 150 = 432 ⇒ 3P = 282 ⇒ P = 94.

∴ The correct answer is 94.

Alternate Method

Given:

Upstream Speed (U) = 216 km ÷ 6 hours = 36 kmph

Stream Speed (s) = 25% of Downstream Speed (D) = D/4

Formula Used:

Downstream Speed (D) = v + s | Upstream Speed (U) = v − s

D − U = 2s | Time = Distance ÷ Speed

Calculations:

⇒ D − 36 = 2 × (D/4)

⇒ D − 36 = D/2

⇒ D/2 = 36 ⇒ D = 72 kmph

⇒ Total Time = (Distance_down/D) + (Distance_up/U)

⇒ 6 = (P + 6)/72 + (P + 72)/36

⇒ 6 × 72 = (P + 6) + 2(P + 72)

⇒ 432 = P + 6 + 2P + 144

⇒ 432 = 3P + 150

⇒ 3P = 282

⇒ P = 94

∴ The correct answer is 94.

Additional Information

Relative Speeds in Water

Speed of boat in still water v = (D + U)/2 and Speed of stream s = (D − U)/2.

Average Speed of Boat

If a boat travels same distance upstream and downstream, Average Speed = (D × U) / v = (v² − s²) / v.

Shortcut Trick

Let the three consecutive multiples of 12 in Set A be x − 12, x, and x + 12.

The largest number in Set A is x + 12, which is also the smallest number in Set B.

The four consecutive odd numbers in Set B are x + 12, x + 14, x + 16, and x + 18.

The average of the smallest numbers in Set A (x − 12) and Set B (x + 12) is [(x − 12) + (x + 12)] ÷ 2 = x = 42.

The 2^nd smallest number in Set A is x = 42.

The 2^nd largest number in Set B is x + 16 = 42 + 16 = 58.

Required Difference = 58 − 42 = 16.

∴ The correct answer is 16.

Alternate Method

Given:

Set A contains 3 consecutive multiples of 12.

Set B contains 4 consecutive odd numbers.

The largest number of Set A is equal to the smallest number of Set B.

The average of the smallest numbers of Set A and Set B is 42.

Calculations:

⇒ Let the three multiples in Set A be A₁, A₂, and A₃.

⇒ Since they are consecutive multiples of 12: A₁ = A₂ − 12 and A₃ = A₂ + 12.

⇒ Let the four consecutive odd numbers in Set B be B₁, B₂, B₃, and B₄.

⇒ Given that B₁ = A₃ ⇒ B₁ = A₂ + 12.

⇒ The subsequent odd numbers are B₂ = B₁ + 2 = A₂ + 14, B₃ = B₁ + 4 = A₂ + 16, B₄ = B₁ + 6 = A₂ + 18.

⇒ The smallest number in Set A is A₁ = A₂ − 12.

⇒ The smallest number in Set B is B₁ = A₂ + 12.

⇒ Average of smallest numbers = (A₁ + B₁) ÷ 2 = 42

⇒ [(A₂ − 12) + (A₂ + 12)] ÷ 2 = 42

⇒ 2 × A₂ ÷ 2 = 42 ⇒ A₂ = 42

⇒ The 2^nd smallest number in Set A is A₂ = 42.

⇒ The 2^nd largest number in Set B is B₃ = A₂ + 16 = 42 + 16 = 58.

⇒ Required Difference = B₃ − A₂ = 58 − 42 = 16.

∴ The correct answer is 16.

Additional Information

Consecutive Multiples

Consecutive multiples of a number n have a constant difference of n between any two successive terms.

Consecutive Odd Numbers

Consecutive odd numbers always have a common difference of 2. For example, if the first term is y, the next terms are y + 2, y + 4, and so on.

Averages of Symmetric Sets

The average of two numbers positioned symmetrically around a central value x (like x − d and x + d) is always equal to the central value x.