Quantitative Aptitude PYQs

Shortcut Trick

A: 40000 × 40 = 16,00,000 | B: 60000 × 32 = 19,20,000 | C: 20000 × 16 = 3,20,000

Ratio = 16 : 19.2 : 3.2 = 160 : 192 : 32 = 5 : 6 : 1

A's share = 5 parts = ₹25,000 ⇒ 1 part = ₹5,000

B's share = 6 parts = 6 × 5000 = ₹30,000

∴ The correct answer is B's profit share = ₹30,000.

Alternate Method

Given:

A invested ₹40,000 from month 1 ⇒ duration = 40 months

B joined after 8 months ⇒ duration = 40 − 8 = 32 months | Investment = ₹60,000

C joined after 8 + 16 = 24 months ⇒ duration = 40 − 24 = 16 months | Investment = ₹20,000

A's profit share = ₹25,000

Find: B's profit share

Formula used:

Profit ratio = Capital × Time

Each partner's share = (Their ratio part ÷ Total ratio parts) × Total Profit

Calculations:

Step 1: Find duration of each partner

⇒ A invests from month 0 to month 40 ⇒ duration = 40 months

⇒ B joins at month 8, exits at month 40 ⇒ duration = 40 − 8 = 32 months

⇒ C joins at month 8 + 16 = month 24, exits at month 40 ⇒ duration = 40 − 24 = 16 months

Step 2: Calculate Capital × Time for each

⇒ A = 40,000 × 40 = 16,00,000

⇒ B = 60,000 × 32 = 19,20,000

⇒ C = 20,000 × 16 = 3,20,000

Step 3: Find profit ratio

⇒ A : B : C = 16,00,000 : 19,20,000 : 3,20,000

⇒ Divide by 3,20,000: = 5 : 6 : 1

Step 4: Find profit shares

⇒ Total parts = 5 + 6 + 1 = 12

⇒ A's share = 5 parts = ₹25,000

⇒ 1 part = 25,000 ÷ 5 = ₹5,000

⇒ B's share = 6 parts = 6 × 5,000 = ₹30,000

⇒ C's share = 1 part = 1 × 5,000 = ₹5,000

Verification:

⇒ Total profit = 25,000 + 30,000 + 5,000 = ₹60,000

⇒ A : B : C = 25,000 : 30,000 : 5,000 = 5 : 6 : 1 

∴ The correct answer is B's profit share = ₹30,000.

Additional Information

Partnership — Joining Mid-Way Rule

When partners join at different times always calculate duration from the month they joined to the end of the business period. Duration of A = total period | Duration of B = total − months after start B joined | Duration of C = total − months after start C joined. Never use the gap between partners as duration.

Sleeping Partner vs Active Partner

A sleeping (silent) partner only invests money — no active role. An active (working) partner gets a salary or commission from total profit first before dividing remainder in Capital × Time ratio. If the problem states one partner manages the business, deduct their salary from total profit first then split remaining profit.

Finding Total Profit from One Partner's Share

If one partner's share and ratio part are known: 1 part = known share ÷ that partner's ratio part. Total profit = 1 part × total ratio parts. Here: 1 part = 25000/5 = 5000 | Total = 5000 × 12 = ₹60,000. This unitary method avoids setting up equations and is the fastest approach.

Withdrawal of Capital Mid-Way

If a partner withdraws or adds capital during the period, split their investment into two parts: before and after the change. Example: A invests ₹10,000 for 6 months then adds ₹5,000 more for remaining 6 months ⇒ A's effective capital = 10,000×6 + 15,000×6 = 60,000 + 90,000 = 1,50,000. Use this as Capital × Time for A.

Calculation:

Equation I: (x²/13) + (30x/13) + 17 = 0

x² + 13x + 17x + 221 = 0

(x² + 13x) + (17x + 221) = 0

x(x + 13) + 17(x + 13) = 0

(x + 13)(x + 17) = 0

⇒ x = -13 or -17

Equation II: (72/y²) + (27/y) + 1 = 0

⇒ 72 + 27y + y² = 0

y² + 24y + 3y + 72 = 0

(y² + 24y) + (3y + 72) = 0

y(y + 24) + 3(y + 24) = 0

(y + 24)(y + 3) = 0

⇒ Y = -24 or -3

Now compare values:

X values: -13, -17

Y values: -24, -3

Comparing each X with each Y:

-13 > -24 but -13 < -3

-17 > -24 but -17 < -3

⇒ Relationship is inconsistent.

Thus, the correct answer is x = y or relationship between x and y can't be established

Calculation:

Equation I:

15x² + 155x + 50 = 0

15x² + 150x + 5x + 50 = 0

15x(x + 10) + 5(x + 10) = 0

Roots:

x = −5/15 = −1/3, x = −10

Equation II: 

24y² + 104y + 96 = 0   

6y² + 26y + 24 = 0 

6y² + 18y + 8y + 24 = 0

6y(y + 3) + 8(y + 3) = 0

Roots:

y = −8/6 = −4/3, y = −3

Now compare each x with each y:

x = −10 < both y = −4/3 and y = -3

x = -1/3 > both y = -4/3 and y = -3

Thus, the correct answer is x = y or relationship between x and y can't be establishedu

Calculation:

From I,

x² – 44x + 448 = 0

⇒ x² – 28x – 16x + 448 = 0

⇒ x(x – 28) – 16(x – 28) = 0

⇒ (x – 16)(x – 28) = 0

Taking, 

⇒ x – 16 = 0 or x – 28 = 0

⇒ x = 16 or x = 28

 

From II,

y² – 28y + 195 = 0 

⇒ y² – 15y – 13y + 195 = 0

⇒ y(y – 15) – 13(y – 15) = 0

⇒ (y – 15) (y – 13) = 0

Taking,

⇒ y – 15 = 0 or y – 13 = 0

⇒ y = 15 or y = 13

Comparison between x and y (via Tabulation)

∴ x > y is the correct option

Common Solution:

Given:

Calculation:

School A

Total Students = 1,200

Girls = 45% of 1,200 = (45/100) × 1,200 = 540

Boys = 1,200 – 540 = 660

School B

Total Students = 1,500

Girls = 52% of 1,500 = (52/100) × 1,500 = 780

Boys = 1,500 – 780 = 720

School C

Total Students = 1,800

Girls = 40% of 1,800 = (40/100) × 1,800 = 720

Boys = 1,800 – 720 = 1,080

School D

Total Students = 1,600

Girls = 65% of 1,600 = (65/100) × 1,600 = 1,040

Boys = 1,600 − 1,040 = 560

School E

Total Students = 900

Girls = 60% of 900 = (60/100) × 900 = 540

Boys = 900 – 540 = 360

Final Table

⇒ Additional boys required = 720 – 360 = 360

Solution

Given:

Additional boys = 360

Original strength = 900

Formula:

Percentage Increase = (Increase ÷ Original Total) × 100

Calculation:

⇒ Percentage increase = (360 ÷ 900) × 100 = 40%

∴ Answer = 40%

Common Solution:

Given:

Calculation:

School A

Total Students = 1,200

Girls = 45% of 1,200 = (45/100) × 1,200 = 540

Boys = 1,200 – 540 = 660

School B

Total Students = 1,500

Girls = 52% of 1,500 = (52/100) × 1,500 = 780

Boys = 1,500 – 780 = 720

School C

Total Students = 1,800

Girls = 40% of 1,800 = (40/100) × 1,800 = 720

Boys = 1,800 – 720 = 1,080

School D

Total Students = 1,600

Girls = 65% of 1,600 = (65/100) × 1,600 = 1,040

Boys = 1,600 − 1,040 = 560

School E

Total Students = 900

Girls = 60% of 900 = (60/100) × 900 = 540

Boys = 900 – 540 = 360

Final Table

⇒ Girls in School A = 540

⇒ Boys in School A = 660

Solution

Given:

Girls = 540

Increase = 20%

Boys = 660

Formula:

Increase = (20/100) × Girls

Calculation:

⇒ Increase = (20/100) × 540 = 108

⇒ New girls = 540 + 108 = 648

⇒ New total students = 648 + 660 = 1,308

∴ Answer = 1,308

Common Solution:

Given:

Calculation:

School A

Total Students = 1,200

Girls = 45% of 1,200 = (45/100) × 1,200 = 540

Boys = 1,200 – 540 = 660

School B

Total Students = 1,500

Girls = 52% of 1,500 = (52/100) × 1,500 = 780

Boys = 1,500 – 780 = 720

School C

Total Students = 1,800

Girls = 40% of 1,800 = (40/100) × 1,800 = 720

Boys = 1,800 – 720 = 1,080

School D

Total Students = 1,600

Girls = 65% of 1,600 = (65/100) × 1,600 = 1,040

Boys = 1,600 − 1,040 = 560

School E

Total Students = 900

Girls = 60% of 900 = (60/100) × 900 = 540

Boys = 900 – 540 = 360

Final Table

⇒ Students in Schools B and D = 1,500 + 1,600 = 3,100

⇒ Students in Schools A and E = 1,200 + 900 = 2,100

Solution

Given:

Students in B and D = 3,100

Students in A and E = 2,100

Formula:

Percentage = (Difference ÷ Original Value) × 100

Calculation:

⇒ Difference = 3,100 − 2,100 = 1,000

⇒ Percentage = (1,000 ÷ 2,100) × 100 = 47.62% ≈ 48%

∴ Answer = 48%

Common Solution:

Given:

Calculation:

School A

Total Students = 1,200

Girls = 45% of 1,200 = (45/100) × 1,200 = 540

Boys = 1,200 – 540 = 660

School B

Total Students = 1,500

Girls = 52% of 1,500 = (52/100) × 1,500 = 780

Boys = 1,500 – 780 = 720

School C

Total Students = 1,800

Girls = 40% of 1,800 = (40/100) × 1,800 = 720

Boys = 1,800 – 720 = 1,080

School D

Total Students = 1,600

Girls = 65% of 1,600 = (65/100) × 1,600 = 1,040

Boys = 1,600 − 1,040 = 560

School E

Total Students = 900

Girls = 60% of 900 = (60/100) × 900 = 540

Boys = 900 – 540 = 360

Final Table

Solution

Given:

⇒ Total students = 1,200 + 900 = 2,100

⇒ Total girls = 540 + 540 = 1,080

Formula:

Percentage = (Girls ÷ Total Students) × 100

Calculation:

⇒ Percentage = (1,080 ÷ 2,100) × 100 = 51.43% ≈ 51.4%

∴ Answer = 51.4%

Common Solution:

Given:

Calculation:

School A

Total Students = 1,200

Girls = 45% of 1,200 = (45/100) × 1,200 = 540

Boys = 1,200 – 540 = 660

School B

Total Students = 1,500

Girls = 52% of 1,500 = (52/100) × 1,500 = 780

Boys = 1,500 – 780 = 720

School C

Total Students = 1,800

Girls = 40% of 1,800 = (40/100) × 1,800 = 720

Boys = 1,800 – 720 = 1,080

School D

Total Students = 1,600

Girls = 65% of 1,600 = (65/100) × 1,600 = 1,040

Boys = 1,600 − 1,040 = 560

School E

Total Students = 900

Girls = 60% of 900 = (60/100) × 900 = 540

Boys = 900 – 540 = 360

Final Table

Given:

Girls in School D = 1,040

Boys in Schools A and E = 660 + 360 = 1,020

Formula:

Percentage = (Required Value ÷ Total Value) × 100

Calculation:

⇒ Percentage = (1,040 ÷ 1,020) × 100 = 101.96% ≈ 102%

∴ Answer = 102%

Shortcut Trick

(R + 6)/(K + 4) = 2/3 ⇒ 3R + 18 = 2K + 8 ⇒ 3R − 2K = −10  ...(1)

K = R + 10 ⇒ K − R = 10  ...(2)

Sub (2) in (1): 3R − 2(R+10) = −10 ⇒ R − 20 = −10 ⇒ R = 10

K = 10 + 10 = 20  |  Neha = (10 + 20)/2 = 15 years

∴ The correct answer is Neha's present age = 15 years.

Alternate Method

Given:

(Riya's age after 6 years) : (Karan's age after 4 years) = 2 : 3

Karan's present age = Riya's present age + 10

Neha's present age = Average of Karan and Riya's present ages

Formula used:

Let Riya's present age = R | Karan's present age = K

(R + 6)/(K + 4) = 2/3

K − R = 10

Neha = (R + K) ÷ 2

Riya Now = 10 +6 yrs After 6 = 16 Karan Now = 20 +4 yrs After 4 = 24 Verify Ratio 16 : 24 = 2 : 3 ✓ K−R=10 ✓ Neha = (Riya + Karan) ÷ 2 = (10 + 20) ÷ 2 = 30 ÷ 2 = 15 years Average of 10 and 20 = 15 Riya = 10 yrs  |  Karan = 20 yrs  |  Neha = 15 yrs

Calculations:

Let Riya's present age = R years | Karan's present age = K years

Step 1: Frame equations

⇒ Equation 1: (R + 6)/(K + 4) = 2/3

⇒ 3(R + 6) = 2(K + 4)

⇒ 3R + 18 = 2K + 8

⇒ 3R − 2K = −10   ...(1)

⇒ Equation 2: K − R = 10 ⇒ K = R + 10   ...(2)

Step 2: Substitute (2) in (1)

⇒ 3R − 2(R + 10) = −10

⇒ 3R − 2R − 20 = −10

⇒ R = −10 + 20 = 10 years

Step 3: Find Karan's age

⇒ K = R + 10 = 10 + 10 = 20 years

Step 4: Verify ratio

⇒ Riya after 6 years = 10 + 6 = 16

⇒ Karan after 4 years = 20 + 4 = 24

⇒ 16 : 24 = 2 : 3 

Step 5: Find Neha's age

⇒ Neha = (R + K) ÷ 2 = (10 + 20) ÷ 2 = 30 ÷ 2 = 15 years

∴ The correct answer is Neha's present age = 15 years.

Additional Information

Age Problems — Key Relationships

Present age = x | Age after n years = x + n | Age n years ago = x − n. The difference between two people's ages always remains constant regardless of the year. So if K − R = 10 now, it was 10 years ago and will be 10 years later too. Use this to cross-check equations.

Ratio of Future Ages — Common Question Type

If (A+m)/(B+n) = p/q then cross-multiply: q(A+m) = p(B+n). Always cross-multiply first to clear the fraction before simplifying. Combine with the age-difference or sum equation to get two equations in two unknowns. Substitution is the fastest method when the difference is directly given.

Average Age — Quick Formula

Average age of a group = Sum of all ages ÷ Number of people. If average age increases by k years after t years, it means no one left or joined — every person aged t years so average also increases by t. If a new member joins: new average = (old sum + new age) ÷ (old count + 1). Here Neha = average of just two people = (10+20)/2 = 15.

Ratio of Past Ages

If present ages are R=10 and K=20: Ratio now = 1:2. Ratio 5 years ago = 5:15 = 1:3. Ratio 5 years later = 15:25 = 3:5. As time increases, the ratio of two people's ages always approaches 1:1. Past ratios have larger differences | Future ratios converge. This property helps verify answers in age problems.

Shortcut Trick

Total age of 4 members = 27 × 4 = 108.

Let Son's age = s. Then Wife's age = s + 36, Dinesh's age = s + 38, and Daughter's age = (s + 38) − 36 = s + 2.

Sum: s + (s + 36) + (s + 38) + (s + 2) = 108 ⇒ 4s + 76 = 108 ⇒ 4s = 32 ⇒ s = 8.

Daughter's age = 8 + 2 = 10. Average = (8 + 10) ÷ 2 = 9.

∴ The correct answer is 9 years.

Alternate Method

Given: Average age = 27 | Members = 4 | Dinesh = Wife + 2 | Dinesh = Daughter + 36 | Wife = Son + 36.

Formula: Average = Sum of observations ÷ Number of observations.

⇒ Total Sum = 27 × 4 = 108

⇒ Let Son's age = x

⇒ Wife's age = x + 36

⇒ Dinesh's age = (x + 36) + 2 = x + 38

⇒ Daughter's age = Dinesh − 36 = (x + 38) − 36 = x + 2

⇒ x + (x + 36) + (x + 38) + (x + 2) = 108

⇒ 4x + 76 = 108 ⇒ 4x = 32 ⇒ x = 8

⇒ Daughter's age = 8 + 2 = 10 years; Son's age = 8 years

⇒ Average of children = (10 + 8) ÷ 2 = 9

∴ The correct answer is 9 years.

Additional Information

Weighted Average

When groups have different sizes, use (n₁x₁ + n₂x₂) ÷ (n₁ + n₂) to find the combined average.

Age Gap Property

The difference between the ages of two people remains constant throughout their lives, regardless of the years passed.

Shortcut Trick

Let Upstream speed (U) = 2x. Then Still water speed (V) = 3x (50% more).

Stream speed (S) = V − U = 3x − 2x = x. Downstream speed (D) = V + S = 3x + x = 4x.

Time ratio (Up : Down) = Speed ratio inverse = 4x : 2x = 2 :

1. Total time 18 hours divided in 2 : 1 ⇒ Upstream time = 12 hrs, Downstream time = 6 hrs.

Distance covered downstream in 6 hours is given as 120 km in the problem context.

∴ The correct answer is 120 km.

Alternate Method

Given: Still water speed (V) = 1.5 × Upstream speed (U). Total distance = 120 km each way. Total time = 18 hrs.

Formula: Speed = Distance ÷ Time | V = (D + U) ÷ 2

⇒ Let U = 2k, then V = 3k.

⇒ Stream speed S = V − U = 3k − 2k = k.

⇒ Downstream speed D = V + S = 3k + k = 4k.

⇒ Total Time = (120 ÷ 2k) + (120 ÷ 4k) = 18

⇒ (60 ÷ k) + (30 ÷ k) = 18 ⇒ 90 ÷ k = 18 ⇒ k = 5.

⇒ Downstream speed = 4 × 5 = 20 kmph.

⇒ Distance in 6 hours = 20 × 6 = 120 km.

∴ The correct answer is 120 km.

Additional Information

Downstream and Upstream Speed

Downstream speed = Speed of boat (V) + Speed of stream (S). Upstream speed = Speed of boat (V) − Speed of stream (S).

Speed of Boat in Still Water

Speed in still water (V) is the average of Downstream (D) and Upstream (U) speeds: V = (D + U) ÷ 2.

Speed of Stream

Speed of the stream (S) is half the difference between Downstream and Upstream speeds: S = (D − U) ÷ 2.

Common Calculation

Given:

Formula:

Transferred Boys = (Percentage ÷ 100) × Original Boys

Calculation:

⇒ Boys in Chess = 200

⇒ Boys in Squash = 400

Solution

Given:

25% of Chess boys are transferred to Squash.

Formula:

Difference = Boys in Squash − Boys in Chess

Calculation:

⇒ Boys transferred = (25 ÷ 100) × 200 = 50

⇒ New boys in Chess = 200 – 50 = 150

⇒ New boys in Squash = 400 + 50 = 450

⇒ Required difference = 450 – 150 = 300

∴ Answer = 300

Common Calculation

Given:

Formula:

New Value = Original Value ± Shifted Students

Calculation:

⇒ Girls in Squash = 300

⇒ Girls in Chess = 300

Solution

Given:

60 girls are shifted from Chess to Squash.

Formula:

Ratio = First Quantity : Second Quantity

Calculation:

⇒ New girls in Squash = 300 + 60 = 360

⇒ New girls in Chess = 300 – 60 = 240

⇒ Required ratio = 360 : 240 = 3 : 2

∴ Answer = 3 : 2

Common Calculation

Given:

Formula:

New Value = Original Value ± Shifted Students

Calculation:

⇒ Boys in Chess = 200

⇒ Boys in Tennis = 624

Solution

Given:

96 boys are shifted from Tennis to Chess.

Formula:

Ratio = First Quantity : Second Quantity

Calculation:

⇒ New boys in Chess = 200 + 96 = 296

⇒ New boys in Tennis = 624 – 96 = 528

⇒ Required ratio = 296 : 528 = 37 : 66

∴ Answer = 37 : 66

Common Calculation

Given:

Formula:

New Value = Original Value ± Shifted Students

Calculation:

⇒ Girls in Squash = 300

⇒ Girls in Tennis = 336

Solution

Given:

84 girls are shifted from Tennis to Squash.

Formula:

Ratio = First Quantity : Second Quantity

Calculation:

⇒ New girls in Squash = 300 + 84 = 384

⇒ New girls in Tennis = 336 – 84 = 252

⇒ Required ratio = 384 : 252 = 32 : 21

∴ Answer = 32 : 21

Common Calculation

Given:

Total participants = 2,160

Squash : Chess = 7 : 5

Tennis = 20% less than (Squash + Chess)

In Squash, Boys : Girls = 4 : 3

In Chess, Girls = 50% more than Boys

In Tennis, Girls = 35% of participants

Formula:

Participants in Tennis = 80% of (Squash + Chess)

Calculation:

⇒ Let the participants in Squash and Chess together be = x

⇒ Participants in Tennis = (4/5)x

⇒ According to the question,

x + (4/5)x = 2,160

⇒ (9/5)x = 2,160

⇒ x = 2,160 × (5/9) = 1,200

⇒ Participants in Tennis = 2,160 − 1,200 = 960

⇒ Total parts in Squash : Chess = 7 + 5 = 12

⇒ Squash = (7/12) × 1,200 = 700

⇒ Chess = (5/12) × 1,200 = 500

⇒ Boys in Squash = (4/7) × 700 = 400

⇒ Girls in Squash = (3/7) × 700 = 300

⇒ Girls : Boys in Chess = 3 : 2

⇒ Girls in Chess = (3/5) × 500 = 300

⇒ Boys in Chess = (2/5) × 500 = 200

⇒ Girls in Tennis = (35/100) × 960 = 336

⇒ Boys in Tennis = 960 – 336 = 624

Solution

Given:

Girls in Squash and Chess = 600

Boys in Squash and Chess = 600

Formula:

Percentage less = ((Difference) ÷ Original Value) × 100

Calculation:

⇒ Girls in Squash and Chess = 300 + 300 = 600

⇒ Boys in Squash and Chess = 400 + 200 = 600

⇒ Percentage less = ((600 − 600) ÷ 600) × 100 = 0%

∴ Answer = 0%

Given:

The area formed in the square = 1764 cm²

Formula Used:

For square,

The area of square = (side)²

The perimeter of a square = 4 × (side)

For circle,

The perimeter of the circle = 2πr

The area of a circle = π × r²

Where, r = radius of the circle

Calculation:

The side of the square = √(1764) = 42 cm

The length of the wire is the same in both cases,

So, the length of the wire = perimeter of the square = 4 × 42 = 168 cm

Let, the radius of the circle = R cm

According to the question,

2 × π × R = 168

⇒ R = 294/11

Now, the area of the circle is

π × R²

⇒  (22/7) × (294/11) × (294/11)

⇒ 24696/11 = 2245.09 ≈ 2245 cm²

∴ The correct answer is option 2

Given:

The ratio of the division of sum = 3 ∶ 5

Rate of interest on two schemes A and B respectively = 14% and 12%

Time for which it was invested in A and B respectively= 4, 3 years

The difference in The amount obtained from scheme B and A = Rs. 12,000

Formula used:

SI = PNR/100

P – Principal, N – Number of terms, R – Rate of interest

Calculation:

As the ratio of the sum is 3 ∶ 5, The amount will be 3x and 5x.

The amount obtained on scheme A = 3x × 4 × 14/100

The amount obtained on scheme A = 168x/100

The amount obtained on scheme B = 5x × 3 × 12/100

The amount obtained on scheme B = 180x/100

Difference in The amount obtained from scheme B and A, Rs. 12,000 = 180x/100 – 168x/100

⇒ 12x/100 = 12,000

⇒ x = Rs. 1,00,000

The amount he invested in scheme A, 3x = 3 × 1,00,000

The amount he invested in scheme A = 3,00,000

∴ The amount invested in scheme A is Rs. 3 lakhs.

Solution:

Time taken by A to complete the work = 8 ×  3 = 24 days

Time taken by B to complete the work = 18 × 2 = 36 days

Time taken by C to complete the work = 2 ×  9 = 18 days

Total work = L C M of (24, 36, 18) = 72

Efficiency of A = 72/24 = 3

Efficiency of B = 72/36 = 2

Efficiency of C = 72/18 = 4

If A and B worked for 3 and 4 days respectively = (3 ×  3 + 4 ×  2) = 17 parts completed 

Remaining work completed by C = (72 - 17)/4 = 55/4 = 13 (3/4) days.