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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.