Quantitative Aptitude PYQs

Shortcut Trick

Ratio of Lengths = (Ratio of Speed) × (Ratio of Time)

L_A : L_B = (4 × 5) : (7 × 3) = 20 : 21

Total units = 20 + 21 = 41 units → 410 m

Difference = 21 − 20 = 1 unit → (410 ÷ 41) × 1 = 10 m

∴ The correct answer is 10 m.

Alternate Method

Given: Sum of lengths (L_A + L_B) = 410 m; Ratio of speeds (S_A : S_B) = 4 : 7; Ratio of time (T_A : T_B) = 5 : 3.

Formula: Length of train = Speed × Time (to cross a pole)

⇒ Let S_A = 4x, S_B = 7x and T_A = 5y, T_B = 3y

⇒ L_A = 4x × 5y = 20xy

⇒ L_B = 7x × 3y = 21xy

⇒ L_A + L_B = 20xy + 21xy = 41xy

⇒ 41xy = 410 ⇒ xy = 10

⇒ Difference = L_B − L_A = 21xy − 20xy = xy

⇒ Difference = 10 m

∴ The correct answer is 10 m.

Additional Information

Relative Speed

When two trains move in the opposite direction, relative speed = S₁ + S₂. In the same direction, relative speed = |S₁ − S₂|.

Crossing Objects

To cross a pole/man, distance = length of train. To cross a platform/bridge, distance = length of train + length of platform.

Unit Conversion

To convert kmph to m/s, multiply by 5/18. To convert m/s to kmph, multiply by 18/5.

Shortcut Trick

Ratio of Speed of Boat (u) to Speed of Stream (v) = 5 : 2

Downstream speed (D) = 5 + 2 = 7 units; Upstream speed (U) = 5 − 2 = 3 units

Time difference = 42/3x − 42/7x = 4 ⇒ 14/x − 6/x = 4 ⇒ 8/x = 4 ⇒ x = 2

Speed of boat in still water = 5 × 2 = 10 km/hr

∴ The correct answer is 10 km/hr.

Alternate Method

Given: Ratio u : v = 5 : 2. Distance (D) = 42 km. Time difference = 4 hours.

Formula Used: Time = Distance ÷ Speed; Downstream Speed = u + v; Upstream Speed = u − v.

⇒ Let Speed of boat = 5x and Speed of stream = 2x

⇒ Upstream Speed = 3x; Downstream Speed = 7x

⇒ Time Upstream − Time Downstream = 4 hours

⇒ 42/3x − 42/7x = 4

⇒ 14/x − 6/x = 4

⇒ 8/x = 4 ⇒ x = 2

⇒ Speed of boat = 5 × 2 = 10 km/hr

∴ The correct answer is 10 km/hr.

Additional Information

Speed in Still Water

It is the average of downstream and upstream speeds: u = (D + U) / 2.

Speed of Current

It is half the difference between downstream and upstream speeds: v = (D − U) / 2.

Relative Speed in Streams

When moving downstream, speeds add up; when moving upstream, the stream speed is subtracted from the boat speed.

Shortcut Trick

Initial Ratio R : P : S = 7 : 6 : 9 (Total parts = 22)

Final Ratio R : P : S = 5 : 3 : 3 (Multiply by 2 to equate Puneet's balls) → 10 : 6 : 6 (Total parts = 22)

Change in Sweety's balls = 9 − 6 = 3 units | Change in Rinku's balls = 10 − 7 = 3 units

3 units = 15 balls ⇒ 1 unit = 5 balls

Initial Rinku's balls = 7 units = 7 × 5 = 35 balls.

∴ The correct answer is 35 balls.

Alternate Method

Given: Initial ratio of balls with Rinku, Puneet and Sweety = 7 : 6 : 9.

Formula: Ratio = Value ÷ Common Factor (x)

⇒ Let balls be 7x, 6x, and 9x.

⇒ According to question: (7x + 15) / 6x / (9x − 15) = 5 / 3 / 3

⇒ Comparing Puneet and Sweety: 6x / (9x − 15) = 3 / 3

⇒ 6x = 9x − 15

⇒ 3x = 15 ⇒ x = 5

⇒ Initial balls with Rinku = 7x = 7 × 5 = 35.

∴ The correct answer is 35 balls.

Additional Information

Ratio Constancy Principle

When items are transferred between two people in a group, the total sum of items in the group remains constant. Always equate the total parts of ratios to simplify calculations.

Compound Ratio

If two ratios are a : b and c : d, then the compound ratio is (a × c) : (b × d).

Shortcut Trick

Let A's present age be 1 unit.

Mother's present age = 6 units.

Sum of Mother and Father after 5 years = 80 years → Present sum = 80 − 10 = 70 years.

Since Father is 10 years older, 2 × Mother + 10 = 70 ⇒ Mother = 30 years.

6 units = 30 years ⇒ 1 unit = 5 years.

∴ The correct answer is 5 years.

Alternate Method

Given:

Let A's present age = x

Mother's age (M) = 6x

Father's age (F) = M + 10 = 6x + 10

Calculations:

⇒ Sum of ages of Mother and Father after 5 years = (M + 5) + (F + 5) = 80

⇒ M + F + 10 = 80

⇒ M + F = 70

⇒ 6x + (6x + 10) = 70

⇒ 12x + 10 = 70

⇒ 12x = 60

⇒ x = 5

∴ The correct answer is 5 years.

Additional Information

Age Sum Property

If the sum of ages of 'n' people after 't' years is S, their present sum is S − (n × t).

Ratio Method for Ages

The difference between the ages of two persons remains constant at all points in time.

Base Change in Ages

Always define all members in terms of a single variable (usually the youngest person) to simplify linear equations.

Shortcut Trick

Total Loss = Rs. 3000. Ratio of loss shared by A and B is 2 :

1. Total units of loss = 2 + 1 = 3 units.

3 units = Rs. 3000 ⇒ 1 unit = Rs. 1000.

Share of A = 2 units = 2 × 1000 = Rs. 2000. ∴ The correct answer is Rs. 2000.

Alternate Method

Given → Total Loss = Rs. 3000, Ratio of Loss (A:B) = 2:1.
Formulas: Share of X = (Ratio of X ÷ Sum of Ratios) × Total Value.

⇒ Let the loss shared by B be x.
⇒ Then, the loss shared by A = 2x.
⇒ Total Loss = 2x + x = 3x.
⇒ 3x = 3000 ⇒ x = 1000.
⇒ Share of A = 2x = 2 × 1000 = Rs. 2000. ∴ The correct answer is Rs. 2000.

Additional Information

Partnership Ratios

In a partnership, profit or loss is generally divided in the ratio of Investment × Time unless a specific condition is mentioned.

Conditional Profit Sharing

Agreements can specify different ratios for Profit and Loss as per the contract between partners.

General solution:

 

AMU college

Number females in AMU college in section A  = 6a

Number of males in AMU college in section A = 3b

Number of females in AMU college in section B = 5a

Number of males in AMU college in section B = 4b

Total males in in AMU college = (3b + 4b) = 7b

Total females in AMU college = (6a + 5a) = 11a

7b - 11a = 50 .............

(1) Total number of students in section B = (5a + 4b) 

Total number of students in section A = (6a + 3b)

(5a + 4b) - (6a + 3b) = 10

-a + b = 10

(b - a) = 10 ................ (2) Solving 1 and 2 we get 

7b - 11a = 50

7b - 7a = 70

-4a = - 20

a = 5

b = 15

Number females in AMU college in section A  = 6a = 6 ×  5 = 30

Number of males in AMU college in section A = 3b = 3 ×  15 = 45

Number of females in AMU college in section B = 5a = 5 ×  5 = 25

Number of males in AMU college in section B = 4b = 4 ×  15 = 60

MMH college

Number females in MMH college in section A  = 5c

Number of males in MMH college in section A = d

Number of females in MMH college in section B = 4c

Number of males in MMH college in section B = 3d

Total males in in MMH college = (d + 3d) = 4d

Total females in MMH college = (5c + 4c) = 9c

9c - 4d = 30 ............. (3) Total number of students in section B = (4c + 3d) 

Total number of students in section A = (5c + d)

(4c + 3d) - (5c + d) = 20

-c + 2d = 20

(2d - c) = 20 ............. (4) Solving 3 and 4 we get 

9c - 4d = 30

9 × [(2d - c) = 20]

9c - 4d = 30

18d - 9c = 180

14d = 210

d = 15

c = 10

Number females in MMH college in section A  = 5c = 5 ×  10 = 50

Number of males in MMH college in section A = d = 15

Number of females in MMH college in section B = 4c = 4 ×  10 = 40

Number of males in MMH college in section B = 3d = 3 ×  15 = 45

LPU college

Number females in LPU college in section A  = 2e

Number of males in LPU college in section A = 4f

Number of females in LPU college in section B = 3e

Number of males in LPU college in section B = 3f

Total males in in LPU college = (4f + 3f) = 7f

Total females in LPU college = (3e + 2e) = 5e

5e - 7f = 55 ............. (5) Total number of students in section B = (3e + 3f) 

Total number of students in section A = (2e + 4f)

(3e + 3f) - (2e + 4f) = 15

e - f = 15 ............. (6) Solving 5 and 6 we get 

Multiply equation (6) by 5 and subtract from equation (5) we get

-2f = - 20

f = 10

e = 25 

Number females in LPU college in section A  = 2e = 2 ×  25 = 50

Number of males in LPU college in section A = 4f = 4 ×  10 = 40

Number of females in LPU college in section B = 3e = 3 ×  25 = 75

Number of males in LPU college in section B = 3f = 3 ×  10 = 30

 

 

Solution:

Number of males in section B from college AMU = 60

Total number of males in section B from all the colleges = (60 + 45 + 30) 

Required percentage = 60/135 × 100 = 44.44%

General solution:

 

AMU college

Number females in AMU college in section A  = 6a

Number of males in AMU college in section A = 3b

Number of females in AMU college in section B = 5a

Number of males in AMU college in section B = 4b

Total males in in AMU college = (3b + 4b) = 7b

Total females in AMU college = (6a + 5a) = 11a

7b - 11a = 50 .............

(1) Total number of students in section B = (5a + 4b) 

Total number of students in section A = (6a + 3b)

(5a + 4b) - (6a + 3b) = 10

-a + b = 10

(b - a) = 10 ................ (2) Solving 1 and 2 we get 

7b - 11a = 50

7b - 7a = 70

-4a = - 20

a = 5

b = 15

Number females in AMU college in section A  = 6a = 6 ×  5 = 30

Number of males in AMU college in section A = 3b = 3 ×  15 = 45

Number of females in AMU college in section B = 5a = 5 ×  5 = 25

Number of males in AMU college in section B = 4b = 4 ×  15 = 60

MMH college

Number females in MMH college in section A  = 5c

Number of males in MMH college in section A = d

Number of females in MMH college in section B = 4c

Number of males in MMH college in section B = 3d

Total males in in MMH college = (d + 3d) = 4d

Total females in MMH college = (5c + 4c) = 9c

9c - 4d = 30 ............. (3) Total number of students in section B = (4c + 3d) 

Total number of students in section A = (5c + d)

(4c + 3d) - (5c + d) = 20

-c + 2d = 20

(2d - c) = 20 ............. (4) Solving 3 and 4 we get 

9c - 4d = 30

9 × [(2d - c) = 20]

9c - 4d = 30

18d - 9c = 180

14d = 210

d = 15

c = 10

Number females in MMH college in section A  = 5c = 5 ×  10 = 50

Number of males in MMH college in section A = d = 15

Number of females in MMH college in section B = 4c = 4 ×  10 = 40

Number of males in MMH college in section B = 3d = 3 ×  15 = 45

LPU college

Number females in LPU college in section A  = 2e

Number of males in LPU college in section A = 4f

Number of females in LPU college in section B = 3e

Number of males in LPU college in section B = 3f

Total males in in LPU college = (4f + 3f) = 7f

Total females in LPU college = (3e + 2e) = 5e

5e - 7f = 55 ............. (5) Total number of students in section B = (3e + 3f) 

Total number of students in section A = (2e + 4f)

(3e + 3f) - (2e + 4f) = 15

e - f = 15 ............. (6) Solving 5 and 6 we get 

Multiply equation (6) by 5 and subtract from equation (5) we get

-2f = - 20

f = 10

e = 25 

Number females in LPU college in section A  = 2e = 2 ×  25 = 50

Number of males in LPU college in section A = 4f = 4 ×  10 = 40

Number of females in LPU college in section B = 3e = 3 ×  25 = 75

Number of males in LPU college in section B = 3f = 3 ×  10 = 30

 

 

Solution:

Total number of male students in colleges AMU, MMH, LPU and PQR in section A = 45 × 4 = 180 

Number of male students in college PQR in section A = 180 - (45 + 15 + 40) = 80

Total number of female students in colleges AMU, MMH, LPU and PQR in section A = 40 × 4 = 160 

Number of female students in college PQR in section A = 160 - (30 + 50 +50) = 30

Required total= (80 + 30) = 110

General solution:

 

AMU college

Number females in AMU college in section A  = 6a

Number of males in AMU college in section A = 3b

Number of females in AMU college in section B = 5a

Number of males in AMU college in section B = 4b

Total males in in AMU college = (3b + 4b) = 7b

Total females in AMU college = (6a + 5a) = 11a

7b - 11a = 50 .............

(1) Total number of students in section B = (5a + 4b) 

Total number of students in section A = (6a + 3b)

(5a + 4b) - (6a + 3b) = 10

-a + b = 10

(b - a) = 10 ................ (2) Solving 1 and 2 we get 

7b - 11a = 50

7b - 7a = 70

-4a = - 20

a = 5

b = 15

Number females in AMU college in section A  = 6a = 6 ×  5 = 30

Number of males in AMU college in section A = 3b = 3 ×  15 = 45

Number of females in AMU college in section B = 5a = 5 ×  5 = 25

Number of males in AMU college in section B = 4b = 4 ×  15 = 60

MMH college

Number females in MMH college in section A  = 5c

Number of males in MMH college in section A = d

Number of females in MMH college in section B = 4c

Number of males in MMH college in section B = 3d

Total males in in MMH college = (d + 3d) = 4d

Total females in MMH college = (5c + 4c) = 9c

9c - 4d = 30 ............. (3) Total number of students in section B = (4c + 3d) 

Total number of students in section A = (5c + d)

(4c + 3d) - (5c + d) = 20

-c + 2d = 20

(2d - c) = 20 ............. (4) Solving 3 and 4 we get 

9c - 4d = 30

9 × [(2d - c) = 20]

9c - 4d = 30

18d - 9c = 180

14d = 210

d = 15

c = 10

Number females in MMH college in section A  = 5c = 5 ×  10 = 50

Number of males in MMH college in section A = d = 15

Number of females in MMH college in section B = 4c = 4 ×  10 = 40

Number of males in MMH college in section B = 3d = 3 ×  15 = 45

LPU college

Number females in LPU college in section A  = 2e

Number of males in LPU college in section A = 4f

Number of females in LPU college in section B = 3e

Number of males in LPU college in section B = 3f

Total males in in LPU college = (4f + 3f) = 7f

Total females in LPU college = (3e + 2e) = 5e

5e - 7f = 55 ............. (5) Total number of students in section B = (3e + 3f) 

Total number of students in section A = (2e + 4f)

(3e + 3f) - (2e + 4f) = 15

e - f = 15 ............. (6) Solving 5 and 6 we get 

Multiply equation (6) by 5 and subtract from equation (5) we get

-2f = - 20

f = 10

e = 25 

Number females in LPU college in section A  = 2e = 2 ×  25 = 50

Number of males in LPU college in section A = 4f = 4 ×  10 = 40

Number of females in LPU college in section B = 3e = 3 ×  25 = 75

Number of males in LPU college in section B = 3f = 3 ×  10 = 30

 

 

Solution:

Total number of students in section A to colleges AMH and MMH = (45 + 15 + 30 + 50) = 140

Total number of female students in all the colleges to the section B = (25 + 40 + 75) = 140

Required ratio = 1: 1

General solution:

 

AMU college

Number females in AMU college in section A  = 6a

Number of males in AMU college in section A = 3b

Number of females in AMU college in section B = 5a

Number of males in AMU college in section B = 4b

Total males in in AMU college = (3b + 4b) = 7b

Total females in AMU college = (6a + 5a) = 11a

7b - 11a = 50 .............

(1) Total number of students in section B = (5a + 4b) 

Total number of students in section A = (6a + 3b)

(5a + 4b) - (6a + 3b) = 10

-a + b = 10

(b - a) = 10 ................ (2) Solving 1 and 2 we get 

7b - 11a = 50

7b - 7a = 70

-4a = - 20

a = 5

b = 15

Number females in AMU college in section A  = 6a = 6 ×  5 = 30

Number of males in AMU college in section A = 3b = 3 ×  15 = 45

Number of females in AMU college in section B = 5a = 5 ×  5 = 25

Number of males in AMU college in section B = 4b = 4 ×  15 = 60

MMH college

Number females in MMH college in section A  = 5c

Number of males in MMH college in section A = d

Number of females in MMH college in section B = 4c

Number of males in MMH college in section B = 3d

Total males in in MMH college = (d + 3d) = 4d

Total females in MMH college = (5c + 4c) = 9c

9c - 4d = 30 ............. (3) Total number of students in section B = (4c + 3d) 

Total number of students in section A = (5c + d)

(4c + 3d) - (5c + d) = 20

-c + 2d = 20

(2d - c) = 20 ............. (4) Solving 3 and 4 we get 

9c - 4d = 30

9 × [(2d - c) = 20]

9c - 4d = 30

18d - 9c = 180

14d = 210

d = 15

c = 10

Number females in MMH college in section A  = 5c = 5 ×  10 = 50

Number of males in MMH college in section A = d = 15

Number of females in MMH college in section B = 4c = 4 ×  10 = 40

Number of males in MMH college in section B = 3d = 3 ×  15 = 45

LPU college

Number females in LPU college in section A  = 2e

Number of males in LPU college in section A = 4f

Number of females in LPU college in section B = 3e

Number of males in LPU college in section B = 3f

Total males in in LPU college = (4f + 3f) = 7f

Total females in LPU college = (3e + 2e) = 5e

5e - 7f = 55 ............. (5) Total number of students in section B = (3e + 3f) 

Total number of students in section A = (2e + 4f)

(3e + 3f) - (2e + 4f) = 15

e - f = 15 ............. (6) Solving 5 and 6 we get 

Multiply equation (6) by 5 and subtract from equation (5) we get

-2f = - 20

f = 10

e = 25 

Number females in LPU college in section A  = 2e = 2 ×  25 = 50

Number of males in LPU college in section A = 4f = 4 ×  10 = 40

Number of females in LPU college in section B = 3e = 3 ×  25 = 75

Number of males in LPU college in section B = 3f = 3 ×  10 = 30

 

 

Solution:

Total number of students in section B in MMH college = (40 + 45) = 85

Number of male students in section C in LPU = 85/17 × 10 = 50

Total male students in all the 3 three sections of LPU college = (50 + 40 + 30) = 120

General solution:

 

AMU college

Number females in AMU college in section A  = 6a

Number of males in AMU college in section A = 3b

Number of females in AMU college in section B = 5a

Number of males in AMU college in section B = 4b

Total males in in AMU college = (3b + 4b) = 7b

Total females in AMU college = (6a + 5a) = 11a

7b - 11a = 50 .............

(1) Total number of students in section B = (5a + 4b) 

Total number of students in section A = (6a + 3b)

(5a + 4b) - (6a + 3b) = 10

-a + b = 10

(b - a) = 10 ................ (2) Solving 1 and 2 we get 

7b - 11a = 50

7b - 7a = 70

-4a = - 20

a = 5

b = 15

Number females in AMU college in section A  = 6a = 6 ×  5 = 30

Number of males in AMU college in section A = 3b = 3 ×  15 = 45

Number of females in AMU college in section B = 5a = 5 ×  5 = 25

Number of males in AMU college in section B = 4b = 4 ×  15 = 60

MMH college

Number females in MMH college in section A  = 5c

Number of males in MMH college in section A = d

Number of females in MMH college in section B = 4c

Number of males in MMH college in section B = 3d

Total males in in MMH college = (d + 3d) = 4d

Total females in MMH college = (5c + 4c) = 9c

9c - 4d = 30 ............. (3) Total number of students in section B = (4c + 3d) 

Total number of students in section A = (5c + d)

(4c + 3d) - (5c + d) = 20

-c + 2d = 20

(2d - c) = 20 ............. (4) Solving 3 and 4 we get 

9c - 4d = 30

9 × [(2d - c) = 20]

9c - 4d = 30

18d - 9c = 180

14d = 210

d = 15

c = 10

Number females in MMH college in section A  = 5c = 5 ×  10 = 50

Number of males in MMH college in section A = d = 15

Number of females in MMH college in section B = 4c = 4 ×  10 = 40

Number of males in MMH college in section B = 3d = 3 ×  15 = 45

LPU college

Number females in LPU college in section A  = 2e

Number of males in LPU college in section A = 4f

Number of females in LPU college in section B = 3e

Number of males in LPU college in section B = 3f

Total males in in LPU college = (4f + 3f) = 7f

Total females in LPU college = (3e + 2e) = 5e

5e - 7f = 55 ............. (5) Total number of students in section B = (3e + 3f) 

Total number of students in section A = (2e + 4f)

(3e + 3f) - (2e + 4f) = 15

e - f = 15 ............. (6) Solving 5 and 6 we get 

Multiply equation (6) by 5 and subtract from equation (5) we get

-2f = - 20

f = 10

e = 25 

Number females in LPU college in section A  = 2e = 2 ×  25 = 50

Number of males in LPU college in section A = 4f = 4 ×  10 = 40

Number of females in LPU college in section B = 3e = 3 ×  25 = 75

Number of males in LPU college in section B = 3f = 3 ×  10 = 30

 

 

Solution:

Total number of male students in college AMU and MMH = (45 + 60 + 15 + 45) = 165

Total number of female students in college MMH and LPU = (50 + 40 + 50 + 75) = 215

Required difference = (215 - 165) = 50

General solution:

 

AMU college

Number females in AMU college in section A  = 6a

Number of males in AMU college in section A = 3b

Number of females in AMU college in section B = 5a

Number of males in AMU college in section B = 4b

Total males in in AMU college = (3b + 4b) = 7b

Total females in AMU college = (6a + 5a) = 11a

7b - 11a = 50 .............

(1) Total number of students in section B = (5a + 4b) 

Total number of students in section A = (6a + 3b)

(5a + 4b) - (6a + 3b) = 10

-a + b = 10

(b - a) = 10 ................ (2) Solving 1 and 2 we get 

7b - 11a = 50

7b - 7a = 70

-4a = - 20

a = 5

b = 15

Number females in AMU college in section A  = 6a = 6 ×  5 = 30

Number of males in AMU college in section A = 3b = 3 ×  15 = 45

Number of females in AMU college in section B = 5a = 5 ×  5 = 25

Number of males in AMU college in section B = 4b = 4 ×  15 = 60

MMH college

Number females in MMH college in section A  = 5c

Number of males in MMH college in section A = d

Number of females in MMH college in section B = 4c

Number of males in MMH college in section B = 3d

Total males in in MMH college = (d + 3d) = 4d

Total females in MMH college = (5c + 4c) = 9c

9c - 4d = 30 ............. (3) Total number of students in section B = (4c + 3d) 

Total number of students in section A = (5c + d)

(4c + 3d) - (5c + d) = 20

-c + 2d = 20

(2d - c) = 20 ............. (4) Solving 3 and 4 we get 

9c - 4d = 30

9 × [(2d - c) = 20]

9c - 4d = 30

18d - 9c = 180

14d = 210

d = 15

c = 10

Number females in MMH college in section A  = 5c = 5 ×  10 = 50

Number of males in MMH college in section A = d = 15

Number of females in MMH college in section B = 4c = 4 ×  10 = 40

Number of males in MMH college in section B = 3d = 3 ×  15 = 45

LPU college

Number females in LPU college in section A  = 2e

Number of males in LPU college in section A = 4f

Number of females in LPU college in section B = 3e

Number of males in LPU college in section B = 3f

Total males in in LPU college = (4f + 3f) = 7f

Total females in LPU college = (3e + 2e) = 5e

5e - 7f = 55 ............. (5) Total number of students in section B = (3e + 3f) 

Total number of students in section A = (2e + 4f)

(3e + 3f) - (2e + 4f) = 15

e - f = 15 ............. (6) Solving 5 and 6 we get 

Multiply equation (6) by 5 and subtract from equation (5) we get

-2f = - 20

f = 10

e = 25 

Number females in LPU college in section A  = 2e = 2 ×  25 = 50

Number of males in LPU college in section A = 4f = 4 ×  10 = 40

Number of females in LPU college in section B = 3e = 3 ×  25 = 75

Number of males in LPU college in section B = 3f = 3 ×  10 = 30

 

 

Solution:

Total males in MMH college = 60

Total females in MMH college = 90

150 - (x + x + 10) = 40

150 - 2x + 10 = 40

-2x = -100

x = 50

Shortcut Trick

Ratio of Speed of Boat (u) to Stream (v) = 5 :

1. Ratio of Downstream Speed (u+v) to Upstream Speed (u-v) = (5+1) : (5-1) = 6 : 4 = 3 : 2.

Since distance is constant, Time Ratio is inverse of Speed Ratio = 2 : 3.

Difference in time ratio parts = 3 − 2 = 1 unit → 2 hours. Therefore, Upstream Time = 3 units = 6 hours.

Upstream Speed = 120 km ÷ 6 hours = 20 km/hr. Since (u − v) is 4 parts → 1 part = 5. Still water speed (5 parts) = 25 km/hr.

∴ The correct answer is 25 km/hr.

Alternate Method

Given: Ratio of boat speed (u) to stream speed (v) = 5 :

1. Distance (D) = 120 km. Time difference = 2 hours.

Formula Used: Time = Distance ÷ Speed; Upstream Speed = u − v; Downstream Speed = u + v.

⇒ Let Speed of Boat = 5x and Speed of Stream = x.

⇒ Downstream Speed = 6x and Upstream Speed = 4x.

⇒ [120 ÷ 4x] − [120 ÷ 6x] = 2

⇒ 30/x − 20/x = 2

⇒ 10/x = 2

⇒ x = 5 km/hr

⇒ Speed of boat in still water = 5x = 5 × 5 = 25 km/hr.

∴ The correct answer is 25 km/hr.

Additional Information

Still Water Speed formula

Speed of Boat in Still Water (u) = (Downstream Speed + Upstream Speed) ÷ 2.

Stream Speed formula

Speed of Current/Stream (v) = (Downstream Speed − Upstream Speed) ÷ 2.

Distance in T hours difference

Distance = [ (u² − v²) × Time Difference ] ÷ 2v.

Shortcut Trick

For Compound Interest, 44% in 2 years means the effective rate is 20% (since 20 + 20 + (20×20)/100 = 44%).

Given 2R = 20% ⇒ R = 10%.

For Simple Interest, Total interest = 40% of Sum.

R × Time = 40% ⇒ 10 × 2t = 40 ⇒ 2t = 4 ⇒ t = 2.

∴ The correct answer is 2 years.

Alternate Method

Given: Simple Interest (SI) = 40% of P for 2t years at R%; Compound Interest (CI) = 44% of P for 2 years at 2R%.

Formula Used: SI = (P×R×T)/100; Effective CI rate for 2 years = x + y + (xy/100).

⇒ Let Sum (P) = 100.

⇒ CI Rate (2R) effective for 2 years: 2R + 2R + (2R × 2R)/100 = 44.

⇒ 4R + 4R²/100 = 44 ⇒ R² + 100R − 1100 = 0.

⇒ (R + 110)(R − 10) = 0 ⇒ R = 10% (as Rate cannot be negative).

⇒ Now for SI: Interest = (P × R × Time)/100.

⇒ 40 = (100 × 10 × 2t)/100.

⇒ 40 = 20t.

⇒ t = 2 years.

∴ The correct answer is 2 years.

Additional Information

Effective Rate of Interest

For compound interest of 2 years, the net percentage change formula (x + y + xy/100) quickly gives the interest percentage of the principal.

Simple Interest vs Compound Interest

Simple interest is calculated only on the principal, whereas compound interest is calculated on the principal plus the accumulated interest of previous periods.

Rule of 72

A quick way to estimate the time required to double an investment: Time ≈ 72 ÷ Interest Rate.

Shortcut Trick

Profit ratio is calculated as Investment × Time.

Arun profit : Nisha profit = 9 : 10.

Arun total investment units = 3000 × 12 = 36000.

Nisha total investment units = (2000 × 4) + (2000 + P) × 8 = 8000 + 16000 + 8P.

Equating ratios: 36000 ÷ (24000 + 8P) = 9 ÷ 10 ⇒ 40000 = 24000 + 8P.

8P = 16000 ⇒ P = 2000.

∴ The correct answer is Rs. 2000.

Alternate Method

Given → Initial Investments: Nisha = Rs. 2000, Arun = Rs. 3000. Time = 12 months.
Formulas: Profit Share = Investment × Time period | Ratio of Profits = (I₁ × T₁) : (I₂ × T₂)

⇒ Total investment-months for Arun = 3000 × 12 = 36000.
⇒ Total investment-months for Nisha = (2000 × 4) + (2000 + P) × 8 = 8000 + 16000 + 8P = 24000 + 8P.
⇒ According to question, Arun Profit = (9/10) × Nisha Profit.
⇒ Arun Profit / Nisha Profit = 9 / 10.
⇒ 36000 / (24000 + 8P) = 9 / 10.
⇒ 4000 / (24000 + 8P) = 1 / 10.
⇒ 40000 = 24000 + 8P ⇒ 8P = 16000.
⇒ P = 2000.

∴ The correct answer is Rs. 2000.

Additional Information

Partnership Basics

In a simple partnership, profits are shared in the ratio of investments if the time period is same. In a compound partnership, profits are shared in the ratio of the product of capital and time.

Ratio Proportionality

If A and B invest C₁ and C₂ for T₁ and T₂ respectively, then Profit A : Profit B = (C₁ × T₁) : (C₂ × T₂).

Follow BODMAS rule to solve this question, as per the order given below.

Step - 1: Parts of an equation enclosed in 'Brackets' must be solved first, and following BODMAS rule in the bracket -

Step - 2: Any mathematical 'Of' or 'Exponent' must be solved next.

Step - 3: Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated.

Step - 4: Last but not the least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

Since, we need to find out the approximate value, we can write these values to their nearest integers.

Given expression is

(71.98 + 143.9 × 3.99) ÷ 17.95 + 27.98 = ?

⇒ (72 + 144 × 4) ÷ 18 + 28 = ?

⇒ (72 + 576) ÷ 18 + 28 = ?

⇒ 648 ÷ 18 + 28 = ?

⇒ 36 + 28 = ?

∴ ? = 64

Follow BODMAS rule to solve this question, as per the order given below.

Step - 1: Parts of an equation enclosed in 'Brackets' must be solved first, and following BODMAS rule in the bracket -

Step - 2: Any mathematical 'Of' or 'Exponent' must be solved next.

Step - 3: Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated.

Step - 4: Last but not the least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

Since, we need to find out the approximate value, we can write these values to their nearest integers.

Given expression is

39.99% of (124.95 ÷ 25.05 × 3.94 + 15.05) = ?

⇒ 40% of (125 ÷ 25 × 4 + 15) = ?

⇒ 40% of (125 ÷ 25 × 4 + 15) = ?

⇒ 40% of  (5 × 4 + 15) = ?

⇒ 40% of (20 + 15) = ?

⇒ 40% of 35 = ?

⇒ ? = 14

Follow BODMAS rule to solve this question, as per the order given below.

Step - 1: Parts of an equation enclosed in 'Brackets' must be solved first, and following BODMAS rule in the bracket -

Step - 2: Any mathematical 'Of' or 'Exponent' must be solved next.

Step - 3: Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated.

Step - 4: Last but not the least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

Since, we need to find out the approximate value, we can write these values to their nearest integers.

Given expression is

6.97² + 161.92 ÷ 2.99² – 6.99 × 8.99 = ?

⇒ 72 + 162 ÷ 32 – 7 × 9 = ?

⇒ 49 + 162 ÷ 9 – 7 × 9 = ? 

⇒ 49 + 18 – 63 = ?

⇒ 67 – 63 = ?

⇒ ? = 4

Follow BODMAS rule to solve this question, as per the order given below.

Step - 1: Parts of an equation enclosed in 'Brackets' must be solved first, and following BODMAS rule in the bracket -

Step - 2: Any mathematical 'Of' or 'Exponent' must be solved next.

Step - 3: Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated.

Step - 4: Last but not the least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

Since, we need to find out the approximate value, we can write these values to their nearest integers.

Given expression is

79.98% of 599.98 + 192.02 ÷ 31.92 = √35.96 × ?

⇒ 80% of 600 + 192 ÷ 32 = √36 × ?

⇒ 480 + 192 ÷ 32 = 6 × ?

⇒ 480 + 6 = 6 × ?

⇒ 486 = 6 × ?

⇒ 81 = ?

Follow BODMAS rule to solve this question, as per the order given below.

Step - 1: Parts of an equation enclosed in 'Brackets' must be solved first, and following BODMAS rule in the bracket -

Step - 2: Any mathematical 'Of' or 'Exponent' must be solved next.

Step - 3: Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated.

Step - 4: Last but not the least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

Since we need to find out the approximate value, we can write these values to their nearest integers.

Given expression is

11.98 – [25.96 – {1.98 + 4.95 × (6.06 – 2.93)}] = ?

⇒ 12 – [26 – {2 + 5 × (6 – 3)}] = ?

⇒ 12 – [26 – {2 + 5 × 3}] = ?

⇒ 12 – [26 – 17] = ?

⇒ 12 – 9 = ?

∴ ? = 3

Shortcut Trick

Let the Cost Price (CP) of the bulb be B and the fan be F.

Ratio of Selling Prices: B × 135% = F × 60%.

Ratio of Cost Prices: B / F = 60 / 135 = 4 / 9.

Difference in ratio parts = 9 − 4 = 5 units.

According to the question, 5 units = Rs. 250 ⇒ 1 unit = Rs. 50. Cost of bulb = 4 units = 4 × 50 = Rs. 200. ∴ The correct answer is Rs. 200.

Alternate Method

Given: CP_fan = CP_bulb + 250, Profit%_bulb = 35%, Loss%_fan = 40%, SP_bulb = SP_fan.

Formulas: SP = CP × (100 + P%) ÷ 100 | SP = CP × (100 − L%) ÷ 100.

Let CP of bulb = x. Then CP of fan = x + 250.

⇒ SP of bulb = x × (135/100) = 1.35x.

⇒ SP of fan = (x + 250) × (60/100) = 0.6(x + 250).

⇒ 1.35x = 0.6x + 150.

⇒ 1.35x − 0.6x = 150 ⇒ 0.75x = 150.

⇒ x = 150 ÷ 0.75 = 200.

∴ The correct answer is Rs. 200.

Additional Information

Cost Price and Selling Price Relationship

When the selling prices are equal, the ratio of their cost prices is inversely proportional to their profit/loss multipliers: CP₁/CP₂ = (100 ± R₂)/(100 ± R₁).

Profit and Loss Percentage

Profit or Loss is always calculated on the Cost Price (CP) unless specifically stated otherwise in the problem.