Quantitative Aptitude PYQs

General Solution:

Let:

GA : Questions given to A

GB : Questions given to B

GC : Questions given to C

AA : Questions attempted by A

AB : Questions attempted by B

AC : Questions attempted by C

B attempted 50% of the questions he was given:

AB = 0.5 × GB

A was given 20 fewer questions than B:

GA = GB - 20

C was given 130 questions:

GC = 130

Combined questions attempted by A and C is 145:

AA + AC = 145

Total questions given to all three is twice the total questions they attempted:

GA + GB + GC = 2 × (AA + AB + AC)

Number of questions attempted by B and C is the same:

AB = AC

From Equation 1:

AB = 0.5 × GB

From Equation 3:

GC = 130

From Equation 2:

GA = GB - 20

From Equation 4:

AA + AC = 145

From Equation 5:

GA + GB + GC = 2 × (AA + AB + AC)

Substitute GA = GB - 20 and GC = 130 into Equation 5:

GB - 20 + GB + 130 = 2 × (AA + AB + AC)

Simplify:

2GB + 110 = 2 × (AA + AB + AC)

GB + 55 = AA + AB + AC

From Equation 4:

AA + AC = 145

From Equation 6:

AB = AC

Substitute AB = AC into GB + 55 = AA + AB + AC:

GB + 55 = AA + AB + AC

GB + 55 = AA + 2AC

From Equation 4:

AA = 145 - AC

Substitute AA = 145 - AC into GB + 55 = AA + 2AC:

GB + 55 = 145 - AC + 2AC

Simplify:

GB + 55 = 145 + AC

GB = 90 + AC

From Equation 1:

AB = 0.5 × GB

From Equation 6:

AB = AC

So:

AC = 0.5 × GB

Substitute AC = 0.5 × GB into GB = 90 + AC:

GB = 90 + 0.5 × GB

Subtract 0.5 × GB from both sides:

0.5 × GB = 90

GB = 180

Now, find AC:

AC = 0.5 × GB = 0.5 × 180 = 90

From Equation 6:

AB = AC = 90

From Equation 4:

AA = 145 - AC = 145 - 90 = 55

From Equation 2:

GA = GB - 20 = 180 - 20 = 160

From Equation 3:

GC = 130

Thus,

 

Calculations:

Not Attempted = Given − Attempted

For A:

Not Attempted = 160 − 55 = 105

For B:

Not Attempted = 180 − 90 = 90

For C:

Not Attempted = 130 − 90 = 40

Now, sum the not attempted questions for A, B, and C:

Total Not Attempted = 105 + 90 + 40 = 235

Thus, the total number of questions not attempted by A, B, and C together is 235.

General Solution:

Let:

GA : Questions given to A

GB : Questions given to B

GC : Questions given to C

AA : Questions attempted by A

AB : Questions attempted by B

AC : Questions attempted by C

B attempted 50% of the questions he was given:

AB = 0.5 × GB

A was given 20 fewer questions than B:

GA = GB - 20

C was given 130 questions:

GC = 130

Combined questions attempted by A and C is 145:

AA + AC = 145

Total questions given to all three is twice the total questions they attempted:

GA + GB + GC = 2 × (AA + AB + AC)

Number of questions attempted by B and C is the same:

AB = AC

From Equation 1:

AB = 0.5 × GB

From Equation 3:

GC = 130

From Equation 2:

GA = GB - 20

From Equation 4:

AA + AC = 145

From Equation 5:

GA + GB + GC = 2 × (AA + AB + AC)

Substitute GA = GB - 20 and GC = 130 into Equation 5:

GB - 20 + GB + 130 = 2 × (AA + AB + AC)

Simplify:

2GB + 110 = 2 × (AA + AB + AC)

GB + 55 = AA + AB + AC

From Equation 4:

AA + AC = 145

From Equation 6:

AB = AC

Substitute AB = AC into GB + 55 = AA + AB + AC:

GB + 55 = AA + AB + AC

GB + 55 = AA + 2AC

From Equation 4:

AA = 145 - AC

Substitute AA = 145 - AC into GB + 55 = AA + 2AC:

GB + 55 = 145 - AC + 2AC

Simplify:

GB + 55 = 145 + AC

GB = 90 + AC

From Equation 1:

AB = 0.5 × GB

From Equation 6:

AB = AC

So:

AC = 0.5 × GB

Substitute AC = 0.5 × GB into GB = 90 + AC:

GB = 90 + 0.5 × GB

Subtract 0.5 × GB from both sides:

0.5 × GB = 90

GB = 180

Now, find AC:

AC = 0.5 × GB = 0.5 × 180 = 90

From Equation 6:

AB = AC = 90

From Equation 4:

AA = 145 - AC = 145 - 90 = 55

From Equation 2:

GA = GB - 20 = 180 - 20 = 160

From Equation 3:

GC = 130

Thus,

 

Calculations:

C was given G_C = 130 questions.

It is stated that D was given 20% less than C. So:

G_D = G_C − 0.20 × G_C

G_D = 130 − 0.20 × 130

G_D = 130 − 26

G_D = 104

It is stated that D attempted 75% of the total questions given to D. So:

A_D = 0.75 × G_D

A_D = 0.75 × 104

A_D = 78

Thus, the number of questions attempted by D is 78.

General Solution:

Let:

GA : Questions given to A

GB : Questions given to B

GC : Questions given to C

AA : Questions attempted by A

AB : Questions attempted by B

AC : Questions attempted by C

B attempted 50% of the questions he was given:

AB = 0.5 × GB

A was given 20 fewer questions than B:

GA = GB - 20

C was given 130 questions:

GC = 130

Combined questions attempted by A and C is 145:

AA + AC = 145

Total questions given to all three is twice the total questions they attempted:

GA + GB + GC = 2 × (AA + AB + AC)

Number of questions attempted by B and C is the same:

AB = AC

From Equation 1:

AB = 0.5 × GB

From Equation 3:

GC = 130

From Equation 2:

GA = GB - 20

From Equation 4:

AA + AC = 145

From Equation 5:

GA + GB + GC = 2 × (AA + AB + AC)

Substitute GA = GB - 20 and GC = 130 into Equation 5:

GB - 20 + GB + 130 = 2 × (AA + AB + AC)

Simplify:

2GB + 110 = 2 × (AA + AB + AC)

GB + 55 = AA + AB + AC

From Equation 4:

AA + AC = 145

From Equation 6:

AB = AC

Substitute AB = AC into GB + 55 = AA + AB + AC:

GB + 55 = AA + AB + AC

GB + 55 = AA + 2AC

From Equation 4:

AA = 145 - AC

Substitute AA = 145 - AC into GB + 55 = AA + 2AC:

GB + 55 = 145 - AC + 2AC

Simplify:

GB + 55 = 145 + AC

GB = 90 + AC

From Equation 1:

AB = 0.5 × GB

From Equation 6:

AB = AC

So:

AC = 0.5 × GB

Substitute AC = 0.5 × GB into GB = 90 + AC:

GB = 90 + 0.5 × GB

Subtract 0.5 × GB from both sides:

0.5 × GB = 90

GB = 180

Now, find AC:

AC = 0.5 × GB = 0.5 × 180 = 90

From Equation 6:

AB = AC = 90

From Equation 4:

AA = 145 - AC = 145 - 90 = 55

From Equation 2:

GA = GB - 20 = 180 - 20 = 160

From Equation 3:

GC = 130

Thus,

 

Calculations:

Let:

x: Number of correct answers by C 

y: Number of incorrect answers by C

C attempted 90 questions:

 

x + y = 90

C scored 70 marks:

x - (1/4)y = 70

From x + y = 90, we can write:

x = 90 - y

Substitute x = 90 - y into the second equation:

(90 - y) - (1/4)y = 70

90 - (5/4)y = 70

(5/4)y = 20

y = 16

Thus, the number of incorrect questions answered by C is 16.

Shortcut Trick

Total Work = LCM(40, 80, 10) = 80 units.

Efficiency of A = 2, B = 1, C = 8.

Combined Efficiency = 2 + 1 + 8 = 11 units/day.

Time taken = 80 ÷ 11 days.

∴ The correct answer is 80/11 days.

Alternate Method

Given:

Time taken by A = 40 days

Time taken by B = 80 days

Time taken by C = 10 days

Formula Used:

Total Work = Time × Efficiency

Time = Total Work ÷ Combined Efficiency

Calculations:

⇒ Total Efficiency of (A + B + C) = 2 + 1 + 8 = 11 units/day

⇒ Total Work = 80 units

⇒ Time taken by all three together = Total Work ÷ Total Efficiency

⇒ Time = 80/11 days

∴ The correct answer is 80/11 days.

Additional Information

Work and Efficiency

Efficiency is inversely proportional to the time taken to complete a work. If a person completes work in 'n' days, their 1-day work is 1/n.

Combined Work Formula

For three people with times x, y, and z, the combined time is (x × y × z) ÷ (xy + yz + zx).

Pipes and Cisterns

Similar logic applies where filling pipes have positive efficiency and leakage/outlet pipes have negative efficiency.

Shortcut Trick

Total Distance = (5 × 2) + (8 × 1) = 10 + 8 = 18 km.

Total Time = 2 + 1 = 3 hours.

Average Speed = Total Distance ÷ Total Time = 18 ÷ 3 = 6 km/h.

∴ The correct answer is 6 km/hr.

Alternate Method

Given:

Speed of A = 5 km/h, Time of A = 2 hours.

Speed of B = 8 km/h, Time of B = 1 hour.

Formula Used:

Average Speed = Total Distance ÷ Total Time

Calculations:

⇒ Distance covered by A = Speed × Time = 5 × 2 = 10 km.

⇒ Distance covered by B = Speed × Time = 8 × 1 = 8 km.

⇒ Total distance = 10 + 8 = 18 km.

⇒ Total time taken = 2 + 1 = 3 hours.

⇒ Average Speed = 18 ÷ 3 = 6 km/h.

∴ The correct answer is 6 km/hr.

Additional Information

Weighted Average Speed

When different speeds are maintained for different durations, Average Speed = (s₁t₁ + s₂t₂) ÷ (t₁ + t₂).

Equal Distance Case

If a person travels equal distances at speeds x and y, the average speed is given by the Harmonic Mean: 2xy ÷ (x + y).

Relative Speed

In the same direction, relative speed = s₁ − s₂. In the opposite direction, relative speed = s₁ + s₂.

Shortcut Trick

Rate of Interest = 12% per annum.

Difference in Time = 8 years − 3 years = 5 years.

Difference in Interest % = 12% × 5 = 60%.

Given 60% of Sum = ₹2,880 → Sum = 2880 ÷ 0.6 = ₹4,800.

∴ The correct answer is ₹4800.

Alternate Method

Given: Rate (R) = 12%, Time₁ (T₁) = 3 years, Time₂ (T₂) = 8 years, Difference in SI = ₹2,880.

Formula Used: Simple Interest (SI) = (P × R × T) ÷ 100

⇒ [(P × 12 × 8) ÷ 100] − [(P × 12 × 3) ÷ 100] = 2880

⇒ (96P − 36P) ÷ 100 = 2880

⇒ 60P ÷ 100 = 2880

⇒ 0.6P = 2880

⇒ P = 2880 ÷ 0.6 = 4800

∴ The correct answer is ₹4800.

Additional Information

Simple Interest vs Amount

The total Amount (A) is the sum of Principal (P) and Interest (I). A = P + (P × R × T)/100.

Difference in SI for same Principal

If R is constant, Difference in SI = P × R × (T₂ − T₁) ÷ 100. If T is constant, Difference in SI = P × (R₂ − R₁) × T ÷ 100.

Rule of 72

A quick way to estimate how long it takes to double your money: Time ≈ 72 ÷ Interest Rate (used mainly for Compound Interest).

Concept Used

Using BODMAS 

Calculation:

⇒ 1/8 of 248 × 18/6 + ? = 744/8 + 125

⇒ 31 × 3 + ? = 93 + 125

⇒ 93 + ? = 218

⇒ ? = 218 – 93

⇒ ? = 125

∴ The ? is  125

Given:

(1/3) of 570 + (1/8) of 6464 - √81 = ?

Concept Used:

Follow the BODMAS rule according to the table given below:

Calculation:

(1/3) of 570 + (1/8) of 6464 - √81 = ?

⇒ (1/3 × 570) + (1/8 × 6464) - 9 = ?

⇒ 190 + 808 – 9 = ?

⇒ 998 – 9 = ?

⇒ ? = 989

∴ The value of ‘?’ Is 989.

Given:

(√729 ÷ 9 + 13 × 4)/5 = ?

Rule:

Calculation:

(√729 ÷ 9 + 13 × 4)/5 = ?

⇒ (27 ÷ 9 + 13 × 4)/5 = ?

⇒ (3 + 13 × 4)/5 = ?

⇒ (3 + 52)/5 = ?

⇒ ? = 55/5

∴ The required answer is 11.

Hint

√ 729 = 27

Convert the data into mathematical form,

⇒ 630 × (4/7) × (60/100) = ?

⇒ 90 × 4 × 0.6 = ?                                      

⇒ ? = 216

Given:

60% of 360 + 56 ÷ 8 = ? + 5 × 4

Concept used:

Follow the BODMAS rule according to the table given below:

Calculation:

60% of 360 + 56 ÷ 8 = ? + 5 × 4

⇒ (60/100) × 360 + 7 = ? + 20

⇒ (3/5) × 360 + 7 – 20 = ?

⇒ ? = 216 – 13

⇒ ? = 203

∴ 203 will come in place of the question mark (?).

Concept:

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. 

Calculation:

Applying BODMAS:

Taking square both sides

⇒ (2/9 of 405 ÷ 15 + 23)² = x

⇒ (90 ÷ 15 + 23)² = x

⇒ (6 + 23)2 = x

⇒ (29)² = x

⇒ x = 841

√(5/8 × 32% of 1200 ÷ 16 + 34)

Applying BODMAS Rule;

⇒ √(5/8 × (32/100) × 1200 ÷ 16 + 34)

⇒ √(5/8 × 384 ÷ 16 + 34)

⇒ √(5/8 × 24 + 34)

⇒ √(15 + 34)

⇒ √49

⇒ 7

Given:

125% of 560 + 22% of 450 = ?

Concept Used:

Follow the BODMAS rule according to the table given below:

Calculation:

125% of 560 + 22% of 450 = ?

⇒ 125/100 × 560 + 22/100 × 450

⇒ 5/4 × 560 + 11/50 × 450

⇒ 700 + 99

⇒ 799

∴ The value of ? is 799.

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 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 least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated 

 25 × 4 ÷ 5 = 80 ÷ 20 × ?

⇒ 25 × (4/5) = 4 × ?

⇒ 5 × 4 = 4 × ?

∴ ? = 5

Concept used:

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

⇒ 153 ÷ 17 × (9)² – 53 = (?)²

⇒ 9 × (9)² – 53 = (?)²

⇒ 729 – 53 = (?)²

⇒ (±26)² = (?)²

∴ ? = ±26

Shortcut Trick

Upstream speed (u) = 60 km ÷ 4 hours = 15 kmph.

Stream speed (v) = Still water speed − Upstream speed = 18 − 15 = 3 kmph.

Downstream speed (d) = Still water speed + Stream speed = 18 + 3 = 21 kmph.

Downstream distance (3 hr) = 21 × 3 = 63 km.

Difference = 173 − 63 = 110 km.

∴ The correct answer is 110 km.

Alternate Method

Given: Speed of boat in still water (x) = 18 kmph, Upstream distance = 60 km, Time = 4 hr.

Formula Used: Speed = Distance ÷ Time; Upstream speed = x − y; Downstream speed = x + y.

⇒ Upstream Speed = 60 ÷ 4 = 15 kmph.

⇒ Let Stream Speed be y kmph.

⇒ 18 − y = 15

⇒ y = 3 kmph.

⇒ Downstream Speed = 18 + 3 = 21 kmph.

⇒ Distance covered in 3 hours downstream = 21 × 3 = 63 km.

⇒ Required Difference = 173 − 63 = 110 km.

∴ The correct answer is 110 km.

Additional Information

Still Water Speed Formula

Speed of boat in still water (x) = (Downstream speed + Upstream speed) ÷ 2.

Stream Speed Formula

Speed of current/stream (y) = (Downstream speed − Upstream speed) ÷ 2.

Average Speed of Boat

Average speed for a round trip = (Downstream Speed × Upstream Speed) ÷ Still Water Speed.

The total number of mobile covers sold by B.

Total = March + April + May = 120 + 99 + 102 = 321

The average is calculated as:

Average = Total / Number of months

Here, the number of months = 3.

So:

Average = 321 / 3 = 107

Thus, the average number of mobile covers sold by B in all the given months is 107.

A sold 72 mobile covers in April.

Each mobile cover is sold for Rs. 50.

Total revenue = Number of covers sold × Price per cover

Total revenue =72 × 50 = 3600 .

Thus, the total revenue generated by A in April is Rs. 3600.

C in April: 65 mobile covers.

B in March: 120 mobile covers.

The formula to calculate the percentage is:

Percentage = Part / Whole × 100

Here:

Part = Number of mobile covers sold by C in April = 65.

Whole = Number of mobile covers sold by B in March = 120.

Substitute the values:

Percentage = 65 / 120 × 100

65 / 120 = 0.5417

Percentage = 0.5417 × 100 = 54.17%

Thus, the number of mobile covers sold by C in April is 54.17% of the number of mobile covers sold by B in March.