Quantitative Aptitude PYQs

Given:

Number of boys of:

A = 600

B = 450

C = 750

D = 700

E = 800

Formula used:

Average = (Sum of Boys)/Number of school

Calculation:

Average = (Sum of Boys)/Number of school

⇒ Average = (600 + 450 + 750 + 700 + 800)/5

⇒ Average = 3300/5

⇒ Average = 660

∴ The average of boys in schools A, B, C, D and E is 660.

Given:

∠A = 66°

Formula used:

Angle between exterior angle bisector = 90° – 1/2 × Angle of opposite vertex

∠BOC = 90° – ∠A/2

Calculation:

∠BOC = 90° – ∠A/2

⇒ ∠BOC = 90° – 66°/2

⇒ ∠BOC = 90° – 33°

⇒ ∠BOC = 57°

∴ ∠BOC is equal to 57°.

Given:

Number of cars with speed less than 60 km/h = 50 + 70 = 120

Number of cars with speed more than 80 km/h = 60 + 45 = 105

Concept used:

Using the concept of ratio

Calculation:

Required ratio = Number of cars with speed less than 60 km/h ∶ Number of cars with speed more than 80 km/h

⇒ Required ratio = 120 ∶ 105

⇒ Required ratio = 8 ∶ 7

∴ The required ratio is 8 ∶ 7.

Given:

Diameter = 42 cm

Number of revolution = 12

Concept used:

Circumference = 2πr

Total distance = Number of revolution × Circumference

Calculation:

Radius = Diameter/2

⇒ Radius = 42/2

⇒ Radius = 21

Circumference = 2πr

⇒ Circumference = 2 × 22/7 × 21

⇒ Circumference = 2 × 22 × 3

⇒ Circumference = 132

Total distance = Number of revolution × Circumference

⇒ Total distance = 12 × 132

⇒ Total distance = 1584 cm

⇒ Total distance = 1584/100 m

⇒ Total distance = 15.84 m

∴ The wheel go 15.84 m.

Given:

The successive discounts are 8%, 15% and 12%.

Concept used:

Using the concept of percentage.

Calculation:

Let the initial value be 100.

Discounted value = (100 – 8)% of (100 – 15)% of (100 – 12)% of 100

⇒ Discounted value = 92% of 85% of 88% of 100

⇒ Discounted value = 92/100 × 85/100 × 88/100 × 100

⇒ Discounted value = 68.816

Equivalent discount = (100 – Discounted value)/Initial value × 100

⇒ Equivalent discount = (100 – 68.816)/100 × 100

⇒ Equivalent discount = 31.184%

∴ A single discount equivalent to three successive discounts is 31.184%.

Alternate Method 8% = 8/100 = 2/25

15% = 15/100 = 3/20

12% = 12/100 = 3/25

25 × 20 × 25 = 12500

23 × 17 × 22 = 8602 

Discount % = \(\frac{12500-8602}{12500}\) × 100 = 389800/12500 = 31.184 %

 ∴ A single discount equivalent to three successive discounts is 31.184%. 

Shortcut Trick 

x + y -\(\frac{x \times y}{100}\)

8 + 15 - ​\(\frac{8 \times 15}{100}\)= 23 - 1.2 = 21.8​21.8 + 12 -\(\frac{21.8 \times 12}{100}\)= 33.8 - 2.616 = 31.184 %

Given:

\(\frac {0.325 \times 0.325 + 0.175 \times 0.175 + 25 \times 0.00455}{5 \times 0.0065 \times 3.25 - 7 \times 0.175 \times 0.025} - \frac {0.5}{1.5}\) 

Calculations:

\(\frac {0.325 \times 0.325 + 0.175 \times 0.175 + 25 \times 0.00455}{5 \times 0.0065 \times 3.25 - 7 \times 0.175 \times 0.025} - \frac {0.5}{1.5}\)

\(⇒ \frac {325 \times 325 + 175 \times 175 + 25 0\times 455}{5 \times 65 \times 325 - 7 \times 175 \times 25} - \frac {0.5}{1.5}\)

\(⇒\frac {(625)(13 \times 13 + 7 \times 7 + 2 \times 91)}{(625)(1 \times 13 \times 13 - 7 \times 7 \times 1)} - \frac {0.5}{1.5}\)

\(⇒\frac {169 + 49 + 182}{169 - 49 } - \frac {0.5}{1.5}\)

\(⇒\frac {400}{120 } - \frac {0.5}{1.5}\)

\(⇒\frac {10}{3 } - \frac {1}{3}\)

\(⇒\frac {9}{3 }\)

⇒ 3

∴ The value of \(\frac {0.325 \times 0.325 + 0.175 \times 0.175 + 25 \times 0.00455}{5 \times 0.0065 \times 3.25 - 7 \times 0.175 \times 0.025} - \frac {0.5}{1.5}\) is 3.

Given:

Length = 2 × Breadth

Breadth = 12m

Cost of levelling = Rs. 100/m²

Formula used:

Area of rectangle = Length × Breadth 

Area of semi-circle = 1/2 × πr²

Calculation:

Length = 2 × Breadth

⇒ Length = 2 × 12

⇒ Length = 24 m

Area of rectangle = Length × Breadth 

⇒ Area of rectangle = 12 × 24

⇒ Area of rectangle = 288 m²

Area of semi-circle on larger side = 1/2 × π × 12²

⇒ Area of semi-circle on larger side = 72π m²

Area of semi-circle on smaller side = 1/2 × π × 62

⇒ Area of semi-circle on smaller side = 18π m²

Total area of lawn = Area of rectangle + 2 × Area of semi-circle on larger side + 2 × Area of semi-circle on smaller side

⇒ Total area of lawn = 288 + (2 × 72π ) + (2 × 18π)

⇒ Total area of lawn = 288 + 2π(72 + 18)

⇒ Total area of lawn = 288 + 2 × 3.14 × 90

⇒ Total area of lawn = 288 + 565.2

⇒ Total area of lawn = 853.2 m²

Total cost of levelling = Total area of lawn × 100

⇒ Total cost of levelling = 853.2 × 100

⇒ Total cost of levelling = Rs. 85320

∴ The total cost of levelling the entire lawn is Rs. 85320.

Alternate Method 

Length = 2 × Breadth

⇒ Length = 2 × 12

⇒ Length = 24 m

Area of rectangle = Length × Breadth 

⇒ 12 × 24 = 288 m2

Area of semi-circle on larger side = 1/2 × π × 122

⇒ 72π m2

Area of semi-circle on smaller side = 1/2 × π × 62

⇒  18π m2

Total area of lawn = Area of rectangle + 2 × Area of semi-circle on larger side + 2 × Area of semi-circle on smaller side

⇒ 288 + (2 × 72π ) + (2 × 18π)

⇒ 288 + 2π(72 + 18)

⇒ 288 + 2 × 3.14 × 90

⇒ 288 + 565.2

⇒ 853.2 m2

Total cost of levelling = Total area of lawn × 100

⇒ 853.2 × 100

⇒ 85320

Given:

x2 - 5x + 1 = 0

Formula used:

(a + b)3 = a3 + b3 + 3ab(a + b)

Calculations:

x2 - 5x + 1 = 0

⇒ x(x - 5 + 1/x) = 0

⇒ (x - 5 + 1/x) = 0

⇒ (x + 1/x) = 5

Taking cube both sides,

⇒ (x + 1/x)³ = (5)³

⇒ x³ + 1/x³ + 3(x)(1/x)(x + 1/x) = 125

⇒ x³ + 1/x³ + 3(5) = 125

⇒ x3 + 1/x3 + 15 = 125

⇒ x3 + 1/x3 = 125 - 15

⇒ x3 + 1/x3 = 110

(x⁴ + 1/x²)/(x² + 1)

Dividing the numerator and denominator by x,

(1/x)(x4 + 1/x2)/(x2 + 1)(1/x)

⇒ (x³ + 1/x³)/(x + 1/x)

⇒ 110/5

⇒ 22

∴ The value of \(\left(x^4 + \frac 1 {x^2}\right) \div \left(x^2 + 1\right)\) is 22.

Given:

Percentage of employees in department A

Formula used:

Central angle = 360° × (Percentage of employees in department A)/100%

Calculation:

Central angle = 360° × (Percentage of employees in department A)/100%

⇒ Central angle = 360° × 20%/100%

⇒ Central angle = 360°/5

⇒ Central angle = 72° 

∴ Central angel of the sector representing the number of employees in department A is 72°.

Shortcut Trick 

⇒ 100% = 360°

⇒ 1% = 360 / 100

∴ 20% = (360 / 100) × 20

⇒ 72°

Given:

X can complete the work in 3 days.

Y can do three times the same work in 8 days.

Z does five times the same work in 12 days.

Working hours per day = 6 hours

Formula used:

Efficiency = Total work/Total Time

Calculations:

Let the total work be 1 unit.

Time to complete 1 unit by X alone = 3 × 6 = 18 hours

Time to complete the 3 units by Y alone =  8 × 6 = 48 hours

Time to complete the 1 unit by Y alone = 48/3 = 16 hours

Time to complete the 5 units by Z alone =  12 × 6 = 72 hours

Time to complete the 1 unit by Z alone = 72/5 hours

Efficiency of X = 1/18 units/hour

Efficiency of Y = 1/16 units/hour

Efficiency of Z = 5/72 units/hour

Combined efficiency of X, Y and Z = (1/18) + (1/16) + (5/72)

⇒ (8 + 9 + 10)/144

⇒ 27/144 units/hour

Time to complete total work by X, Y and Z together

⇒ 1/(27/144)

⇒ 144/27 hours

⇒ 5 hours 20 minutes

∴ They can complete the work in 5 hours 20 minutes.

Given:

Total number of cars 

50 + 70 + 85 + 90 + 60 + 45 = 400

Total number of cars with speeds less than 70 km/h

50 + 70 + 85 = 205

Formula used:

Percentage = (Part value/Original value) × 100

Calculation:

Percentage of cars running with a speed less than 70 km/h

⇒ (205/400) × 100

⇒ 51.25%

∴ The percentage value is 51.25%.

Given:

Total money = Rs. 8,200

A's share = Rs. 500 more than B

C's share = Rs. 300 more than A

Calculations:

Let share of A, B and C are A, B and C respectively.

A = 500 + B

⇒ B = A - 500

C = 300 + A

⇒ C = 300 + A

A + B + C = 8,200

⇒ A + A - 500 + 300 + A = 8,200

⇒ 3A - 200 = 8,200

⇒ 3A = 8,400

⇒ A = 2,800

∴ A's share was Rs. 2,800

Given:

Height of Asha's mud fort = AB = 9 cm

Height of Suman's mud fort = DE = 16 cm

Distance between mud forts = BD = 24 cm

AB = CD

AC = BD

Calculations:

From the above figure,

DE = CD + CE

⇒ DE = AB + CE

⇒ CE = DE - AB

⇒ CE = 16 - 9

⇒ CE = 7 cm

In ΔACE, By using Pythagoras theorem,

(AE)² = (AC)² + (CE)²

⇒ (AE)² = (24)² + (7)²

⇒ (AE)2 = 576 + 49

⇒ (AE)2 = 625

⇒ AE = √625

⇒ AE = 25 cm

∴ The fort tops are 25 cm far from each other.

Given:

Rate of interest = 17%

Time = 2 years

The difference between CI and SI = Rs. 433.50

Formula used:

Simple interest = P × R% × T

Calculation:

Let the principal be Rs.

x. SI = P × R% × T

⇒ SI = x × 17% × 2

⇒ SI = x × 17/100 × 2

⇒ SI = 34x/100

⇒ SI = 0.34x

In the case of CI:

1^st year CI = P × R% × T

⇒ 1^st year CI = x × 17% × 1

⇒ 1^st year CI = 17x/100

⇒ 1^st year CI = 0.17x

Amount after 1^st year = P + CI 

⇒ Amount = x + 0.17x

⇒ Amount = 1.17x

2^nd year CI = 1.17x × 17% × 1

⇒ 2^nd year CI = 0.1989x

Total CI = 1^st year CI + 2^nd year CI

⇒ CI = 0.17x + 0.1989x

⇒ CI = 0.3689x

Required difference = CI – SI

⇒ 433.50 = 0.3689x – 0.34x

⇒ 433.50 = 0.0289x

⇒ x = 433.50/0.0289

⇒ x = Rs. 15000

∴ The sum is Rs. 15000.

Given:

Total employees = 450

Number of employees working in department B = 14%

Calculations:

Number of employees working in department B = 450 × 14%

⇒ 63

∴ The number of employees working in department B is 63.

Given:

x² + 8y² – 12y – 4xy + 9 = 0

Concept used:

(a – b)² = a² + b² – 2ab

Calculation:

x² + 8y² – 12y – 4xy + 9 = 0

⇒ x² + 4y² + 4y² – 12y – 4xy + 9 = 0

⇒ (x2 + 4y2 – 4xy) + (4y2 – 12y + 9) = 0 

⇒ (x – 2y)² + (2y – 3)² = 0

Either (2y – 3)² = 0

⇒ (2y – 3) = 0

⇒ 2y = 3

⇒ y = 3/2

Or (x – 2y)2 = 0

⇒ (x – 2y) = 0

⇒ x = 2 × 3/2

⇒ x = 3

(7x – 8y) = 7 × 3 – 8 × 3/2

⇒ (7x – 8y) = 21 – 12

⇒ (7x – 8y) = 9

∴ The value of (7x – 8y) is 9.

Given:

Rate = 11%

Rate = \(17 \frac 1 2%\)

Increase in Amount = Rs. 1,071.20

Formula used:

(i) SI = PRT/100

Where, P = principal

R = rate

T = time

(ii) A = SI + P

Where, A = Amount

SI = simple interest

Calculations:

A = SI + P

⇒ A = PRT/100 + P

⇒ A = P{(PRT/100) + 1}

For R = 11%

⇒ A₁ = P{(11/100) + 1}

For R = \(17 \frac 1 2%\)

⇒ A₂ = P{(35/200) + 1}

A₂ - A₁ = P{(35/200) + 1} - P{(11/100) + 1}

⇒ P{35/200 + 1 - 11/100 - 1}

⇒ P{35/200 - 11/100}

⇒ P{(35 - 22)/200}

⇒ 13P/200 = 1,071.20

⇒ P = 214240/13

⇒ 16,480

∴ The principal amount invested is Rs.16,480.

Given:

Distance travelled = 300 km

Time is taken by outward journey = Time taken for return journey + 5 hrs

Speed of returning journey = Speed of going journey + 10 km/h

Concept used:

Average speed = (Total distance traveled)/(Total time is taken)

Calculation:

Let the speed of the outward journey be x km/h.

Speed of returning journey = (x + 10) km/h

Time is taken by outward journey = 300/x hrs

Time is taken by returning journey = 300/(x + 10) hrs

According to the question, 

Time is taken by outward journey = Time taken for return journey + 5 hrs

⇒ 300/x = 300/(x + 10) + 5

⇒ 300/x – 300/(x + 10) = 5

⇒ (300x + 3000 – 300x)/(x² + 10x) = 5

⇒ 3000 = 5x² + 50x

⇒ x² + 10x – 600 = 0

⇒ x² + 30x – 20x – 600 = 0

⇒ x(x + 30) – 20(x + 30) = 0

⇒ (x + 30)(x – 20) = 0

⇒ (x + 30) = 0 or (x – 20) = 0

⇒ x = -30 or x = 20

Speed of outward journey = 20 hrs

Time is taken by outward journey = 300/x hrs

⇒ Outward journey time = 300/20

⇒ Outward journey time = 15 hrs

Time is taken by outward journey = Time taken for return journey + 5 hrs

⇒ 15 = Returning time + 5

⇒ Returning time = 10 hrs

Average speed = (Total distance traveled)/(Total time is taken)

⇒ Average speed = (300 + 300)/(15 + 10)

⇒ Average speed = 600/25

⇒ Average speed = 24 km/h

∴ The average speed for the entire journey is 24 km/h.

Alternate Method Let speed of the outward journey = s₁ = x km/h.

Speed of returning journey = s₂ = (x + 10) km/h

Distance = \(\frac{s_1× s_2}{s_2 - s_1}\) × (Difference in time)

300 = \(\frac{x× (x+10)}{(x + 10)-x}\) × 5

⇒ 600 = x(x + 10)

⇒ 600 = 20 × (20 + 10)

By Comparing

x = 20

Average speed = \(\frac{2\times 20 \times30}{20 + 30}\) = 1200/50 = 24 km/h.

Given:

A and B can do a piece of work in 36 days

B and C can do the same work in 60 days

A and C can do the same work in 45 days

Concept used:

Total days of work = Work/(One day work)

Calculation:

1 day work of (A + B) = 1/36           …………..(1)

1 day work of (B + C) = 1/60           …………..(2)

1 day work of (A + C) = 1/45           …………..(3)

Adding (1), (2), and (3)

1 day work of 2(A + B + C) = 1/36 + 1/60 + 1/45

⇒ 1/36 + 1/60 + 1/45

⇒ (5 + 3 + 4)/180

⇒ 12/180

⇒ 1/15

⇒ 1 day work of (A + B + C) = 1/30

1 day work of A = 1 day work of (A + B + C) – 1 day work of (B + C)

⇒ 1 day work of A = 1/30 – 1/60

⇒ (2 – 1)/60

⇒ 1/60

Total number of days completed by A = Work/(1 day work of A)

⇒ Total number of days completed by A = 1/(1/60)

⇒ Total number of days completed by A = 60 days

∴ Total number of days completed by A is 60 days.

Given:

sec3x = cosec(3x - 45°)

Concept used:

cosec(90 - θ) = secθ 

Calculation:

sec3x = cosec(3x - 45°)

⇒ cosec(90° - 3x) = cosec(3x - 45°)

⇒ 90 - 3x = 3x - 45°

⇒ 6x = 135° 

⇒ x = 22.5° 

∴ The value of x is 22.5°.