Quantitative Aptitude PYQs

Given:

Monthly salary = Rs. 50,000

E = 60% of income

T = 20% of E

C = 15% of T

After salary got raised by 40%,

E = 60% of income

T = 30% of E

C = 20% of T

Calculations:

Let the initial salary be 1000x.

E = 60% of 1000x

⇒ 600x

T = 20% of 600x

⇒ 120x

C = 15% of 120x

⇒ 18x

Total expense = 600x + 120x + 18x

⇒ 738x

Initial savings = 1000x - 738x

⇒ 262x

After salary got raised by 40%,

New salary = 1000x + 1000x × 40%

⇒ 1400x

New E = 60% of 1400x

⇒ 840x

New T = 30% of 840x

⇒ 252x

New C = 20% of 252x

⇒ 50.4x

New Total expense = 840x + 252x + 50.4x

⇒ 1142.4x

New savings = 1400x - 1142.4x

⇒ 257.6x

Difference in savings = 262x - 257.6x

⇒ 4.4x

∵ 1000x = Rs. 50,000

⇒ x = 50

Difference in savings = 4.4x

⇒ 4.4 × 50

⇒ Rs. 220

∴ The difference between the two savings is Rs. 220.

Given:

\(3 \frac 1 3 ÷ 2 \frac 1 2 \rm;of;1 \frac 3 5 + \left(\frac 3 8 + \frac 1 7 × 1 \frac 3 4\right)\)

Concept Used:

Calculation:

\(3 \frac 1 3 ÷ 2 \frac 1 2 \rm;of;1 \frac 3 5 + \left(\frac 3 8 + \frac 1 7 × 1 \frac 3 4\right)\)

⇒ (10/3) ÷ [(5/2) × (8/5)] + [(3/8) + (1/7) × (7/4)]

⇒ (10/3) ÷ 4 + [(3/8) + (1/7) × (7/4)]

⇒ (5/6) + [(3/8) + (1/4)]

⇒ (5/6) + (5/8)

⇒ 70/48

⇒ 35/24

∴ The value is 35/24.

Given:

Pipe A takes 5 hours less than pipe B to fill a cistern.

Time to fill the tank by pipe A and B together = 6 hours

Formula used:

Efficiency = Total work/total time

Concept used:

Let time to fill the whole tank by pipe B alone be x hours.

So, time to fill the whole tank by pipe A will be (x - 5) hours.

Let capacity of total tank be 6x(x - 5) units.

Efficiency of pipe A = 6x(x - 5)/(x - 5)

⇒ 6x units/hour

Efficiency of pipe B = 6x(x - 5)/x

⇒ 6(x - 5) units/hour

Efficiency of A and B together = 6x(x - 5)/6

⇒ x(x - 5) units/hour

Equating combined efficiency,

6x + 6(x - 5) = x(x - 5)

⇒ 6x + 6x - 30 = x² - 5x

⇒ 12x - 30 = x² - 5x

⇒ x² - 5x - 12x + 30 = 0

⇒ x² - 17x + 30 = 0

⇒ (x - 15)(x - 2) = 0

Taking (x - 15) = 0

⇒ x = 15 hours

∴ Pipe B will fill the cistern in 15 hours.

Shortcut Trick Let total work = x(x-5)

 

Total efficiency = x + x - 5 = 2x - 5

Time to fill the tank = x(x - 5)/(2x -5) = 6

⇒ x² - 5x = 12x - 30

⇒ x² - 17x + 30 =0

⇒ (x - 15)(x - 2) = 0

∴ x = 15, 2

∴ Pipe B will fill the cistern in 15 hours.

Given:

Length of shadow = 18 m

cos θ = 12/13

Concept used:

In a right angled triangle,

Cos θ = Base/Hypotenuse

Calculations:

Let height of pole be h.

In ΔABC,

cos θ = BC/AC = 12/13

Let BC and AC be 12x and 13x respectively.

12x = 18

⇒ x = 3/2

By pythagoras theorem,

(AC)² = (AB)² + (BC)²

⇒ (13x)2 = (h)2 + (12x)2

⇒ 169x² = (h)2 + 144x²

⇒ 169x2 - 144x2 = (h)2

⇒ 25x² = (h)2

⇒ h = 5x

⇒ h = 5 × (3/2)

⇒ h = 7.5 m

∴ The height of the pole is 7.5 m

Given:

Dimensions of parallel sides of trapezium = 27 cm and 13 cm

Height of trapezium = 7 cm

Formula used:

(i) Area of trapezium = {(Sum of parallel sides)/2} × height

(ii) 1 cm = 0.01m

Calculations:

27 cm = 0.27 m

13 cm = 0.13 m

7 cm = 0.07 m

Area of trapezium = {(0.27 + 0.13)/2} × 0.07

⇒ (0.40/2) × 0.07

⇒ 0.20 × 0.07

⇒ 0.014 m²

∴ The area of trapezium is 0.014 m²

Given:

\(\left( {5\frac{1}{4} \div ;\frac{3}{7};of;\frac{1}{2}} \right) \div ;\left( {5\frac{1}{9} - 7\frac{7}{8} \div 9\frac{9}{{20}}} \right) \times ;\frac{{11}}{{21}} - ;\left( {5 \div 2;of;\frac{1}{2}} \right)\)

Concept used:

Calculation:

\(\left( {5\frac{1}{4} \div ;\frac{3}{7};of;\frac{1}{2}} \right) \div ;\left( {5\frac{1}{9} - 7\frac{7}{8} \div 9\frac{9}{{20}}} \right) \times ;\frac{{11}}{{21}} - ;\left( {5 \div 2;of;\frac{1}{2}} \right)\)

\( \Rightarrow \left( {\frac{{21}}{4} \div ;\frac{3}{7};of;\frac{1}{2}} \right) \div ;\left( {\frac{{46}}{9} - \frac{{63}}{8} \div \frac{{189}}{{20}}} \right) \times ;\frac{{11}}{{21}} - ;\left( {5 \div 2;of;\frac{1}{2}} \right)\)

\( \Rightarrow \left( {\frac{{21}}{4} \div ;\frac{3}{{14}};;} \right) \div ;\left( {\frac{{46}}{9} - \frac{{63}}{8} \times \frac{{20}}{{189}}} \right) \times ;\frac{{11}}{{21}} - ;\left( {5 \div 1} \right)\)

\( \Rightarrow \left( {\frac{{21}}{4}; \times ;\frac{{14}}{3};;} \right) \div ;\left( {\frac{{46}}{9} - \frac{{63}}{8} \times \frac{{20}}{{189}}} \right) \times ;\frac{{11}}{{21}} - 5\)

\(\Rightarrow ;\frac{{49}}{2}; \div ;\left( {\frac{{46}}{9} - \frac{5}{6}} \right) \times ;\frac{{11}}{{21}} - 5\)

\( \Rightarrow ;\frac{{49}}{2}; \div ;\left( {\frac{{92 - 15}}{{18}}} \right) \times ;\frac{{11}}{{21}} - 5\)

\( \Rightarrow ;\frac{{49}}{2} \div ;\frac{{77}}{{18}}; \times ;\frac{{11}}{{21}} - 5\)

\( \Rightarrow ;\frac{{49}}{2} \times \frac{{18}}{{77}}; \times ;\frac{{11}}{{21}} - 5\)

⇒ 3 – 5

⇒ -2

∴ The value is -2.

Given:

The number of cars with speed between 70 - 80 km/h = 90

The number of cars with speed between 50 - 60 km/h = 70

Formula used:

Percentage increased = (greater value - smaller value)/smaller value × 100

Calculation:

Percentage increased = (greater value - smaller value)/smaller value × 100

⇒ Percentage increased = (90 - 70)/70 × 100

⇒ Percentage increased = 20/70 × 100

⇒ Percentage increased = 28.6%

∴ The percentage increase is 28.6%.

Given:

Initial salary = Rs. 8100

Increased salary = Rs. 9000

Formula used:

Percentage increased = (Initial salary – Increased salary)/Initial salary × 100

Calculation:

Initial salary = Rs. 8100

Increased salary = Rs. 9000

Percentage increased = (Initial salary – Increased salary)/Initial salary × 100

⇒ Required percent = (9000 – 8100)/8100 × 100

⇒ Required percent = 900/8100 × 100

⇒ Required percent = 1/9 × 100

⇒ Required percent = 100/9%

⇒ Required percent = \(11\frac{1}{9}% \)

∴ The percentage increase in his salary is \(11\frac{1}{9}% \).

Given:

Investment of A  = Rs. 92,500

Investment of B = Rs. 1,12,500

B's share in profit = Rs. 9,000

Concept used:

If time of investment is equal for A and B then, ratio of profit earned by A and B is same as the ratio of investment done.

Calculations:

Ratio of profit = 92,500 : 1,12,500

⇒ 37 : 45

Let total profit be y.

B's share in the profit = {45/(45 + 37)} × y

⇒ (45/82) × y = 9,000

⇒ y = 7,38,000/45

⇒ y = 16,400

∴ The total profit earned by them together is Rs. 16,400

Given:

x + y + z = 19, xyz = 216 and xy + yz + zx = 114

Formula used:

(x + y + z)² = x² + y² + z² + 2(xy + yz + zx)

x³ + y³ + z³ - 3xyz = (x + y + z)[x2 + y2 + z² - (xy + yz + zx)]

Calculation:

x + y + z = 19

squaring both side

⇒ (x + y + z)² = 19²

⇒ x2 + y2 + z2 + 2(xy + yz + zx) = 361

⇒ x2 + y2 + z2 = 361 - 2 × 114

⇒ x2 + y2 + z2 = 133

x³ + y3 + z3 - 3xyz = (x + y + z)[x2 + y2 + z2 - (xy + yz + zx)]

⇒ x3 + y3 + z3 = 19(133 - 114) + 648

⇒ x3 + y3 + z3 = 19 × 19 + 648

⇒ x3 + y3 + z3 = 1009

x3 + y3 + z3 + xyz

⇒ 1009 + 216

⇒ 1225

∴ The value is 1225.

Given:

Percentage distribution of employees in an organization

Department A = 20%

Department B = 14%

Department C = 16%

Department D = 8%

Department E = 20% 

Department F = 22%

Total employees in the organization = 450 = 100%

Calculations:

Number of employees in Department D = (8 × 450)/100 = 36

75% of employees of Department D are male 

So, 25% of employees of Department D are female

Number of female Employees in Department D = (25 × 36)/100 = 9

∴ The number of female employees in Department D is 9.

Given:

Number of cars between 40-50 speed = 50 

Number of cars between 50-60 speed = 70

Number of cars between 60-70 speed = 85

Number of cars between 70-80 speed = 90

Number of cars between 80-90 speed = 60

Number of cars between 90-100 speed = 45

Calculations: 

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

Total  number of cars with speed more than 70 km/h = 90 + 60 + 45 = 195 km/h

The ratio of the number of cars with a speed less than 60 km/h to the number of cars with speed more than 70 km/h = 120/195

⇒ 8 ∶ 13

∴ The ratio of the number of cars with a speed less than 60 km/h to the number of cars with speed more than 70 km/h is 8 ∶ 13.

Given:

Distance covered by bus and car = 60 km.

Time taken by bus to cover the distance = 1 hour 30 minutes.

Time taken by car to cover the same distance = 45 minutes.

Formula used:

Speed = Distance/time

Calculation:

Time taken by bus = 1hr 30 min.

⇒ T = (1 + 30/60) = 3/2 hr.

Time taken by car to cover the same distance = 45 minutes.

⇒ T = 45/60 hr = 3/4 hr.

According to question:

Speed of car ∶ Speed of bus 

⇒ [60/(3/4)] ∶ [60/(3/2)] 

⇒ 80 ∶ 40

⇒ 2 ∶ 1

∴ The ratio of the speed of car and bus is 2 : 1

Alternate Method

Concept used:

When the distance is constant, then the speed is inversely proportional to time.

Calculation:

Time covered by bus = 1hr 30 min.

⇒ T2 = (1 + 30/60) hr.

⇒ T₂ = 3/2 hr.

again,

Time covered by car = 45 min

⇒ T₁ = 45/60 hr.

⇒  T1 = 3/4 hr.

S₁ ∶ S₂ = (1/T₁) ∶ (1/T₂)

⇒ 1/(3/4) ∶ 1/(3/2)

⇒ 4/3 ∶ 2/3 = 2 ∶ 1

∴ The ratio of the speed of car and bus is 2 : 1

Given:

Amount invested by A =Rs. 92,500

Amount invested by B = Rs. 1,12,500

B's share in profit = Rs. 9,000

Concept used:

Divide the total profit according to investment ratio.

Calculation:

Let the profit earned by A be Rs. X

According to the question,

(92500/112500) = (X/9000)

⇒ X = (92500 × 9000)/112500

⇒ X =  37 × 200

⇒ X = 7400

∴ The profit earned by A is Rs. 7400

Given:

Employees working in department F = 22%

Total percent population = 100%

Calculation:

Out of 100%, 22% of Employees work in the department 'F'

Total number of Employees in Department 'F' = (22/100) × 450 = 99

∴ Total number of Employees in Department 'F' is 99

Given:

Distance covered by the person in a single journey = 300km

Time taken by him for the outward journey = 5 hours more than the time taken for the return journey.

The return speed is 10km/h more than the speed of going.

Formula used:

Distance covered = Speed × Time taken

Calculation:

Let the speed of outward journey be ‘x’ km/h

Hence, the speed of return journey will be (x + 10) km/h

So, time taken in return journey = 300/(x + 10) hours

Time taken in outward journey = (300/x) hours

According to the question,

⇒ (300/x) – 300/(x + 10) = 5

⇒ [300(x + 10) – 300x]/x(x + 10) = 5

⇒ [300x + 3000 – 300x] = 5x(x + 10)

⇒ 3000 = 5x² + 50x

⇒ 5x² + 50x – 3000 = 0

⇒ x² + 10x – 600 = 0

⇒ x² + 30x – 20x – 600 = 0

⇒ x(x + 30) – 20(x + 30) = 0

⇒ (x – 20)(x + 30) = 0

⇒ x = 20, -30

⇒ x = 20km/h as negative speed is not possible.

∴ The speed for the outward journey is 20km/h

Shortcut Trick Concept:

[(Speed 1 × Speed 2) / Speed difference] × Time difference = Total distance

Calculation:

Let the speed of outward journey be ‘x’ km/h

Hence, the speed of return journey will be (x + 10) km/h

So, 

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

⇒ \(x \times (x+10) = 600\)

⇒ \(x \times (x+10) = 20 \times 30\)

⇒ x = 20

∴ The speed for the outward journey is 20km/h

Given:

Length of shadow = 36m.

Let the height of pole = (x) m.

Formula Used:

sec2θ = tan2θ + 1

tanθ = P/B 

Where, 

P = Perpendicular 

B =Base

Calculations:

secθ = 13/12  

⇒ sec2θ = tan2θ + 1

⇒ (13/12)2 = tan2θ + 1

⇒ tan2θ = 1 - (13/12)2  

⇒ tan2θ = 1 - 169/144

⇒ tan2θ = 25/144

⇒ tanθ = 5/12  

tanθ = P/B 

tanθ = x/36

⇒ 5/12 = x/36

⇒ x = 36 × (5/12)

⇒ x = 15

∴ The height of pole is 15 m.

Given:

∠APB = 130°

Formula Used:

Sum of angles of quadrilateral = 360° 

Sum of angles of triangle = 180° 

Concept Used:

The radius of a circle always perpendicular to the tangent of the circle

In a triangle opposite angle of two equal sides always equal 

Calculation:

∠OAP = ∠OBP = 90°       ----(Radius of a circle always perpendicular to the tangent of the circle)

Now consider □OAPB,

∠OAP + ∠APB + ∠PBO + ∠BOA = 360°

⇒ 90° + 130° + 90° + ∠BOA = 360°

⇒ 310° + ∠BOA = 360°

⇒ ∠BOA = 50°    

In ΔOAB,

OA = OB (Radius of the circle)

So, ∠OAB = ∠OBA      ----(In a triangle opposite angle of two equal side always equal )

⇒ ∠OAB + ∠ABO + ∠BOA = 180°

⇒ 2∠OAB + 50° = 180°

⇒ 2∠OAB = 130°

⇒ ∠OAB = 65°

∴ The value of ∠OAB is 65°        

Shortcut Trick

∠OAB = ∠OBA  = (1/2) × ∠APB 

⇒ ∠OAB = (1/2) × 130 = 65°

∴ The value of ∠OAB is 65°

Given:

Initial salary = Rs. 7000

Increased salary = Rs. 12,000

Calculation:

Increase in salary = Rs. 12000 – Rs. 7000 = Rs. 5000

Increase % in salary = (5000/7000) × 100 

\(⇒ 71\frac{3}{7}\)%

∴ The percentage increase in the salary is \(71\frac{3}{7}\)%

Shortcut Trick Increase % in salary = (5000/7000) × 100 

⇒ \(71\frac{3}{7}\)%

Given:

BO and CO is the angle bisector of ∠CBD and ∠BCE respectively

∠ BOC = 57° 

Concept used:

BO and CO is the angle bisector of ∠CBD and ∠BCE respectively.

Then ∠BOC = 90° – (∠BAC/2)

Calculations:

∠BOC = 57° 

⇒ ∠BOC = 90° – (∠BAC/2)

⇒ ∠BAC/2 = 90 – 57

⇒ ∠BAC/2 = 33° 

⇒ ∠BAC = 66° 

∴ The value of ∠BAC is 66° 

Shortcut Trick  ∠BOC = 90° – (∠BAC/2)

⇒ ∠BAC/2 = 90 – 57

⇒ ∠BAC/2 = 33° 

⇒ ∠BAC = 66°