Quantitative Aptitude PYQs

Given:

Percentage increase in the price of diesel = 26%

Percentage increase in total expenditure = 15%

Concept used:

P × C = E

Where P is Price, C is Consumption and E is expenditure

Calculation:

Let initial price be P1, consumption be C1, and expenditure be E1 

P1 × C1 = E1 

⇒ C1 = E1/P1

Let new price be P2, new quantity consumed be C2, and new expenditure be E2 

P₂ = P₁ + 26% of P₁ 

⇒ P2 = 1.26P₁ 

E₂ = E₁ + 15% of E₁ 

⇒ E2 = 1.15E₁ 

As, P₂ × C₂ = E₂ 

⇒ 1.26P1 × C₂ = 1.15E1 

⇒ C₂ = (1.15E₁)/(1.26P₁)

⇒ C2 = 0.9126 × E₁/P₁ 

⇒ C₂ = 0.9126C₁ 

Decrease in consumption = C₁ – C₂ 

⇒ Decrease in consumption = C₁ – 0.9126C1 = 0.0874C₁ 

Percentage decrease in consumption = (Decrease in consumption/Initial consumption) × 100

⇒ Percentage decrease in consumption = (0.0874C₁/C₁) × 100

∴ The percentage decrease in consumption is 8.7% (correct to 1 decimal place)

Shortcut Trick 26% = 26/100

 

 

∴ Consumption decrease = 1 - (115/126) = (126 - 115)/126 = 11/126

∴ The percentage decrease in consumption is 8.7%

Given:

We have to find the value of 1 - 3 ÷ 6 of 2 + (4 ÷ 4 of \(\frac{1}{4}\)) ÷ 8 + (4 × 8 ÷ \(\frac{1}{4}\)) × \(\frac{1}{8}\)

Concept Used:

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

Calculation:

1 - 3 ÷ 6 of 2 + (4 ÷ 4 of ) ÷ 8 + (4 × 8 ÷ \(\frac{1}{4}\)) × \(\frac{1}{8}\)

⇒ 1 - 3 ÷ 12 + (4 ÷ 1) ÷ 8 + (4 × 8 × 4) × 1/8

⇒ 1 - 1/4 + 4 ÷ 8 + 16

⇒ 1 - 1/4 + 1/2 + 16

⇒ 69/4

∴ The required value is 69/4

Concept used:

Profit% = (Profit/cost price) × 100

Calculation:

Let the cost price be 'X'.

⇒ 108X/100 - 84X/100 = 660

⇒ 24X = 66000

⇒ X = 2750

When article sold at Rs. 3080

⇒ Profit = 3080 - 2750 = 330

Profit%

⇒ (330/2750) × 100

⇒ 12%

∴ The profit% is 12%.

Given:

A takes 2 hours 30 minutes more than B to walk 40 km.

If A doubles his speed, then he can make it in 1 hour less than B.

Formula used:

Speed = Distance/Time

Calculations:

Let the initial speeds of A and B be x and y respectively and the time taken by A and B be t_A and t_B respectively.

t_A = 40/x

t_B = 40/y

t_A - t_{B =} 40/x - 40/y

⇒ 40/x - 40/y = 2.5      ----(i)

After A doubles his speed,

⇒ 40/y - 40/2x = 1      ----(ii)

On adding equation (i) and (ii),

40/x - 40/2x = 2.5 + 1

⇒ 40/x - 20/x = 3.5

⇒ 20/x = 3.5

⇒ x = 40/7 km/h

From equation (ii),

40/y - 40/2x = 1

⇒ 40/y - 40/2(40/7) = 1

⇒ 40/y - 7/2 = 1

⇒ 40/y = 1 + 7/2

⇒ 40/y = 9/2

⇒ y = 80/9 km/h

tA = 40/x

⇒ tA = 40/(40/7)

⇒ tA = 7 hours

tB = 40/y

⇒ 40/(80/9)

⇒ 9/2 hours

Average time of A and B = {7 + (9/2)}/2

⇒ 23/4 hours

⇒ 5 hours 45 minutes

∴ The average time taken by A and B to walk a 40 km distance is 5 hours 45 minutes. 

Shortcut Trick 

Speed ratio of A      Old   :  New 

Speed ratio of A       1            2

Time Ratio of A         2            1 

Here, time difference is 1 unit.

And according to question time difference is 2.5 + 1 = 3.5 hr

​⇒ 1 unit = 3.5 hr

​⇒ 2 unit = 7 hr 

So, the time taken by A is = 7 h

​⇒ Time taken by B is = 7 - 2.5 = 4.5 h

​⇒ Average time taken = (7 + 4.5)/2 = 5.75 hr 

​⇒ 5.75 hr = 5 hour and 45 minutes 

∴ The average time taken by A and B to walk a 40 km distance is 5 hours 45 minutes.

Given:

Height in the interval 165-175 = 26

Total student = 70

Concept used:

Percentage = Given interval/Total student × 100

Calculation:

Percentage of student = (Number of student in 165-175)/(Total student) × 100

⇒ Percentage of student = 26/70 × 100

⇒ Percentage of student = 37.14%

∴ The percentage of student whose height is in the class interval 165-175 is 37%.

Given:

A, B and C started a business investing amounts of Rs. 13,750, Rs. 16,250 and Rs. 18,750 respectively.

Concept Used:

In the partnership business profit is distributed in the ratio of their Investment.

Calculation:

Ratio of investment of A, B and C is 13750 : 16250 : 18750

⇒ 11 : 13 : 15

B's share is 13 units → 5200

⇒ 1 unit → 400

Difference in the profit earned by A and C is (15 - 11) = 4 unit

1 unit → Rs. 400

⇒ 4 unit → Rs. 400 × 4 = Rs. 1600

∴ The difference in the profit earned by A and C is Rs. 1600.

Given:

The person invested Rs. 2,000 on simple interest for 7 years and obtain a total amount of Rs. 3,400 on a certain annual rate of interest.

Concept Used:

If P be the principal and r be the rate of interest then interest after t year will be (prt/100)

Amount (A) = Principal (P) + Interest

Calculation:

Here,

Principal (P) = Rs. 2000

Time (t) = 7 years

Amount after 7 years (A) = 3400

Interest = A - P

⇒ 3400 - 2000

⇒ 1400

Let, the rate of interest be r%

Accordingly,

(2000 × r × 7)/100 = 1400

⇒ r = 10

∴ The rate of interest was 10%

Alternate Method SI in 7 years = 3400 - 2000 = 1400

∴ SI in 1 year = 1400/7 = 200

Rate(%) = \(\frac{SI}{P}\times100\) = \(\frac{200}{2000}\times100\) = 10 %

∴ The rate of interest was 10%

Given,1433 × 1433 × 1422 × 1425

Looking at the unit digits of the terms, the last two terms contain digits 2 and 5 respectively.

So, when we multiply all the terms, the last digit will be zero, as 2 × 5 = 10.

The product will be completely divisible by 10. Thus, the remainder will be 0.

Alternate Method

Given:

1433 × 1433 × 1422 × 1425

Concept used:

Divide individually to get a smaller number than 10.

Calculation:

(1433 × 1433 × 1422 × 1425) ÷ 10

⇒ (1433/10) × (1433/10) × (1422/10) × (1425/10)

⇒ (3 × 3 × 2 × 5)/10

⇒ (9/10) × (10/10) 

The remainder of the above terms are 9 and 0. Then,

⇒ 9 × 0

⇒ 0

∴ The remainder is 0.

Given:

1% = Rs. 5000

In 2005,

Strawberry = 18.90%, Butterscotch = 10.80%

In 2015,

Strawberry = 20.50%, Butterscotch = 16.60%

Calculation:

Total sales in 2005

⇒ 18.90 + 10.80

⇒ 29.70%

Total sales in 2015

⇒ 20.50 + 16.60

⇒ 37.10%

Percentage Increase

⇒ 37.10 - 29.70

⇒ 7.4%

⇒ 7.4 × 5000

⇒ 37000

∴ The percentage increase from 2005 to 2015 is Rs. 37000.

Given:

Time is taken by A to fill the tank = 12 hrs

Time is taken by B to fill the tank = 18 hrs

Concept used:

Total time = Volume/Volume filled in 1 hour

Calculation:

Volume filled by A in 1 hr = 1/12

Volume filled by B in 1 hr = 1/18

Volume filled by both A and B in 1 hr = 1/12 + 1/18

⇒ Volume filled by both A + B in 1 hr = (3 + 2)/36

⇒ Volume filled by both (A + B) in 1 hr = 5/36

Total Time to fill the tank = (Total volume of the tank)/(Volume filled by both (A + B) in 1 hr)

⇒ Total Time to fill the tank = 1/(5/36)

⇒ Total Time to fill the tank = 36/5 hr

⇒ Total Time to fill the tank = 7 hours 12 minutes

∴ The total time to the filled the empty tank is 7 hours 12 minutes.

 Alternate Method

Efficiency of A is 3

Efficiency of B is 2

Total efficiency of A & B is 3 + 2 = 5

So total time taken to fill the tank = 36/5 = 7¹/₅ hour = 7 hour 12 minutes

Given:

Students in School A in 2008 = 640

Students in School B in 2008 = 550

Students in School A in 2009 = 800

Students in School B in 2009 = 820

Students in School A in 2010 = 500

Students in School B in 2010 = 600

Students in School A in 2011 = 700

Students in School B in 2011 = 750

Students in School A in 2012 = 900

Students in School B in 2012 = 500

Students in School A in 2013 = 750

Students in School B in 2013 = 480

Calculations:

Absolute difference of the students from schools A to B in 2008 = 640 - 550

⇒ 90

Absolute difference of the students from schools A to B in 2009 = 820 - 800

⇒ 20

Absolute difference of the students from schools A to B in 2010 = 600 - 500

⇒ 100

Absolute difference of the students from schools A to B in 2011 = 750 - 700

⇒ 50

Absolute difference of the students from schools A to B in 2012 = 900 - 500

⇒ 400

Absolute difference of the students from schools A to B in 2013 = 750 - 480

⇒ 270

∴ In 2012, the absolute difference of the students from schools A to B taken together is maximum.

Given:

Seven years ago the ratio of ages of A and B = 4 : 5

After 8 years from present the ratio of ages of A and B = 9 : 10

Calculation:

Seven years ago the ratio of ages of A and B

⇒ 4x : 5x

Ratio of Present ages of A and B

⇒ 4x + 7 : 5x + 7

After 8 years the ratio of ages of A and B

⇒ (4x + 7 + 8) : (5x + 7 + 8) = 9 : 10

⇒ 40x + 150 = 45x + 135

⇒ 5x = 15

⇒ x = 3

Difference of present ages of A and B

⇒ (5 × 3 + 7) - (4 × 3 + 7)

⇒ 3 years

∴ The difference of present ages of A and B is 3 years.

Given:

∠APB = 142°

Concept used:

Radius makes 90° to the point of contact of tangent.

Line joining centre and meeting point of two tangents bisects the angle.

Line joining centre makes 90° to the chord.

Calculation:

∠APO = 1/2 × ∠APB

⇒ ∠APO = 1/2 × 142°

⇒ ∠APO = 71°

∠OAP = 90°

In ΔAMP,

∠OAP + ∠APO + ∠POA = 180°                [Angle sum property]

⇒ 90° + 71° + ∠PAM = 180°

⇒ ∠POA = 180° – 161°

⇒ ∠POA = 19°

∠OAP = 90°

⇒ ∠OAB + AMO+ MOA = 180

⇒ ∠OAB = 71°

∴ ∠OAB is equal to 71°.

Given:

The efficiency of A = 2 × Efficiency of B 

Total time is taken to do work together = 13 days

Concept used:

Time = 1/Efficiency

Calculation:

Let the time taken by A to complete work alone be x days. 

The time is taken by B to complete work alone is x days. 

1 day work of A = 1/x

1 day work of B = 1/2x

1 day work of A and B together = 1/13

⇒ 1 day work of A + B = 1/x + 1/2x

⇒ 1/13 = (2 + 1)/2x

⇒ 2x = 39

⇒ x = 39/2 days

⇒ x = \(19 \frac{1}{2}\) days

∴ The time taken by A alone to finish the work is \(19 \frac{1}{2}\) days.

Shortcut Trick

 

Total Work = 3 × 13 = 39

Total time is taken by A = Total work/Efficiency of A

⇒ 39/2 = 19\(\frac{1}{2}\)

∴ The time taken by A alone to finish the work is 19\(\frac{1}{2}\)  days

Given:

Decreasing 110 by x% gives the same result as increasing 50 by x%.

Calculations:

110 × (100 - x)/100 = 50 × (100 + x)/100

⇒ 11 × (100 - x) = 5 × (100 + x)

⇒ 1100 - 11x = 500 + 5x

⇒ 16x = 600

⇒ x = 600/16

⇒ x = 37.5

x% of 650 = 37.5% × 650

⇒ 243.75

(x - 10)% of 780 = (37.5 - 10)% × 780

⇒ 27.5% × 780

⇒ 214.50

Percentage difference = {(243.75 - 214.50)/214.50} × 100

⇒ (29.25/214.50) × 100

⇒ 13.63% ≈ 14%

∴ x% of 650 is 14% more than (x - 10)% of 780.

Given:

(5√5x3 – 3√3y3) ÷ (√5x – √3y) = (Ax2 + By2 + Cxy)

Identity used:

(a³ – b³) = (a – b) × (a² + b² + ab)

Calculation:

(5√5x3 – 3√3y3) = (√5x)³ – (√3y)³ 

(√5x)3 – (√3y)3 = (√5x – √3y) × (5x² + 3y² + √15xy)

As, (5√5x3 – 3√3y3) ÷ (√5x – √3y) = (Ax2 + By2 + Cxy)

⇒ [(√5x – √3y) × (5x2 + 3y2 + √15xy)] ÷ (√5x – √3y) = Ax2 + By2 + Cxy

⇒ 5x2 + 3y2 + √15xy = Ax2 + By2 + Cxy

Comparing both sides we'll get

A = 5, B = 3 and C = √15

According to question we have to find 3A – B – √15C

⇒ 3A – B – √15C = 3 × 5 – 3 – √15 × √15

∴ 3A – B – √15C is –3

Given:

Speed of train x = 74 km/h

Speed of train y = 52 km/h

Train x crosses train y in opposite direction in 12 seconds.

Length of train y = 2/3 of length of train x

Concept used:

(i) If speed of body A is x and speed of body B is y then, Relative speed of two bodies moving in opposite direction = (x + y)

(ii) 1 km/h = 5/18 m/s

Formula used:

Relative speed = Total length/Time taken to cross

Calculations:

Relative speed = Total length/Time taken to cross

⇒ (74 + 52)(5/18) = Total length/12

⇒ (126)(5/18) = Total length/12

⇒ Total length = 35 × 12

⇒ Total length = 420 m

Let the length of train x be 3a.

So, length of train y = 2/3 × 3a = 2a

⇒ 2a + 3a = 420

⇒ 5a = 420

⇒ a = 84

Length of train y = 2a

= 2 × 84

= 168 m

∴ The length of train y is 168 m.

Given:

X⁴ + 1/X⁴ = 194

Concept used:

(a + b)² = a² + b² + 2ab

Calculation:

X4 + 1/X4 = 194

⇒ (X²)² + 1/(X²)² = 194

⇒ (X2)2 + 1/(X2)² + 2 = 194 + 2                     [Adding 2 both sides]

⇒ (X2)2 + 1/(X2)2 + 2 × X² × 1/X² = 196

⇒ (X² + 1/X²)² = 196

⇒ X2 + 1/X² = √196

⇒ X2 + 1/X2 = 14

⇒ X2 + 1/X² + 2 = 14 + 2                                [Adding 2 both sides]

⇒ X2 + 1/X² + 2 × X × 1/X = 16

⇒ (X + 1/X)² = 16

⇒ X + 1/X = √16

⇒ X + 1/X = 4

X + 1/X + 2 = 4 + 2

⇒ X + 1/X + 2 = 6

∴ The value of X + 1/X + 2 is 6.

Shortcut Trick Concept: 

X4 + 1/X4 = a

⇒ X2 + 1/X2 = √(a + 2)

Calculation:

X4 + 1/X4 = 194

⇒ X2 + 1/X2 = √196

⇒ X2 + 1/X2 = 14

⇒ X + 1/X = √(14+2)

⇒ X + 1/X = 4

⇒ X + 1/X + 2 = 6

∴ The value of X + 1/X + 2 is 6.

Given:

The circle is inscribed in a triangle ABC. It touches side AB, BC and AC at points R, P and Q, respectively.

AQ = 2.6 cm, PC = 2.7 cm, BR = 3 cm

Concept Used:

The length of two tangent drawn on a circle from an outside point are same.

Calculation:

AQ and AR are two tangents on the circle from point A

⇒ AQ = AR = 2.6 cm

BP and BR are two tangents on the circle from point B

⇒ BR = BP = 3 cm

CP and CQ are two tangents on the circle from point C

⇒ CP = CQ = 2.7 cm

The perimeter of the triangle is AB + BC + CA

⇒ AR + RB + BP + PC + CQ + QA

⇒ 2.6 + 3 + 3 + 2.7 + 2.7 + 2.6

⇒ 16.6

∴ The perimeter of the triangle is 16.6 cm.

Given:

Students from school B in the year 2008, 2009, 2010, 2011, 2012 and 2013 is 550, 820, 600, 750, 500 and 480 respectively.

Students from school A in the year 2008, 2009, 2010, 2011, 2012 and 2013 is 640, 800, 500, 700, 500 and 750 respectively.

Concept Used:

Average = sum of all entry/Total number of entry

Calculation:

The average of total students from school B is (550 + 820 + 600 + 750 + 500 + 480)/6

⇒ 3700/6

The average of total students from school A is (640 + 800 + 500 + 700 + 900 + 750)/6

⇒ 4290/6

The required ratio is 3700/6 : 4290/6

⇒ 370 : 429

∴ The ratio of the average of total students from school B to the average of the total student from school A is 370 : 429