Quantitative Aptitude PYQs

Given:

 The successive discounts of 20%, 10%, and 5% respectively

Formula used:

 Total discount = [x + y – (x y) ÷ 100] %

Calculation

First, we take 20% and 10%

Total discount = [20 + 10 – (20 × 10) ÷ 100] %

Total discount = 30 – 2 = 28 %

Now we take 28% and 5 %

Total discount = [28 + 5 – (28 × 5) ÷ 100] %

Total discount 33 – 1.40 = 31.6%

Hence, overall discount = 31.6%

Given:

The simple Interest (SI) = Rs 500, Time = 5 years, and Rate = 5%

Formula used:

We know the formula,

SI = (P × R × T) ÷ 100

Where P = principle, T = time, and R = Rate

Calculation

We know that formula,

SI = (P × R × T) ÷ 100

Putting the value,

500 = (5 × 5 × P) ÷ 100

P = (500 × 100) ÷ 25

P = 20 × 100 = 2000

Hence, Principle = Rs 2000

Shortcut TrickIn case of Compounded annually questions,

The difference between compound interest and simple interest is given as:

Principal amount = Difference × (100/Rate)²

⇒ Principal amount = 72 × (100/12)²

⇒ Principal amount = (72 × 25 × 25)/9 = 5000

∴ The correct answer is 5000 Rs.

Alternate Method

Formula used:

We know that formula, A = P(1 + r/100)²

CI = A – P and SI = PRT/ 100

Calculation       

Let principle amount = Rs x

We know that, A = P(1 + r/ 100)^t

A = x (1 + 12/100)²

A = x (1.12)²

A = 1.2544 x

CI = A – P

CI = 1.2544x – x = 0.2544x ……..(1)

SI = PRT/100

SI = x × 12 × 2 / 100

SI = 0.24x …….(2)

According to question,

CI – SI = 72

0.2544x – 0.24x = 72

X = 720000/144 = 5000

Hence, the value principle = Rs 5000

Given:

 The distance travels by trains 600 km and 900 km

Taking time 5 hours and 10 hours

Formula used:

 Average speed of the trains = total distance ÷ total time

Calculation

Total distance = 600 + 900 = 1500 km

Total time = 10 + 5 = 15 hours

Now average of the trains = 1500 ÷ 15 = 100 km/hours

Hence, average of the trains = 100 km/hours

Given:

 Walking at 4/5 of his usual speed, a man is 12 minutes late for his office

Calculation

Let the usual time be x minute

Walking at 4/5^th of his usual speed, he will take time 5/4 of the usual time

Now, 5/4^th of the usual time = usual time + 12 minute

5/4 x = x + 12                     

5/4x – x = 12

1/4 x = 12

X = 48 minute

Hence, he usually time take = 48 minute

Shortcut Trick

 

⇒ 1 unit = 12 min. [given]

⇒ usual time = 4 unit = 4 × 12

⇒ usual time = 48 min.

∴ The correct answer is option (1).

Given:

 The cost price of 100 mangoes is equal to the selling price of 80 mangoes

Calculation

Let selling price = cost price

According to question,

100 × CP = 80 × SP

CP ÷ SP = 80/100

CP ÷ SP = 4/5

Profit percentage = (5 – 4) ÷ 4 × 100

Profit percentage = 1 ÷ 4 × 100

Profit percentage = 25%

Hence, profit percentage = 25%

Shortcut TrickAfter selling 80 mangoes the seller recovers the cost price,

Hence, Profit = 100 - 80 = 20

Cost price will be 80 because that is total number of mangoes seller sold

Profit % = (Profit × 100)/Cost

Profit % = (20 × 100)/80 = 25

∴ The correct answer is 25%.

Shortcut Trick 

X : Y = 3 : 4

X : Z = 5 : 4

⇒ X : Y : Z = 15 : 20 : 12

⇒ % Less = (8/20) × 100 = 40%

∴ Z is 40% cheaper than Y

Alternate Method 

Given:

 X sells his goods 25% cheaper than Y and 25% dearer than Z

Calculation

Let price to y goods = x

Then price of y goods = x – 25 ÷ 100 = x – x/4 = 3x/4 …..

(1) Let price of x sells = y + 25y/100 = y + y/4 = 5y/4 ….(2)

From the equation 1 and 2

⇒ 3x/4 = 5y/4

⇒ 3x= 5y

⇒ Y = 0.6x

Then y = 60% of x

Percentage of Z’s good cheaper than y = 100 – 60 = 40%

∴ Z is 40% cheaper than Y

Alternate Method

Let the price of y be 100

As X is 25% cheaper, rate of x = 100 - 25% of 100 = 75

Ax X is 25% dearer than Z

Price of X = Z + 25% of Z

⇒ 75 = (5/4) × Z

⇒ Z = 60

Z is 100 - 60 = 40 cheaper than Y

⇒ Z is (40/100) × 100 = 40% cheaper than Y

∴ Z is 40% cheaper than Y

Shortcut Trick

                          before             after

   liquid                 1                     1

   water                 3                     4

⇒ Quantity of added water = (80/4) × 1 = 20 litres

∴ The correct answer is 20 litres.

Alternate Method 

Given:

 In a mixture of 80 liters of a liquid and water, 25% of the mixture is the liquid

Calculation

Liquid amount = 80 × 25 ÷ 100

Liquid amount = 20 liters

Let x be the amount of water be added

According to question,

⇒ 20 = 20% of (80 + x)

⇒ 20 = 20 ÷ 100 (80 + x)

⇒ 100 = 80 + x

⇒ X = 20 liters

∴ The correct answer is 20 litres.

Shortcut Trick 

(20 × 20) / 20 = ( 16 × D₂) / 16

⇒ D₂ = 20 days

∴ The correct answer is 20 days.

Alternate Method 

Given:

 if 20 persons can clean 20 floors in 20 days

Formula used:

We know that, M₁ D₁ / W₁ = M₂ D₂ / W₂

Calculation

⇒ D₂ = 20 × 20 × 16 / 20 × 16

⇒ D₂ = 20 days

∴ The correct answer is 20 days.

Given:

the work done by (x - 1) men in (x + 1) days : work done by (x + 2) men in (x - 1) days = 9 : 10

Calculation       

According to question,

(x - 1)(x + 1)/(x + 2) (x - 1) = 9/10

⇒ (x + 1)/(x + 2) = 9/ 10

⇒ 10x + 10 = 9x +18

⇒ x = 8

∴ The value of x is 8.

Given:

The expressions 9x² + 8kx + 16 = 0

Calculation       

For real and equal roots D = 0

We know that, b² – 4ac = 0

Put the value, b = 8k, a = 9 and c = 16

b² – 4ac = 0

(8k)² – 4 × 16 × 9 = 0

64k² – 576 = 0

K² = 576/64 = 9

K = -3, +3

Hence, the value of k = -3, +3

Given:

Given that x + 1/ x = 8

Calculation       

The expression x + 1/ x = 8

Squaring both sides

x² + 1/x² = 64 – 2 = 62

∴ The value of x² + 1/x² is 62.

Given:

The equation 4x² - 2kx + 3k = 0 has equal roots

Calculation       

The equation 4x² - 2kx + 3k = 0

Here, a = 4, b = -2k, and c = 3k

D = b² – 4ac = 0

Put the value,

(-2k)² – 4 × 4 × 3K = 0

4k² – 48k = 0

4k (k – 12) = 0

K = 0

And (k - 12) = 0

K = 12

Hence, the value of K = 0, 12

Shortcut Trick

Sum of roots = -b / a 

-(-6) / p = 6

⇒ p = 1

Product of roots = c / a

⇒ q / p = 6

⇒ q = 6

⇒ p + q = 7

Therefore the correct answer is 7.

Alternate Method 

Given:

 The expression px² – 6x + q

Sum of root = 6 and product of root = 6

Formula used:

Sum of root = - b/a and product of root = c/a

Calculation

The expression px² – 6x + q

Here, a = p, b = -6 and c = q

Sum of roots = - b/a = - (- 6)/p

6 = 6/p

⇒ p = 1

Now, product of roots = c/a = q/p

6 = q/1 (put the value of p)

Then q = 6

Hence, (p + q) = 6 + 1 = 7   

Therefore the correct answer is 7.

Calculation:

Let, 

P(x) = 4x3 + 12x2 - x - 3

⇒ P(x) = 4x²(x + 3) - 1(x + 3)

⇒ P(x) = (4x² - 1)(x + 3)

We know that, (a² - b²) = (a + b)(a - b)

⇒ P(x) = (2x - 1)(2x + 1)(x + 3)

We can see that,  (2x - 1) & (2x + 1) are factor of polynomial. ​

∴ 4x3 + 12x2 - x - 3 is divisible by both (2x + 1) and (2x - 1)

Given:

The value if 3 is added to both the numerator and the denominator of all the fractions

Calculation       

We check all option 2/3, 3/4, 4/5, 5/6

(1) 2/3 added 3 both numerator and denominator

= (2 + 3)/(3 + 3) = 5/6

Change = 5/6 – 2/3 = (5-4)/6 = 1/6  

(2) 3/4 added 3 both numerator and denominator

(3 + 3)/(4 + 3) = 6/7

Change = 6/7 – 3/4 = (24 – 21)/ 28 = 3/28

(3) 4/5 added 3 both numerator and denominator

(4 + 3)/(5 + 3) = 7/8

Change = 7/8 – 4/5 = (35 – 32)/ 40 = 3/ 40

(4) 5/6 added 3 both numerator and denominator

(5 + 3)/(6 + 3) = 8/9

Change = 8/9 – 5/6 = (16 – 15)/18 = 1/18

Hence, required minimum change in 5/6

Given:

Let the average score of a class of boys and girls in an examination be p. The ratio of boys and girls in the class is 3 ∶ 1. If the average score of the boys is (p + 1),

Calculation       

Let the number of boys be 3x and number of girls x

Let the average score of the girls be q

Sum of score of boys = 3x (p + 1)

Sum of score of girls = qx

Sum of score of the whole class = (3x + x)p = 4xp

According to question,

3x (p+1) + qx = 4xp

3xp + 3x + qx = 4xp

x(3 + q) = xp

3 + q = p

Then, q = (p – 3)

Hence, average score of the girls = (p - 3) 

Shortcut Trick 

⇒ (p + 1 - p) / (p - x) = 1 / 3

⇒ p - x = 3

⇒ x = p - 3

∴ The average score of girls is (p - 3).

Given:

The incomes of A, B, and C are in the ratio 7 ∶ 9 ∶ 12 and their expenditures are in the ratio 8 ∶ 9 ∶ 15. If A's saving is one-fourth of his income

Calculation

Let the income of A, B, C be 7I, 9I, and 12I respectively and their expenditure be 8E, 9E, and 15E respectively

Now A’s saving = ¼ 7E = 7I - 8E

8E = 7I – 7/4 I

8E = 21/4 I

E = 21/32 I

Expenditure, of A, B, and C are 21/32 × 8I, 21/32 × 9I, and 21/32 × 15I

Saving of A, B, and C are respectively

7I – 168I/32, 9I – 189I/32, 12I – 315I/32

224I – 168I = 56I, 288I -189I = 99I, 384I – 315I = 69I

Now, 56, 99, 69

Then ratio of saving A : B : C = 56 : 99 : 69

Hence, Saving ratio = 56 : 99 : 69

Given:

 A train 200 m long passes a platform 100 m long in 10 seconds

Formula used:

Speed of the train = total distance ÷ time taken

Calculation

Total distance covered in 10 second = platform length + length of train

Total distance covered in 10 second = 200 +100 = 300 meters

Speed of the trains = 300 ÷ 10

Speed of the trains = 30 m/s

Hence, speed of the trains = 30 m/s

Shortcut Trick

For a telescoping series of the form 1/(1×2) + 1/(2×3) + ... + 1/(n(n+1)), the sum is always n/(n+1).

Given: Sum = 99/100

⇒ n/(n+1) = 99/100

By comparing the numerator and denominator directly:

∴ The correct answer is 99.

Alternate Method

Given: 1/(1×2) + 1/(2×3) + 1/(3×4) + ... + 1/(n(n+1)) = 99/100

Method: Using Partial Fraction Decomposition, we know that 1/(r(r+1)) = 1/r − 1/(r+1).

Applying this to each term of the series:

⇒ (1/1 − 1/2) + (1/2 − 1/3) + (1/3 − 1/4) + ... + (1/n − 1/(n+1)) = 99/100

⇒ 1 − 1/2 + 1/2 − 1/3 + 1/3 − 1/4 + ... + 1/n − 1/(n+1) = 99/100

Observing the series, all intermediate terms cancel out (telescoping property):

⇒ 1 − 1/(n+1) = 99/100

⇒ (n + 1 − 1) / (n + 1) = 99/100

⇒ n / (n + 1) = 99/100

⇒ 100n = 99(n + 1)

⇒ 100n = 99n + 99

⇒ 100n − 99n = 99

⇒ n = 99

∴ The correct answer is 99.

Additional Information

General Formula for Telescoping Series

For a series where terms are 1/(n(n+d)), the sum S_n = (1/d) × [1/(first factor) − 1/(last factor)]. In this case, d = 1.

Sum of Infinite Telescoping Series

If the series goes to infinity, the sum S_∞ = 1/d × (1/a₁) because the term 1/(n+d) approaches zero as n tends to infinity.

Property of Partial Fractions

1/[n(n+1)(n+2)] = (1/2) × [1/(n(n+1)) − 1/((n+1)(n+2))], which is used to solve higher-order telescoping sums.