Quantitative Aptitude PYQs

Given:

Ramchandran gave his wife = 40%

He gave his 3 daughters and 3 sons = 10% each

Amount spent on shopping = 1/2 of remaining

The amount deposited in the bank = Rs. 24,000

Calculation:

Let the initial amount with Ramchandran be Rs.

x. ∴ Required equation according to the question:       

⇒ (1/2) × [100 – (3 × 20)]% of (100 – 40)% of x = 24000.

⇒ 1/2 × (40/100) × (60/100) × x = 24000

⇒ 3x/25 = 24000

⇒ 3x = 24000 × 25

⇒ x = 8000 × 25

⇒ x = Rs. 2,00,000

∴ Initially Ramchandran had Rs. 2,00,000.

Alternate Method

Let, Ramchandran has 100x rupees initially.

⇒ Wife's share = 40% = 40x rupees.

Remaining Amount = 100x - 40x = 60x.

10% of remaining amount = 6x

⇒ share of 3 daughter and 3 son = 6x × 6 = 36x

Remaining Amount = 60x - 36x = 24x

⇒ Amount Spent on Shopping = (1/2) of remaining = 24x/2 = 12x

Remaining Amount = 24x - 12x = 12x = 24000 (Given)

⇒ x = 24000/12 = 2000

⇒ Initial value = 100x = 100 × 2000 = 200000

∴ Initially Ramchandran had Rs. 2,00,000

Solution:

A and B together can finish a task in 12.5 days.

B and C together can do it in 18.75 days.

A and C together can do it in 15 days.

D having only 40% of C's efficiency.

Total work = Time × Efficiency

A + B = 12.5 = 25/2 days

B + C = 18.75 = 75/4 days

A + C = 15 days

Total work = LCM(25/2, 75/4, 15) = 75 units

Efficiency of A + B = 75/(25/2) = 6

Efficiency of B + C = 75/(75/4) = 4

Efficiency of A + C = 75/15 = 5

Efficiency of A + B + C = 15/2 = 7.5

Efficiency of C = 7.5 - 6 = 1.5

Efficiency of D = 1.5 × 0.4 = 0.6

Time taken by A, B, C and D to complete the task collectively = 75/8.1

= 750/81 = 250/27

= 9(7/27)

Hence, option(4) is correct.

Given:

A group of X+5 students has a average weight of 12 kg

X students with an average of 8 kg is removed and X+3 students with an average of 14 kg is added then, the new average is increased by 3 kg

Formula used:

Total Sum = Average × Number of terms

Calculation:

New Average = 12kg + 3 kg

= 15 kg

Latter number of students = X + 5 - X + X + 3

= X + 8

According to the question,

\(\frac{(X+5)\times12 - (X) 8 + (X+3) \times 14}{X+8}=15\)

\(\frac{12X+60 - 8X + 14X+42}{X+8}=15\)

18X + 102 = 15(X + 8)

18X + 102 = 15X + 120

18X - 15X = 120 - 102

3X = 18

X = 18/3

X = 6

Answer is 6.

⇒ 37.79 × 3.98 + 36.05 × ? = 50.40 × 1.99 + 7.01 × 27.88

⇒ 38 × 4 + 36 × ? = 50 × 2 + 7 × 28

Using BODMAS rule,

⇒ 152 + 36 × ? = 100 + 196

⇒ 36 × ? = 144

⇒ ? = 4

(7.9899 × 7.002) + √1024.78 + (1.99086)³ = ? ÷ 1/ 5.7865

(8 × 7) + 32 + (2)³ = ? × 5.7865

(56 + 32 + 8) / 5.7865 = 96/6 = 16

Concept used:

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:

12.99% of 6699.99 + (32.001 + 39.98) ÷ 8.99 × 8.001 = 999.001 – x

⇒ 13% of 6700 + (32 + 40) ÷ 9 × 8 = 999 – x

⇒ (13/100) × 6700 + 72 ÷ 9 × 8 = 999 – x

⇒ 871 + 8 × 8 = 999 – x

⇒ 935 = 999 – x

⇒ x = 999 – 935

⇒ x = 64

∴ The approximate value of x is 64

Concept used:

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.

Calculations:

√(47.99 + 15.990) ÷ 3.89 × 40.89 – 35.35 + x^1/2 = (8.22)²

⇒ √(48 + 16) ÷ 4 × 41 – 35 + x^1/2 ≈ (8)²

⇒ √64 ÷ 4 × 41 – 35 + x^1/2 = 64

⇒ 8 ÷ 4 × 41 – 35 + x^1/2 = 64

⇒ 2 × 41 – 35 + x^1/2 = 64

⇒ 82 – 35 + x^1/2 = 64

⇒ x^1/2 = 64 – 47

⇒ x^1/2 = 17

⇒ x = (17)²

⇒ x = 289

∴ The Value of x is 289

Rules of Approximation:

1. If a number has digits to the right of the decimal less than 5, then just drop the digits to the right of the decimal. The number so obtained will be the approximated value.
2. If a number has digits to the right of the decimal more than 5, then just drop the digits to the right of the decimal and raise the remaining number by '1'.The number so obtained will be the approximated value.

Concept Used:       

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

Calculation:

{(784)^1/2 × 4} + (29 × 13) = (?)² – (8 × 5)

⇒ {28 × 4} + (29 × 13) = (?)² – (8 × 5)

⇒ {28 × 4} + 377 = (?)² – 40

⇒ 112 + 377 = (?)² – 40

⇒ 489 = (?)² – 40

⇒ (?)² = 489 + 40

⇒ (?)² = 529

∴ ? = 23

Given :

Average age of P, Q, and R = 90 years

Average age of P, Q, R and S = 86 years

Average age of Q, R, S and T = 84 years

Formula Used :

Average = Sum of observation/No. of observation

Shortcut Trick

The age of S = Sum of age of P, Q, R, S - Sum of age of P, Q, R

The age of S = 86 × 4 - 90 × 3 = 74 years

The age of T = 74 + 6 = 80 years

The age of P = Sum of age of P, Q, R, S, T - Sum of age of Q, R, S, T

The age of P = 86 × 4 + 80 - 84 × 4 = 424 - 336 = 88 years

∴ Age of P is 88 Years

Alternate Method

Calculation :

Average age of P, Q, and R = 90 years

⇒ (P + Q + R)/3 = 90

⇒ P + Q + R = 270

Similarly, Average age of P, Q, R and S = 86 years

⇒ (P + Q + R + S)/4 = 86

⇒ P + Q + R + S = 344

⇒ Age of P, Q, and R + Age of S = 344

⇒ 270 + S = 344

⇒ S = 74 years

If the age of S is 74 year

Then the Age of T = 74 + 6 = 80 years

Similarly, Average age of Q, R, S and T = 84 years

⇒ (Q + R + S + T)/4 = 84

⇒ Q + R + S + T = 84 × 4

⇒ Q + R + S + T = 336

⇒ Q + R + 74 + 80 = 336 (Putting the value of S, and T)

⇒ Q + R = 182

⇒ Age of P, Q, and R = 270

⇒ P + Q + R = 270 (Putting the value of Q + R from above)

⇒ P + 182 = 270

⇒ P = 88 years

∴ Age of P is 88 Years.

Given:

The ratio of water and milk in a mixture is 2 : 3.

Calculation:

Let the initial quantity of water and milk in the mixture be 2x and 3x litres respectively.

After adding 'a + 20' litres of water and 'a' litres of milk, the total quantity of water and milk in the mixture becomes (2x + a + 20) litres and (3x + a) litres respectively.

The ratio of water and milk in the new mixture is 7 : 8.

Therefore,

⇒ (2x + a + 20) / (3x + a) = 7 / 8

⇒ 8(2x + a + 20) = 7(3x + a)

⇒ 16x + 8a + 160 = 21x + 7a

⇒ 5x = a + 160

⇒ x = (a + 160) / 5

Now, half of the solution is removed.

Therefore, the remaining quantity of water and milk in the mixture is (2x + a + 20) / 2 litres and (3x + a) / 2 litres respectively.

The remaining mixture contains 10 litres more milk than water.

Therefore,

(3x + a) / 2 - (2x + a + 20) / 2 = 10

⇒ 3x + a - 2x - a - 20 = 20

⇒ x = 40

Therefore, the initial quantity of water in the mixture = 2x = 80 litres.

And, the initial quantity of milk in the mixture = 3x = 120 litres.

Therefore, the initial quantity of mixture = 80 + 120 = 200 litres.

Hence option (3) is correct.

General solution:

 

Company A:

Number of two wheelers and four wheelers manufactured = 400

Number of two wheelers and four wheelers sold = 250

Number of two wheelers and four wheelers unsold = (400 - 250) = 150

 

Company B:

Number of two wheelers and four wheelers manufactured = 450

Number of two wheelers and four wheelers sold = 100

Number of two wheelers and four wheelers unsold = (450 - 100) = 350

 

Company C:

Number of two wheelers and four wheelers manufactured = 500

Number of two wheelers and four wheelers sold = 200

Number of two wheelers and four wheelers unsold = (500 - 200) = 300

 

Company D:

Number of two wheelers and four wheelers manufactured = 550

Number of two wheelers and four wheelers sold = 300

Number of two wheelers and four wheelers unsold = (550 - 300) = 250

 

Company E:

Number of two wheelers and four wheelers manufactured = 550

Number of two wheelers and four wheelers sold = 350

Number of two wheelers and four wheelers unsold = (550 - 350) = 200

 

 

Solution:

Average of two wheelers and four wheelers sold = (250 + 100 + 200 + 300 + 350)/5 = 240

Average of two wheelers and four wheelers unsold = (150 + 350 + 300 + 250 + 200)/5 = 250

Required difference = (250 - 240) = 10

General solution:

 

Company A:

Number of two wheelers and four wheelers manufactured = 400

Number of two wheelers and four wheelers sold = 250

Number of two wheelers and four wheelers unsold = (400 - 250) = 150

 

Company B:

Number of two wheelers and four wheelers manufactured = 450

Number of two wheelers and four wheelers sold = 100

Number of two wheelers and four wheelers unsold = (450 - 100) = 350

 

Company C:

Number of two wheelers and four wheelers manufactured = 500

Number of two wheelers and four wheelers sold = 200

Number of two wheelers and four wheelers unsold = (500 - 200) = 300

 

Company D:

Number of two wheelers and four wheelers manufactured = 550

Number of two wheelers and four wheelers sold = 300

Number of two wheelers and four wheelers unsold = (550 - 300) = 250

 

Company E:

Number of two wheelers and four wheelers manufactured = 550

Number of two wheelers and four wheelers sold = 350

Number of two wheelers and four wheelers unsold = (550 - 350) = 200

 

 

Solution:

Number of four wheelers sold in company E = 350/7 ×  4 = 200

Number of two wheelers sold in company E = (350 - 200) = 150

Number of maruti four wheelers sold = 200 ×  30/100 = 60

Number of Tata four wheelers sold = (200 - 60) = 140

Number of  KTM two wheelers sold = 150 ×  (40/100) = 60

Required total = (140 + 60) = 200

General solution:

 

Company A:

Number of two wheelers and four wheelers manufactured = 400

Number of two wheelers and four wheelers sold = 250

Number of two wheelers and four wheelers unsold = (400 - 250) = 150

 

Company B:

Number of two wheelers and four wheelers manufactured = 450

Number of two wheelers and four wheelers sold = 100

Number of two wheelers and four wheelers unsold = (450 - 100) = 350

 

Company C:

Number of two wheelers and four wheelers manufactured = 500

Number of two wheelers and four wheelers sold = 200

Number of two wheelers and four wheelers unsold = (500 - 200) = 300

 

Company D:

Number of two wheelers and four wheelers manufactured = 550

Number of two wheelers and four wheelers sold = 300

Number of two wheelers and four wheelers unsold = (550 - 300) = 250

 

Company E:

Number of two wheelers and four wheelers manufactured = 550

Number of two wheelers and four wheelers sold = 350

Number of two wheelers and four wheelers unsold = (550 - 350) = 200

 

 

Solution:

Number of four wheelers unsold in company D = 70/100 ×  250 = 175

Number of two wheelers unsold in company D = (250 - 175) = 75

Number of four wheelers unsold in company E = (175 + 5) = 180

Number of two wheelers unsold in company E = (200 - 180) = 20

Required sum = (75 + 20) = 95

General solution:

 

Company A:

Number of two wheelers and four wheelers manufactured = 400

Number of two wheelers and four wheelers sold = 250

Number of two wheelers and four wheelers unsold = (400 - 250) = 150

 

Company B:

Number of two wheelers and four wheelers manufactured = 450

Number of two wheelers and four wheelers sold = 100

Number of two wheelers and four wheelers unsold = (450 - 100) = 350

 

Company C:

Number of two wheelers and four wheelers manufactured = 500

Number of two wheelers and four wheelers sold = 200

Number of two wheelers and four wheelers unsold = (500 - 200) = 300

 

Company D:

Number of two wheelers and four wheelers manufactured = 550

Number of two wheelers and four wheelers sold = 300

Number of two wheelers and four wheelers unsold = (550 - 300) = 250

 

Company E:

Number of two wheelers and four wheelers manufactured = 550

Number of two wheelers and four wheelers sold = 350

Number of two wheelers and four wheelers unsold = (550 - 350) = 200

 

 

Solution:

Number of vehicles sold to males  by company B = 100/5 ×  3= 60

Number of vehicles  sold to females  by company B  = (100 - 60) = 40

Number of vehicles sold to males by company C  = 200/5 ×  4 = 160

Number of vehicles sold to females by company C = (200 - 160) = 40

Required difference = (160 + 60): (80) = 220: 80 = 11: 4

General solution:

 

Company A:

Number of two wheelers and four wheelers manufactured = 400

Number of two wheelers and four wheelers sold = 250

Number of two wheelers and four wheelers unsold = (400 - 250) = 150

 

Company B:

Number of two wheelers and four wheelers manufactured = 450

Number of two wheelers and four wheelers sold = 100

Number of two wheelers and four wheelers unsold = (450 - 100) = 350

 

Company C:

Number of two wheelers and four wheelers manufactured = 500

Number of two wheelers and four wheelers sold = 200

Number of two wheelers and four wheelers unsold = (500 - 200) = 300

 

Company D:

Number of two wheelers and four wheelers manufactured = 550

Number of two wheelers and four wheelers sold = 300

Number of two wheelers and four wheelers unsold = (550 - 300) = 250

 

Company E:

Number of two wheelers and four wheelers manufactured = 550

Number of two wheelers and four wheelers sold = 350

Number of two wheelers and four wheelers unsold = (550 - 350) = 200

 

Solution:

Number of two wheelers sold in company A = 250 ×  (30/100) = 75

Number of four wheelers sold in company A = (250 - 75) = 175

Number of four wheelers unsold in company A = 150 ×  40/100 = 60

Number of two wheelers unsold in company A = (150 - 60) = 90

Number of two wheelers manufactured = (90 + 75) = 165

Number of four wheelers manufactured = (175 + 60) = 235

Required difference = (235 - 165) = 70

Shortcut Trick

Ratio of Height (h) to Radius (r) = 2 :

1. Let h = 2x and r = x.

Total Surface Area (TSA) = 2πr(r + h) = 2 × (22/7) × x × (x + 2x) = 924.

⇒ (44/7) × 3x² = 924 ⇒ x² = 49 ⇒ x = 7 cm.

Volume = πr²h = (22/7) × 7² × 14 = 22 × 7 × 14 = 2156 cm³.

∴ The correct answer is 2156 cm³.

Alternate Method

Given: Ratio of height to radius = 2 : 1, TSA = 924 cm².

Formula Used: TSA = 2πr(h + r); Volume = πr²h

⇒ Let Radius (r) = x and Height (h) = 2x.

⇒ 2 × (22/7) × x × (2x + x) = 924

⇒ 2 × (22/7) × 3x² = 924

⇒ (132/7) × x² = 924

⇒ x² = (924 × 7) ÷ 132

⇒ x² = 7 × 7 = 49 ⇒ x = 7 cm.

⇒ Radius (r) = 7 cm, Height (h) = 14 cm.

⇒ Volume = (22/7) × 7 × 7 × 14

⇒ Volume = 22 × 7 × 14 = 2156 cm³.

∴ The correct answer is 2156 cm³.

Additional Information

Curved Surface Area of Cylinder

The area of the side surface of the cylinder is given by CSA = 2πrh.

Volume of Hollow Cylinder

If R is outer radius and r is inner radius, Volume = π(R² − r²)h.

Base Area of Cylinder

The base of a right circular cylinder is a circle, so Base Area = πr².

Shortcut Trick

Approximate Total mixture = 80 litres, Water = 20%, Milk = 80%.

Mixture remaining after 25% removal = 80 × 0.75 = 60 litres.

Remaining Milk = 80% of 60 = 48 litres; Remaining Water = 20% of 60 = 12 litres.

Final ratio 12:5 means 12 units = 48 litres ⇒ 1 unit = 4 litres.

Required Water = 5 units = 5 × 4 = 20 litres. Added Water = 20 − 12 = 8 litres.

∴ The correct answer is 8 litres.

Alternate Method

Given: Total Mixture ≈ 80 litres, Water ≈ 20.13% ≈ 20%, Removal ≈ 25%.

Step 1: Calculate initial quantities in litres.

⇒ Total mixture = 80 litres.

⇒ Water = 20% of 80 = 16 litres.

⇒ Milk = 80 − 16 = 64 litres.

Step 2: Calculate remaining quantities after taking out 25% of mixture.

⇒ Milk remaining = 64 − (25% of 64) = 64 − 16 = 48 litres.

⇒ Water remaining = 16 − (25% of 16) = 16 − 4 = 12 litres.

Step 3: Let x be the quantity of water added.

⇒ Milk : (Water + x) = 12 : 5

⇒ 48 / (12 + x) = 12 / 5

⇒ 4 / (12 + x) = 1 / 5

⇒ 20 = 12 + x

⇒ x = 20 − 12 = 8 litres.

∴ The correct answer is 8 litres.

Additional Information

Mixture Removal Property

When a portion of a mixture is removed, the ratio of components in the remaining mixture stays the same as the initial ratio.

Mixtures & Alligations

For replacement problems, Quantity left = Initial Quantity × (1 − Removed/Total)ⁿ, where n is the number of operations.

Approximation Rule

In banking exams, values like 79.99 or 20.13% should be rounded to the nearest whole number (80 and 20%) to speed up calculations.

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.

Since, we need to find out the approximate value, we can write these values to their nearest integers.

Given expression is

12.98 + 417.98 = 34.98 × 19.98 - ?2 - 99.99

13 + 418 = 35 × 20 - ?2 - 100

⇒ 13 + 418 = 700 - ?2 - 100

⇒ 431 = 700 - ?2 - 100

⇒ ?2 = 700 - 431 - 100

⇒ ? = √169 = 13

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.

Since, we need to find out the approximate value, we can write these values to their nearest integers.

Given expression is

44.99% of 399.99 + 49.99% of 299.99 = ? + 69.99% of 199.99

⇒ 45% of 400 + 50% of 300 = ? + 70% of 200

⇒ 180 + 150 = ? + 140

⇒ 330 = ? + 140

∴ ? = 190

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.

Since, we need to find out the approximate value, we can write these values to their nearest integers.

Given expression is

[12.935 ÷ 12.984] × [13.033 ÷ 13.136] = 12.83(? – 4)

⇒ [135 ÷ 134] × [133 ÷ 136] = 13(? – 4)

⇒ 13 × 13-3 = 13(? – 4)

⇒ 13-2 = 13(? – 4)

⇒ ? - 4 = -2

⇒ ? = 2