Quantitative Aptitude PYQs

Common Calculation

Let number of items manufactured on Monday is 5x and number of items manufactured on Tuesday is 2x in company A.

So, ATQ, 5x – 2x = 45

Or, 3x = 45

Or, x = 45/3 = 15

So, Number of items manufactured on Monday is 5 × 15 = 75

And number of items manufactured on Tuesday is 2 × 15 = 30

Similarly, we can calculate others values also which is given in the table.

 

Calculation

Items manufactured Monday = 105

Items manufactured Tuesday = 135

→ Average = [105 + 135] /2 = 240/2 = 120

Common Calculation

Let number of items manufactured on Monday is 5x and number of items manufactured on Tuesday is 2x in company A.

So, ATQ, 5x – 2x = 45

Or, 3x = 45

Or, x = 45/3 = 15

So, Number of items manufactured on Monday is 5 × 15 = 75

And number of items manufactured on Tuesday is 2 × 15 = 30

Similarly, we can calculate others values also which is given in the table.

 

Calculation

Monday: 40% items sold

→ Items Unsold = 60% of 120 = 0.6 × 120 = 72

Tuesday: 60% items sold

→ Unsold = 40% of 150 = 0.4 × 150 = 60

Total items unsold = 72 + 60 = 132

Common Calculation

Let number of items manufactured on Monday is 5x and number of items manufactured on Tuesday is 2x in company A.

So, ATQ, 5x – 2x = 45

Or, 3x = 45

Or, x = 45/3 = 15

So, Number of items manufactured on Monday is 5 × 15 = 75

And number of items manufactured on Tuesday is 2 × 15 = 30

Similarly, we can calculate others values also which is given in the table.

 

Calculation

Company F:

Number of items manufactured on Monday by E = 240

→ Number of items manufactured on Monday by F

= 240 + [0.3 × 240] = 240 + 72 = 312

Number of items manufactured on Tuesday by F

= Monday + 68 = 312 + 68 = 380

Common Calculation

Let number of items manufactured on Monday is 5x and number of items manufactured on Tuesday is 2x in company A.

So, ATQ, 5x – 2x = 45

Or, 3x = 45

Or, x = 45/3 = 15

So, Number of items manufactured on Monday is 5 × 15 = 75

And number of items manufactured on Tuesday is 2 × 15 = 30

Similarly, we can calculate others values also which is given in the table.

 

Calculation

Monday total:

B = 105, C = 75

→ Total = 105 + 75 = 180

Tuesday total:

B = 135, C = 50

→ Total = 135 + 50 = 185

So, ratio = 180 : 185

Simplify: divide by 5 → 36:37

So, x = 36, y = 37:

So, 7x + 8y = [7 × 36] + [8 × 37] = 252 + 296 = 548

Common Calculation

Let number of items manufactured on Monday is 5x and number of items manufactured on Tuesday is 2x in company A.

So, ATQ, 5x – 2x = 45

Or, 3x = 45

Or, x = 45/3 = 15

So, Number of items manufactured on Monday is 5 × 15 = 75

And number of items manufactured on Tuesday is 2 × 15 = 30

Similarly, we can calculate others values also which is given in the table.

 

Calculation

Total number of items manufactured on Monday = 75

→ 20% defective = 0.2 × 75 = 15

→ Non-defective = 75 – 15 = 60

Now, non-defective on Tuesday = 40% of 60 = 24

Total number of items manufactured on Tuesday = 30

→ Defective = 30 – 24 = 6

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.

Concept used:

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

Calculation:

⇒ 813.5 × 7291.5 ÷  95.5 = ?

⇒ 9^{2 × 3.5} × 9^{3 × 1.5} ÷ 9^5.5 = ?

⇒ 9^{7 + 4.5 - 5.5} = ?

⇒ 9⁶ = ?

∴ The value of ? is 96

Concept used:

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

Calculation:

⇒ 6/11 of 5/7 of 462 - 30 = ?

⇒ 6/11 ×  5/7 × 462 - 30 = ?

⇒ 180 - 30 = ?

⇒ ? = 150

∴ The value of ? is 150.

Given:

(13/11) × 143 + (12)3 ÷ 8 + ? = (20)²

Concept used:

Follow the BODMAS rule according to the table given below:

Calculation:

(13/11) × 143 + (12)3 ÷ 8 + ? = (20)²

⇒ 169 + 1728 ÷ 8 + ? = 400

⇒ 169 + 216 + ? = 400

⇒ ? = 400 - 385

⇒ ? = 15

∴ 15 should come in place of the question mark (?).

Given:

(3/7) × 343 - 25% of 160 + 62 = ? - 1

Concept used:

Follow the BODMAS rule according to the table given below:

Calculation:

(3/7) × 343 - 25% of 160 + 62 = ? - 1

⇒ 147 - (25/100) × 160 + 36 = ? - 1

⇒ ? = 147 - 40 + 36 + 1

⇒ ? = 184 - 40

⇒ ? = 144

∴ 144 should come in place of the question mark (?).

Given:

(625 × 25) ÷ 625 - 10 = ?

Concept Used:

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

Calculation:

(625 × 25) ÷ 625 - 10 = ?

⇒ 25 - 10 = ?

⇒ ? = 15

∴ The required value is 15