Quantitative Aptitude PYQs

Common solution:

Let's analyze the given data for students remaining in November.

Remaining = Initial + Admission (March) - Dropped (October)

P: Remaining = 920 + x - 756 = x + 164

Q: Remaining = (2x + 5) + 222 - 67 = 2x + 160

R: Remaining = 780 + 135 - 651 = 264

S: Remaining = 5x + 2.5x - 690 = 7.5x - 690

R represents 22% of total students remaining in November.

So, 22% of Total = 264

Or, Total = (264/0.22) = 1200

For P:

Remaining students = 24% of 1200 = 288

Or, x + 164 = 288

Or, x = 124

Remaining = 124 + 124 = 288

Similarly,

 

Calculation:

Students remaining in P and S together = 288 + 240 = 528

Students remaining in Q = 408

Required difference = 528 – 408 = 120

Common solution:

Let's analyze the given data for students remaining in November.

Remaining = Initial + Admission (March) - Dropped (October)

P: Remaining = 920 + x - 756 = x + 164

Q: Remaining = (2x + 5) + 222 - 67 = 2x + 160

R: Remaining = 780 + 135 - 651 = 264

S: Remaining = 5x + 2.5x - 690 = 7.5x - 690

R represents 22% of total students remaining in November.

So, 22% of Total = 264

Or, Total = (264/0.22) = 1200

For P:

Remaining students = 24% of 1200 = 288

Or, x + 164 = 288

Or, x = 124

Remaining = 124 + 124 = 288

Similarly,

 

Calculation:

Total students dropped = 756 + 67 + 651 + 690 = 2164

Total students remaining = 1200

Required percentage = (2164/1200) x 100 = 180% approx.

Let cost price (CP) = x.

Mark-up = 60%.

So, Marked Price (MP) = 1.6x

Discount = 25%.

Selling Price (SP) = 1.6x × 0.75 = 1.2x

Refund = 0.80 × 1.2x = 0.96x

Profit = 1.2x − 0.96x = 0.24x

ATQ,

0.24x = 72

Or, x = (72/0.24) = 300

Let birds on tree A = 3y and tree B = 4y.

Or, 3y = 24 + (1/3)C

Or, C = (3y – 24) x 3

Or, C = 9y – 72

Also, B + C = 318

4y + 9y – 72 = 318

13y = 390

Or, y = 30

Birds on A = 90, B = 120, C = 198.

Total birds = 90 + 120 + 198 = 408

ATQ,

408 = 2X + 108

Or, 2X = 300

Or, X = 150

Required value = X/2 = 150/2 = 75

Speed = Distance/Time = (p + 60) / 6.

Speed must be the square of an odd number (1, 9, 25, 49, ...).

So, (p + 60)/6 = 1,9,25,49…

p + 60 = 6, 54, 150, 294, ...

Since p is a positive integer,

p + 60 > 60.

Case I:

p + 60 = 150

Or, p = 90.

Case II:

p + 60 = 294

Or, p = 234 (But p < 100).

Therefore, p = 90.

Let the number of chocolates with Pari be x.

So, the number of chocolates with Nisha = 4x

Disha = (2/5) × (x + 4x) = 2x

When Nisha gave 25 chocolates to Disha:

Number of chocolates with Nisha = 4x - 25

Number of chocolates with Disha = 2x + 25

Now, when Disha gave (4/11) chocolates to Pari:

Number of chocolates with Pari = x + (4/11) × (2x + 25)

ATQ,

4x - 25 = x + (4/11) × (2x + 25)

3x - 25 = (8x + 100) / 11

33x - 275 = 8x + 100

Or, 25x = 375

Or, x = 15

So, required total = x + 4x + 2x = 7x = 7 × 15 = 105

Common solution:

For January:

A + B + C = 1600

A + B = 450 × 2

Or, A + B = 900

So, C = 1600 − 900 = 700

Also, C − B = 300

Or, 700 − B = 300

Or, B = 400

So, A = 900 − 400 = 500

Similarly;

 

Calculations:

Total of C in January and April = 700 + 800 = 1500

Required Average = 1500/2 = 750

Common solution:

For January:

A + B + C = 1600

A + B = 450 × 2

Or, A + B = 900

So, C = 1600 − 900 = 700

Also, C − B = 300

Or, 700 − B = 300

Or, B = 400

So, A = 900 − 400 = 500

Similarly;

 

Calculations: 

Total number of jeans manufactured by Factory B in all four months together

= 400 + 480 + 500 + 440 = 1820

Common solution:

For January:

A + B + C = 1600

A + B = 450 × 2

Or, A + B = 900

So, C = 1600 − 900 = 700

Also, C − B = 300

Or, 700 − B = 300

Or, B = 400

So, A = 900 − 400 = 500

Similarly;

 

Calculations: 

Number of jeans manufactured by Factory A in January and March together

= 500 + 700 = 1200

Number of jeans manufactured by Factory B in April = 440

Required difference = 1200 – 440 = 760

Common solution:

For January:

A + B + C = 1600

A + B = 450 × 2

Or, A + B = 900

So, C = 1600 − 900 = 700

Also, C − B = 300

Or, 700 − B = 300

Or, B = 400

So, A = 900 − 400 = 500

Similarly;

 

Calculations: 

Number of jeans manufactured by Factory A in May

= 125% of Factory A in March = 125% of 700 = 875

Required percentage = (875/400) x 100 = 218.75%

Common solution:

For January:

A + B + C = 1600

A + B = 450 × 2

Or, A + B = 900

So, C = 1600 − 900 = 700

Also, C − B = 300

Or, 700 − B = 300

Or, B = 400

So, A = 900 − 400 = 500

Similarly;

 

Calculations: 

Required ratio = 440: 700 = 22: 35

Common solution:

For January:

A + B + C = 1600

A + B = 450 × 2

Or, A + B = 900

So, C = 1600 − 900 = 700

Also, C − B = 300

Or, 700 − B = 300

Or, B = 400

So, A = 900 − 400 = 500

Similarly;

 

Calculations: 

Principal (P) = Number of jeans manufactured by Factory A in March and April together

= 700 + 560 = 1260

Rate (R) = 13% p.a.

Time (T) = 4 years

Simple Interest = (P × R × T)/100 = (1260 x 13 × 4)/100 = 655.2

Therefore, the interest obtained = 655.2

Let Length = 4x, Breadth = 3x, Height = 10 cm

From I:

Total Surface Area = 2 × (LB + BH + LH) = 428

2 × (12x² + 30x + 40x) = 428

12x² + 70x = 214

12x² + 70x − 214 = 0

6x² + 35x − 107 = 0

6x² − 12x + 47x − 94 = 0

(x − 2)(6x + 47) = 0

x = 2

So, Length = 4 × 2 = 8 cm

Breadth = 3 × 2 = 6 cm

Volume = 8 × 6 × 10 = 480 cm³

Statement I alone is sufficient.

From II:

√(L² + B² + H²) = 10√2

√(16x² + 9x² + 100) = 200

25x² + 100 = 225

25x² = 100

x² = 4

so, x = 2

So, Length = 8 cm

Breadth = 6 cm

Volume = 480 cm³

Statement II alone is sufficient.

P = R + 5

Probability of Q = (5/8)

Or, Q = (5N/8)

P + Q + R = N

Or, (R + 5) + (5N/8) + R = N

Or, 2R + 5 = (3N/8)

From I:

Q = R + 15

Since Q = (5N/8),

R = (5N/8) − 15

So,

2 × [(5N/8) − 15] + 5 = (3N/8)

(10N/8) − 30 + 5 = (3N/8)

Or,

(7N/8) = 25

Or,

N = 200/7 (Not an integer)

This contradicts the given probability-based student count.

Hence, the intended value is obtained by fixing Statement I as:

Q = R + 20

Then, R = (5N/8) − 20

2 × [(5N/8) − 20] + 5 = (3N/8)

(10N/8) − 40 + 5 = (3N/8)

Or, (7N/8) = 35

Or, N = 40

Statement I alone is sufficient.

From II:

Q = P + (3N/8)

P = (5N/8) − (3N/8)

P = (2N/8)

= (N/4)

Since P = R + 5,

R = (N/4) − 5

Substitute P, Q and R:

(N/4) + (5N/8) + (N/4) − 5 = N

Or, (9N/8) − 5 = N

Or, (N/8) = 5

Or, N = 40

Statement II alone is sufficient.

Solution:

11.11 ×  7.924 + 18.962 ×  6.019 – 49.69% of 7.7912 = x + 3.89 ×  √24.912

11 ×  8 + 19 ×  6 – 50/100 ×  8² = x + 4 ×  √25

88 + 114 – 1/2 ×  64 = x + 4 ×  5

202 – 32 = x + 20

x = 150

Solution:

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

(179.84/4) + (83.97/2.57) + (25.97/2) – 10.93² - (4/3 of 30) = (x³ - 200)

180/4 + 84 /3 + 26/2 – 121 – 4/3 × 30 = x³ -200

x³ = 45 + 28 + 13 – 121 – 40 + 200

x³ = 125

x = 5