Quantitative Aptitude PYQs

Calculations:

i. x² - 36x + 320 = 0

⇒ x² - 16x - 20x + 320 = 0

⇒ (x - 16) (x - 20) = 0

⇒ x = 16, 20

 

ii. y² - 28y + 192 = 0

⇒ y² - 12y - 16y + 192 = 0

⇒ (y - 12) (y - 16) = 0

⇒ y = 12, 16

 

 

Hence, x ≥ y.

Calculation:

Using Statement I only:

Let present ages be 8k (father) and 5k (son).

[8k – 12] / [5k – 12] = 13/7

From this we can get value k from value of k we get age of father and son. From age of Father and Son we can get ratio after 5 years.

Using Statement II only:

Let son = s, father = s + 24.

Given product: s(s + 24) = 2560

⇒ s2 + 24s - 2560 = 0.

from here we can also get value of s. from value of s we age of son and father also.

Conclusion:

either I or II alone is sufficient.

Calculation:

Statement I analysis:

Let digits be t (tens) and u (units). t + u = 21 and t × u = 110.

So, statement I alone is not sufficient

Statement II analysis:

t + u = 5. When reversed, new number = 10u + t is 27 less than original: 10t + u - (10u + t) = 27

⇒ 9t - 9u = 27

⇒ t - u = 3.

Solve t + u = 5 and t - u = 3

⇒ add: 2t = 8

⇒ t = 4

⇒ u = 1.

Original number = 41.

Conclusion:

Statement II alone is sufficient.

Calculation

Given:

• Average granted in three countries (A, B, C) = 15000 → Total granted = 15000 × 3 = 45000

• Granted B = 6000, Granted C = 9000 → Granted A = 45000 - (6000 + 9000) = 30000

• In A, males are 60% of granted, females 40%

Calculation:

Female granted in A = 0.4 × 30000 = 12000

Answer: Female visas granted in A = 12,000.

Calculation

Given:

• Ratio of male granted in C : D = 3 : 4

• Female granted in D = female granted in B

• Visas granted in D = 60% of total applications in D

• From table: Granted in C = 9000, males in C = 70% of 9000 = 6300

• From B: granted B = 6000, males in B = 60% of 6000 = 3600, so females in B = 2400 (this equals female granted in D)

Let total applications in D = x. Then granted in D = 0.6x.

Female granted in D = 2400 → Male granted in D = 0.6x - 2400

Using male ratio C:D = 3:4 → 6300 : (0.6x - 2400) = 3 : 4

So, 6300 / (0.6x - 2400) = 3 / 4

3(0.6x - 2400) = 25200

1.8x - 7200 = 25200

1.8x = 32400 → x = 18000

Granted in D = 0.6 × 18000 = 10800

Rejected in D = 18000 - 10800 = 7200

Answer: Rejected applications in D = 7200.

Calculation

Given:

• Female visas granted in A = 640

• Difference between male _granted and female _rejected = 600

• From table: 80% of applications from A are granted, and 60% of granted are male.

Calculation:

Since males are 60% of granted, females are 40% of granted.

Female granted = 640 = 40% of total granted

→ Total granted in A = 640 / 0.4 = 1600

Male granted = 60% of 1600 = 960

Let female _rejected = f _rej.

Given: male _granted – female _rejected = 600

→ 960 – f _rej = 600

→ f _rej = 360

Total applications in A = total _granted / 0.8 = 1600 / 0.8 = 2000

Total rejected = 2000 - 1600 = 400

male _rejected = total _rejected – female _rejected = 400 - 360 = 40

Calculation

Given:

• Visas granted from B = 6000

• % of visas granted to males = 60% → Male granted = 60% of 6000 = 3600

• Ratio of male:female rejected visas in B = 2:3

• Difference between male & female rejected visas = 1600

Calculation:

Let male _rejected = 2k, female _rejected = 3k.

Difference = k = 1600

→ k = 1600.

Male rejected = 2k = 3200

Female rejected = 3k = 4800

Total rejected = 3200 + 4800 = 8000

Total applications = granted + rejected = 6000 + 8000 = 14000

Answer: 14,000 applications received from country B.

Calculation

Total visas granted from C = 9000.

Males granted from C = 70% of 9000 = 0.70 × 9000 = 6300.

Females granted from C = 9000 − 6300 = 2700

Among males, HB:HD = 7:5.

HB (males) = [7/12] × 6300 = 3675.

HD (males) = [5/12] × 6300 = 2625

HD (females) = one-fifth of HD (males) = [1/5] × 2625 = 525

So HB (females) = total females − HD (females) = 2700 – 525 = 2175

Total HB = 3675 + 2175 = 5850

Total HD = 2625 + 525 = 3150

Ratio HB : HD = 5850:3150 = 13 : 7.

Solution :

 Concept used :

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

 Step 1: Parts of an equation enclosed in ' Brackets ' must be solved first and in the bracket,

 Step 2: Any mathematical ' of ' or 'exponent' must be solved next,

 Step 3: Next, The part of the equation that contains 'Division' and 'multiplication' are calculated,

 Step 4: Last but not least, the parts of the equation that contains 'Addition ' and 'Subtraction' should be calculated

 Given :

⇒ 125.12% of 59.920 - 5.103 × 16.1456 + 48.09% of 24.99 + 39.95 of 9.989% = ?

Approximate the number to its nearest digit 

⇒ 125% of 60 - 5× 16 + 48% of 25 + 40 of 10% = ?

⇒ 125 × 60 /100 - 80 + 48 × 25 / 100 + 40 × 10 /100 = ?

⇒ 75 - 80 + 12 + 4 = ?

⇒ 91 - 80 = ?

⇒ ? = 11

Hence 11 is the correct answer.

Concept used :

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

Step 1: Parts of an equation enclosed in ' Brackets ' must be solved first and in the bracket,

Step 2: Any mathematical ' of ' or 'exponent' must be solved next,

Step 3: Next, The part of the equation that contains 'Division' and 'multiplication' are calculated,

Step 4: Last but not least, the parts of the equation that contains 'Addition ' and 'Subtraction' should be calculated.

In approximation, we used the concept that if the digit after the decimal is 5 or more than 5  we increase the digit before the decimal by 1 but if the digit after the decimal is less than 5 we will not increase the digit before the decimal by 1.

Calculations:

16.09% of 350.01 + 3.25 x (?) = (14.99)2 - (27.01)2/3 - 11.8² + 83.11/1.34

(16/100) × 350 + 3(?)  = (15)² - (27)^2/3 - 12² + 83/1

56 + 3(?) = 225 - 9 - 144 + 83

3(?) = 216 - 56 - 144 + 83

3(?) =  99

(?) = 33  

The approximate value of (?) we get is 33.

Concept used :

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

Step 1: Parts of an equation enclosed in ' Brackets ' must be solved first and in the bracket,

Step 2: Any mathematical ' of ' or 'exponent' must be solved next,

Step 3: Next, The part of the equation that contains 'Division' and 'multiplication' are calculated,

Step 4: Last but not least, the parts of the equation that contains 'Addition ' and 'Subtraction' should be calculated.

In approximation, we used the concept that if the digit after the decimal is 5 or more than 5  we increase the digit before the decimal by 1 but if the digit after the decimal is less than 5 we will not increase the digit before the decimal by 1.

Calculations:

120.09 % of 34.92 + (?)  =  (13.66)2 - (6.8547 × 9.12345) + 96.4561/12.3789

(120/100) 35 + (?) = (14)²  - (7 × 9)  + 8 

(12/10) × 35 + (?) = 196 - 63 + 8

(?) = 196 - 63 + 8 - 42 

(?) = 99

The approximate value we get is 99.

Concept used :

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

Step 1: Parts of an equation enclosed in ' Brackets ' must be solved first and in the bracket,

Step 2: Any mathematical ' of ' or 'exponent' must be solved next,

Step 3: Next, The part of the equation that contains 'Division' and 'multiplication' are calculated,

Step 4: Last but not least, the parts of the equation that contains 'Addition ' and 'Subtraction' should be calculated.

In approximation, we used the concept that if the digit after the decimal is 5 or more than 5  we increase the digit before the decimal by 1 but if the digit after the decimal is less than 5 we will not increase the digit before the decimal by 1.

Calculations:

 [(728.990)1/3 + (1330.887)1/3] = (?) + 5.789  + (88.456/11.123) + 7.01

 ⇒ (729)^1/3 + (1331)^1/3 =  (?) + 6 + (88/11) + 7

 ⇒ 9 + 11 = (?) + 6 + 8 + 7

 ⇒  (?) = -1

The approximate value of (?) is -1.

Calculation:

Total marbles = x + (x+10) + (x+20) = 3x + 30.

P(white) - P(black) = 1/6.

P(white) = (x + 10)/ (3x + 30). P(black) = x/(3x + 30).

Difference = [(x + 10) - x] / (3x + 30) = 10/(3x + 30) = 1/6.

Therefore 3x + 30 = 60

⇒ 3x = 30

⇒ x = 10.

Yellow marbles = x + 20 = 10 + 20 = 30.

Quantity I = 30,

Quantity II = 40

⇒ Quantity II

Quantity II > Quantity I.

Calculation:

From second equation: (6b - a)/7 = 4

⇒ 6b - a = 28

⇒ a = 6b - 28.

Substitute into first: 4b + 2a = 24

⇒ 4b + 2(6b - 28) = 24

⇒ 4b + 12b - 56 = 24

⇒ 16b = 80

⇒ b = 5.

Then a = 6(5) - 28 = 30 - 28 = 2.

2a = 4.

So, Quantity I = 4

Quantity II = 5

Quantity II > Quantity I.

Calculation

Four years ago

[X−4]/[Y−4] = 5/7

Or, 7(X−4) = 5(Y−4)

Or, 7X – 28 = 5Y – 20

Or, 7X = 5Y + 8 ------(1)

[X + 12] [Y + 12] = 9/11

Or, 11(X + 12) = 9(Y + 12)

Or, 11X + 132 = 9Y + 108

11X = 9Y − 24 -------(2)

After solving above equation, we get Y = 32 and X = 24

Quantity I:

[X – (4Y/8)] = X – (Y/2) = 24 – 16 = 8

Quantity II:

Or, 0.2Y + 0.3X = 0.2(32) + 0.3(24) = 6.4 + 7.2 = 13.6

Quantity II > Quantity I.

Calculation

So , x2 – Px + 84 = 0

Roots: a and b

Given: a – b = 5

Sum: a + b=P

Product: ab=84

a=P + 52, b = P − 52

a⋅b = [P + 52]⋅[P – 52] = P2 – 254 = 84  

P2 – 25 = 336  

⟹   P2 =361 

 ⟹   P = 19

Quantity I: 2P= 38

Quantity II: b2  – a + 1 = 72 – 12 + 1 = 49 − 11=38

✅ Quantity I = Quantity II = 38

Shortcut Trick

Sum of P + Q + R = 16 × 3 = 48.

Value of S = (Q + R)/2 + 9.

Required value = 2S + P = 2[(Q + R)/2 + 9] + P = (Q + R) + 18 + P.

Rearranging, we get (P + Q + R) + 18 = 48 + 18 = 66.

∴ The correct answer is 66.

Alternate Method

Given:

Average of P, Q, R = 16

S = (Average of Q, R) + 9

Calculations:

⇒ (P + Q + R) ÷ 3 = 16

⇒ P + Q + R = 48 ...

(i) ⇒ S = (Q + R)/2 + 9

⇒ 2S = Q + R + 18 ...

(ii) ⇒ Required value = 2S + P

⇒ Substituting 2S from (ii): (Q + R + 18) + P

⇒ Reordering: (P + Q + R) + 18

⇒ From (i), P + Q + R = 48

⇒ 48 + 18 = 66

∴ The correct answer is 66.

Additional Information

Average Concept

The average is the sum of observations divided by the total number of observations. Sum = Average × Number of terms.

Algebraic Substitution

When solving for a combined expression like 2S + P, look for ways to substitute entire blocks of variables (like P + Q + R) to avoid solving for individual unknowns.

Weighted Average

If two groups have averages A₁ and A₂ with n₁ and n₂ terms, the combined average is (n₁A₁ + n₂A₂) ÷ (n₁ + n₂).

Common Calculation

Students who did not pass:

A: 180 − 90 = 90

B: 450 − 180 = 270

C: 270 − 162 = 108

 

Calculation:

Total students in C = 270

→ F = 270 × 1.2 =324

Ratio boys:girls = 14:13

→ total parts = 27

→ one part = 12

Boys in F = 14 × 12 = 168, Girls in F = 12 × 13 = 132

Girls in E = 193 – 132 = 61

Cannot calculate exact difference in boys without total students in E.

Answer: Not enough information to determine exact difference.

Common Calculation

Students who did not pass:

A: 180 − 90 = 90

B: 450 − 180 = 270

C: 270 − 162 = 108

Calculation:

Exam A: Appeared = 180, Passed = 90, Not passed = 90

Exam B: Appeared = 450, Passed = 180, Not passed = 270

Exam C: Appeared = 270, Passed = 162, Not passed = 108

Statement I: Not passed B = 270 → multiple of 9 ✅

Statement II: Passed A+B = 90 + 180 = 270, Passed B + C = 180 + 162 = 342  → 270<342 ❌

Statement III: 50% of not passed A = 0.5 × 90 = 45, 1/6 of not passed B = 270/6 = 45 → 45=45 ✅

Answer: Statements I and III are correct.

Common Calculation

Students who did not pass:

A: 180 − 90 = 90

B: 450 − 180 = 270

C: 270 − 162 = 108

Calculation:

Not passed A = 90, B = 270, C = 108

Average = (90 + 270 + 108)/3 = 468/3 = 156

Answer: 156 students