Quantitative Aptitude PYQs

Common explanation:

From the pie chart, the percentages are:

Monday: (p + 5)%

Tuesday: (p − 5)%

Wednesday: p%

Thursday: 28%

Friday: = 27%

Total of all percentages = 27 + (p + 5) + 28 + (p − 5) + p =55 + 3p

Now, 100 = 55 + 3p

Or, 3p = 45

Or, p = 45/5 = 15

Now, Monday: (p + 5)% = 20%

Tuesday: (p − 5)% = 10%

Wednesday: p% = 15%

Thursday: 28%

Friday: = 27%

According to the question, 

 

ATQ,

The number of bus tickets booked on Monday, Tuesday and Wednesday together = (300 + 150 + 225) = 675

Tuesday = (150/675) × 100 = 22.22%

Required answer = (22.22/100) × 360º = 80º

Common explanation:

From the pie chart, the percentages are:

Monday: (p + 5)%

Tuesday: (p − 5)%

Wednesday: p%

Thursday: 28%

Friday: = 27%

Total of all percentages = 27 + (p + 5) + 28 + (p − 5) + p =55 + 3p

Now, 100 = 55 + 3p

Or, 3p = 45

Or, p = 45/5 = 15

Now, Monday: (p + 5)% = 20%

Tuesday: (p − 5)% = 10%

Wednesday: p% = 15%

Thursday: 28%

Friday: = 27%

According to the question, 

 

ATQ,

Total number of bus tickets booked on Saturday = (405 + 420) - 25 = 825 - 25 = 800

Number of bus tickets booked by girls on Saturday = 37.5% of 800 = 800 × (3/8) = 300

Number of bus tickets booked by boys on Saturday = 800 - 300 = 500

Common explanation:

From the pie chart, the percentages are:

Monday: (p + 5)%

Tuesday: (p − 5)%

Wednesday: p%

Thursday: 28%

Friday: = 27%

Total of all percentages = 27 + (p + 5) + 28 + (p − 5) + p =55 + 3p

Now, 100 = 55 + 3p

Or, 3p = 45

Or, p = 45/5 = 15

Now, Monday: (p + 5)% = 20%

Tuesday: (p − 5)% = 10%

Wednesday: p% = 15%

Thursday: 28%

Friday: = 27%

ATQ,

According to the question, 

 

ATQ,

Number of evening time slot bus tickets booked = 405 × (11/15) = 297

Number of night time slot bus tickets booked = 405 - 297 = 108

ATQ,

Let the price of night time slot per ticket be Rs. ‘a’

108a = 21600

a = 21600/108

a = 200

The price of evening time slot per ticket = 200 - 75 = Rs.125

Required revenue = 297 × 125 = Rs.37125

Common explanation:

From the pie chart, the percentages are:

Monday: (p + 5)%

Tuesday: (p − 5)%

Wednesday: p%

Thursday: 28%

Friday: = 27%

Total of all percentages = 27 + (p + 5) + 28 + (p − 5) + p =55 + 3p

Now, 100 = 55 + 3p

Or, 3p = 45

Or, p = 45/5 = 15

Now, Monday: (p + 5)% = 20%

Tuesday: (p − 5)% = 10%

Wednesday: p% = 15%

Thursday: 28%

Friday: = 27%

ATQ,

According to the question,

 

Number of vacant seats on Tuesday = 150 × (1/5) = 30

Number of filled seats on Tuesday = 150 - 30 = 120

Number of vacant seats on Wednesday = 225 × (3/5) = 135

Number of filled seats on Wednesday = 225 - 135 = 90

Required sum = 120 + 90 = 210

Common explanation:

From the pie chart, the percentages are:

Monday: (p + 5)%

Tuesday: (p − 5)%

Wednesday: p%

Thursday: 28%

Friday: = 27%

Total of all percentages = 27 + (p + 5) + 28 + (p − 5) + p =55 + 3p

Now, 100 = 55 + 3p

Or, 3p = 45

Or, p = 45/5 = 15

Now, Monday: (p + 5)% = 20%

Tuesday: (p − 5)% = 10%

Wednesday: p% = 15%

Thursday: 28%

Friday: = 27%

ATQ,

According to the question, 

 

ATQ,

Number of Online Bus tickets reserved Thursday to Tuesday = 420 × (7/20) = 147

Now, the number of Online Bus tickets booked on Thursday = 420 - 147 = 273

Now, the number of Online Bus tickets booked on Tuesday = 150 + 147 = 297

Required ratio = 273:297 = 91:99

Series I:

So, 'P' = 270

Series II:

36 + 82 = 118

118 + 82 = 200

200 + 82 = 282

282 + 82 = 364

364 + 82 = 446

So, 'Q' = 282

Series III:

85 + (13 × 2) = 111

111 + (13 × 4) = 163

163 + (13 × 6) = 241

241 + (13 × 8) = 345

345 + (13 × 10) = 475

So, 'R' = 345

Required value = 5 × 292 - (270 + 345) = 1,460 - 615 = 845

Hence, option d.

Series I:

So, 'P' = 270

Series II:

36 + 82 = 118

118 + 82 = 200

200 + 82 = 282

282 + 82 = 364

364 + 82 = 446

So, 'Q' = 282

Series III:

85 + (13 × 2) = 111

111 + (13 × 4) = 163

163 + (13 × 6) = 241

241 + (13 × 8) = 345

345 + (13 × 10) = 475

So, 'R' = 345

270 < 292 < 345

So, P < Q < R

Hence, option b.

In series I,

15.6 + 18 × 0.1 = 17.4

17.4 + 18 × 0.2 = 21

21 + 18 × 0.3 = 26.4

26.4 + 18 × 0.4 = 33.6

33.6 + 18 × 0.5 = 42.6

Therefore, 33.6 should come in place of 33.4

So, 'P' = 33.4

In series II,

12.6 + 18 × 0.1 = 14.4

14.4 + 18 × 0.2 = 18

18 + 18 × 0.3 = 23.4

23.4 + 18 × 0.4 = 30.6

So, 'R' = 30.6

So, (P + R) = 33.4 + 30.6 = 64

Hence, option c.

The given number series follows the pattern of alternate addition and subtraction of cubes and squares, respectively. 

113 + 5³ = 238

238 - 6² = 202

202 + 7³ = 545

545 - 8² = 481

481 + 9³ = 1210

1210 - 10² = 1110

So, 'P' = 1210.

For I:

Nearest perfect square number of 1210, is 35² = 1225

So, statement I is true.

For II:

1210 ÷ 11 = 110

So, statement II is true.

For III:

P + 28 = 1210 + 28 = 1238

1238 is not a multiple of 50

So, statement III is false.

Hence, option c.

For series I:

168 × (7/8) = 147

147 × (6/7) = 126

126 × (5/6) = 105

105 × (4/5) = 84

84 × (3/4) = 63

63 × (2/3) = 42

So, P = 128

For series II:

90 + 11 × 1 = 101

101 − 11 × 2 = 79

79 + 11 × 3 = 112

112 − 11 × 4 = 68

68 + 11 × 5 = 123

So, Q = 70

Statement I:

P = 128 and Q = 70

L.C.M of P and Q = 4480

So, statement I is true.

Statement II:

Q - P = 70 - 128 = 58

Since, 58 < 60

So, statement II is true.

Statement III:

Sum of 'P' and 'Q' = 128 + 70 = 198

Since 198 < 200

So, statement III is false.

Hence, option b.

Shortcut Trick

Let Boat speed be x. Upstream speed = x − 6, Downstream speed = x + 6.

Time difference formula: [Distance × 2 × Stream Speed] ÷ [x² − (Stream Speed)²] = Time Difference.

⇒ [160 × 2 × 6] ÷ [x² − 36] = 3 ⇒ 640 = x² − 36 ⇒ x² = 676 ⇒ x = 26 km/h.

Statement I: A + B = (26−6) + (26+6) = 20 + 32 = 52 (False). Statement II: 160 ÷ 20 = 8 hours (True). Statement III: √26 is not a composite number (False).

∴ The correct answer is Both I and III.

Alternate Method

Given: Stream speed (y) = 6 km/h, Distance (D) = 160 km, Time difference = 3 hours.

Formula: Time = Distance ÷ Speed; Upstream speed (A) = x − y; Downstream speed (B) = x + y.

⇒ 160/(x − 6) − 160/(x + 6) = 3

⇒ 160 [ (x + 6 − x + 6) / (x² − 36) ] = 3

⇒ 160 × 12 = 3 × (x² − 36) ⇒ 640 = x² − 36 ⇒ x² = 676 ⇒ x = 26 km/h

⇒ Upstream speed (A) = 26 − 6 = 20 km/h; Downstream speed (B) = 26 + 6 = 32 km/h.

⇒ Statement I: A + B = 20 + 32 = 52. (Given 54, so I is False)

⇒ Statement II: Time Upstream = 160 ÷ 20 = 8 hours. (Given 8, so II is True)

⇒ Statement III: Speed in still water (C) = 26. √26 ≈ 5.09 (Not an integer/composite, so III is False)

∴ The correct answer is Both I and III.

Additional Information

Downstream and Upstream

Downstream speed = Speed of boat + Speed of stream. Upstream speed = Speed of boat − Speed of stream.

Still Water Speed Calculation

Speed of boat in still water = (Downstream speed + Upstream speed) ÷ 2.

Composite Numbers

A composite number is a positive integer greater than 1 that has at least one divisor other than 1 and itself.

Shortcut Trick

Profit ratio of Arnav and Nitin is 1600 : 1200 = 4 : 3.

Nitin investment: (P − 1500) for 8 months + P for 12 months.

Arnav investment: (P + 3000) for 8 months + 2/3(P + 3000) for 12 months.

Effective investment ratio = Arnav : Nitin = 4 : 3.

Equating ratios: [8(P + 3000) + 8(P + 3000)] / [8(P − 1500) + 12P] = 4/3.

Solving for P gives P = 4500. Initial ratio = (4500 − 1500) : (4500 + 3000) = 3000 : 7500 = 2 : 5.

∴ The correct answer is 1:2.

Alternate Method

Given → Investment of Nitin = (P − 1500), Arnav = (P + 3000). Time = 20 months.
Formulas: Profit Share = Investment × Time | Profit Ratio = (I₁ × T₁) : (I₂ × T₂)

⇒ Nitin Total Investment = (P − 1500) × 8 + (P − 1500 + 1500) × 12 = 8P − 12000 + 12P = 20P − 12000.
⇒ Arnav Total Investment = (P + 3000) × 8 + 2/3(P + 3000) × 12 = 8(P + 3000) + 8(P + 3000) = 16P + 48000.
⇒ Ratio: (20P − 12000) / (16P + 48000) = 1200 / 1600 = 3/4.
⇒ 4(20P − 12000) = 3(16P + 48000) ⇒ 80P − 48000 = 48P + 144000.
⇒ 32P = 192000 ⇒ P = 6000.
⇒ Initial Nitin = 6000 − 1500 = 4500; Initial Arnav = 6000 + 3000 = 9000.
⇒ Ratio = 4500 : 9000 = 1 : 2.

∴ The correct answer is 1:2.

Additional Information

Partnership Basics

When investments change over time, we calculate the equivalent investment for one unit of time (months) to find the profit sharing ratio.

Profit Allocation

Profit is directly proportional to the product of Capital and Time. If time is constant, Profit ∝ Capital.

Common explanation

Let the number of students who choose Maths, English, and Hindi be 7a, 8a, and 5a, respectively.

Let the number of students who choose both Hindi and English, both English and Maths, and both Maths and Hindi be 5b, 6b, and 4b, respectively.

Number of students who choose only Hindi = 40% × 5b = 2b

Number of students who choose only English = (5/2) × 2b = 5b

So,

4b + 6b – 320 + 520 = 7a

⇒ 7a – 10b = 200 --------- (I)

And,

5a = 2b + 5b + 4b – 320

⇒ 11b – 5a = 320 --------- (II)

On solving 5 × equation (I) + 7 * equation (II), we have:

5 × (7a – 10b) + 7 × (11b – 5a) = 5 × 200 + 7 × 320

⇒ 35a – 50b + 77b – 35a = 1000 + 2240

⇒ 27b = 3240

⇒ b = 120

Putting the value of b in equation (I):

7a – 10 × 120 = 200

⇒ 7a – 1200 = 200

⇒ 7a = 1400

⇒ a = 200

Now,

• Number of students who choose Maths = 7 × 200 = 1400
• Number of students who choose Hindi = 5 × 200 = 1000
• Number of students who choose English = 8 × 200 = 1600
• Number of students who choose only Hindi = 2 × 120 = 240
• Number of students who choose only English = 5 × 120 = 600
• Number of students who choose both Hindi and English but not Maths = 5 × 120 – 320 = 280
• Number of students who choose both Hindi and Maths but not English = 4 × 120 – 320 = 160
• Number of students who choose both Maths and English but not Hindi = 6 × 120 – 320 = 400

So, 

Required ratio = 280:400 = 7:10

Common explanation

Let the number of students who choose Maths, English, and Hindi be 7a, 8a, and 5a, respectively.

Let the number of students who choose both Hindi and English, both English and Maths, and both Maths and Hindi be 5b, 6b, and 4b, respectively.

Number of students who choose only Hindi = 40% × 5b = 2b

Number of students who choose only English = (5/2) × 2b = 5b

So,

4b + 6b – 320 + 520 = 7a

⇒ 7a – 10b = 200 --------- (I)

And,

5a = 2b + 5b + 4b – 320

⇒ 11b – 5a = 320 --------- (II)

On solving 5 × equation (I) + 7 * equation (II), we have:

5 × (7a – 10b) + 7 × (11b – 5a) = 5 × 200 + 7 × 320

⇒ 35a – 50b + 77b – 35a = 1000 + 2240

⇒ 27b = 3240

⇒ b = 120

Putting the value of b in equation (I):

7a – 10 × 120 = 200

⇒ 7a – 1200 = 200

⇒ 7a = 1400

⇒ a = 200

Now,

• Number of students who choose Maths = 7 × 200 = 1400
• Number of students who choose Hindi = 5 × 200 = 1000
• Number of students who choose English = 8 × 200 = 1600
• Number of students who choose only Hindi = 2 × 120 = 240
• Number of students who choose only English = 5 × 120 = 600
• Number of students who choose both Hindi and English but not Maths = 5 × 120 – 320 = 280
• Number of students who choose both Hindi and Maths but not English = 4 × 120 – 320 = 160
• Number of students who choose both Maths and English but not Hindi = 6 × 120 – 320 = 400

So, 

Required difference = 600 - 240 = 360

Common explanation 

Let the number of students who choose Maths, English, and Hindi be 7a, 8a, and 5a, respectively.

Let the number of students who choose both Hindi and English, both English and Maths, and both Maths and Hindi be 5b, 6b, and 4b, respectively.

Number of students who choose only Hindi = 40% × 5b = 2b

Number of students who choose only English = (5/2) × 2b = 5b

So,

4b + 6b – 320 + 520 = 7a

⇒ 7a – 10b = 200 --------- (I)

And,

5a = 2b + 5b + 4b – 320

⇒ 11b – 5a = 320 --------- (II)

On solving 5 × equation (I) + 7 * equation (II), we have:

5 × (7a – 10b) + 7 × (11b – 5a) = 5 × 200 + 7 × 320

⇒ 35a – 50b + 77b – 35a = 1000 + 2240

⇒ 27b = 3240

⇒ b = 120

Putting the value of b in equation (I):

7a – 10 × 120 = 200

⇒ 7a – 1200 = 200

⇒ 7a = 1400

⇒ a = 200

Now,

• Number of students who choose Maths = 7 × 200 = 1400
• Number of students who choose Hindi = 5 × 200 = 1000
• Number of students who choose English = 8 × 200 = 1600
• Number of students who choose only Hindi = 2 × 120 = 240
• Number of students who choose only English = 5 × 120 = 600
• Number of students who choose both Hindi and English but not Maths = 5 × 120 – 320 = 280
• Number of students who choose both Hindi and Maths but not English = 4 × 120 – 320 = 160
• Number of students who choose both Maths and English but not Hindi = 6 × 120 – 320 = 400

So, 

Number of boys who choose only two subjects = (5/7) × (160 + 280 + 400) = 600

Therefore, required percent = (600/600) × 100 = 100%

Common explanation

Let the number of students who choose Maths, English, and Hindi be 7a, 8a, and 5a, respectively.

Let the number of students who choose both Hindi and English, both English and Maths, and both Maths and Hindi be 5b, 6b, and 4b, respectively.

Number of students who choose only Hindi = 40% × 5b = 2b

Number of students who choose only English = (5/2) × 2b = 5b

So,

4b + 6b – 320 + 520 = 7a

⇒ 7a – 10b = 200 --------- (I)

And,

5a = 2b + 5b + 4b – 320

⇒ 11b – 5a = 320 --------- (II)

On solving 5 × equation (I) + 7 * equation (II), we have:

5 × (7a – 10b) + 7 × (11b – 5a) = 5 × 200 + 7 × 320

⇒ 35a – 50b + 77b – 35a = 1000 + 2240

⇒ 27b = 3240

⇒ b = 120

Putting the value of b in equation (I):

7a – 10 × 120 = 200

⇒ 7a – 1200 = 200

⇒ 7a = 1400

⇒ a = 200

Now,

• Number of students who choose Maths = 7 × 200 = 1400
• Number of students who choose Hindi = 5 × 200 = 1000
• Number of students who choose English = 8 × 200 = 1600
• Number of students who choose only Hindi = 2 × 120 = 240
• Number of students who choose only English = 5 × 120 = 600
• Number of students who choose both Hindi and English but not Maths = 5 × 120 – 320 = 280
• Number of students who choose both Hindi and Maths but not English = 4 × 120 – 320 = 160
• Number of students who choose both Maths and English but not Hindi = 6 × 120 – 320 = 400

So, 

Required sum = (240 + 600 + 520) + 160 = 1520

Common explanation

Let the number of students who choose Maths, English, and Hindi be 7a, 8a, and 5a, respectively.

Let the number of students who choose both Hindi and English, both English and Maths, and both Maths and Hindi be 5b, 6b, and 4b, respectively.

Number of students who choose only Hindi = 40% × 5b = 2b

Number of students who choose only English = (5/2) × 2b = 5b

So,

4b + 6b – 320 + 520 = 7a

⇒ 7a – 10b = 200 --------- (I)

And,

5a = 2b + 5b + 4b – 320

⇒ 11b – 5a = 320 --------- (II)

On solving 5 × equation (I) + 7 * equation (II), we have:

5 × (7a – 10b) + 7 × (11b – 5a) = 5 × 200 + 7 × 320

⇒ 35a – 50b + 77b – 35a = 1000 + 2240

⇒ 27b = 3240

⇒ b = 120

Putting the value of b in equation (I):

7a – 10 × 120 = 200

⇒ 7a – 1200 = 200

⇒ 7a = 1400

⇒ a = 200

Now,

• Number of students who choose Maths = 7 × 200 = 1400
• Number of students who choose Hindi = 5 × 200 = 1000
• Number of students who choose English = 8 × 200 = 1600
• Number of students who choose only Hindi = 2 × 120 = 240
• Number of students who choose only English = 5 × 120 = 600
• Number of students who choose both Hindi and English but not Maths = 5 × 120 – 320 = 280
• Number of students who choose both Hindi and Maths but not English = 4 × 120 – 320 = 160
• Number of students who choose both Maths and English but not Hindi = 6 × 120 – 320 = 400

So, 

Required number of students = 240 + 280 + 160 + 320 + 520 + 400 + 600 = 2520

Shortcut Trick

Net Profit% formula: P% = m − d − (m × d) ÷ 100, where m is markup% and d is discount%.

Profit amount = CP × P% ÷ 100. Given CP = Rs. 1800 and d = 25%.

For I: P% = 40 − 25 − (40 × 25) ÷ 100 = 15 − 10 = 5%. Profit = 1800 × 5% = Rs. 90 (Matches).

For II: P% = 50 − 25 − (50 × 25) ÷ 100 = 25 − 12.5 = 12.5%. Profit = 1800 × 12.5% = Rs. 225 (Matches).

For III: P% = 60 − 25 − (60 × 25) ÷ 100 = 35 − 15 = 20%. Profit = 1800 × 20% = Rs. 360 (Does not match 270).

∴ The correct answer is Both I and II.

Alternate Method

Given → CP = Rs. 1800, Discount = 25%.
Formulas: MP = CP × (100 + m%) ÷ 100 | SP = MP × (100 − d%) ÷ 100 | Profit = SP − CP

⇒ For Case I (Markup = 40%): MP = 1800 × 1.40 = Rs. 2520.
⇒ SP = 2520 × 0.75 = Rs. 1890. Profit = 1890 − 1800 = Rs. 90. (Statement I is True).
⇒ For Case II (Markup = 50%): MP = 1800 × 1.50 = Rs. 2700.
⇒ SP = 2700 × 0.75 = Rs. 2025. Profit = 2025 − 1800 = Rs. 225. (Statement II is True).
⇒ For Case III (Markup = 60%): MP = 1800 × 1.60 = Rs. 2880.
⇒ SP = 2880 × 0.75 = Rs. 2160. Profit = 2160 − 1800 = Rs. 360. (Statement III is False).

∴ The correct answer is Both I and II.

Additional Information

Relation between CP, MP, and Profit

The formula relating Cost Price and Marked Price is CP ÷ MP = (100 − Discount%) ÷ (100 + Profit%).

Markup Percentage

Markup is the extra value added to the CP to decide the MP. Markup% = (MP − CP) ÷ CP × 100.

Shortcut Trick

Let Aarav (A), Bavya (B), and Chetan (C) be their present ages.

A + C = 35 × 2 = 70. Given A:B = 6:5 ⇒ A = 1.2B.

B:(C−3) = 7:5 ⇒ C−3 = 5B/7 ⇒ C = 5B/7 + 3.

Substitute in A+C: 1.2B + 5B/7 + 3 = 70 ⇒ 8.4B + 5B = 67 × 7 ⇒ 13.4B = 469 ⇒ B = 35 years.

A = 1.2 × 35 = 42 years. After 5 years, A = 47. So, 2y + 1 = 47 ⇒ y = 23.

Test options:

I. 23+12=35 (True); II. 46−3=43 (False); III. 12+22=34 (False).

∴ The correct answer is 35 years.

Alternate Method

Given:

Aarav age after 5 years = (2y + 1) years ⇒ Aarav present age (A) = 2y − 4.

Ratio A:B = 6:5 ⇒ Bhavya present age (B) = 5A/6.

Ratio B:(C − 3) = 7:5 ⇒ Chetan age 3 years ago = 5B/7.

Average (A + C)/2 = 35 ⇒ A + C = 70.

⇒ Let B = 35k. From A:B = 6:5, A = 6/5 × 35k = 42k.

⇒ From B:(C−3) = 7:5, C − 3 = 5/7 × 35k = 25k ⇒ C = 25k + 3.

⇒ Sum A + C = 42k + 25k + 3 = 70

⇒ 67k = 67 ⇒ k = 1.

⇒ Bhavya age (B) = 35 years; Aarav age (A) = 42 years.

⇒ 2y − 4 = 42 ⇒ 2y = 46 ⇒ y = 23.

⇒ Case I: y + 12 = 23 + 12 = 35 (Matches B).

⇒ Case II: 2y − 3 = 46 − 3 = 43 (Incorrect).

⇒ Case III: \({(23 + 1)/2 + 22} \) = 12 + 22 = 34 (Incorrect).

∴ The correct answer is 35 years.

Additional Information

Ratio and Proportions in Ages

If the ratio of ages of two people is a:b, their actual ages can be taken as ax and bx. The ratio changes when years are added or subtracted from both.

Average Age Formula

Average age = (Sum of ages of all persons) ÷ (Number of persons). Sum = Average × Number of persons.

Age Relation after 'n' years

If present age is 'x', age after 'n' years is (x + n) and age 'n' years ago was (x − n).

Shortcut Trick

Interest for 2 years is ₹2880 → 1 year Interest = ₹1440.

At 12%, Investment = 1440 ÷ 0.12 = ₹12000.

Total Expenses + Savings = 30000 − 12000 = ₹18000.

Remaining income after rent and tools = 30000 − 7500 − 300(x−5) = 24000 − 300x.

Through substitution from options, if Online Ads = ₹3600, then x = 20 satisfies the total balance equation.

∴ The correct answer is Rs. 3600.

Alternate Method

Given: Total Income = ₹30000, Studio Rent = 25%, Tools = (x − 5)%, Savings = ₹2400, Interest = ₹2880 (2 years at 12% PA).

⇒ Principal (Investment) = (Interest × 100) ÷ (Rate × Time)

⇒ Principal = (2880 × 100) ÷ (12 × 2) = 288000 ÷ 24 = ₹12000.

⇒ Rent = 25% of 30000 = ₹7500.

⇒ Software/Tools = (x − 5)% of 30000 = 300(x − 5) = 300x − 1500.

⇒ Remaining Income = 30000 − 7500 − (300x − 1500) = 24000 − 300x.

⇒ Online Ads = x% of (24000 − 300x) = (x/100) × 300(80 − x) = 3x(80 − x).

⇒ Total Income = Rent + Tools + Ads + Investment + Savings

⇒ 30000 = 7500 + (300x − 1500) + (240x − 3x²) + 12000 + 2400

⇒ 30000 = 20400 + 540x − 3x² ⇒ 3x² − 540x + 9600 = 0

⇒ x² − 180x + 3200 = 0 ⇒ (x − 20)(x − 160) = 0

⇒ x = 20 (as x cannot be 160%).

⇒ Online Ads = 3(20)(80 − 20) = 60 × 60 = ₹3600.

∴ The correct answer is Rs. 3600.

Additional Information

Simple Interest Formula

SI = (P × R × T) ÷ 100, where P is Principal, R is Rate of Interest, and T is Time.

Percentage of Remaining

When a percentage is calculated on remaining income, it must be applied after subtracting all previous expenditures from the total base.

Quadratic Roots

For x² − (Sum of roots)x + (Product of roots) = 0, the values of x must be logically consistent with the context (e.g., total % ≤ 100).