Quantitative Aptitude PYQs

Let the total number of students in IIM A be 100a and in IIM B be 100b.

For IIM A:

Number of students in Engineering = 60a

Number of students in Commerce = 18a

Number of students in Arts = 12a

Number of students in Others = 10a

For IIM B:

Number of students in Engineering = 55b

Number of students in Commerce = 20b

Number of students in Arts = 15b

Number of students in Others = 10b

Let total intake in IIM A = 100a and in IIM B = 100b.

Given, total students in IIM A = 600

100a = 600

So, a = 6

ATQ, Engineering students in IIM A are 80 less than the total intake of students in IIM B.

60a = 100b – 80

Substitute (a = 6):

60 x 6 = 100b – 80

360 = 100b – 80

100b = 440

So, b = 4.4

Required sum of Commerce students = 18a + 20b

= 18 x 6 + 20 x 4.4

= 108 + 88 = 196

Let the total number of students in IIM A be 100a and in IIM B be 100b.

For IIM A:

Number of students in Engineering = 60a

Number of students in Commerce = 18a

Number of students in Arts = 12a

Number of students in Others = 10a

For IIM B:

Number of students in Engineering = 55b

Number of students in Commerce = 20b

Number of students in Arts = 15b

Number of students in Others = 10b

ATQ, {60a - 55b}/{18a – 20b} = {9/1}

Or, 60a – 55b = 9(18a – 20b)

Or, 60a – 55b = 162a – 180b

Or, 125b = 102a

Or, {a/b} = {125/102}

Ratio of total intake =100a:100b = a: b =125:102

Copper in X kg of alloy P = 5X/8 kg

Zinc in X kg of alloy P = 3X/8 kg

Copper in 9 kg of alloy Q = 9 × (7/9) = 7 kg

Zinc in 9 kg of alloy Q = 9 × (2/9) = 2 kg

Copper in 15 kg of alloy R = 15 × (4/5) = 12 kg

Zinc in 15 kg of alloy R = 15 × (1/5) = 3 kg

Total copper in final mixture = (5X/8) + 7 + 12 = (5X/8) + 19

Total zinc in final mixture = (3X/8) + 2 + 3 = (3X/8) + 5

ATQ, [(5X/8) + 19] ÷ [(3X/8) + 5] = 37/14

Or, (70X/8) + 266 = (111X/8) + 185

Or, 81 = 41X/8

Or, X = 648/41 ≈ 15.8 kg

So, X ≈ 15.8 kg

Side of equilateral triangular field = (24/3) = 8 m

Side of square field = 8 × (5/3) = 40/3 m

Area of triangular field = (√3/4) × 8² = (1.732/4) × 64 = 27.712 sq.m

Area of square field = (40/3)² = 1600/9 = 177.778 sq.m

Required difference = 177.778 − 27.712 = 150.066 ≈ 150 sq.m

Let total number of students = 100x

Participated in all clubs = 10x

Participated in none = 10x

Remaining = 100x − 20x = 80x

Participated in only 1 club = 25% of 80x = 20x

Participated in 4 clubs = 20% of 80x = 16x

Remaining students (participated in 2 clubs) = 80x − (20x + 16x) = 44x

ATQ, 44x = 264

x = (264/44) = 6

Total number of students = 100 × 6 = 600

Common solution:

Let the number of engineers in city A, B and C be 11x, 8x and 5x respectively.

Number of doctors in city A = 8x + 120

Number of doctors in city B = 8x + 5x = 13x

Number of doctors in city C = 5x + M

Given,

Number of doctors in city B = 2600

13x = 2600

So, x = 200

So, Engineers in A = 2200

Engineers in B = 1600

Engineers in C = 1000

Number of doctors in A = 8 × 200 + 120 = 1720

Given, Doctors in A = 5(M − 60)

1720 = 5(M – 60)

344 = M – 60

M = 404

Therefore, Doctors in C = 1000 + 404 = 1404

 

Calculations:

Total doctors and engineers in city A = 1720 + 2200 = 3920

Common solution:

Let the number of engineers in city A, B and C be 11x, 8x and 5x respectively.

Number of doctors in city A = 8x + 120

Number of doctors in city B = 8x + 5x = 13x

Number of doctors in city C = 5x + M

Given,

Number of doctors in city B = 2600

13x = 2600

So, x = 200

So, Engineers in A = 2200

Engineers in B = 1600

Engineers in C = 1000

Number of doctors in A = 8 × 200 + 120 = 1720

Given, Doctors in A = 5(M − 60)

1720 = 5(M – 60)

344 = M – 60

M = 404

Therefore, Doctors in C = 1000 + 404 = 1404

 

Calculations: 

Required ratio = 2600:2200 = 13:11

Common solution:

Let the number of engineers in city A, B and C be 11x, 8x and 5x respectively.

Number of doctors in city A = 8x + 120

Number of doctors in city B = 8x + 5x = 13x

Number of doctors in city C = 5x + M

Given,

Number of doctors in city B = 2600

13x = 2600

So, x = 200

So, Engineers in A = 2200

Engineers in B = 1600

Engineers in C = 1000

Number of doctors in A = 8 × 200 + 120 = 1720

Given, Doctors in A = 5(M − 60)

1720 = 5(M – 60)

344 = M – 60

M = 404

Therefore, Doctors in C = 1000 + 404 = 1404

 

Calculations: 

Required percentage more = (1720 – 1000)/1000 x 100 = 72%

Common solution:

Let the number of engineers in city A, B and C be 11x, 8x and 5x respectively.

Number of doctors in city A = 8x + 120

Number of doctors in city B = 8x + 5x = 13x

Number of doctors in city C = 5x + M

Given,

Number of doctors in city B = 2600

13x = 2600

So, x = 200

So, Engineers in A = 2200

Engineers in B = 1600

Engineers in C = 1000

Number of doctors in A = 8 × 200 + 120 = 1720

Given, Doctors in A = 5(M − 60)

1720 = 5(M – 60)

344 = M – 60

M = 404

Therefore, Doctors in C = 1000 + 404 = 1404

 

Calculations: 

Doctors in city S = 1404 x 125% = 1755

Engineers in city S = 1000 – (404 + 50) = 546

Total doctors and engineers in city S = 1755 + 546 = 2301

Common solution:

Let the number of engineers in city A, B and C be 11x, 8x and 5x respectively.

Number of doctors in city A = 8x + 120

Number of doctors in city B = 8x + 5x = 13x

Number of doctors in city C = 5x + M

Given,

Number of doctors in city B = 2600

13x = 2600

So, x = 200

So, Engineers in A = 2200

Engineers in B = 1600

Engineers in C = 1000

Number of doctors in A = 8 × 200 + 120 = 1720

Given, Doctors in A = 5(M − 60)

1720 = 5(M – 60)

344 = M – 60

M = 404

Therefore, Doctors in C = 1000 + 404 = 1404

 

Calculations: 

Doctors in city Q and R together = 2600 + 1404 = 4004

Required difference = 4004 – 1600 = 2404

Amount (A) = Rs. 10580

Rate = 15% p.a.

Time = 2 years

We know,

A = P(1 + R/100)²

10580 = P(115/100)²

10580 = P x 529/400

 P = 20 x 400 = Rs. 8000

168 - 26 = 142

142 - 24 = 118

118 - 22 = 96

96 - 20 = 76

76 - 18 = 58

Hence, option D.

20 x 1 - 2 = 18

18 x 2 - 4 = 32

32 x 3 - 6 = 90

90 x 4 - 8 = 352

352 x 5 - 10 = 1750

Hence, option B.