# SBI Clerk Prelims 2020 Quantitative Aptitude Questions (Day-17)

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Inequality

Directions (Q. 1 – 5): Following questions have two quantities as Quantity I and Quantity II. You have to determine the relationship between them and give answer as,

a) Quantity I > Quantity II

b) Quantity I ≥ Quantity II

c) Quantity I < Quantity II

d) Quantity I ≤ Quantity II

e) Quantity I = Quantity II

1) The average weight of A, B and C is 36 kg and the average weight of A and B is 39 kg. A’s weight is 15 kg more than C.

Quantity I: Average weight of A and C?

Quantity II: Weight of B?

2) Quantity I: The respective ratio of the present ages of Komal and Harish is 5: 6. After twelve years, respective ratio of their ages will be 13: 15. Present age of Neeta is 15% more than the present age of Komal. Find the average of the present ages of Harish and Neeta?

Quantity II: 47

3) Quantity I: Perimeter of the rectangle is 108 cm and the ratio of the length to breadth of the rectangle is 5: 4. What is area of the rectangle?

Quantity II: Length of the rectangle is 8 cm more than the breadth of the rectangle and the perimeter of the rectangle is 20 cm less the circumference of the circle whose radius is 14 cm. What is the area of the rectangle?

4) An urn contains stones of five different colours viz. 4 green, some yellow, 6 red, 7 blue and some black stones. Probability of drawing one green stone from the urn is 1/6 and probability of drawing one black stone from the urn is 1/12.

Quantity I: Find the probability of drawing 1 green, 1 yellow and 1 black stone from the urn.

Quantity II: Find the probability of drawing 1 red and 1 blue stone from the urn.

5) A, B and C together started the business with the investments of Rs. 6000, Rs. 7000 and Rs. 12000 respectively. The ratio of the investment period of A, B and C is 5 : 6 : 4, and the total profit is Rs. 20000?

Quantity I: B’s profit share?

Quantity II: Average profit share of A and C?

Caselet

Directions (6 – 10): Study the following information carefully and answer the questions given below.

A total of x number of students appeared in some competitive exams like CAT, GMAT and NET. The following diagram shows the number of students who passed these three exams. None of the students failed in all the exams together. 6) How many students passed in GMAT and NET but failed in CAT?

a) x/2

b) x/4

c) x/6

d) x/8

e) None of these

7) What is the difference between the number of students who passed in GMAT and the number of students who passed in CAT? (Note: The number of students who passed in all the three exams is 5).

a) 35

b) 40

c) 45

d) 50

e) None of these

8) How many students passed in exactly one exam, if the number of students who passed in exactly two exams is 35?

a) 20

b) 80

c) 40

d) 60

e) None of these

9) The number of students who passed in at least two exams is what percent of the total number of students?

a) 33.33 %

b) 12.24 %

c) 38.88 %

d) 42.24 %

e) None of these

10) The number of students who passed only NET exam is what percentage more than the number of students who passed in all the three exams?

a) 100 %

b) 300 %

c) 200 %

d) 500 %

e) None of these

Directions (1-5) :

A + B + C = 36 * 3 = 108

A + B = 78

C = 108 – 78 = 30 kg

A = 30 + 15 = 45 kg

From quantity I,

Average weight of A and C = (30 + 45)/2 = 37.5

From quantity II,

B = 78 – 45 = 33 kg

Quantity I > Quantity II

Quantity I:

Let, present ages of Komal and Harish be 5k years and 6k years respectively.

(5k + 12)/(6k + 12) = 13/15

=> 15 x (5k + 12) = 13 x (6k + 12)

=> 75k + 180 = 78k + 156

=> 78k – 75k = 180 – 156

=> 3k = 24

=> k = 24/3

=> k = 8

Present age of Komal = 5 x 8 = 40 years

Present age of Harish = 6 x 8 = 48 years

Present age of Neeta = 40 x 115/100 = 46 years

Required average = (48 + 46)/2 = 94/2 = 47 years

Quantity II:

47

Quantity I = Quantity II

From quantity I,

Perimeter of the rectangle = 2 * (l + b)

108 = 2 * (9x)

x = 6 cm

Area of the rectangle = 30 * 24 = 720 cm2

From quantity II,

Circumference of the circle = 2 * 22/7 * 14 = 88 cm

Perimeter of the rectangle = 88 – 20 = 68 cm

2 * (b + 8 + b) = 68

2b = 26

b = 13 cm

Length of the rectangle = 13 + 8 = 21 cm

Area of the rectangle = 13 * 21 = 273 cm2

Quantity I > Quantity II

Green = 4

Red = 6

Blue = 7

Let, number of yellow stones = a

And number of black stones = b

Total = 4 + 6 + 7 + a + b = 17 + a + b

4/(17 + a + b) = 1/6

=> 24 = 17 + a + b

=> a + b = 24 – 17

=> a + b = 7 —— (i)

b/(17 + a + b) = 1/12

=> 12b = 17 + a + b

=> 12b – b – a = 17

=> 11b – a = 17 ——- (ii)

a + b + 11b – a = 7 + 17

=> 12b = 24

=> b = 24/12

=> b = 2

From (i)

a + 2 = 7

=> a = 7 – 2

=> a = 5

Total = 17 + 5 + 2 = 24

Quantity I:

Required probability = (4c1 x 5c1 x 2c1)/24c3

= (4 x 5 x 2)/2024

= 5/253

Quantity II:

Required probability = (6c1 x 7c1)/24c2

= 7/46

Quantity I < Quantity II

Profit ratio = (6000 * 5): (7000 * 6): (12000 * 4)

= 5: 7: 8

From quantity I,

B’s profit share = 7/20 * 20000 = Rs. 7000

From quantity II,

Average of A and C = (13/20 * 20000)/2 = Rs. 6500

Quantity I > Quantity II

Directions (6-10) :

x/12 + x/12 + x/3 + x/12 + x/24 + x/4 + A = x

=> 3 * x/12 + x/3 + x/4 + x/24 + A = x

=> A = x/8 = Number of students passed in GMAT and NET but failed in CAT.

Number of students passed in GMAT

= x/12 + x/3 + x/24 + x/8 = 14x/24 = 7x/12

Number of students passed in CAT = x/12 + x/12 + x/12 + x/24 = 7x/24

Required difference = 7x/12 – 7x/24 = 7x/24

Given, No. of students passed in all three exams = 5

Hence, x/24 = 5 => x = 120

Required difference = 7x/24 = 7 * 120/24 = 35

Number of students who passed in exactly two exams

x/12 + x/12 + x/8 = 7x/24 = 35

x = 120

Number of students who passed in exactly one exam

= x/12 + x/3 + x/4 = 2x/3

= 2 * 120/3 = 80

Number of students who passed in at least two exams

= x/12 + x/12 + x/8 + x/24

= 8x/24 = x/3

Required percentage = [(x/3)/x] * 100 = 33.33 %

Number of students passed only in NET = x/4

Number of students passed in all three exams = x/24

Required percentage

= {[(x/4) – (x/24)] / (x/24)} * 100

= 500 %

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