SBI Clerk Mains Quantitative Aptitude (Day-45)

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Dear Aspirants, Our IBPS Guide team is providing new series of Quantitative Aptitude Questions for SBI Clerk Mains 2020 so the aspirants can practice it on a daily basis. These questions are framed by our skilled experts after understanding your needs thoroughly. Aspirants can practice these new series questions daily to familiarize with the exact exam pattern and make your preparation effective.

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Directions (01 – 05): Each question contains a statement followed by Quantity I, II and III. Read the information clearly and answer your questions accordingly.

A) >

B) <

C)=

D) 

E) 

For example:

Quantity I = 200

Quantity II = 300

Quantity III = 100

Hence, Quantity I < Quantity II > Quantity III

1)

Quantity I: Ratio of the radius of the sphere to cone is 1:2. If the total surface area of the sphere is 154 cm2 and the total surface area of the cone is 704 cm2, then what is the height of cone?

Quantity II: Ratio of the side of the cube to length of the cuboid is 2:3. If the surface area of the cube is 384 cm2 and the height of the cuboid is 16 cm, diagonal of the cuboid is 25 cm, then find the base of the cuboid?

Quantity III: If the area of the rhombus is 384 cm2 and one of the diagonal is 24 cm, then what is the side of the rhombus?

A) A, B

B) B, C

C) B, A

D) E, B

E) B, D

2)

Quantity I: Rahul invests Rs.x in a scheme in simple interest at 15% per annum for 6 years. If he earns the interest is Rs.2880 and he invests the same amount in a compound interest scheme at the rate of 15% per annum for 2 years, then find the compound amount received by Rahul after 2 years?

Quantity II: Rahul invests Rs.x in scheme A which offer simple interest at the rate of 12% per annum for 5 years and also he invests Rs.2500 in scheme B which offer simple interest at the rate of 10% per annum for 6 years. If he gets the interest by scheme A is Rs.900 more than that of the interest gets by scheme B, and then find the value of x?

Quantity III: Rs.3999

A) A, D

B) B, C

C) A, A

D) B, B

E) E, A

3)

Quantity I: Average ages of A, B and C is 36 years. Ratio of the ages of A to B is 2: 3 and the ratio of the ages of C to D after 5 years is 2: 1. C is 19 years elder than D. What is A’s age after 5 years?

Quantity II: Product of the ages of A and B is 1176 years. The ratio of the ages of A to B is 2:3 and C is 5 years elder than A. What is C’s present age?

Quantity III: Sum of the ages of A and his father is 60 years and A’s mother age is 25 when he is born. A’s father is 5 years older than his mother. What is A’s mother’s present age?

A) A, B

B) B, C

C) B, A

D) E, B

E) B, D

4)

Quantity I: If the marked price of the article is 25% more than that of the cost price of the article and the shopkeeper offers a discount of 15% on marked price of the article. If the cost price of price of the article is Rs.Rs.2400, then what is the profit percentage of the article?

Quantity II: If the cost price of the laptop is Rs.960 and the shopkeeper offer a discount of x% while he earned the profit of 25%. If the marked price of the laptop is Rs.540 more than that of cost price of the laptop, then find the value of x?

Quantity III: The profit earned by selling article for Rs.1200 is equal to the loss incurred when the same article sold for Rs.800. If marked price of the article is Rs.500 more than that of the cost price of the article, then in how much percentage the marked price more than that of cost price of the article?

A) A, D

B) B, C

C) A, A

D) B, B

E) E, A

5)

Quantity I: A can complete half of the work in 9 days and A and B together can complete the 75% of the work in 9 days. C is 20% more efficiency than B. In how many days A, B and C together can complete the half of the work?

Quantity II: A is 20% less efficiency than B and B is 50% more efficiency than C. If C alone completed the work in 24 days, in how many days A, B and C together can complete half of the work?

Quantity III: 3(1/3) days

A) A, B

B) B, C

C) B, A

D) E, B

E) B, D

Caselet

Directions (06 – 10): Study the following information carefully and answer the questions given below.

In four boxes A, B, C and D contains three different colours balls – red, yellow and blue.

A: The number of yellow balls is equal to the total number of yellow balls from box B and C together. The total number of red balls from A is 30 more than that of the blue balls from A.

B: The number of red balls is 20 more than that of number of blue balls. The number of blue balls is 10 less than that of number of yellow balls. Total number of balls from box B is 150.

C: The ratio of the total number of balls from C to B is 8:5 and ratio of the total number of balls from C to A is 2:3. The ratio of the number of red balls from C to the number of red balls from B is 3:2. Ratio of the number of blue balls from C to number of yellow balls from C is 7:8.

D: Ratio of the number of red balls to yellow balls is 3:7 and the number of yellow balls is 40% more than that of the number of blue balls. Total number of balls from all the boxes together is 1050.

6) The number of total number of balls from C is what percent of the total number of balls from D?

A) 60%

B) 70%

C) 80%

D) 90%

E) None of these

7) What is the total number of blue balls from all the boxes together?

A) 290

B) 380

C) 370

D) 310

E) None of these

8) Total number of red balls from all the boxes together is what percent of the total number of yellow balls from all the boxes together?

A) 80%

B) 70%

C) 85%

D) 75%

E) None of these

9) What is the difference between the numbers of yellow balls from D and C?

A) 40

B) 60

C) 70

D) 50

E) None of these

10) Number of Blue balls from A is approximately what percent more or less than that of the number of blue balls from C?

A) 41%

B) 43%

C) 45%

D) 47%

E) 49%

Answers :

Directions (1-5) :

1) Answer: A

From quantity I,

TSA of sphere = 4 * 22/7 * r * r

154 = 4 * 22/7 * r * r

Radius of the sphere = 3.5 cm

Radius of the cone = 2 * 3.5 = 7 cm

TSA of cone = 22/7 * r * (l + r)

704 = 22/7 * 7 * (l + 7)

32 = l + 7

L = 25 cm

Height of the cone = √(l2 – r2)

= √(252 – 72)

= 24 cm

From quantity II,

SA of cube = 6a2 = 384

a = 8 cm

Length of the cuboid = 3/2 * 8 = 12 cm

Diagonal of cuboid = √(l2 + b2 + h2)

25 = √162 + 122 + b2

625 = 256 + 144 + b2

B2 = 225

B = 15 cm

From quantity III,

Area of the rhombus = ½ * (d1 * d2)

384 = ½ * 24 * d2

d2= 32 cm

Side of the rhombus = √[(d1/2)2 + (d2/2)2]

= √(122 + 162)

= 20 cm

24 > 15 < 20

2) Answer: C

From quantity I,

SI = P * N * R/100

x * 15 * 6 /100 = 2880

x = 3200

CA = P * (1 + R/100)2

CA = 3200 * (1 + 15/100)2

CA = Rs.4232

From quantity II,

x * 12 * 5/100 – 2500 * 10 * 6/100 = 900

60x = 90000 + 150000

x = 4000

From quantity III,

Ra.3999

4232 > 4000 > 3999

3) Answer: A

From quantity I,

A + B + C = 36 * 3 = 108

2x – x = 19

x = 19

C’s age after 5 years = 2 * 19 = 38

C’s present age = 38 – 5 = 33

A + B = 108 – 33 = 75

5x = 75

x = 15 years

A’s age = 15 * 2 = 30

A’s age after 5 years = 30 + 5 = 35 years

From quantity II,

A * B = 1176

2x * 3x = 1176

x = 14

A’s age = 14 * 2 = 28

C’s age = 28 + 5 = 33 years

From quantity III,

A + F = 60 ——–(1)

A + 25 = M

M – A = 25 ——–(2)

F – M = 5 ———(3)

(2) + (3)

F – A = 30 ———(4)

(1) + (4)

2F = 90

F = 45

M = 45 – 5 = 40 years

35 > 33 < 40

4) Answer: D

From quantity I,

CP = 2400

MP= 2400 * 125/100 = 3000

3000 * 85/100 = 2400 * (100 + x)/100

106.25 = 100 + x

x = 6.25%

From quantity II,

CP = 960

MP = 540 + 960 = 1500

1500 * (100 – x)/100 = 960 * 125/100

100 – x = 80

x = 20%

From quantity III,

SP – CP = CP – SP

1200 – CP = CP-800

2CP = 2000

CP = 1000

MP = 1000 + 500 = 1500

Required percentage = (1500 – 1000)/1000 * 100 = 50%

6.25 < 20 < 50

5) Answer: A

From quantity I,

A = 1/18

A + B = 4/3 * 9 = 12 days

B = 1/12 – 1/18

B = 1/36

C = 36 * 5/6 = 30

A + B + C = 1/18 + 1/36 + 1/30

= 10 + 5 + 6/180

= 7/60

Half of the work completed in = 60/7 * ½ = 30/7 days

From quantity II,

C = 1/24

B = 24 * 100/150 = 16

A = 5/4 * 16 = 20

A + B + C=1/24 + 1/16 + 1/20

=10 + 15 + 12/240

=37/240

Half of the work complete in = 240/37 * ½ =120/37

From quantity III,

3(1/3) days

30/7 > 120/37 < 3(1/3)

Directions (06 – 10):

Total number of balls = 150

Number of red balls = x

Number of blue balls = x – 20

Number of yellow balls = x – 20 + 10

x + x – 20 + x – 20 + 10 = 150

3x = 180

x = 60

Number of red balls = 60

Number of blue balls = 60 – 20 = 40

Number of yellow balls = 60 – 20 + 10 = 50

Total number of balls from C = 8/5 * 150 = 240

Total number of balls from A = 3/2 * 240 = 360

Number of red balls from C = 3/2 * 60 = 90

Number of blue and yellow balls from C = 240 – 90 = 150

Number of blue balls from C= 7/15 * 150 = 70

Number of yellow balls from C =8/15 * 150 = 80

Number of yellow balls from A = 80 + 50 = 130

Number of red and blue balls from A = 360 – 130 = 230

Number of red balls from A = 100 + 30 = 130

Number of blue balls from A = 100

Total number of balls = 1050

Total Number of balls from D = 1050 – 240 – 360 – 150 = 300

Number of red balls from D = 3y

Number of yellow balls from D = 7y

Number of blue balls from D = 7y * 100/140 = 5y

3y + 7y + 5y = 300

15y = 300

y = 20

Number of red balls from D = 3 * 20 = 60

Number of yellow balls from D = 7 * 20 = 140

Number of blue balls from D = 5 * 20 = 100

6) Answer: C

Required percentage = 240/300 * 100 = 80%

7) Answer: D

Required total = 100 + 40 + 70 + 100 = 310

8) Answer: C

Required percentage = [(130 + 60 + 90 + 60)/(130 + 50 + 80 + 140)] * 100

= 85%

9) Answer: B

Difference = 140 – 80 = 60

10) Answer: B

Required percentage = [(100 – 70)/70] * 100 = 43%

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