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Quantitative Aptitude DI

Here we are providing new series of Quantitative Aptitude Questions for upcoming exams, so the aspirants can practice it on a daily basis.

Study the following data carefully and answer the questions:

There are total five cuboid A, B, C, D and E. Stacked bar graph given below shows the ratio of length, width and height respectively of those cuboids.

Example: In stacked bar graph, length, width and height of cuboid A is showing as 8, 4 and 3 respectively.So, the ratio of length, width and height of cuboid A is 8: 4: 3 respectively.

Pie chart given below shows the distribution (in degree) of total surface area of those six cuboids A, B, C, D and E.

Sum of total surface area of all the six cuboids together = 24000 cm2

1) What is the difference between volume of a cylinder whose base radius and height is equal to the width and height respectively of cuboid A and volume of a cone whose base radius and height is equal to the length and height respectively of cuboid A?

a) 1750π cm3

b) 1500π cm3

c)2000π cm3

d) 1000π cm3

e) None of these

2) If cuboids C and E are solid, these two cuboids are melted and 88% of total volume is casted into a hemisphere, then what is the total surface area of that hemisphere? (Use 3√21 = 2.8 and ∏ = 22/7)

a) 4928 cm2

b) 7392 cm2

c) 9856 cm2

d) 6570 cm2

e) None of these

3) If the cuboid E is filled with water up to height of 16.8 cm and a solid cylinder is put inside this cuboid in such a way that it is completely immersed in to water. Base radius and height of cylinder is 7 cm and 25 cm respectively, then what is the height of water in the cuboid E after the cylinder is immersed?

a) 18 cm

b) 17.2 cm

c) 19 cm

d) 18.5 cm

e) None of these

4) There are ‘m’ cuboidal boxes having dimension 5 cm * 4 cm * 2 cm that can be kept inside cuboid B and ‘n’ cuboidal boxes having dimension 6 cm * 5 cm * 1 cm that can be kept inside cuboid C respectively, then what is the maximum possible value of sum of ‘m’ and ‘n’?

a) 900

b) 1200

c) 800

d) 750

e) None of these

5) If cuboids D and E solid, cuboid D is melted and 15.4% of its total volume is used to cast ‘m’ number of cones of base radius 14 cm and height 3 cm. Cuboid E is melted and 83.16% of its total volume is used to cast ‘n’ number of cylinders of base radius 7 cm and height 21 cm, then what is the ratio of ‘m’ to ‘n’?

a) 4: 5

b) 6: 7

c) 3: 4

d) 2: 3

e) None of these

Answers :

1) Answer: C

Ratio of length, width and height respectively of cuboid A = 8: 4: 3

Total surface area of cuboid A = 2 * (8a * 4a + 4a * 3a + 8a * 3a) = 24000 * (51/360) = 3400 cm2

136a2 = 3400

a = 5

Ratio of length, width and height respectively of cuboid B = 4: 4: 1

Total surface area of cuboid B = 2 * (4b * 4b + 4b * b + 4b * b) = 24000 * (72/360) = 4800 cm2

48b2 = 4800

b = 10

Ratio of length, width and height respectively of cuboid C = 6: 5: 4

Total surface area of cuboid C = 2 * (6c * 5c + 5c * 4c + 6c * 4c) = 24000 * (55.5/360) = 3700 cm2

148c2 = 3700

c = 5

Ratio of length, width and height respectively of cuboid D = 4: 3: 2

Total surface area of cuboid D = 2 * (4d * 3d + 3d * 2d + 4d * 2d) = 24000 * (78/360) = 5200 cm2

52d2 = 5200

d = 10

Ratio of length, width and height respectively of cuboid E = 10: 7: 4

Total surface area of cuboid E = 2 * (10e * 7e + 7e * 4e + 10e * 4e) = 24000 * (103.5/360) = 6900 cm2

276e2 = 6900

e = 5

Base radius and height and height of cylinder is 20 cm and 15 cm respectively.

Base radius and height and height of cone is 40 cm and 15 cm respectively.

Volume of cylinder = π * (20)2 * 15 = 6000π cm3

Volume of cone = (π/3) * (40)2 * 15 = 8000π cm3

Required difference = 8000π – 6000π = 2000π cm3

2) Answer: B

Total volume of cuboids C and E together = (30 * 25 * 20) + (50 * 35 * 20) = 15000 + 35000 = 50000 cm3

Volume of hemisphere = (2/3) * (22/7) * R3 = 88% of 50000 = 44000

R3 = (44000 * 7 * 3)/(2 * 22)

R3 = 1000 * 21

R = 10 * 2.8

R = 28 cm

Total surface area of that hemisphere = 3 * ∏ * R2 = 3 * (22/7) * 282 = 7392 cm2

3) Answer: C

Volume of cylinder = ∏ * 72 * 25 = (22/7) * 49 * 25 = 3850 cm3

Base area of cuboid E = Length * Width = 50 * 35 = 1750 cm2

Increment in the height of water in cuboid E = 3850/1750 = 2.2 cm

Height of water in the cuboid E after the cylinder is immersed = 16.8 + 2.2 = 19 cm

4) Answer: A

Maximum number of ‘m’ cuboids that can be kept inside cuboid B = Volume of cuboid B/Volume of each small cuboid = m = (40 * 40 * 10)/(5 * 4 * 2) = 8 * 10 * 5 = 400

Maximum number of ‘n’ cuboids that can be kept inside cuboid C = Volume of cuboid C/Volume of each small cuboid = n = (30 * 25 * 20)/(6 * 5 * 1) = 5 * 5 * 20 = 500

Maximum possible value of sum of ‘m’ and ‘n’ = m + n = 400 + 500 = 900

5) Answer: D

Volume of cuboid D = (40 * 30 * 20) = 24000 cm3

Volume of cuboid E = (50 * 35 * 20) = 35000 cm3

Volume used to cast ‘m’ number of cones = 15.4% of 24000 = 3696 cm3

m * (1/3) * (22/7) * 142 * 3 = 3696

m = (3696 * 7 * 3)/(22 * 142 * 3)

m = 6

Volume used to cast ‘n’ number of cylinders = 83.16% of 35000 = 29106 cm3

n * (22/7) * 72 * 21 = 29106

n = (29106 * 7)/(22 * 72 * 21)

n = 9

Required ratio = m: n = 6: 9 = 2: 3