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Find the total surface area of a right circular cylinder whose volume is 4928 cubic cm and surface area is 704 square cm.

Correct

Explanation:
ᴨ r^2 h = 4928, 2ᴨrh = 704
Divide both equations, r = 14
TSA of cylinder = surface area + 2ᴨ r^2 = 704 + 2ᴨ(14^2)

Incorrect

Explanation:
ᴨ r^2 h = 4928, 2ᴨrh = 704
Divide both equations, r = 14
TSA of cylinder = surface area + 2ᴨ r^2 = 704 + 2ᴨ(14^2)

Question 7 of 10

7. Question

1 points

Category: Quantitative Aptitude

A can do thrice as much work as B can complete in a day. C can complete half as much work as A and B do together can complete in a day. If C alone can complete a work in 33 days, then in how many days A alone would complete the work?

Correct

Explanation:
A does half as much work as B does in a day. So if A does work in 3x days, then B in x days.
So A+B completes 1/x + 1/3x = 4/3x work in a day, or completes a work in 3x/4 days.
Now C does half as much work as A and B do together in a day. So if A+B in 3x/4 days, then C in 2 * 3x/4 = 3x/2 days
So 3x/2 = 33, solve, x = 22 days, so A in 3*22 days

Incorrect

Explanation:
A does half as much work as B does in a day. So if A does work in 3x days, then B in x days.
So A+B completes 1/x + 1/3x = 4/3x work in a day, or completes a work in 3x/4 days.
Now C does half as much work as A and B do together in a day. So if A+B in 3x/4 days, then C in 2 * 3x/4 = 3x/2 days
So 3x/2 = 33, solve, x = 22 days, so A in 3*22 days

Question 8 of 10

8. Question

1 points

Category: Quantitative Aptitude

A box contains 2 blue, 3 green and 5 red balls. If three balls are drawn at random, what is the probability that exactly 2 balls are same in color?

Correct

Explanation:
Total balls = 10
Case1: 2 blue, 1 green or red
So prob. = ^{2}C_{2}×^{8}C_{1/10C}_{3} = 1/15
Case2: 2 green, 1 blue or red
So prob. = ^{3}C_{2}×^{7}C_{1/10C}_{3} = 21/120
Case3: 2 red, 1 green or blue
So prob. = ^{5}C_{2}×^{5}C_{1/10C}_{3} = 5/12
Add all cases

Incorrect

Explanation:
Total balls = 10
Case1: 2 blue, 1 green or red
So prob. = ^{2}C_{2}×^{8}C_{1/10C}_{3} = 1/15
Case2: 2 green, 1 blue or red
So prob. = ^{3}C_{2}×^{7}C_{1/10C}_{3} = 21/120
Case3: 2 red, 1 green or blue
So prob. = ^{5}C_{2}×^{5}C_{1/10C}_{3} = 5/12
Add all cases

Question 9 of 10

9. Question

1 points

Category: Quantitative Aptitude

Two alloys contain copper and zinc in the ratio of 2 : 3 and 3 : 5 respectively. 25 kg of first alloy, 40 kg of second alloy and some quantity of pure copper are melted together. If in final alloy the ratio of copper and zinc is 11 : 8, find the amount of pure zinc that is melted.

Correct

Explanation:
In 1st alloy copper = (2/5)*25 = 10kg and zinc = 25-10 = 15
in 2nd alloy copper = (3/8)*40 = 15kg and zinc = 40-15 = 25
So, (10+15+x)/(15+25) = 11/8
Solve for x, x = 30

Incorrect

Explanation:
In 1st alloy copper = (2/5)*25 = 10kg and zinc = 25-10 = 15
in 2nd alloy copper = (3/8)*40 = 15kg and zinc = 40-15 = 25
So, (10+15+x)/(15+25) = 11/8
Solve for x, x = 30

Question 10 of 10

10. Question

1 points

Category: Quantitative Aptitude

The ratio of the ages of A and B is 8 : 11. C’s present age is 3 less than B’s age plus half of A’s age. If the ratio of C’s present age and B’s age 3 years hence is 7 : 6, find the total of the ages of all of three.

Correct

Explanation:
A/B = 8/11
8x and 11x
So C’s present age = (11x + 8x/2) – 3 = 15x – 3
So (15x-3)/(11x+3) = 7/6
Solve, x = 3
So total = 8x+11x+15x-3 = 34x – 3

Incorrect

Explanation:
A/B = 8/11
8x and 11x
So C’s present age = (11x + 8x/2) – 3 = 15x – 3
So (15x-3)/(11x+3) = 7/6
Solve, x = 3
So total = 8x+11x+15x-3 = 34x – 3