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Question 1 of 10
1. Question
1 pointsCategory: Quantitative AptitudeDirections for Questions 15: Study the following charts and answer the questions that follow:
Arun and Bimal started doing the work. After 2 days they both left, and Chandra joined the work. He completed his part of the work. Now the remaining work completed by Farhan in 7 days. In how many days can Farhan complete whole work?
Correct
Solution: Arun does 20% work in 4 days. So 100% work in 100*4/20= 20 days, Similarly Bimal can complete 100% in 100*3/10= 30 days. They worked for two days, so did (1/20+ 1/30)*2 = 1/6 work, Now Chanra completed 25% = ¼ work. Therefore, remaining work = 1 (1/6 + ¼) = 7/12, Therefore Farhan completed 7/12 work in 7 days, so will complete the whole work in 12 days.
Incorrect
Solution: Arun does 20% work in 4 days. So 100% work in 100*4/20= 20 days, Similarly Bimal can complete 100% in 100*3/10= 30 days. They worked for two days, so did (1/20+ 1/30)*2 = 1/6 work, Now Chanra completed 25% = ¼ work. Therefore, remaining work = 1 (1/6 + ¼) = 7/12, Therefore Farhan completed 7/12 work in 7 days, so will complete the whole work in 12 days.

Question 2 of 10
2. Question
1 pointsCategory: Quantitative AptitudeDirections for Questions 15: Study the following charts and answer the questions that follow:
Girish who can complete the whole work in 30 days replaced Arun and did Arun’s part of the work. He left and then Bimal also worked for the same number of days as Girish. If remaining work was completed by Manish who can do complete work in onefourth the number of days in which Ella can complete the whole work, then how many days was the whole wok completed?
Correct
Answer: d) 18 days
Solution: Arun’s part of work = 20 %= 1/5. So, Girish did 1/5 of work, and whole work in 30 days, so 1/5 work in 1/5*30 = 6 days, Now B also worked for 6 days. Bimal can complete whole work in 30 days, so in 6 days, completed 6/30= 1/5 of work. Now remaining work = 1 (1/5 + 1/5) = 3/5 (After Girish and Bimal did work). Now Ella can complete whole work in (100*6/15) = 40 days. So Manish can complete work in 10 days. So completed 3/5*10= 6 days. So total number of days = 6=6=6 = 18 daysIncorrect
Answer: d) 18 days
Solution: Arun’s part of work = 20 %= 1/5. So, Girish did 1/5 of work, and whole work in 30 days, so 1/5 work in 1/5*30 = 6 days, Now B also worked for 6 days. Bimal can complete whole work in 30 days, so in 6 days, completed 6/30= 1/5 of work. Now remaining work = 1 (1/5 + 1/5) = 3/5 (After Girish and Bimal did work). Now Ella can complete whole work in (100*6/15) = 40 days. So Manish can complete work in 10 days. So completed 3/5*10= 6 days. So total number of days = 6=6=6 = 18 days 
Question 3 of 10
3. Question
1 pointsCategory: Quantitative AptitudeDirections for Questions 15: Study the following charts and answer the questions that follow:
All five people decided to complete work in less number of days. So they divided the work equally among themselves. In how many days will the work be completed this way?( They all worked individually)
Correct
Answer: b) 27 4/5 days
Solution: 5 people equally divided the work so each did 1/5 work , So Arun does 1/5 work in 4 days (found earlier), Similarly bimal does 1/10 work in 3 days, so 1/5 work in 6 days, Similarly Chandra does ¼ work in 6 days , so 1/5 work in 24/5 days, Similarly Dev does 3/10 work in 7.5 days, so 1/5 work in 5 days, similarly Ella does 3/20 wrk in 6 days, so 1/5 work in 8 days. So complete work in = 4+6+24/5 + 5 + 8 = 27.8 days.Incorrect
Answer: b) 27 4/5 days
Solution: 5 people equally divided the work so each did 1/5 work , So Arun does 1/5 work in 4 days (found earlier), Similarly bimal does 1/10 work in 3 days, so 1/5 work in 6 days, Similarly Chandra does ¼ work in 6 days , so 1/5 work in 24/5 days, Similarly Dev does 3/10 work in 7.5 days, so 1/5 work in 5 days, similarly Ella does 3/20 wrk in 6 days, so 1/5 work in 8 days. So complete work in = 4+6+24/5 + 5 + 8 = 27.8 days. 
Question 4 of 10
4. Question
1 pointsCategory: Quantitative AptitudeDirections for Questions 15: Study the following charts and answer the questions that follow:
Piyush is 20% more efficient than Bimal and Ramesh is 60% more efficient than Chandra. They worked together for 5 days and left the work, after which the remaining work was completed by Chandra and Ella together. Had all worked together, in how many less days they would have completed the work?
Correct
Answer: a) 6 3/13 days
Solution: Piyush and Bimal efficiency ratio = 120: 100= 6:5 respectively, so no of days taken= 5:6 Since 6 = 30days, so 1 = 5, Piyush can complete work in 25 days. Similarly Ramesh and Chandra efficiency ratio is 160: 100 = 8 : 5. So no of days taken = 5 : 8 . Since 8= 24 days, so 1 = 3 days, so 5 = 15, Ramesh can complete work in 15 days. They worked for 5 days, so (1/25+ 1/15)*5 = 8/15 work, Remaining 7/15 by Chandra and Ella , So ( 1/24 + 1/40)* X= 7/15, Therefore X = 7 days. Total = 5+ 7= 12 days. Now had they worked together: (1/25 + 1/15 + 1/24 + 1/40) = 13/75, So 75/13 days. So less days = 12 – 75/13= 81/13= 6 3/13 days.Incorrect
Answer: a) 6 3/13 days
Solution: Piyush and Bimal efficiency ratio = 120: 100= 6:5 respectively, so no of days taken= 5:6 Since 6 = 30days, so 1 = 5, Piyush can complete work in 25 days. Similarly Ramesh and Chandra efficiency ratio is 160: 100 = 8 : 5. So no of days taken = 5 : 8 . Since 8= 24 days, so 1 = 3 days, so 5 = 15, Ramesh can complete work in 15 days. They worked for 5 days, so (1/25+ 1/15)*5 = 8/15 work, Remaining 7/15 by Chandra and Ella , So ( 1/24 + 1/40)* X= 7/15, Therefore X = 7 days. Total = 5+ 7= 12 days. Now had they worked together: (1/25 + 1/15 + 1/24 + 1/40) = 13/75, So 75/13 days. So less days = 12 – 75/13= 81/13= 6 3/13 days. 
Question 5 of 10
5. Question
1 pointsCategory: Quantitative AptitudeDirections for Questions 15: Study the following charts and answer the questions that follow:
Jayant can complete the whole work in number of days equal to the average to the average of number of days in which Arun and Bimal can complete the work. Jayant, Chandra, Dev and ella all started the work and after 5 days they were replaced by Arun and Bimal. Arun and Bimal completed the remaining work in how many days?
Correct
Answer: a) 3 1/5 days
Solution: Arun completes work in 20 days, Bimal in 30 days. So Jayant completes in (20 + 30)/2 = 25 days. Jayant, Chandra, Dev and Ella all started work: (1/25 +1/24 + 1/25 + 1/40) = 11/75. Worked for 5 days so did 11/75*5= 11/15 of work. Remaining work = 4/15, ( 1/20 + 1/30)*X = 4/15, solving X = 48/15 days = 3 1/5 days.Incorrect
Answer: a) 3 1/5 days
Solution: Arun completes work in 20 days, Bimal in 30 days. So Jayant completes in (20 + 30)/2 = 25 days. Jayant, Chandra, Dev and Ella all started work: (1/25 +1/24 + 1/25 + 1/40) = 11/75. Worked for 5 days so did 11/75*5= 11/15 of work. Remaining work = 4/15, ( 1/20 + 1/30)*X = 4/15, solving X = 48/15 days = 3 1/5 days. 
Question 6 of 10
6. Question
1 pointsCategory: Quantitative AptitudeDirection for Question 68: In the following question, two statements are numbered as I and II. On solving these statements you get Quantity I and Quantity II respectively. Solve for both the statements individually and mark the correct answer.
Quantity I: Three metal cubes with edges 6 cm, 8 cm and 10 cm respectively are melted together and formed into a single cube. Find the side of the resulting cube?
Quantity II: The curves surface area of a cone is 2376 square cm and its slant height is 18 cm. Find the diameter.Correct
Answer: b) Quantity I < Quantity II Solution: Quantity I, the total volume will remain the same, let the side of the resulting cube be = a. Then, 63 + 83 + 103 = a3 =3√1728 = 12 cm. Quantity II, Curved Surface Area = πrl, Radius= πrl/πl= 2376/(3.14*18)= 42 cm. Therefore, diameter is = 2*Radius = 2 * 42 = 84 cm
Incorrect
Answer: b) Quantity I < Quantity II Solution: Quantity I, the total volume will remain the same, let the side of the resulting cube be = a. Then, 63 + 83 + 103 = a3 =3√1728 = 12 cm. Quantity II, Curved Surface Area = πrl, Radius= πrl/πl= 2376/(3.14*18)= 42 cm. Therefore, diameter is = 2*Radius = 2 * 42 = 84 cm

Question 7 of 10
7. Question
1 pointsCategory: Quantitative AptitudeDirection for Question 68: In the following question, two statements are numbered as I and II. On solving these statements you get Quantity I and Quantity II respectively. Solve for both the statements individually and mark the correct answer.
Quantity I: The profit earned when an article is sold for Rs.800 is 20 times the loss incurred when it is sold for Rs. 275. At what price should the article be sold if it is desired to make a profit of 25%.
Quantity II: In the town of Andher Nagari Chaupat Raja, shopkeepers have to buy and sell goods in the range of Rs. 500 to Rs. 999. A shopkeeper in such a town decided not to buy or sell goods for amounts that contain the digit 9 or for amounts that add up to 13 or are a multiple of 13. What is the maximum possible profit he can earn?Correct
Answer: b) Quantity I < Quantity II Solution: Quantity I,The interpretation of the first statement is the if the loss at 275 is L, the profit at 800 is 20L. Thus, 21L = 800 – 275= 525, therefore L= 25. Thus cost price of article is Rs.300. To get a profit of 25%, the selling price should be 1.25* 300 = 375. Quantity II, He would buy at 500 and sell at Rs.888 to get a profit of 388.
Incorrect
Answer: b) Quantity I < Quantity II Solution: Quantity I,The interpretation of the first statement is the if the loss at 275 is L, the profit at 800 is 20L. Thus, 21L = 800 – 275= 525, therefore L= 25. Thus cost price of article is Rs.300. To get a profit of 25%, the selling price should be 1.25* 300 = 375. Quantity II, He would buy at 500 and sell at Rs.888 to get a profit of 388.

Question 8 of 10
8. Question
1 pointsCategory: Quantitative AptitudeDirection for Question 68: In the following question, two statements are numbered as I and II. On solving these statements you get Quantity I and Quantity II respectively. Solve for both the statements individually and mark the correct answer.
Quantity I: In the famous Bhojpur Islands, there are four men for every three women and five children for every three men. How many children are there in the Island if it has 531 women?
Quantity II : King Sheru had ordered the distribution of apples according to the following plan: for every 20 apples the elephant gets, the zebra should get 13 apples and the deer should get 8 apples. Now his servant Shambha jackal is in a fix. Can you help him by telling how much should he give to the elephant if there were 820 Apples in total?Correct
Answer: a) Quantity I > Quantity II
Solution: Quantity I, Women : Men= 3:4, Man : Children = 3 :5, therefore women : men : children = 9 : 12 : 20. In the ratio, 9 implies 531 women, thus 20 would imply 1180 children. Quantity II, Ratio of distribution between elephant, zebra, deer is 20 : 13 : 8 respectively. So the elephant should get (20/41)*820 = 400Incorrect
Answer: a) Quantity I > Quantity II
Solution: Quantity I, Women : Men= 3:4, Man : Children = 3 :5, therefore women : men : children = 9 : 12 : 20. In the ratio, 9 implies 531 women, thus 20 would imply 1180 children. Quantity II, Ratio of distribution between elephant, zebra, deer is 20 : 13 : 8 respectively. So the elephant should get (20/41)*820 = 400 
Question 9 of 10
9. Question
1 pointsCategory: Quantitative AptitudeThree equal cubes are placed adjacently in a row. Find the ratio of total surface area of the new cuboid to that of the sum of the surface areas of the three cubes.
Correct
Answer: a) 7 : 9
Solution: Let the side of the cube be = a units, Total surface area of 3 cubes = 3* 6a2 = 18 a2 . Total surface area of cuboid = 18a2 – 4 a2 = 14 a2. Hence ratio = 7 : 9Incorrect
Answer: a) 7 : 9
Solution: Let the side of the cube be = a units, Total surface area of 3 cubes = 3* 6a2 = 18 a2 . Total surface area of cuboid = 18a2 – 4 a2 = 14 a2. Hence ratio = 7 : 9 
Question 10 of 10
10. Question
1 pointsCategory: Quantitative AptitudeOn his deathbed, Mr. Kalu called upon his three sons and told them to distribute all his assets worth Rs. 5,25,000 in the ratio of 1/15 : 1/21 : 1/35 amongst themselves. Find the biggest share amongst the three portions.
Correct
Answer: b) 2,45,000
Solution: The ratio of distribution will be: 21*35 : 15*35 : 15*21 = 7 : 5 : 3 The biggest will be worth: 7*525000/15= 245000Incorrect
Answer: b) 2,45,000
Solution: The ratio of distribution will be: 21*35 : 15*35 : 15*21 = 7 : 5 : 3 The biggest will be worth: 7*525000/15= 245000
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