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Question 1 of 10
1. Question
1 pointsCategory: Quantitative AptitudeDirections for questions 15: Read the following figures and answer the questions that follow:
A, B, C, D and E are five persons employed to complete a job X. Line graph shows the data regarding the time taken by these persons to complete the job X. Table 2 shows the actual time for which every one of them worked on the job X.
Note 1: All the persons worked on the job X for ‘whole number’ days. Note 2: Two jobs Y and Z are similar to job X and require same effort as required by job X.If C worked on job Y with 5/4 times his given efficiency and was assisted by B every 3rd day, then find the time taken to complete the job Y?
Correct
Solution: Initially C took 15 days to complete the job, with increased efficiency he finishes the job in 15*4/5 = 12 days. C’s one day work = 1/12. C’s three day work + B’s assistance on third day = 3/12+1/12 = 4/12 =1/3. In three days 1/3 of the work is complete. Hence, in order to complete the work 3*3 = 9 days are required.
Incorrect
Solution: Initially C took 15 days to complete the job, with increased efficiency he finishes the job in 15*4/5 = 12 days. C’s one day work = 1/12. C’s three day work + B’s assistance on third day = 3/12+1/12 = 4/12 =1/3. In three days 1/3 of the work is complete. Hence, in order to complete the work 3*3 = 9 days are required.

Question 2 of 10
2. Question
1 pointsCategory: Quantitative AptitudeDirections for questions 15: Read the following figures and answer the questions that follow:
A, B, C, D and E are five persons employed to complete a job X. Line graph shows the data regarding the time taken by these persons to complete the job X. Table 2 shows the actual time for which every one of them worked on the job X.
Note 1: All the persons worked on the job X for ‘whole number’ days. Note 2: Two jobs Y and Z are similar to job X and require same effort as required by job X.If the ratio of number of days for which B and D worked on job X is 4 : 3, then find the difference between number of days for which B and D worked ?
Correct
Solution: Since the no of days each work on job is given in the table except for B and D. Hence work completed by A, C and E = 2/10 + 3/15 + 2/20 = ½ . wok remaining = ½. Now, Remaining work done by B and D in 5x and 3x days i.e. 4x/12 + 3x/18 = ½, Solving we get x =1, hence difference = 43 = 1 Day
Incorrect
Solution: Since the no of days each work on job is given in the table except for B and D. Hence work completed by A, C and E = 2/10 + 3/15 + 2/20 = ½ . wok remaining = ½. Now, Remaining work done by B and D in 5x and 3x days i.e. 4x/12 + 3x/18 = ½, Solving we get x =1, hence difference = 43 = 1 Day

Question 3 of 10
3. Question
1 pointsCategory: Quantitative AptitudeDirections for questions 15: Read the following figures and answer the questions that follow:
A, B, C, D and E are five persons employed to complete a job X. Line graph shows the data regarding the time taken by these persons to complete the job X. Table 2 shows the actual time for which every one of them worked on the job X.
Note 1: All the persons worked on the job X for ‘whole number’ days. Note 2: Two jobs Y and Z are similar to job X and require same effort as required by job X.If A, C and E worked on job Z for 2 days each and the remaining job was done by B and D. If the ratio of no. days for which B and D worked is in ratio 20 : 21, then find the number of days for which B worked ?
Correct
Solution: Work done by A, C and E for two days each = 2/10 + 2/15 + 2/20 = 13/30. Work remaining = 17/30. Now, Remaining work done by B and D in 20x and 21x days i.e. 20x/12 + 21x/18 = 17/30, solving 102x= 17*36/30, x = 1/5 Hence B takes 20*1/5 = 4 days
Incorrect
Solution: Work done by A, C and E for two days each = 2/10 + 2/15 + 2/20 = 13/30. Work remaining = 17/30. Now, Remaining work done by B and D in 20x and 21x days i.e. 20x/12 + 21x/18 = 17/30, solving 102x= 17*36/30, x = 1/5 Hence B takes 20*1/5 = 4 days

Question 4 of 10
4. Question
1 pointsCategory: Quantitative AptitudeDirections for questions 15: Read the following figures and answer the questions that follow:
A, B, C, D and E are five persons employed to complete a job X. Line graph shows the data regarding the time taken by these persons to complete the job X. Table 2 shows the actual time for which every one of them worked on the job X.
Note 1: All the persons worked on the job X for ‘whole number’ days. Note 2: Two jobs Y and Z are similar to job X and require same effort as required by job X.E worked on job ‘Z’ for 5 days and the remaining job was completed by A, B and C who worked on alternate days starting with A followed by B and C in that order. Find the no. of days B worked for?
Correct
Solution: E‘s five day work completes 5*1/20 =1/4th work. Work remaining = 3/4th. Work done by A, B and C in three days = (1/10 + 1/12 +1/15)= ¼ . Therefore if in three days ¼ is done, ¾ will be done in 3*3 = 9 days. Hence B worked for 3 days.
Incorrect
Solution: E‘s five day work completes 5*1/20 =1/4th work. Work remaining = 3/4th. Work done by A, B and C in three days = (1/10 + 1/12 +1/15)= ¼ . Therefore if in three days ¼ is done, ¾ will be done in 3*3 = 9 days. Hence B worked for 3 days.

Question 5 of 10
5. Question
1 pointsCategory: Quantitative AptitudeDirections for questions 15: Read the following figures and answer the questions that follow:
A, B, C, D and E are five persons employed to complete a job X. Line graph shows the data regarding the time taken by these persons to complete the job X. Table 2 shows the actual time for which every one of them worked on the job X.
Note 1: All the persons worked on the job X for ‘whole number’ days. Note 2: Two jobs Y and Z are similar to job X and require same effort as required by job X.A and C worked on job Y working alternatively for 10 days. B and D then worked together for ‘x ’ days. If 1/36 of the job was still remained, then find the value of ‘x ’ ?
Correct
Solution: As per given condition (1/10 + 1/12)*5 + (x/12+x/18) = (11/36) i.e. work done. Solving, 5x/36 = 35/36 – 5/6, x = 1 day
Incorrect
Solution: As per given condition (1/10 + 1/12)*5 + (x/12+x/18) = (11/36) i.e. work done. Solving, 5x/36 = 35/36 – 5/6, x = 1 day

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: A, B, C rent a pasture. A puts 10 oxen for 7 months, B puts 12 oxen for 5 months and C puts 15 oxen for 3 months for grazing. If the rent of the pasture is Rs. 175, how much must C pay as his share of rent?
Quantity II: If the simple interest is 10.5% annually and compound interest is 10% annually. Find the difference between the interests after 3 years on a sum of Rs. 1000.
Correct
Solution: Quantity I: A:B:C = [ 10 x 7 ] : [ 12 x 5 ] : ( 15 x 3 ) = 70 : 60 : 45 = 14 : 12 : 9 . Therefore, C’s Rent = Rs. (175 x 9/35) = Rs. 45. Quantity II: At 10% compound interest the interest in three years would be 33.1 % = Rs.331 At 10.5% simple interest the interest in 3 years would be 31.5% = Rs. 315. Therefore difference = 331315 = Rs 16. Hence Quantity I > Quantity II
Incorrect
Solution: Quantity I: A:B:C = [ 10 x 7 ] : [ 12 x 5 ] : ( 15 x 3 ) = 70 : 60 : 45 = 14 : 12 : 9 . Therefore, C’s Rent = Rs. (175 x 9/35) = Rs. 45. Quantity II: At 10% compound interest the interest in three years would be 33.1 % = Rs.331 At 10.5% simple interest the interest in 3 years would be 31.5% = Rs. 315. Therefore difference = 331315 = Rs 16. Hence Quantity I > Quantity II

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: Find the area of trapezium whose parallel sides are 20 cm and 18 cm long, and the distance between them is 15 cm.
Quantity II: Find the area of parallelogram with base 24 cm and height 16 cm.Correct
Solution: Quantity I: Area of a trapezium = 1/2 (sum of parallel sides) * (perpendicular distance between them) = 1/2 (20 + 18) * (15) = 285 cm2. Area of a parallelogram = base * height = 24 * 16 = 384 cm2. Area of a parallelogram = base * height = 24 * 16 = 384 cm2. Hence, Quantity I < Quantity II
Incorrect
Solution: Quantity I: Area of a trapezium = 1/2 (sum of parallel sides) * (perpendicular distance between them) = 1/2 (20 + 18) * (15) = 285 cm2. Area of a parallelogram = base * height = 24 * 16 = 384 cm2. Area of a parallelogram = base * height = 24 * 16 = 384 cm2. Hence, Quantity I < Quantity II

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 covering a distance of 30 km, Abhay takes 2 hours more than Sameer. If Abhay doubles his speed, then he would take 1 hour less than Sameer. Abhay’s speed is:
Quantity II: Robert is travelling on his cycle and has calculated to reach point A at 2 P.M. if he travels at 10 kmph, he will reach there at 12 noon if he travels at 15 kmph. At what speed must he travel to reach A at 1 P.M.?
Correct
Solution: Quantity I: Let Abhay’s speed be x km/hr. Then, 30/x – 30/2x = 3, i.e. 6x = 30, Therefore x= 5kmpr. Quantity II, Let the distance travelled by x km.
Then x/10 – x/15 = 2, 3x 2x =60, therefore x= 60 km. Time taken to travel 60 km at 10 km/hr = 6 hr. So, Robert started before 2 P.M. i.e. 8 A.M. Therefore required speed = 60/5 kmph= 12 kmph. Hence, Quantity I < Quantity II.Incorrect
Solution: Quantity I: Let Abhay’s speed be x km/hr. Then, 30/x – 30/2x = 3, i.e. 6x = 30, Therefore x= 5kmpr. Quantity II, Let the distance travelled by x km.
Then x/10 – x/15 = 2, 3x 2x =60, therefore x= 60 km. Time taken to travel 60 km at 10 km/hr = 6 hr. So, Robert started before 2 P.M. i.e. 8 A.M. Therefore required speed = 60/5 kmph= 12 kmph. Hence, Quantity I < Quantity II. 
Question 9 of 10
9. Question
1 pointsCategory: Quantitative AptitudeA pair of fair dice are thrown. Given that the sum of the dice is less than or equal to 4, find the probability that only one dice shows two.
Correct
Solution: The possible outcomes are : (1,1); (1,2); (2,1); (2,2); (3,1); (1,3). Out of six cases, in two cases there is exactly one ‘2’. Thus the probability is 2/6=1/3.
Incorrect
Solution: The possible outcomes are : (1,1); (1,2); (2,1); (2,2); (3,1); (1,3). Out of six cases, in two cases there is exactly one ‘2’. Thus the probability is 2/6=1/3.

Question 10 of 10
10. Question
1 pointsCategory: Quantitative AptitudeIn business, A and C invested amounts in the ratio 2:1, whereas the ratio between amounts invested by A and B was 3:2, If Rs 157300 was their profit, how much amount did B receive?
Correct
Solution: A:B = 3:2 = 6:4, A:C = 2:1 = 6:3 Therefore A:B:C = 6:4:3, B share = (4/13)*157300 = 48400
Incorrect
Solution: A:B = 3:2 = 6:4, A:C = 2:1 = 6:3 Therefore A:B:C = 6:4:3, B share = (4/13)*157300 = 48400
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