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Question 1 of 10
1. Question
1 pointsCategory: Quantitative AptitudeDirections (Q15). Sumit is a clerk in government office and typed certain number of words in four different months of 2018 i.e. January, February, March and April. The days of the month are categorized into odd number days and even number days with respect to their date.
 In January, he typed 1350 words per odd day. No words were typed on even day.
 In February, he typed 1320 words per odd day and 1400 words per even day.
 In March, he typed 2000 words per odd day and 1500 words per even day. In this month, he was on a leave of 5 days (3 odd days and 2 even days)
 In April, he typed 1600 words per even day. Average number of words typed in this month is 1900 words per day.
Total number of words typed by Sumit on odd days in April are:
Correct
Answer 2) 33000
Explanation
Total number of words typed in April = 1900 × 30 = 57000
Total number of words typed on even days in April = 1600 × 15 = 24000
Required number of words = 57000 – 24000 = 33000Incorrect
Answer 2) 33000
Explanation
Total number of words typed in April = 1900 × 30 = 57000
Total number of words typed on even days in April = 1600 × 15 = 24000
Required number of words = 57000 – 24000 = 33000 
Question 2 of 10
2. Question
1 pointsCategory: Quantitative AptitudeDirections (Q15). Sumit is a clerk in government office and typed certain number of words in four different months of 2018 i.e. January, February, March and April. The days of the month are categorized into odd number days and even number days with respect to their date.
 In January, he typed 1350 words per odd day. No words were typed on even day.
 In February, he typed 1320 words per odd day and 1400 words per even day.
 In March, he typed 2000 words per odd day and 1500 words per even day. In this month, he was on a leave of 5 days (3 odd days and 2 even days)
 In April, he typed 1600 words per even day. Average number of words typed in this month is 1900 words per day.
In February, total number of words typed on even days are how much more than the total number of words typed on odd days?
Correct
Answer 5) 1120
Explanation
Required difference = 1400 × 14 – 1320 × 14 = 14 × (1400 – 1320) = 14 × 80 = 1120 (there are 28 days in February, 14 odd days and 14 even days)Incorrect
Answer 5) 1120
Explanation
Required difference = 1400 × 14 – 1320 × 14 = 14 × (1400 – 1320) = 14 × 80 = 1120 (there are 28 days in February, 14 odd days and 14 even days) 
Question 3 of 10
3. Question
1 pointsCategory: Quantitative AptitudeDirections (Q15). Sumit is a clerk in government office and typed certain number of words in four different months of 2018 i.e. January, February, March and April. The days of the month are categorized into odd number days and even number days with respect to their date.
 In January, he typed 1350 words per odd day. No words were typed on even day.
 In February, he typed 1320 words per odd day and 1400 words per even day.
 In March, he typed 2000 words per odd day and 1500 words per even day. In this month, he was on a leave of 5 days (3 odd days and 2 even days)
 In April, he typed 1600 words per even day. Average number of words typed in this month is 1900 words per day.
Average number of words typed per day in January are (approximately):
Correct
Answer 3) 697
Explanation
Total number of words typed in January = 16 × 1350 = 21600 (total number of days in March are 31 and number of odd days are 16)
Required average = 21600/31 = 697 (approx.)Incorrect
Answer 3) 697
Explanation
Total number of words typed in January = 16 × 1350 = 21600 (total number of days in March are 31 and number of odd days are 16)
Required average = 21600/31 = 697 (approx.) 
Question 4 of 10
4. Question
1 pointsCategory: Quantitative AptitudeDirections (Q15). Sumit is a clerk in government office and typed certain number of words in four different months of 2018 i.e. January, February, March and April. The days of the month are categorized into odd number days and even number days with respect to their date.
 In January, he typed 1350 words per odd day. No words were typed on even day.
 In February, he typed 1320 words per odd day and 1400 words per even day.
 In March, he typed 2000 words per odd day and 1500 words per even day. In this month, he was on a leave of 5 days (3 odd days and 2 even days)
 In April, he typed 1600 words per even day. Average number of words typed in this month is 1900 words per day.
In March, total number of words typed on odd days are how much more than the number of words typed on even days?
Correct
Answer 4) 6500
Explanation
Total number of words typed on odd days in March = 2000 × (16 – 3) = 2000 × 13 = 26000
Total number of words typed on even days in March = 1500 × (15 – 2) = 1500 × 13 = 19500
Required difference = 26000 – 19500 = 6500Incorrect
Answer 4) 6500
Explanation
Total number of words typed on odd days in March = 2000 × (16 – 3) = 2000 × 13 = 26000
Total number of words typed on even days in March = 1500 × (15 – 2) = 1500 × 13 = 19500
Required difference = 26000 – 19500 = 6500 
Question 5 of 10
5. Question
1 pointsCategory: Quantitative AptitudeDirections (Q15). Sumit is a clerk in government office and typed certain number of words in four different months of 2018 i.e. January, February, March and April. The days of the month are categorized into odd number days and even number days with respect to their date.
 In January, he typed 1350 words per odd day. No words were typed on even day.
 In February, he typed 1320 words per odd day and 1400 words per even day.
 In March, he typed 2000 words per odd day and 1500 words per even day. In this month, he was on a leave of 5 days (3 odd days and 2 even days)
 In April, he typed 1600 words per even day. Average number of words typed in this month is 1900 words per day.
Total number of words typed by Sumit in May are 9970 more than the total number of words typed by him in February. Average number of words typed per day by him in May are:
Correct
Answer 4) 1550
Explanation
Total number of words typed in February = 1320 × 14 + 1400 × 14 = 18480 + 19600 = 38080
Total number of words typed by him in May = 38080 + 9970 = 48050
Required average = 48050/31 = 1550Incorrect
Answer 4) 1550
Explanation
Total number of words typed in February = 1320 × 14 + 1400 × 14 = 18480 + 19600 = 38080
Total number of words typed by him in May = 38080 + 9970 = 48050
Required average = 48050/31 = 1550 
Question 6 of 10
6. Question
1 pointsCategory: Quantitative AptitudeDirections (Q610). Six friends A, B, C, D, E and F cover different amount of distance in their respective cars which is shown by a pie chart (in percentage). The table below shows the average speeds of their respective cars. Few values are missing in the table. Based on this information, answer the following questions.
Time taken by A to travel his share of distance is (in minutes):
Correct
Answer 4) 245
Explanation
Distance travelled by A = 21% of 1400 = 294 km
Time taken = 294/72 × 60 = 245 minutesIncorrect
Answer 4) 245
Explanation
Distance travelled by A = 21% of 1400 = 294 km
Time taken = 294/72 × 60 = 245 minutes 
Question 7 of 10
7. Question
1 pointsCategory: Quantitative AptitudeDirections (Q610). Six friends A, B, C, D, E and F cover different amount of distance in their respective cars which is shown by a pie chart (in percentage). The table below shows the average speeds of their respective cars. Few values are missing in the table. Based on this information, answer the following questions.
The average speed of C is 20% less the average speed of B. The time taken by C to travel his share of distance is approximately how much more than the time taken by B to travel his share of distance?Correct
Answer 3) 1.77 hours
Explanation
Average speed of C = 80% of 75 = 60 km/h
Distance travelled by C = 18% of 1400 = 252 km
Time taken by C = 252/60 = 4.2 hours
Distance travelled by B = 13% of 1400 = 182 km
Time taken by B = 182/75 = 2.43 hours (approximately)
Required difference = 4.2 – 2.43 = 1.77 hours (approx.)Incorrect
Answer 3) 1.77 hours
Explanation
Average speed of C = 80% of 75 = 60 km/h
Distance travelled by C = 18% of 1400 = 252 km
Time taken by C = 252/60 = 4.2 hours
Distance travelled by B = 13% of 1400 = 182 km
Time taken by B = 182/75 = 2.43 hours (approximately)
Required difference = 4.2 – 2.43 = 1.77 hours (approx.) 
Question 8 of 10
8. Question
1 pointsCategory: Quantitative AptitudeDirections (Q610). Six friends A, B, C, D, E and F cover different amount of distance in their respective cars which is shown by a pie chart (in percentage). The table below shows the average speeds of their respective cars. Few values are missing in the table. Based on this information, answer the following questions.
If the time taken by D to travel his share of distance is 2.1 hours, what is the approximate average speed of D (in m/s)?Correct
Answer 1) 22.2 m/s
Explanation
Distance travelled by D = 12% of 1400 = 168 km
Average speed of D = 168/2.1 = 80 km/h = 80 × 5/18 = 22.2 m/s (approx.)Incorrect
Answer 1) 22.2 m/s
Explanation
Distance travelled by D = 12% of 1400 = 168 km
Average speed of D = 168/2.1 = 80 km/h = 80 × 5/18 = 22.2 m/s (approx.) 
Question 9 of 10
9. Question
1 pointsCategory: Quantitative AptitudeDirections (Q610). Six friends A, B, C, D, E and F cover different amount of distance in their respective cars which is shown by a pie chart (in percentage). The table below shows the average speeds of their respective cars. Few values are missing in the table. Based on this information, answer the following questions.
If E travels with the speed which is 10 km/h less than his usual speed, he will reach his destination 1.68 hrs late. What is the usual average speed of E?Correct
Answer 3) 50 km/h
Explanation
Distance travelled by E = 24% of 1400 = 336 km
Let the usual average speed of E is x km/h
According to question,
336/(x – 10) – 336/x = 1.68
By solving this equation, we have, (try solving it through options, it will take less time)
X = 50 km/hIncorrect
Answer 3) 50 km/h
Explanation
Distance travelled by E = 24% of 1400 = 336 km
Let the usual average speed of E is x km/h
According to question,
336/(x – 10) – 336/x = 1.68
By solving this equation, we have, (try solving it through options, it will take less time)
X = 50 km/h 
Question 10 of 10
10. Question
1 pointsCategory: Quantitative AptitudeDirections (Q610). Six friends A, B, C, D, E and F cover different amount of distance in their respective cars which is shown by a pie chart (in percentage). The table below shows the average speeds of their respective cars. Few values are missing in the table. Based on this information, answer the following questions.
Approximate time taken by F to travel his share of distance is:Correct
Answer 2) 2.5 hours
Explanation
Distance travelled by F = 12% of 1400 = 168 km
Time taken = 168/68 = 2.5 hours (approx.)Incorrect
Answer 2) 2.5 hours
Explanation
Distance travelled by F = 12% of 1400 = 168 km
Time taken = 168/68 = 2.5 hours (approx.)
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