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Question 1 of 10

1. Question

1 points

Category: Quantitative Aptitude

Directions (Q1-5). 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 = 33000

Incorrect

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 points

Category: Quantitative Aptitude

Directions (Q1-5). 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 points

Category: Quantitative Aptitude

Directions (Q1-5). 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 points

Category: Quantitative Aptitude

Directions (Q1-5). 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 = 6500

Incorrect

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 points

Category: Quantitative Aptitude

Directions (Q1-5). 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 = 1550

Incorrect

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 points

Category: Quantitative Aptitude

Directions (Q6-10). 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 minutes

Incorrect

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 points

Category: Quantitative Aptitude

Directions (Q6-10). 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 points

Category: Quantitative Aptitude

Directions (Q6-10). 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 points

Category: Quantitative Aptitude

Directions (Q6-10). 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/h

Incorrect

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 points

Category: Quantitative Aptitude

Directions (Q6-10). 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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