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- Question 1 of 10
##### 1. Question

1 pointsCategory: Quantitative Aptitude**Directions (Q1-5). Six football players A, B, C, D, E and F trains every day and runs some distance every day. Distance run by each every week has been represented by a pie chart in percentage (total distance travelled by them changes every week but percentage of distance travelled by them and their speed don’t change). Their average speed is given in a table, however some values are missing in it.**

**In one of the weeks, B runs for 16 hours. For how much hours, A runs in the same week?**Correct**Answer- 3) 16.8 hours**

Explanation-

Total distance travelled by B in this week = 16 × 8.5 = 136 km

Distance travelled by A in the same week = 136/17 × 21 = 168 km

Time for which A runs = 168/10 = 16.8 hoursIncorrect**Answer- 3) 16.8 hours**

Explanation-

Total distance travelled by B in this week = 16 × 8.5 = 136 km

Distance travelled by A in the same week = 136/17 × 21 = 168 km

Time for which A runs = 168/10 = 16.8 hours - Question 2 of 10
##### 2. Question

1 pointsCategory: Quantitative Aptitude**Directions (Q1-5). Six football players A, B, C, D, E and F trains every day and runs some distance every day. Distance run by each every week has been represented by a pie chart in percentage (total distance travelled by them changes every week but percentage of distance travelled by them and their speed don’t change). Their average speed is given in a table, however some values are missing in it.**

**The total distance travelled by B and E in one of the weeks is 330 km. The time for which E runs in this week is:**Correct**Answer- 5) 40/3 hours**

Explanation-

Distance travelled by E in this week = 330/(17 + 16) × 16 = 160 km

Time taken = 160/12 = 40/3 hoursIncorrect**Answer- 5) 40/3 hours**

Explanation-

Distance travelled by E in this week = 330/(17 + 16) × 16 = 160 km

Time taken = 160/12 = 40/3 hours - Question 3 of 10
##### 3. Question

1 pointsCategory: Quantitative Aptitude**Directions (Q1-5). Six football players A, B, C, D, E and F trains every day and runs some distance every day. Distance run by each every week has been represented by a pie chart in percentage (total distance travelled by them changes every week but percentage of distance travelled by them and their speed don’t change). Their average speed is given in a table, however some values are missing in it.**

**In a week, the total distance ran by C and D together is 351. The total time taken by C to run his distance is 31.2 hours and that by D is 7.8 hours. The speed of D is how much percent more than the speed of C?**Correct**Answer- 3) 100%**

Explanation-

Distance run by C = 351/(24 + 12) × 24 = 234 km

Distance run by D = 351/(24 + 12) × 12 = 117 km

Let the speed of C is x km/h and that of D is y km/h

Then, 234/x = 31.2 , x = 7.5 km/h

Also, 117/y = 7.8, y = 117/7.8 = 15 km/h

Required percentage = (15 – 7.5)/7.5 × 100 = 100%Incorrect**Answer- 3) 100%**

Explanation-

Distance run by C = 351/(24 + 12) × 24 = 234 km

Distance run by D = 351/(24 + 12) × 12 = 117 km

Let the speed of C is x km/h and that of D is y km/h

Then, 234/x = 31.2 , x = 7.5 km/h

Also, 117/y = 7.8, y = 117/7.8 = 15 km/h

Required percentage = (15 – 7.5)/7.5 × 100 = 100% - Question 4 of 10
##### 4. Question

1 pointsCategory: Quantitative Aptitude

**The total distance run by all of the footballers together is 1700 km. The time for which A runs is 22.1 hours more than the time for which F runs. The speed of F is how much percent less than that of A?**Correct**Answer- 2) 25%**

Explanation-

Distance run by A = 21% of 1700 = 357 km

Distance run by F = 10% of 1700 = 170 km

Time for which A runs = 357/10 = 35.7 hours

Time for which F runs = 35.7 – 22.1 = 13.6 hours

Speed of F = 170/13.6 = 12.5 km/h

Required percentage = (12.5 – 10)/10 × 100 = 25%Incorrect**Answer- 2) 25%**

Explanation-

Distance run by A = 21% of 1700 = 357 km

Distance run by F = 10% of 1700 = 170 km

Time for which A runs = 357/10 = 35.7 hours

Time for which F runs = 35.7 – 22.1 = 13.6 hours

Speed of F = 170/13.6 = 12.5 km/h

Required percentage = (12.5 – 10)/10 × 100 = 25% - Question 5 of 10
##### 5. Question

1 pointsCategory: Quantitative Aptitude

**For any total distance run by all the footballers in a week, the time taken by B to run his share of distance will be how much percent more than the time taken by E to run his share of distance?**Correct**Answer- 1) 50%**

Explanation-

Let the total distance ran by all the footballers together is 100 km.

Then, distance run by B = 17 km

Distance run by E = 16 km

Time taken by B to run his distance = 17/8.5 = 2 hours

Time taken by E to run his distance = 16/12 = 4/3 hours

Required percentage = (2 – 4/3)/4/3 × 100 = 1/2 × 100 = 50%Incorrect**Answer- 1) 50%**

Explanation-

Let the total distance ran by all the footballers together is 100 km.

Then, distance run by B = 17 km

Distance run by E = 16 km

Time taken by B to run his distance = 17/8.5 = 2 hours

Time taken by E to run his distance = 16/12 = 4/3 hours

Required percentage = (2 – 4/3)/4/3 × 100 = 1/2 × 100 = 50% - Question 6 of 10
##### 6. Question

1 pointsCategory: Quantitative Aptitude**Directions (Q6-10). There’re six streams i.e. A, B, C, D, E and F having different speeds of stream which is given in the line graph. Based on it answer the following questions.**

**In stream C, a boat travels 36 km upstream and 120 km downstream in total 10 h. The speed of boat in still water is how much percent more than the speed of stream C?**Correct**Answer- 1) 50%**

Explanation-

Let the speed of stream in still water is x km/h.

Then, 36/(x – 12) + 120/(x + 12) = 10

By solving the equation, we have, x = 18 km/h

Required percentage = (18 – 12)/12 × 100 = 50%Incorrect**Answer- 1) 50%**

Explanation-

Let the speed of stream in still water is x km/h.

Then, 36/(x – 12) + 120/(x + 12) = 10

By solving the equation, we have, x = 18 km/h

Required percentage = (18 – 12)/12 × 100 = 50% - Question 7 of 10
##### 7. Question

1 pointsCategory: Quantitative Aptitude**Directions (Q6-10). There’re six streams i.e. A, B, C, D, E and F having different speeds of stream which is given in the line graph. Based on it answer the following questions.**

**In stream F, a boat travels downstream and then returns to the original point in 18 hours. If the speed of the boat in still water is 12 km/h, the total distance travelled by boat is:**Correct**Answer- 2) 66 km**

Explanation-

Let the distance travelled by boat downstream is x km

According to question,

x/(12 + 10) + x/(12 – 10) = 18

x/22 + x/2 = 18

12x/22 = 18

x = 18 × 22/12 = 33 km

total distance travelled = 33 × 2 = 66 kmIncorrect**Answer- 2) 66 km**

Explanation-

Let the distance travelled by boat downstream is x km

According to question,

x/(12 + 10) + x/(12 – 10) = 18

x/22 + x/2 = 18

12x/22 = 18

x = 18 × 22/12 = 33 km

total distance travelled = 33 × 2 = 66 km - Question 8 of 10
##### 8. Question

1 pointsCategory: Quantitative Aptitude**Directions (Q6-10). There’re six streams i.e. A, B, C, D, E and F having different speeds of stream which is given in the line graph. Based on it answer the following questions.**

**Two boats x and y travels in streams D and E respectively. Boat x travels downstream and boat y travels upstream. Time taken by x to travel 92 km is 4 hours and time taken by y to travel 102 km is 6 hours. The speed of boat x in still water is how much percent less than that of boat y?**Correct**Answer- 5) 28%**

Explanation-

Let the speeds of boats x and y in still water are m and n respectively.

According to question, we have,

92/(x + 5) = 4

x + 5 = 23, x = 18 km/h

Also, 102/(y – 8) = 6

y – 8 = 17, y = 25 km/h

required percentage = (25 – 18)/25 × 100 = 28%Incorrect**Answer- 5) 28%**

Explanation-

Let the speeds of boats x and y in still water are m and n respectively.

According to question, we have,

92/(x + 5) = 4

x + 5 = 23, x = 18 km/h

Also, 102/(y – 8) = 6

y – 8 = 17, y = 25 km/h

required percentage = (25 – 18)/25 × 100 = 28% - Question 9 of 10
##### 9. Question

1 pointsCategory: Quantitative Aptitude

**The ratio of speed of a boat while going upstream in stream A to its speed while going downstream in stream B is 2:7. Total time the boat will take to go 72 downstream in stream A and 81 km upstream in stream B is:**Correct**Answer- 1) 12 hrs**

Explanation-

Let the speed of the boat in still water is x km/h

Then,

(x – 9) / (x + 6) = 2/7

7x – 63 = 2x + 12

5x = 75, x = 15 km/h

Required time = 72/(9 + 15) + 81/(15 – 6) = 72/24 + 81/9 = 3 + 9 = 12 hrsIncorrect**Answer- 1) 12 hrs**

Explanation-

Let the speed of the boat in still water is x km/h

Then,

(x – 9) / (x + 6) = 2/7

7x – 63 = 2x + 12

5x = 75, x = 15 km/h

Required time = 72/(9 + 15) + 81/(15 – 6) = 72/24 + 81/9 = 3 + 9 = 12 hrs - Question 10 of 10
##### 10. Question

1 pointsCategory: Quantitative Aptitude

**Two boats x and y travels in stream F downstream and upstream respectively. Both the boats start at the same time from opposite ends. The speed between these opposite ends is 102 km. After how much time they will cross each other if the speeds of boats x and y in still water are 15 km and 18 km respectively?**Correct**Answer- 4) 34/11 hours**

Explanation-

Effective speed of boat x = 15 + 10 = 25 km/h

Effective speed of boat y = 18 – 10 = 8 km/h

Time taken to cross each other = 102/(25 + 8) = 102/33 = 34/11 hoursIncorrect**Answer- 4) 34/11 hours**

Explanation-

Effective speed of boat x = 15 + 10 = 25 km/h

Effective speed of boat y = 18 – 10 = 8 km/h

Time taken to cross each other = 102/(25 + 8) = 102/33 = 34/11 hours

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