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

1 pointsCategory: Quantitative Aptitude**Directions (Q1-5). The data of 8 boats moving in 8 different streams was collected for a day. The line graph gives information about the speed of boats in still water and speed of respective streams in which they are moving. The table below gives information about the total distance travelled by them in the day (both upstream and downstream). Some of the values in the table are missing. Read them carefully and answer the given questions.**

**The boat A goes downstream for 6 hours. The distance travelled by boat D in upstream is 10% less than the distance travelled by boat A in upstream. For how much time does boat D travels in downstream?**Correct**Answer- 1) 6 hours**Distance travelled by boat A in downstream = 18 × 6 = 108 km

Explanation-

Distance travelled by boat A in upstream = 148 – 108 = 40 km

Distance travelled by boat D in upstream = 40 – 10% of 40 = 40 – 4 = 36 km

Distance travelled by boat D in downstream = 156 – 36 = 120 km

Required time = 120/20 = 6 hoursIncorrect**Answer- 1) 6 hours**Distance travelled by boat A in downstream = 18 × 6 = 108 km

Explanation-

Distance travelled by boat A in upstream = 148 – 108 = 40 km

Distance travelled by boat D in upstream = 40 – 10% of 40 = 40 – 4 = 36 km

Distance travelled by boat D in downstream = 156 – 36 = 120 km

Required time = 120/20 = 6 hours - Question 2 of 10
##### 2. Question

1 pointsCategory: Quantitative Aptitude**Directions (Q1-5). The data of 8 boats moving in 8 different streams was collected for a day. The line graph gives information about the speed of boats in still water and speed of respective streams in which they are moving. The table below gives information about the total distance travelled by them in the day (both upstream and downstream). Some of the values in the table are missing. Read them carefully and answer the given questions.**

**What is the total distance travelled by boat C if it goes 3 hours downstream and 16 hours upstream?**Correct**Answer- 3) 112 km**Total distance travelled by boat C = 16 × 3 + 4 × 16 = 48 + 64 = 112 km

Explanation-Incorrect**Answer- 3) 112 km**Total distance travelled by boat C = 16 × 3 + 4 × 16 = 48 + 64 = 112 km

Explanation- - Question 3 of 10
##### 3. Question

1 pointsCategory: Quantitative Aptitude**Directions (Q1-5). The data of 8 boats moving in 8 different streams was collected for a day. The line graph gives information about the speed of boats in still water and speed of respective streams in which they are moving. The table below gives information about the total distance travelled by them in the day (both upstream and downstream). Some of the values in the table are missing. Read them carefully and answer the given questions.**

**Boat F travels in downstream for 4 hours less than the boat E travels downstream and it travels in upstream for 6 hours less than the boat E travels upstream. The distance travelled by boat E in upstream is how much percent less than the distance travelled by boat F in upstream?**Correct**Answer- 3) 58.75%**Let the boat E travels in downstream for x hours and in upstream it travels for y hours.

Explanation-

Then, time for which boat F travels in downstream = (x – 4) h

Time for which boat F travels in upstream = (y – 6) h

We have,

27x + 3y = 222, 9x + y = 74…. (1)

28(x – 4) + 16(y – 6) = 164 or 7x + 4y = 93….. (2)

By solving both the equations we have, x = 7, y = 11

Distance travelled by boat E in upstream = 3 × 11 = 33 km

Distance travelled by boat F in upstream = (11 – 6) × 16 = 80 km

Required percentage = (80 – 33)/80 × 100 = 4700/80 = 58.75 %Incorrect**Answer- 3) 58.75%**Let the boat E travels in downstream for x hours and in upstream it travels for y hours.

Explanation-

Then, time for which boat F travels in downstream = (x – 4) h

Time for which boat F travels in upstream = (y – 6) h

We have,

27x + 3y = 222, 9x + y = 74…. (1)

28(x – 4) + 16(y – 6) = 164 or 7x + 4y = 93….. (2)

By solving both the equations we have, x = 7, y = 11

Distance travelled by boat E in upstream = 3 × 11 = 33 km

Distance travelled by boat F in upstream = (11 – 6) × 16 = 80 km

Required percentage = (80 – 33)/80 × 100 = 4700/80 = 58.75 % - Question 4 of 10
##### 4. Question

1 pointsCategory: Quantitative Aptitude

**Hours for which boat H travels downstream and upstream are two squares of two consecutive numbers (time for which the boat travels downstream is more than that of upstream). The distance travelled by boat H in downstream is how much percent more than the distance travelled by it in upstream?**Correct**Answer- 2) 275%**

**Explanation-**Let the hours for which the boat travels upstream and downstream are x^{2}hours and (x + 1)^{2}hours respectively

Then, we have,

12x^{2}+ 20(x + 1)^{2}= 228

12x^{2}+ 20(x^{2}+ 1 + 2x) = 228

4x^{2}+ 5x – 26 = 0

We have, (x – 2)(4x + 13) = 0

Therefore, x = 2

Distance travelled by boat H in upstream = 12 × 2^{2}= 12 × 4 = 48 km

Distance travelled by boat H in downstream = 20 × (2 + 1)^{2}= 20 × 9 = 180 km

Required percentage = (180 – 48)/48 × 100 = 13200/48 = 275%Incorrect**Answer- 2) 275%**

**Explanation-**Let the hours for which the boat travels upstream and downstream are x^{2}hours and (x + 1)^{2}hours respectively

Then, we have,

12x^{2}+ 20(x + 1)^{2}= 228

12x^{2}+ 20(x^{2}+ 1 + 2x) = 228

4x^{2}+ 5x – 26 = 0

We have, (x – 2)(4x + 13) = 0

Therefore, x = 2

Distance travelled by boat H in upstream = 12 × 2^{2}= 12 × 4 = 48 km

Distance travelled by boat H in downstream = 20 × (2 + 1)^{2}= 20 × 9 = 180 km

Required percentage = (180 – 48)/48 × 100 = 13200/48 = 275% - Question 5 of 10
##### 5. Question

1 pointsCategory: Quantitative Aptitude

**Boat B travels in upstream for 7 hours and boat G travels in downstream for 3 hours. The distance travelled by boat B in upstream is how much percent more than the distance travelled by boat G in downstream?**Correct**Answer- 1) 50/3%**Distance travelled by boat B in upstream = 15 × 7 = 105 km

Explanation-

Distance travelled by boat G in downstream = 30 × 3 = 90 km

Required percentage = (105 – 90)/90 × 100 = 1500/90 = 50/3 %Incorrect**Answer- 1) 50/3%**Distance travelled by boat B in upstream = 15 × 7 = 105 km

Explanation-

Distance travelled by boat G in downstream = 30 × 3 = 90 km

Required percentage = (105 – 90)/90 × 100 = 1500/90 = 50/3 % - Question 6 of 10
##### 6. Question

1 pointsCategory: Quantitative Aptitude**Directions (Q6-10). Three type of liquids P, Q and R are contained in 6 casks A, B, C, D, E and F in different proportions. The line graph below gives information about the percentage of each liquid in the casks. The bar graph below it gives information about the total amount of liquids in four casks A, C, D and E (in liters**

**What is the difference between the average amount of liquid P in casks A, C and D and the average amount of liquid R in the same casks?**Correct**Answer- 4) 1.75 l**Total amount of liquid P in A, C and D = 35% of 105 + 32% of 175 + 25% of 210 = 36.75 + 56 + 52.5 = 145.25 l

Explanation-

Total amount of liquid R in the same casks = 40% of 105 + 20% of 175 + 30% of 210 = 42 + 35 + 63 = 140 l

Required difference = 145.25/3 – 140/3 = 5.25/3 = 1.75 lIncorrect**Answer- 4) 1.75 l**Total amount of liquid P in A, C and D = 35% of 105 + 32% of 175 + 25% of 210 = 36.75 + 56 + 52.5 = 145.25 l

Explanation-

Total amount of liquid R in the same casks = 40% of 105 + 20% of 175 + 30% of 210 = 42 + 35 + 63 = 140 l

Required difference = 145.25/3 – 140/3 = 5.25/3 = 1.75 l - Question 7 of 10
##### 7. Question

1 pointsCategory: Quantitative Aptitude**Directions (Q6-10). Three type of liquids P, Q and R are contained in 6 casks A, B, C, D, E and F in different proportions. The line graph below gives information about the percentage of each liquid in the casks. The bar graph below it gives information about the total amount of liquids in four casks A, C, D and E (in liters**

**40% of the total liquid in A, 50% of the liquid in B and 25% of the liquid in C is drawn and is put in a container. In this container, the total amount of liquid Q is 55.02 l. Before this process, what was the difference between the amount of liquid P in cask B and the amount of liquid R in the same cask?**Correct**Answer- 1) 40.32 l**Let the total amount of liquid in cask C is x l.

Explanation-

Then, according to the question,

40% of 25% of 105 + 50% of 28% of x + 25% of 48% of 175 = 55.02

10.5 + 0.14x + 21 = 55.02

0.14x = 23.52, x = 168 l

Required difference = (48 – 24) % of 168 = 40.32 lIncorrect**Answer- 1) 40.32 l**Let the total amount of liquid in cask C is x l.

Explanation-

Then, according to the question,

40% of 25% of 105 + 50% of 28% of x + 25% of 48% of 175 = 55.02

10.5 + 0.14x + 21 = 55.02

0.14x = 23.52, x = 168 l

Required difference = (48 – 24) % of 168 = 40.32 l - Question 8 of 10
##### 8. Question

1 pointsCategory: Quantitative Aptitude**Directions (Q6-10). Three type of liquids P, Q and R are contained in 6 casks A, B, C, D, E and F in different proportions. The line graph below gives information about the percentage of each liquid in the casks. The bar graph below it gives information about the total amount of liquids in four casks A, C, D and E (in liters**

**60% of the liquid in cask D and 75% of the liquid in cask E is taken out and was mixed in a different vessel. If the cost prizes of liquid P, Q and R are Rs 75, Rs 65 and Rs 90 respectively, what should be the approximate selling price of each liter of the liquid in the vessel in order to gain a profit of 25%?**Correct**Answer- 3) Rs 95**Total amount of liquid P in the vessel after the process = 25% of 60% of 210 + 18% of 75% of 148 = 31.5 + 19.98 = 51.48 l

Explanation-

Total amount of liquid Q = 45% of 60% of 210 + 40% of 75% 148 = 56.7 + 44.4 = 101.1 l

Total amount of liquid R = 30% of 60% of 210 + 42% of 75% of 148 = 37.8 + 46.62 = 84.42 l

Total cost prize = 75 × 51.48 + 65 × 101.1 + 90 × 84.42 = 3861 + 6571.5 + 7597.8 = Rs 18030.3

Selling prize = 125% of 18030.3 = Rs 22537.875

Selling prize per liter = 22537.875 / (51.48 + 101.1 + 84.42) = 22537.875 / 237 = Rs 95 (approx.)Incorrect**Answer- 3) Rs 95**Total amount of liquid P in the vessel after the process = 25% of 60% of 210 + 18% of 75% of 148 = 31.5 + 19.98 = 51.48 l

Explanation-

Total amount of liquid Q = 45% of 60% of 210 + 40% of 75% 148 = 56.7 + 44.4 = 101.1 l

Total amount of liquid R = 30% of 60% of 210 + 42% of 75% of 148 = 37.8 + 46.62 = 84.42 l

Total cost prize = 75 × 51.48 + 65 × 101.1 + 90 × 84.42 = 3861 + 6571.5 + 7597.8 = Rs 18030.3

Selling prize = 125% of 18030.3 = Rs 22537.875

Selling prize per liter = 22537.875 / (51.48 + 101.1 + 84.42) = 22537.875 / 237 = Rs 95 (approx.) - Question 9 of 10
##### 9. Question

1 pointsCategory: Quantitative Aptitude

**The liquids in cask E and F are poured in another cask and are mixed together. If the ratio of total amount of liquid P to the total amount of liquid Q in this cask becomes 2394:3587 respectively, what was the total amount of mixture in the cask F before this process?**Correct**Answer- 2) 192 l**Let the total amount of mixture in the cask F is 100x l.

Explanation-

Total amount of liquid P in new cask after this process = 18% of 148 + 36% of 100x = 26.64 + 36x

Total amount of liquid Q in new cask = 40% of 148 + 44% of 100x = 59.2 + 44x

According to the question,

(26.64 + 36x)/(59.2 + 44x) = 2394/3587

95557.68 + 129132x = 141724.8 + 105336x

23796x = 46167.12

x = 1.92

100x = 192 lIncorrect**Answer- 2) 192 l**Let the total amount of mixture in the cask F is 100x l.

Explanation-

Total amount of liquid P in new cask after this process = 18% of 148 + 36% of 100x = 26.64 + 36x

Total amount of liquid Q in new cask = 40% of 148 + 44% of 100x = 59.2 + 44x

According to the question,

(26.64 + 36x)/(59.2 + 44x) = 2394/3587

95557.68 + 129132x = 141724.8 + 105336x

23796x = 46167.12

x = 1.92

100x = 192 l - Question 10 of 10
##### 10. Question

1 pointsCategory: Quantitative Aptitude

**5 l of the mixture in the cask A was replaced by water. This process was done twice more. What is the approximate amount of water in the mixture after these processes?**Correct**Answer- 1) 14.3 l**We can take the mixture in cask A as one entity.

Explanation-

We have,

amount of water after these processes = 105 – 105[(105-100)3/1053] = 105 – 105 × 8000/9261 = 105 – 90.7029478 = 14.3 l (approx.)Incorrect**Answer- 1) 14.3 l**We can take the mixture in cask A as one entity.

Explanation-

We have,

amount of water after these processes = 105 – 105[(105-100)3/1053] = 105 – 105 × 8000/9261 = 105 – 90.7029478 = 14.3 l (approx.)

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