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

1 pointsCategory: Quantitative Aptitude**Directions (Q1-5). The following line graph shows the time taken by six men i.e. A, B, C, D, E and F to do a work X.**

**In how many days will B and C complete a work which is double the work X?**Correct**Answer- 3) 19.8 days**

Explanation-

B and C will take double the time to complete this work i.e. 36 and 44 days respectively.

LCM of 36 and 44 = 396

396/36 = 11 and 396/44 = 9

Time they will take to compete the work = 396/(11 + 9) = 396/20 = 19.8 daysIncorrect**Answer- 3) 19.8 days**

Explanation-

B and C will take double the time to complete this work i.e. 36 and 44 days respectively.

LCM of 36 and 44 = 396

396/36 = 11 and 396/44 = 9

Time they will take to compete the work = 396/(11 + 9) = 396/20 = 19.8 days - Question 2 of 10
##### 2. Question

1 pointsCategory: Quantitative Aptitude**Directions (Q1-5). The following line graph shows the time taken by six men i.e. A, B, C, D, E and F to do a work X.**

**There’s another person G whose efficiency is 1/3rd of the efficiency of D. In how many days he will complete a piece of work which is 1.5 times of the work X?**Correct**Answer- 1) 216 days**

Explanation-

Since the efficiency of G is 1/3rd of the efficiency of D, time taken by it to complete the work X = 48 × 3 = 144 days

Time taken by it complete a work which will be 1.5 times of X = 1.5 × 144 = 216 daysIncorrect**Answer- 1) 216 days**

Explanation-

Since the efficiency of G is 1/3rd of the efficiency of D, time taken by it to complete the work X = 48 × 3 = 144 days

Time taken by it complete a work which will be 1.5 times of X = 1.5 × 144 = 216 days - Question 3 of 10
##### 3. Question

1 pointsCategory: Quantitative Aptitude**Directions (Q1-5). The following line graph shows the time taken by six men i.e. A, B, C, D, E and F to do a work X.**

**E and F started doing the work X together. However, after 10 days E left the work and the rest of the work was completed by F with the help of H who is half as efficient as A. In approximately how much time the work will be completed?**Correct**Answer- 2) 19 days**

Explanation-

Time taken by H to complete the work X alone = 35 × 2 = 70 days

LCM of 70, 25 and 40 = 1400 = total work

Work completed by H in one day = 1400/70 = 20

Work completed by E in one day = 1400/25 = 56

Work completed by F in one day = 1400/40 = 35

Work completed by E and F in 10 days = (56 + 35) × 10 = 910

Remaining work = 1400 – 910 = 490

Time taken by F and H to complete this work = 490/(35 + 20) = 490/55 = 9 (approx.)

Total time taken for the work to complete = 10 + 9 = 19 daysIncorrect**Answer- 2) 19 days**

Explanation-

Time taken by H to complete the work X alone = 35 × 2 = 70 days

LCM of 70, 25 and 40 = 1400 = total work

Work completed by H in one day = 1400/70 = 20

Work completed by E in one day = 1400/25 = 56

Work completed by F in one day = 1400/40 = 35

Work completed by E and F in 10 days = (56 + 35) × 10 = 910

Remaining work = 1400 – 910 = 490

Time taken by F and H to complete this work = 490/(35 + 20) = 490/55 = 9 (approx.)

Total time taken for the work to complete = 10 + 9 = 19 days - Question 4 of 10
##### 4. Question

1 pointsCategory: Quantitative Aptitude

**In approximately how many days A, B and D can complete a piece of work which is 4 times as much of work X?**Correct**Answer- 3) 38 days**

Explanation-

LCM of 35, 18 and 48 = 5040

Work completed by A in one day = 5040/35 = 144

Work completed by B in one day = 5040/18 = 280

Work completed by D in one day = 5040/48 = 105

Time taken by them to complete the work X together = 5040/(144 + 280 + 105) = 5040/529 = 9.53 days

Time taken by them to complete 4 times as this work = 9.53 × 4 = 38 days (approx.)Incorrect**Answer- 3) 38 days**

Explanation-

LCM of 35, 18 and 48 = 5040

Work completed by A in one day = 5040/35 = 144

Work completed by B in one day = 5040/18 = 280

Work completed by D in one day = 5040/48 = 105

Time taken by them to complete the work X together = 5040/(144 + 280 + 105) = 5040/529 = 9.53 days

Time taken by them to complete 4 times as this work = 9.53 × 4 = 38 days (approx.) - Question 5 of 10
##### 5. Question

1 pointsCategory: Quantitative Aptitude

**A with 1/5th of his efficiency and E with his usual efficiency can complete a work Y in 25 days. The amount of work X is how much percent less than that of Y?**Correct**Answer- 4) 12.5%**

Explanation-

Time taken by A to complete work X with 1/5th of his efficiency = 35 × 5 = 175 days

Time taken by E to complete work X = 25 days

LCM of 175 and 25 = 175

Work completed by A in one day with depleted efficiency = 175/175 = 1

Work completed by E in one day = 175/25 = 7

Time taken by them to complete the work X = 175/8 = 21.875 days

Required percentage = (25 – 21.875)/25 × 100 = 312.5/25 = 12.5%Incorrect**Answer- 4) 12.5%**

Explanation-

Time taken by A to complete work X with 1/5th of his efficiency = 35 × 5 = 175 days

Time taken by E to complete work X = 25 days

LCM of 175 and 25 = 175

Work completed by A in one day with depleted efficiency = 175/175 = 1

Work completed by E in one day = 175/25 = 7

Time taken by them to complete the work X = 175/8 = 21.875 days

Required percentage = (25 – 21.875)/25 × 100 = 312.5/25 = 12.5% - Question 6 of 10
##### 6. Question

1 pointsCategory: Quantitative Aptitude**Directions (Q6-10). The following pie charts show the distribution of alcohol and water in six casks i.e. A, B, C, D, E and F is shown. Total amount of alcohol and water keep changing. However, their percentage distribution is done in same way as shown in the following pie charts.**

**The amount of alcohol in cask C is 35 l. The ratio of total amount of alcohol to the total amount of water is 2:1. The amount of water in cask F is:**Correct**Answer- 1) 28 l**

Explanation-

Total amount of alcohol = 35/10 × 100 = 350 l

Total amount of water = 350/2 = 175 l

Amount of water in cask F = 16% of 175 = 28 lIncorrect**Answer- 1) 28 l**

Explanation-

Total amount of alcohol = 35/10 × 100 = 350 l

Total amount of water = 350/2 = 175 l

Amount of water in cask F = 16% of 175 = 28 l - Question 7 of 10
##### 7. Question

1 pointsCategory: Quantitative Aptitude**Directions (Q6-10). The following pie charts show the distribution of alcohol and water in six casks i.e. A, B, C, D, E and F is shown. Total amount of alcohol and water keep changing. However, their percentage distribution is done in same way as shown in the following pie charts.**

**One day, the ratio of total amount of alcohol to water is 5:2 respectively in all the six casks together. Alcohol in cask E and water in cask A were removed and added in another cask G which already had 73.2 l water in it. After this process, the ratio of alcohol to water becomes 3:10 respectively in the cask G. What is the amount of alcohol in cask B?**Correct**Answer- 5) 32.4 l**

Explanation:

Let the total amount of alcohol and water are 5x and 2x respectively.

Then,

Alcohol in cask E = 14% of 5x = 0.7x

Water in cask A = 15% of 2x = 0.3x

We have,

0.7x/(0.3x + 73.2) = 3/10

7x = 0.9x + 219.6

6.1x = 219.6

x = 219.6/6.1 = 36

Total amount of alcohol = 5 × 36 = 180 l

Amount of alcohol in cask B = 18% of 180 = 32.4 lIncorrect**Answer- 5) 32.4 l**

Explanation:

Let the total amount of alcohol and water are 5x and 2x respectively.

Then,

Alcohol in cask E = 14% of 5x = 0.7x

Water in cask A = 15% of 2x = 0.3x

We have,

0.7x/(0.3x + 73.2) = 3/10

7x = 0.9x + 219.6

6.1x = 219.6

x = 219.6/6.1 = 36

Total amount of alcohol = 5 × 36 = 180 l

Amount of alcohol in cask B = 18% of 180 = 32.4 l - Question 8 of 10
##### 8. Question

1 pointsCategory: Quantitative Aptitude**Directions (Q6-10). The following pie charts show the distribution of alcohol and water in six casks i.e. A, B, C, D, E and F is shown. Total amount of alcohol and water keep changing. However, their percentage distribution is done in same way as shown in the following pie charts.**

**If the total amount of alcohol is 450 l and the amount of alcohol in cask D is 80.5 l more than the amount of water in cask F. What is the amount of water in cask C?**Correct**Answer- 3) 16 l**

Explanation-

Amount of alcohol in cask D = 25% of 450 = 112.5 l

Amount of water in cask F = 112.5 – 80.5 = 32 l

Amount of water in cask C = 32/16 × 8 = 16 lIncorrect**Answer- 3) 16 l**

Explanation-

Amount of alcohol in cask D = 25% of 450 = 112.5 l

Amount of water in cask F = 112.5 – 80.5 = 32 l

Amount of water in cask C = 32/16 × 8 = 16 l - Question 9 of 10
##### 9. Question

1 pointsCategory: Quantitative Aptitude

**The ratio of alcohol to water in cask E is 7:4 respectively. The amount of alcohol in cask C is how much percent more than that of amount of water in the same cask?**Correct**Answer- 5) 275%**

Explanation-

Let the total amount of alcohol is x and total amount of water is y.

Then,

14% of x / 24% of y = 7/4

7x / 12y = 7/4

x/y = 3/1

Therefore, ratio of total amount of alcohol to total amount of water is 3:1.

Now let the total amount of water is 300 l and total amount of water is 100 l.

Then,

Amount of alcohol in cask C = 10% of 300 = 30 l

Amount of water in cask C = 8% of 100 = 8 l

Required percentage = (30 – 8)/8 × 100 = 2200/8 = 275%Incorrect**Answer- 5) 275%**

Explanation-

Let the total amount of alcohol is x and total amount of water is y.

Then,

14% of x / 24% of y = 7/4

7x / 12y = 7/4

x/y = 3/1

Therefore, ratio of total amount of alcohol to total amount of water is 3:1.

Now let the total amount of water is 300 l and total amount of water is 100 l.

Then,

Amount of alcohol in cask C = 10% of 300 = 30 l

Amount of water in cask C = 8% of 100 = 8 l

Required percentage = (30 – 8)/8 × 100 = 2200/8 = 275% - Question 10 of 10
##### 10. Question

1 pointsCategory: Quantitative Aptitude

**One day, the whole mixture of alcohol and water in cask A was poured in the cask B due to which the ratio of alcohol to water in cask B becomes 5:7. On the same day the amount of water in cask B was how much percent more than the amount of alcohol in cask E?**Correct**Answer- 2) 75%**

Explanation-

Let the total amount of alcohol and water are x and y respectively.

Then,

(12% of x + 18% of x) / (15% of y + 21% of y) = 5/7

30x / 36y = 5/7

x/y = 6/7

Now let the total amount of alcohol is 600 l and total amount of water is 700 l

Amount of water is cask B = 21% of 700 = 147 l

Amount of alcohol in cask E = 14% of 600 = 84 l

Required percentage = (147 – 84)/84 × 100 = 6300/84 = 75%Incorrect**Answer- 2) 75%**

Explanation-

Let the total amount of alcohol and water are x and y respectively.

Then,

(12% of x + 18% of x) / (15% of y + 21% of y) = 5/7

30x / 36y = 5/7

x/y = 6/7

Now let the total amount of alcohol is 600 l and total amount of water is 700 l

Amount of water is cask B = 21% of 700 = 147 l

Amount of alcohol in cask E = 14% of 600 = 84 l

Required percentage = (147 – 84)/84 × 100 = 6300/84 = 75%

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