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

1 pointsCategory: Quantitative Aptitude**Directions (Q1-5). Six persons i.e. A, B, C, D, E and F have different efficiencies and therefore take unique number of days to do a piece of work. Time taken by B to complete the work alone is 50% more than that of by A. Ratio of efficiencies of A and C is 5:2 respectively. Time taken by F to do the work alone is equal to the average time taken by A, B and C to do the work (each one alone). Time taken by D to do the work alone is half the time taken by E to do it alone. The sum of the number of days taken by C to complete the work alone and that of by E are 95. Number of days taken by D to complete the work alone are 7 more than that of by A.****Number of days taken by E to complete the work alone are:**Correct**Answer- 3) 50**

Explanation-

Let the number of days taken by A is x.

Then, number of days taken by B to complete the work = 1.5x

Efficiencies of A and C are in ratio 5:2, therefore, the number of days taken by them will be in ratio 2:5.

Number of days taken by C to complete the work = x/2 × 5 = 5x/2

Number of days taken by F to do the work = (x + 1.5x + 5x/2)/3 = 5x/3

Time taken by D to do the work = x + 7

Time taken by E to do the work = 2(x + 7)

We have,

5x/2 + 2(x + 7) = 95

2.5x + 2x = 81

4.5x = 81

x = 81/4.5 = 18

Therefore, the number of days taken by A to complete the work = 18

Number of days taken by B to complete the work = 27

Number of days taken by C to complete the work = 5 × 18/2 = 45

Number of days taken by D to do the work = 18 + 7 = 25

Number of days taken by E to do the work = 2(18 + 7) = 50

Number of days taken by F to do the work = 5 × 18/3 = 30Incorrect**Answer- 3) 50**

Explanation-

Let the number of days taken by A is x.

Then, number of days taken by B to complete the work = 1.5x

Efficiencies of A and C are in ratio 5:2, therefore, the number of days taken by them will be in ratio 2:5.

Number of days taken by C to complete the work = x/2 × 5 = 5x/2

Number of days taken by F to do the work = (x + 1.5x + 5x/2)/3 = 5x/3

Time taken by D to do the work = x + 7

Time taken by E to do the work = 2(x + 7)

We have,

5x/2 + 2(x + 7) = 95

2.5x + 2x = 81

4.5x = 81

x = 81/4.5 = 18

Therefore, the number of days taken by A to complete the work = 18

Number of days taken by B to complete the work = 27

Number of days taken by C to complete the work = 5 × 18/2 = 45

Number of days taken by D to do the work = 18 + 7 = 25

Number of days taken by E to do the work = 2(18 + 7) = 50

Number of days taken by F to do the work = 5 × 18/3 = 30 - Question 2 of 10
##### 2. Question

1 pointsCategory: Quantitative Aptitude**Directions (Q1-5). Six persons i.e. A, B, C, D, E and F have different efficiencies and therefore take unique number of days to do a piece of work. Time taken by B to complete the work alone is 50% more than that of by A. Ratio of efficiencies of A and C is 5:2 respectively. Time taken by F to do the work alone is equal to the average time taken by A, B and C to do the work (each one alone). Time taken by D to do the work alone is half the time taken by E to do it alone. The sum of the number of days taken by C to complete the work alone and that of by E are 95. Number of days taken by D to complete the work alone are 7 more than that of by A.****Number of days taken by F to complete the work are how much more than that of by A?**Correct**Answer- 2) 12**

Explanation-

Required difference = 30 – 18 = 12 (watch the solution of Q1)Incorrect**Answer- 2) 12**

Explanation-

Required difference = 30 – 18 = 12 (watch the solution of Q1) - Question 3 of 10
##### 3. Question

1 pointsCategory: Quantitative Aptitude**Directions (Q1-5). Six persons i.e. A, B, C, D, E and F have different efficiencies and therefore take unique number of days to do a piece of work. Time taken by B to complete the work alone is 50% more than that of by A. Ratio of efficiencies of A and C is 5:2 respectively. Time taken by F to do the work alone is equal to the average time taken by A, B and C to do the work (each one alone). Time taken by D to do the work alone is half the time taken by E to do it alone. The sum of the number of days taken by C to complete the work alone and that of by E are 95. Number of days taken by D to complete the work alone are 7 more than that of by A.****A, B and C started the work together but C left after 5 days. The rest of the work was completed by A and B together. The total time taken by them to complete the work is:**Correct**Answer- 3) 9.6 days**

Explanation-

LCM of 18, 27 and 45 = 270 = total work

Work completed by A in one day = 270/18 = 15

Work completed by B in one day = 270/27 = 10

Work complete by C in one day = 270/45 = 6

Work completed by A, B and C in five days = (15 + 10 + 6) × 5 = 31 × 5 = 155

Remaining work = 270 – 155 = 115

Time taken by A and B to complete the rest of the work = 115/(15 + 10) = 115/25 = 4.6 days

Total time taken to complete the work = 5 + 4.6 = 9.6 daysIncorrect**Answer- 3) 9.6 days**

Explanation-

LCM of 18, 27 and 45 = 270 = total work

Work completed by A in one day = 270/18 = 15

Work completed by B in one day = 270/27 = 10

Work complete by C in one day = 270/45 = 6

Work completed by A, B and C in five days = (15 + 10 + 6) × 5 = 31 × 5 = 155

Remaining work = 270 – 155 = 115

Time taken by A and B to complete the rest of the work = 115/(15 + 10) = 115/25 = 4.6 days

Total time taken to complete the work = 5 + 4.6 = 9.6 days - Question 4 of 10
##### 4. Question

1 pointsCategory: Quantitative Aptitude**D and F work on alternate days starting with D. In how many days the work will be completed?**Correct**Answer- 1) 27.2 days**

Explanation-

LCM of 25 and 30 = 150 = total work

Work done by D in one day = 150/25 = 6

Work done by F in one day = 150/30 = 5

Total work done by them in two days = 6 + 5 = 11

Work done by them in 26 days = 11 × 13 = 143

Work done in 27 days = 143 + 6 = 149

Time taken by F to complete the remaining work = 1/5 = 0.2 day

Total time taken = 27 + 0.2 = 27.2 daysIncorrect**Answer- 1) 27.2 days**

Explanation-

LCM of 25 and 30 = 150 = total work

Work done by D in one day = 150/25 = 6

Work done by F in one day = 150/30 = 5

Total work done by them in two days = 6 + 5 = 11

Work done by them in 26 days = 11 × 13 = 143

Work done in 27 days = 143 + 6 = 149

Time taken by F to complete the remaining work = 1/5 = 0.2 day

Total time taken = 27 + 0.2 = 27.2 days - Question 5 of 10
##### 5. Question

1 pointsCategory: Quantitative Aptitude**There is another person G whose efficiency is 3 times of the efficiency of A. E, F and G take the responsibility of doing the work but they work on alternate days starting with E (E works on the first day, F works on the second day, G works on the third day and so on). In how many days the work will be completed?**Correct**Answer- 5) 14.4 days**

Explanation-

Since the efficiency of G is 3 times as much as of A, the number of days taken by G to complete the work alone = 18/3 = 6 days

LCM of 50, 30 and 6 = 150 = total work

Work done by E in one day = 150/50 = 3

Work done by F in one day = 150/30 = 5

Work done by G in one day = 150/6 = 25

Total work done by them in three days = 3 + 5 + 25 = 33

Work done by them in twelve days = 33 × 4 = 132

Work done on the 13th day = 3

Work done on the 14th day = 5

Remaining work = 150 – (132 + 3 + 5) = 10

Time taken to complete the remaining work = 10/25 = 0.4

Total time taken = 14 + 0.4 = 14.4 daysIncorrect**Answer- 5) 14.4 days**

Explanation-

Since the efficiency of G is 3 times as much as of A, the number of days taken by G to complete the work alone = 18/3 = 6 days

LCM of 50, 30 and 6 = 150 = total work

Work done by E in one day = 150/50 = 3

Work done by F in one day = 150/30 = 5

Work done by G in one day = 150/6 = 25

Total work done by them in three days = 3 + 5 + 25 = 33

Work done by them in twelve days = 33 × 4 = 132

Work done on the 13th day = 3

Work done on the 14th day = 5

Remaining work = 150 – (132 + 3 + 5) = 10

Time taken to complete the remaining work = 10/25 = 0.4

Total time taken = 14 + 0.4 = 14.4 days - Question 6 of 10
##### 6. Question

1 pointsCategory: Quantitative Aptitude**Directions (Q6-10). Compare the value of 2 quantities given in the following questions and choose the correct option (only numerical values)****Quantity 1**– Difference in time taken by A and B to reach their respective destinations (in hours). A started from a point X towards Y. After two hours, B started from Y towards X. By the time A travelled 1/5th of the total distance, B had also travelled the same. The speed of A is 1/3rd of the speed of B.

**Quantity 2**– Distance travelled by the car (in km). The speed of the car reduced to 4/5th of his original speed due to which the time taken by car to reach his destination is increased by 27/16 minutes. The reduced speed of the car is 64 km/h.Correct**Answer- 3) Quantity 1 > Quantity 2**

**Explanation-**

For quantity 1

Let the time taken by A to travel 1/5^{th}of the total distance is x hours

Then, time taken by B to travel 1/5^{th}of the distance = (x – 2) hours

According to question,

x/(x – 2) = 3/1

x = 3x – 6

2x = 6, x = 3 hours

Total time taken by A to reach his destination = 3 × 5 = 15 hours

Total time taken by B to reach his destination = (3 – 2) × 5 =5 hours

Difference of time = 15 – 5 = 10 hours = Quantity 1For quantity 2

Time taken by the car to reach his destination = 27/16 × 1/60 × 5 = 9/64 hours

Distance travelled = 64 × 9/64 = 9 km = Quantity 2

Therefore, Quantity 1 > Quantity 2Incorrect**Answer- 3) Quantity 1 > Quantity 2**

**Explanation-**

For quantity 1

Let the time taken by A to travel 1/5^{th}of the total distance is x hours

Then, time taken by B to travel 1/5^{th}of the distance = (x – 2) hours

According to question,

x/(x – 2) = 3/1

x = 3x – 6

2x = 6, x = 3 hours

Total time taken by A to reach his destination = 3 × 5 = 15 hours

Total time taken by B to reach his destination = (3 – 2) × 5 =5 hours

Difference of time = 15 – 5 = 10 hours = Quantity 1For quantity 2

Time taken by the car to reach his destination = 27/16 × 1/60 × 5 = 9/64 hours

Distance travelled = 64 × 9/64 = 9 km = Quantity 2

Therefore, Quantity 1 > Quantity 2 - Question 7 of 10
##### 7. Question

1 pointsCategory: Quantitative Aptitude**Directions (Q6-10). Compare the value of 2 quantities given in the following questions and choose the correct option (only numerical values)****Quantity 1-**Rate of compound interest per annum at which Rs 20000 becomes Rs 23152.5 in a time period of 9 months when the interest is compounded quarterly.

**Quantity 2-**The rate of compound interest per annum for which a sum of money becomes 9/4 times of itself in 2 years when the interest is compounded annually.Correct**Answer- 4) Quantity 2 > Quantity 1**

**Explanation-**

For quantity 1

We have,

(23152.5 – 20000)/20000 × 100 = 15.7625

Therefore, the rate of interest per annum = 5 × 4 = 20% = Quantity 1For quantity 2

We have,

(9 – 4)/4 × 100 = 500/4 = 125%

Therefore, the rate of interest = 50% = Quantity 2Incorrect**Answer- 4) Quantity 2 > Quantity 1**

**Explanation-**

For quantity 1

We have,

(23152.5 – 20000)/20000 × 100 = 15.7625

Therefore, the rate of interest per annum = 5 × 4 = 20% = Quantity 1For quantity 2

We have,

(9 – 4)/4 × 100 = 500/4 = 125%

Therefore, the rate of interest = 50% = Quantity 2 - Question 8 of 10
##### 8. Question

1 pointsCategory: Quantitative Aptitude**Directions (Q6-10). Compare the value of 2 quantities given in the following questions and choose the correct option (only numerical values)**A sold an article to B at a profit of 17% and B sold it to C at a profit of 25%. C sold it to D at 6/5 of his cost price.

**Quantity 1**– Profit earned by A if cost price of the article for C was Rs 5703.75.

**Quantity 2**– Loss incurred to D if the cost price of the article for A was Rs 2500 and the D sold the article at Rs 3724.5.Correct**Answer- 2) Quantity 1 = Quantity 2**

Explanation-

Let the cost price of the article for A is Rs 100x.

Then, cost price of the article for B = Rs 117x

Cost price of the article for C = 125% of 117x = Rs 146.25x

Cost price of the article for D = 6/5 × 146.25x = Rs 175.5x

For quantity 1,

Profit earned by A = 5703.75/146.25x × 17x = Rs 663

For quantity 2,

Cost price of D = 2500/100x × 175.5x = Rs 4387.5

Profit incurred to D = 4387.5 – 3724.5 = Rs 663

Therefore, quantity 1 = quantity 2Incorrect**Answer- 2) Quantity 1 = Quantity 2**

Explanation-

Let the cost price of the article for A is Rs 100x.

Then, cost price of the article for B = Rs 117x

Cost price of the article for C = 125% of 117x = Rs 146.25x

Cost price of the article for D = 6/5 × 146.25x = Rs 175.5x

For quantity 1,

Profit earned by A = 5703.75/146.25x × 17x = Rs 663

For quantity 2,

Cost price of D = 2500/100x × 175.5x = Rs 4387.5

Profit incurred to D = 4387.5 – 3724.5 = Rs 663

Therefore, quantity 1 = quantity 2 - Question 9 of 10
##### 9. Question

1 pointsCategory: Quantitative AptitudeA box contains total 64 balls. There’re three types of balls i.e. red, yellow and black. Number of black balls are 20% less than that of red balls and number of yellow balls 1/3rd of the number of red balls.

**Quantity 1-**If 30 white balls are added in the box and a ball is picked randomly from it, what is the probability of getting a red ball?

**Quantity 2-**If 15 blue balls are added in the box and a ball is picked randomly from it, what is the probability of getting a black ball?Correct**Answer- 3) Quantity 1 > Quantity 2**

**Explanation-**

Let the number of red balls in the box are x.

Then, number of black balls = 80% of x = 0.8x

Number of yellow balls = x/3

x + 0.8x + x/3 = 64

6.4x/3 = 64

x = 30

Number of red balls = 30

Number of black balls = 0.8 × 30 = 24

Number of yellow balls = 30/3 = 10

For quantity 1

Probability of getting a red ball = 30/(64 + 30) = 30/94 = 15/47 = quantity 1

For quantity 2

Probability of getting a black ball = 24/(64 + 15) = 24/79 = quantity 2

15/47 > 24/79

Therefore, quantity 1 > quantity 2Incorrect**Answer- 3) Quantity 1 > Quantity 2**

**Explanation-**

Let the number of red balls in the box are x.

Then, number of black balls = 80% of x = 0.8x

Number of yellow balls = x/3

x + 0.8x + x/3 = 64

6.4x/3 = 64

x = 30

Number of red balls = 30

Number of black balls = 0.8 × 30 = 24

Number of yellow balls = 30/3 = 10

For quantity 1

Probability of getting a red ball = 30/(64 + 30) = 30/94 = 15/47 = quantity 1

For quantity 2

Probability of getting a black ball = 24/(64 + 15) = 24/79 = quantity 2

15/47 > 24/79

Therefore, quantity 1 > quantity 2 - Question 10 of 10
##### 10. Question

1 pointsCategory: Quantitative AptitudeA, B and C are business partners. A invested Rs 5000 for 12 months, B invested Rs 8000 for 8 months and C invested Rs 12500 for 6 months. Profit earned by B is Rs 1200 more than that of A.

**Quantity 1**– 25% of the average profit earned by all of them.

**Quantity 2**– Difference between the profit earned by B and by C.Correct**Answer- 3) Quantity 1 > Quantity 2**

**Explanation-**

Ratio in which the profit will be divided among A, B and C is,

5000 × 12 : 8000 × 8 : 12500 × 6 = 60:64:75

Total profit earned = 1200/4 × 199 = Rs 59700

Quantity 1 = 25% of 59700/3 = Rs 4975

Quantity 2 = 1200/4 × 11 = Rs 3300

Therefore, Quantity 1 > Quantity 2Incorrect**Answer- 3) Quantity 1 > Quantity 2**

**Explanation-**

Ratio in which the profit will be divided among A, B and C is,

5000 × 12 : 8000 × 8 : 12500 × 6 = 60:64:75

Total profit earned = 1200/4 × 199 = Rs 59700

Quantity 1 = 25% of 59700/3 = Rs 4975

Quantity 2 = 1200/4 × 11 = Rs 3300

Therefore, Quantity 1 > Quantity 2

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