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

1 pointsCategory: Quantitative Aptitude**Directions (1-5) In a school all students play three type of sports (cricket, hockey and tennis). Total number of students who play cricket, hockey and tennis are 548, 557 and 561 respectively. Total students who play both cricket and hockey are 375, total students who play both hockey and tennis are 372 and total students who play both tennis and cricket are 371. Sum of students who play only cricket, only hockey and only tennis is 438. Students who play only tennis are 16 more than student who play only cricket and students who play only cricket are 8 less than students who play only hockey.****How many students are there in the school?**Correct**Answer – 2) 884**

Explanation for this question –

Total students who play sports = C +H + T + x + y + z + p

= 138+146+154+ 39+35+36+ 336 = 884**Complete explanation-**

Let the number of students who play only cricket are a

Number of students who play only hockey will be a+8

Number of students who play only tennis will be a+16

(a+a+8+a+16) = 438

3a+24 = 438

3a = 414

a =138

Only cricket (C) = 138

Only Hockey (H) = 146

Only Tennis (T) = 154

Those who play are three sports = p

Both Cricket and Hockey (x+p) = 375

Both Hockey and Tennis (y+p) = 372

Both Cricket and Tennis (z+p) = 371

548 =C+ x+p+z

548 =138+ 375+z

Z = 35

Z+p = 371

P = 336

x+p = 375

x = 39

y+p = 372

y = 36

Total students who play sports = C +H + T + x + y + z + p

= 138+146+154+ 39+35+36+ 336 = 884Incorrect**Answer – 2) 884**

Explanation for this question –

Total students who play sports = C +H + T + x + y + z + p

= 138+146+154+ 39+35+36+ 336 = 884**Complete explanation-**

Let the number of students who play only cricket are a

Number of students who play only hockey will be a+8

Number of students who play only tennis will be a+16

(a+a+8+a+16) = 438

3a+24 = 438

3a = 414

a =138

Only cricket (C) = 138

Only Hockey (H) = 146

Only Tennis (T) = 154

Those who play are three sports = p

Both Cricket and Hockey (x+p) = 375

Both Hockey and Tennis (y+p) = 372

Both Cricket and Tennis (z+p) = 371

548 =C+ x+p+z

548 =138+ 375+z

Z = 35

Z+p = 371

P = 336

x+p = 375

x = 39

y+p = 372

y = 36

Total students who play sports = C +H + T + x + y + z + p

= 138+146+154+ 39+35+36+ 336 = 884 - Question 2 of 10
##### 2. Question

1 pointsCategory: Quantitative Aptitude**Directions (1-5) In a school all students play three type of sports (cricket, hockey and tennis). Total number of students who play cricket, hockey and tennis are 548, 557 and 561 respectively. Total students who play both cricket and hockey are 375, total students who play both hockey and tennis are 372 and total students who play both tennis and cricket are 371. Sum of students who play only cricket, only hockey and only tennis is 438. Students who play only tennis are 16 more than student who play only cricket and students who play only cricket are 8 less than students who play only hockey.****How many students play any two sports but not all three?**Correct**Answer: 3) 110**

Explanation:

X+y+z = 35+36+39 = 110Incorrect**Answer: 3) 110**

Explanation:

X+y+z = 35+36+39 = 110 - Question 3 of 10
##### 3. Question

1 pointsCategory: Quantitative Aptitude**Directions (1-5) In a school all students play three type of sports (cricket, hockey and tennis). Total number of students who play cricket, hockey and tennis are 548, 557 and 561 respectively. Total students who play both cricket and hockey are 375, total students who play both hockey and tennis are 372 and total students who play both tennis and cricket are 371. Sum of students who play only cricket, only hockey and only tennis is 438. Students who play only tennis are 16 more than student who play only cricket and students who play only cricket are 8 less than students who play only hockey.****How many students play all the three sports?**Correct**Answer – 1) 336**

Explanation–

P=336Incorrect**Answer – 1) 336**

Explanation–

P=336 - Question 4 of 10
##### 4. Question

1 pointsCategory: Quantitative Aptitude**What is the sum of the number of students who play only cricket and students who play both tennis and hockey but not cricket?**Correct**Answer – 3) 174**

Explanation–

Only cricket = 138

Both tennis and hockey but not cricket (y) = 36

Total = 138+36 =174Incorrect**Answer – 3) 174**

Explanation–

Only cricket = 138

Both tennis and hockey but not cricket (y) = 36

Total = 138+36 =174 - Question 5 of 10
##### 5. Question

1 pointsCategory: Quantitative Aptitude**What is the ratio of the sum of the number of students who play only cricket and both cricket and hockey but not tennis to the sum of number of students who play only hockey and both hockey and tennis but not cricket?**Correct**Answer – 1) 177/182**

Explanation:

(138+39)/ (146+36) = 177/182

Let the number of students who play only cricket are a

Number of students who play only hockey will be a+8

Number of students who play only tennis will be a+16

(a+a+8+a+16) = 438

3a+24 = 438

3a = 414

a =138

Only cricket (C) = 138

Only Hockey (H) = 146

Only Tennis (T) = 154

Those who play are three sports = p

Both Cricket and Hockey (x+p) = 375

Both Hockey and Tennis (y+p) = 372

Both Cricket and Tennis (z+p) = 371

548 =C+ x+p+z

548 =138+ 375+z

Z = 35

Z+p = 371

P = 336

x+p = 375

x = 39

y+p = 372

y = 36

Total students who play sports = C +H + T + x + y + z + p

= 138+146+154+ 39+35+36+ 336 = 884Incorrect**Answer – 1) 177/182**

Explanation:

(138+39)/ (146+36) = 177/182

Let the number of students who play only cricket are a

Number of students who play only hockey will be a+8

Number of students who play only tennis will be a+16

(a+a+8+a+16) = 438

3a+24 = 438

3a = 414

a =138

Only cricket (C) = 138

Only Hockey (H) = 146

Only Tennis (T) = 154

Those who play are three sports = p

Both Cricket and Hockey (x+p) = 375

Both Hockey and Tennis (y+p) = 372

Both Cricket and Tennis (z+p) = 371

548 =C+ x+p+z

548 =138+ 375+z

Z = 35

Z+p = 371

P = 336

x+p = 375

x = 39

y+p = 372

y = 36

Total students who play sports = C +H + T + x + y + z + p

= 138+146+154+ 39+35+36+ 336 = 884 - Question 6 of 10
##### 6. Question

1 pointsCategory: Quantitative Aptitude**Anil and Bunty run from Point P to Point Q a distance of 1200 meter at 4m/s and 5m/s respectively. Bunty reaches Y and immediately turns back meeting Anil at point R. What is the distance from Q to R?**Correct**Answer – 3) 400/3 meter**

Explanation –

When they meet the distance covered will be 2 × 1200 = 2400 meter

Time taken = 2400/(4+5) = 800/3 seconds

Distance of R from point P = 4 × 800/3 = 3200/3 meter

Therefore, distance of R from point Q = 1200 – 3200/3 = 400/3 meterIncorrect**Answer – 3) 400/3 meter**

Explanation –

When they meet the distance covered will be 2 × 1200 = 2400 meter

Time taken = 2400/(4+5) = 800/3 seconds

Distance of R from point P = 4 × 800/3 = 3200/3 meter

Therefore, distance of R from point Q = 1200 – 3200/3 = 400/3 meter - Question 7 of 10
##### 7. Question

1 pointsCategory: Quantitative Aptitude**A bullet train takes 1 hour less than an express train to travel 132km between Tokyo and Osaka. If the average speed of the bullet train is 11km/h more than that of the express train, find the average speed of bullet train.**Correct**Answer – 3) 44km/hr**

Explanation –

132/x – 132/x+11 = 1

X^{2}+11x-1452 = 0

X = -44,33

Therefore, speed of express train is 33

Speed of bullet train = 33+11 = 44km/hrIncorrect**Answer – 3) 44km/hr**

Explanation –

132/x – 132/x+11 = 1

X^{2}+11x-1452 = 0

X = -44,33

Therefore, speed of express train is 33

Speed of bullet train = 33+11 = 44km/hr - Question 8 of 10
##### 8. Question

1 pointsCategory: Quantitative Aptitude**Anu and Rishi together can complete a work in 6 days. If the efficiency of Anu is decreased by 20% and efficiency of Rishi in increased by 40% then they completed the same work in 5 days. Find the time taken by Rishi to complete the same work alone with increased efficiency of 40%?**Correct**Answer – 3) 45/7 days**

Explanation –

Let the per day work done by Anu is A

And per day work done by Rishi is R

Work done by Anu in 6 days = 6A and work done by Rishi in 6 days = 6R

Total work done together = 6A+6R

Per day work done by Anu with decreased efficiency of 20% = .8A

And per day work done by Rishi with increased efficiency of 40% = 1.4 R

Now, with changed efficiency of both

Work done by Anu in 5 days = 4A , work done by Rishi in 5days = 7R

Total work done Together = 4A+7R

Total work = Total Work

6A+6R = 4A + 7R

2A = R

Total work in terms of R will be 9R

Time taken by Rishi alone with increased efficiency = 9R/1.4R = 90/14 = 45/7 daysIncorrect**Answer – 3) 45/7 days**

Explanation –

Let the per day work done by Anu is A

And per day work done by Rishi is R

Work done by Anu in 6 days = 6A and work done by Rishi in 6 days = 6R

Total work done together = 6A+6R

Per day work done by Anu with decreased efficiency of 20% = .8A

And per day work done by Rishi with increased efficiency of 40% = 1.4 R

Now, with changed efficiency of both

Work done by Anu in 5 days = 4A , work done by Rishi in 5days = 7R

Total work done Together = 4A+7R

Total work = Total Work

6A+6R = 4A + 7R

2A = R

Total work in terms of R will be 9R

Time taken by Rishi alone with increased efficiency = 9R/1.4R = 90/14 = 45/7 days - Question 9 of 10
##### 9. Question

1 pointsCategory: Quantitative Aptitude**Two pipes P and Q can fill a cistern in 24 hours and 30 hours respectively. Both pipes are opened simultaneously, due to a leakage in cistern it takes 40 minute extra to fill it up. Find the time taken by the leakage to empty the full cistern alone.**Correct**Answer – 4) 280 hours**

Explanation –

40minutes = 2/3 hours

Let the capacity of tank is LCM of 24, 30 = 120units

Capacity of pipe P = 120/24 units = 5units per hour

Capacity of Pipe Q = 120/30units = 4units per hour

Time taken by both the pipes to fill the tank = 120/9 = 40/3 hours

Time taken by both the pipes due to leakage = 40/3 + 2/3 = 14 hours

Capacity of both the pipes with leakage = 120/14 = 60/7 units

Capacity of leakage = 60/7 – 5 -4 = 60/7 – 9 = -3/7units

Hence, time taken by leakage to empty the tank = 120/3/7 =280 hoursIncorrect**Answer – 4) 280 hours**

Explanation –

40minutes = 2/3 hours

Let the capacity of tank is LCM of 24, 30 = 120units

Capacity of pipe P = 120/24 units = 5units per hour

Capacity of Pipe Q = 120/30units = 4units per hour

Time taken by both the pipes to fill the tank = 120/9 = 40/3 hours

Time taken by both the pipes due to leakage = 40/3 + 2/3 = 14 hours

Capacity of both the pipes with leakage = 120/14 = 60/7 units

Capacity of leakage = 60/7 – 5 -4 = 60/7 – 9 = -3/7units

Hence, time taken by leakage to empty the tank = 120/3/7 =280 hours - Question 10 of 10
##### 10. Question

1 pointsCategory: Quantitative Aptitude**Pintu has x number of toffees in his pocket. Out of this, (x-4) are of mango flavour and (x-2) are of orange flavour and. Pintu takes three toffees from his pocket. Find the probability of one being mango flavour and two being orange flavour.**Correct**Answer – 2) 3/5**

Explanation –

x = (x-4) + (x-2)

x = 2x-6

x = 6

Hence, pintu has (6-2) = 4 orange toffees and (6-4) = 2 mango toffees

Probability of one being mango toffee and two being orange flavour = (2C_{1}× 4C_{2})/6C_{3}= 3/5Incorrect**Answer – 2) 3/5**

Explanation –

x = (x-4) + (x-2)

x = 2x-6

x = 6

Hence, pintu has (6-2) = 4 orange toffees and (6-4) = 2 mango toffees

Probability of one being mango toffee and two being orange flavour = (2C_{1}× 4C_{2})/6C_{3}= 3/5

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