Hello Aspirants
Welcome to Online Quant Test in AffairsCloud.com. We are starting Cumulus Course for SBI PO 2019 Prelims Exam and we are creating sample questions in Quant section, this type of Question will be asked in SBI PO 2019.
Click Here to View Cumulus Course: SBI PO 2019 Prelims
SBI PO 2019 Prelims: Reasoning Test – 7.00 PM Every Day
SBI PO 2019 Prelims: Quant Test – 8.00 PM Every Day
SBI PO 2019 Prelims: English Test – 9.00 PM Every Day
Help: Share Our Cumulus Course for SBI PO 2019 Exam Course page to your Friends & FB Groups
SBI PO Previous Year Question Paper
________________________
Quizsummary
0 of 10 questions completed
Questions:
 1
 2
 3
 4
 5
 6
 7
 8
 9
 10
Information
All the Best
You have already completed the quiz before. Hence you can not start it again.
Quiz is loading...
You must sign in or sign up to start the quiz.
You have to finish following quiz, to start this quiz:
Results
0 of 10 questions answered correctly
Your time:
Time has elapsed
You have reached 0 of 0 points, (0)
Categories
 Quantitative Aptitude 0%
 1
 2
 3
 4
 5
 6
 7
 8
 9
 10
 Answered
 Review

Question 1 of 10
1. Question
1 pointsCategory: Quantitative AptitudeDirections (15) 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 = 884Complete 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 = 884Complete 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 AptitudeDirections (15) 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 AptitudeDirections (15) 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 AptitudeDirections (15) 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.
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 AptitudeDirections (15) 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.
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 AptitudeAnil 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 AptitudeA 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}+11x1452 = 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}+11x1452 = 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 AptitudeAnu 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 AptitudeTwo 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 AptitudePintu has x number of toffees in his pocket. Out of this, (x4) are of mango flavour and (x2) 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 = (x4) + (x2)
x = 2x6
x = 6
Hence, pintu has (62) = 4 orange toffees and (64) = 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 = (x4) + (x2)
x = 2x6
x = 6
Hence, pintu has (62) = 4 orange toffees and (64) = 2 mango toffees
Probability of one being mango toffee and two being orange flavour = (2C_{1} × 4C_{2})/6C_{3}= 3/5
________________________
 Click view Questions button to view Explanation
 Note: We are providing unique questions for you to practice well, have a try !!
 Ask your doubt in comment section, AC Mod’s ll clear your doubts in caring way.
 If you find any mistake, please tell us in the comment section.