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Question 1 of 10
1. Question
1 pointsCategory: Quantitative Aptitudewhat is the total numbers of ball faced by N and K together if it is known that ball faced by N is 4/5 of K and matches played by him is 5/4 of K?
Correct
Answer : 4. 5625
Explanation
Matches played by N = 5/4 of K = 5/4*20 = 25,
Total runs scored by N = 90*25 = 2250, total ball faced by him = 2250*100/90 = 2500,
Total ball faced by K = 2500*5/4 = 3125
Total ball faced by them = 3125+2500 = 5625Incorrect
Answer : 4. 5625
Explanation
Matches played by N = 5/4 of K = 5/4*20 = 25,
Total runs scored by N = 90*25 = 2250, total ball faced by him = 2250*100/90 = 2500,
Total ball faced by K = 2500*5/4 = 3125
Total ball faced by them = 3125+2500 = 5625 
Question 2 of 10
2. Question
1 pointsCategory: Quantitative Aptitudeif average runs scored by K is 80 then what will be total balls faced by him?
Correct
Answer : 3. 1250
Explanation
Average runs scored by K = 80, total runs = 80*20 = 1600
Total ball faced = 1600*100/128 = 1250Incorrect
Answer : 3. 1250
Explanation
Average runs scored by K = 80, total runs = 80*20 = 1600
Total ball faced = 1600*100/128 = 1250 
Question 3 of 10
3. Question
1 pointsCategory: Quantitative AptitudeIf average runs scored by L are 75 and matches played were 4/5 of M, then what will by the strike rate of O if it is 3/5 of strike rate of L?
Correct
Answer : 4. 100
Explanation
Average runs scored by L = 75, total runs scored by him = 75*20 = 1500
Then strike rate of him is 1500*100/ 900
Strike rate of O = 3/5 * 1500*100/900 = 100Incorrect
Answer : 4. 100
Explanation
Average runs scored by L = 75, total runs scored by him = 75*20 = 1500
Then strike rate of him is 1500*100/ 900
Strike rate of O = 3/5 * 1500*100/900 = 100 
Question 4 of 10
4. Question
1 pointsCategory: Quantitative Aptitudewhat is the ratio of strike rate of M and O together to L and K together?
Correct
Answer : 5. Data inadequate
Explanation
Data inadequateIncorrect
Answer : 5. Data inadequate
Explanation
Data inadequate 
Question 5 of 10
5. Question
1 pointsCategory: Quantitative Aptitudewhat is the total number of matches played by N players if total ball faced by N is half of the ball faced by O and L together?
Correct
Answer : 1. 12
Explanation
Total ball faced by N = (900+1500)/2 = 1200
Total matches played by him = total runs scored/avg. runs scored
Total runs scored = 90*1200/100=1080,
Total matches played= 1080/90=12Incorrect
Answer : 1. 12
Explanation
Total ball faced by N = (900+1500)/2 = 1200
Total matches played by him = total runs scored/avg. runs scored
Total runs scored = 90*1200/100=1080,
Total matches played= 1080/90=12 
Question 6 of 10
6. Question
1 pointsCategory: Quantitative AptitudeA man was engaged on a job for 40 days on the condition that he will get a wage of Rs. 180 for the day he works, but he will have to pay a fine of Rs. 20 for each day of his absence. If he gets Rs. 5200 at the end of the month, then, he was absent for how many days?
Correct
Answer : 2. 10
Example
No of days absent=x
180(40x)20x = 5200
X= 2000/20 = 10 daysIncorrect
Answer : 2. 10
Example
No of days absent=x
180(40x)20x = 5200
X= 2000/20 = 10 days 
Question 7 of 10
7. Question
1 pointsCategory: Quantitative AptitudeThe compound interest on a certain sum for 2 years at 20% per annum is Rs.880/. The simple Interest on the same sum for double the time at half the rate percent per annum is:
Correct
Answer : 1. Rs.800/
Example
Incorrect
Answer : 1. Rs.800/
Example

Question 8 of 10
8. Question
1 pointsCategory: Quantitative Aptitude4 kg of a metal contains 1/5 copper and rest in Zinc. Another 5 kg of metal contains 1/6 copper and rest in Zinc.The ratio of Copper and Zinc into the mixture of these two metals:
Correct
Answer : 3. 49: 221
Explanation
Copper in 4 kg = 4/5 and Zinc in 4 kg = 4*4/5=16/5 Copper in 5 kg = 5/6 and Zinc in 5 kg = 5*5/6=25/6 Therefore, Copper in mixture = 4/5 + 5/6=49/30 and Zinc in the mixture = 16/5 + 25/6=221/30 Therefore the required ratio = 49: 221Incorrect
Answer : 3. 49: 221
Explanation
Copper in 4 kg = 4/5 and Zinc in 4 kg = 4*4/5=16/5 Copper in 5 kg = 5/6 and Zinc in 5 kg = 5*5/6=25/6 Therefore, Copper in mixture = 4/5 + 5/6=49/30 and Zinc in the mixture = 16/5 + 25/6=221/30 Therefore the required ratio = 49: 221 
Question 9 of 10
9. Question
1 pointsCategory: Quantitative AptitudeThere are three pipes, A, B and C, attached to container. A and B can fill the container alone in 20 and 30 mins, respectively whereas C can empty the container alone in 45 mins. The three pipes are kept opened alone for one minute each in the order A, B and C. The same order is followed subsequently. In how many minutes will the reservoir be full?
Correct
Answer : 4.47
Explanation
Aâ€¦â€¦20 9unit (180/20)
Bâ€¦â€¦30 180 (LCM) 6unit
Câ€¦â€¦45 4unit
1st Minute => A is opened => fills 9 L 2nd Minute => B is opened =>fills another 6 L 3rd Minute => C is opened => empties 4 L Hence every 3 minutes => (9 + 6 â€“ 4 =) 11 litres are filled into the container. So in 45 minutes (11 Ã— 15 =) 165 litres are filled. In the 46th minute A is opened and it fills 9 litres. In the 47th minute B is opened and it fills 6 litres. Hence the container will be full in 47 minutesIncorrect
Answer : 4.47
Explanation
Aâ€¦â€¦20 9unit (180/20)
Bâ€¦â€¦30 180 (LCM) 6unit
Câ€¦â€¦45 4unit
1st Minute => A is opened => fills 9 L 2nd Minute => B is opened =>fills another 6 L 3rd Minute => C is opened => empties 4 L Hence every 3 minutes => (9 + 6 â€“ 4 =) 11 litres are filled into the container. So in 45 minutes (11 Ã— 15 =) 165 litres are filled. In the 46th minute A is opened and it fills 9 litres. In the 47th minute B is opened and it fills 6 litres. Hence the container will be full in 47 minutes 
Question 10 of 10
10. Question
1 pointsCategory: Quantitative AptitudeA man rides his bike 20 km at an average speed of 8 km/hr and again travels 45 km at an average speed of 10 km/hr. What is his average speed for the ride approximately?
Correct
Answer : 3. 9.3
Explanation
Average speed=total distance/total time
Total time=20/8 + 45/10
Avg speed=(20+45)/(20/8 + 45/10) = 65/((200+360)/80) =65*80/560 = 65/7 =9.3km/hrIncorrect
Answer : 3. 9.3
Explanation
Average speed=total distance/total time
Total time=20/8 + 45/10
Avg speed=(20+45)/(20/8 + 45/10) = 65/((200+360)/80) =65*80/560 = 65/7 =9.3km/hr
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