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Question 1 of 10
1. Question
1 pointsCategory: Quantitative AptitudeDirection Q (15): Study the following table carefully to answer the given questions
Data related to performance of 6 Batsman in a tournament is given. Few Values are intentionally left blank.Batsman No of matches Average Runs Total Balls Faced SR P 5 – – 130 Q 30 78 – – R – 37 400 148 S – – – 72 T 28 55 1280 – U – – – 66 I.Strike Rate = [Total Runs Scored/Total Balls Faced]*100
II.None of the batsman stayed not out in any match.What is the number of matches played by batsman R in the tournament?
Correct
Explanation
148 = (37x / 400)* 100
x = 16Incorrect
Explanation
148 = (37x / 400)* 100
x = 16 
Question 2 of 10
2. Question
1 pointsCategory: Quantitative AptitudeDirection Q (15): Study the following table carefully to answer the given questions
Data related to performance of 6 Batsman in a tournament is given. Few Values are intentionally left blank.Batsman No of matches Average Runs Total Balls Faced SR P 5 – – 130 Q 30 78 – – R – 37 400 148 S – – – 72 T 28 55 1280 – U – – – 66 I.Strike Rate = [Total Runs Scored/Total Balls Faced]*100
II.None of the batsman stayed not out in any match.Batsman Q faced equal number of balls in first 15 matches he played in the tournament and last 15 matches he played in the tournament. If his Strike rate in first 15 matches of the tournament is 120 and in last 15 match is 140 respectively, what is the total number of balls faced by him in the tournament?
Correct
Explanation :
(120/100)*(x/2) + (140/100)*(x/2) = 2340
x = 1800Incorrect
Explanation :
(120/100)*(x/2) + (140/100)*(x/2) = 2340
x = 1800 
Question 3 of 10
3. Question
1 pointsCategory: Quantitative AptitudeDirection Q (15): Study the following table carefully to answer the given questions
Data related to performance of 6 Batsman in a tournament is given. Few Values are intentionally left blank.Batsman No of matches Average Runs Total Balls Faced SR P 5 – – 130 Q 30 78 – – R – 37 400 148 S – – – 72 T 28 55 1280 – U – – – 66 I.Strike Rate = [Total Runs Scored/Total Balls Faced]*100
II.None of the batsman stayed not out in any match.In the tournament, the total number of balls faced by Batsman P is 75 less than the total number of runs scored by him. What is the average run scored by Batsman P in the tournament?
Correct
Explanation :
130 = [x/x75]*100
130x 9750= 100x
x = 325
Average = 325/5 = 65Incorrect
Explanation :
130 = [x/x75]*100
130x 9750= 100x
x = 325
Average = 325/5 = 65 
Question 4 of 10
4. Question
1 pointsCategory: Quantitative AptitudeDirection Q (15): Study the following table carefully to answer the given questions
Data related to performance of 6 Batsman in a tournament is given. Few Values are intentionally left blank.Batsman No of matches Average Runs Total Balls Faced SR P 5 – – 130 Q 30 78 – – R – 37 400 148 S – – – 72 T 28 55 1280 – U – – – 66 I.Strike Rate = [Total Runs Scored/Total Balls Faced]*100
II.None of the batsman stayed not out in any match.If the runs scored by T in last 3 matches of the tournament are not considered, his average runs scored in the tournament will decrease by 9. If the runs scored by T in the 26th and 27th match are below 128 and no two scores among these 3 scores are equal, what are the minimum possible runs scored by T in the 28th match?
Correct
Explanation :
Total runs scored = Number of matches played in the tournament * Average Run = 28 * 55 = 1540
Total runs scored(excluding last 3 matches) = 25 * 46(decrease 9 in avg) = 1150
Total runs of last 3 matches = 1540 â€“ 1150 = 390
Average = 390/3 = 130
26th and 27th match are below 128 and no two scores among these 3 scores are equal.
So
Assume 26th = 127
then 27th = 126
and therefore 28th = 137Incorrect
Explanation :
Total runs scored = Number of matches played in the tournament * Average Run = 28 * 55 = 1540
Total runs scored(excluding last 3 matches) = 25 * 46(decrease 9 in avg) = 1150
Total runs of last 3 matches = 1540 â€“ 1150 = 390
Average = 390/3 = 130
26th and 27th match are below 128 and no two scores among these 3 scores are equal.
So
Assume 26th = 127
then 27th = 126
and therefore 28th = 137 
Question 5 of 10
5. Question
1 pointsCategory: Quantitative AptitudeDirection Q (15): Study the following table carefully to answer the given questions
Data related to performance of 6 Batsman in a tournament is given. Few Values are intentionally left blank.Batsman No of matches Average Runs Total Balls Faced SR P 5 – – 130 Q 30 78 – – R – 37 400 148 S – – – 72 T 28 55 1280 – U – – – 66 I.Strike Rate = [Total Runs Scored/Total Balls Faced]*100
II.None of the batsman stayed not out in any match.The respective ratio between total number of balls faced by S and that by U in the tournament is 3:4. Total number of runs scored by U in the tournament is what percent more than the total runs scored by S in the tournament?
Correct
Explanation :
U = 66*4x/100, S = 72*3x/100
U = S*[(100+y)/100] 264x/100 = 216x/100 * [(100+y)/100] y = 200/9%Incorrect
Explanation :
U = 66*4x/100, S = 72*3x/100
U = S*[(100+y)/100] 264x/100 = 216x/100 * [(100+y)/100] y = 200/9% 
Question 6 of 10
6. Question
1 pointsCategory: Quantitative AptitudeDirection:Q (610)Refer to the table and answer the given questions. Some values are intentionally left blank.
Person Type of Interest Principal Amount Years Rate A CI – – 2 2 B SI – – 4 – C CI 21000 – 2 4 D SI – 31860 3 – E CI 10000 – – 4 If the ratio of interest rate of E to that of D is 2:3 then what is the Principal(P) of D?
Correct
Explanation :
R(%) = 4*3/2 = 6
Principal â€“ x
x + (x*3*6/100) = 31860
x = 27000Incorrect
Explanation :
R(%) = 4*3/2 = 6
Principal â€“ x
x + (x*3*6/100) = 31860
x = 27000 
Question 7 of 10
7. Question
1 pointsCategory: Quantitative AptitudeDirection:Q (610)Refer to the table and answer the given questions. Some values are intentionally left blank.
Person Type of Interest Principal Amount Years Rate A CI – – 2 2 B SI – – 4 – C CI 21000 – 2 4 D SI – 31860 3 – E CI 10000 – – 4 If the interest is compounded yearly for three years then what is the amount to be earned by C?
Correct
Explanation :
Amount = P(1 + (R/100)^3)
A = 21000 * 1.04 * 1.04 * 1.04
A = 23622.14Incorrect
Explanation :
Amount = P(1 + (R/100)^3)
A = 21000 * 1.04 * 1.04 * 1.04
A = 23622.14 
Question 8 of 10
8. Question
1 pointsCategory: Quantitative AptitudeDirection:Q (610)Refer to the table and answer the given questions. Some values are intentionally left blank.
Person Type of Interest Principal Amount Years Rate A CI – – 2 2 B SI – – 4 – C CI 21000 – 2 4 D SI – 31860 3 – E CI 10000 – – 4 What is the Simple Interest(SI) of B ? If the ratio of Principal of C to that of B is 4:5 and the rate of interest is 10% more than that of C.
Correct
Explanation :
P = 5/4 * 21000 = 26250
The rate of interest is 10% more than that of C.
R(%) = 4 + (4*(10/100)) = 4.4 %
SI = [26250 * 4.4 * 4]/100 = 4620Incorrect
Explanation :
P = 5/4 * 21000 = 26250
The rate of interest is 10% more than that of C.
R(%) = 4 + (4*(10/100)) = 4.4 %
SI = [26250 * 4.4 * 4]/100 = 4620 
Question 9 of 10
9. Question
1 pointsCategory: Quantitative AptitudeDirection:Q (610)Refer to the table and answer the given questions. Some values are intentionally left blank.
Person Type of Interest Principal Amount Years Rate A CI – – 2 2 B SI – – 4 – C CI 21000 – 2 4 D SI – 31860 3 – E CI 10000 – – 4 If the Principal(P) of A is 20% more than that of E, then What is the amount of A?
Correct
Explanation :
Principal(P) of A = 10000 * 120/100 = 12000
A = P(1 + (R/100)^N) = 12000(1 + (2/100)^2) = 12484.80Incorrect
Explanation :
Principal(P) of A = 10000 * 120/100 = 12000
A = P(1 + (R/100)^N) = 12000(1 + (2/100)^2) = 12484.80 
Question 10 of 10
10. Question
1 pointsCategory: Quantitative AptitudeDirection:Q (610)Refer to the table and answer the given questions. Some values are intentionally left blank.
Person Type of Interest Principal Amount Years Rate A CI – – 2 2 B SI – – 4 – C CI 21000 – 2 4 D SI – 31860 3 – E CI 10000 – – 4 If amount of D equals to four times that of his Principal then what is the Rate of Interest(%)?
Correct
Explanation :
Amount of D = Rs.31860
Principal â€“ x
Amount of D = 4x
4x = 31860
x= 7965
SI = 23895
R = 23895 * 100 / 7965 * 3 = 100%
Else can solve this directlyIncorrect
Explanation :
Amount of D = Rs.31860
Principal â€“ x
Amount of D = 4x
4x = 31860
x= 7965
SI = 23895
R = 23895 * 100 / 7965 * 3 = 100%
Else can solve this directly
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