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Question 1 of 10
1. Question
1 pointsCategory: Quantitative AptitudeA can do a piece of work in the same time in which B and C can together do it. If A and B can together do it in 10 days and C alone in 30 days, then B alone could do it in how many days?
Correct
Explanation:
(A+B+C)’s 1 day’s work: 1/A + 1/B + 1/C = 1/10 + 1/30 = 4/30
Given 1/A = 1/B +1/C
So, 1/A +1/A = 4/30 gives 1/A = 4/60 = 1/15
1/A + 1/B = 1/10, so 1/B = 1/10 – 1/15 = 1/30
So B can do in 30 daysIncorrect
Explanation:
(A+B+C)’s 1 day’s work: 1/A + 1/B + 1/C = 1/10 + 1/30 = 4/30
Given 1/A = 1/B +1/C
So, 1/A +1/A = 4/30 gives 1/A = 4/60 = 1/15
1/A + 1/B = 1/10, so 1/B = 1/10 – 1/15 = 1/30
So B can do in 30 days 
Question 2 of 10
2. Question
1 pointsCategory: Quantitative AptitudeA committee of 5 is to be formed from among 4 girls and 5 boys. What is the probability that the committee will have number of boys less than number of girls?
Correct
Explanation:
Case 1: 4 girls and 1 boys [0 boys cannot be there, because maximum girls are 4]
4C4*5C1/9C5
Case 2: 3 girls and 2 boys
4C3*5C2/9C5
Add both casesIncorrect
Explanation:
Case 1: 4 girls and 1 boys [0 boys cannot be there, because maximum girls are 4]
4C4*5C1/9C5
Case 2: 3 girls and 2 boys
4C3*5C2/9C5
Add both cases 
Question 3 of 10
3. Question
1 pointsCategory: Quantitative AptitudeSolution A has 42 6/7% water and rest milk. If 6 litres of water is added to it, then percentage of water will become 60%. What was the initial weight of solution A?
Correct
Explanation:
42 6/7 = 300/7%
Then
(300/7*100) * x + 6 = (60/100) * (x+6)
Solve, x = 14Incorrect
Explanation:
42 6/7 = 300/7%
Then
(300/7*100) * x + 6 = (60/100) * (x+6)
Solve, x = 14 
Question 4 of 10
4. Question
1 pointsCategory: Quantitative AptitudeThe average age of Anisha and Vinisha is 11 years who are sitting in a group having 11 people. The maximum and minimum average of remaining becomes 12 years and 11 years respectively when anyone of Anisha or Vinisha leaves the group. What is the average of all people in the group?
Correct
Explanation:
Let total age of 9 people except Anisha and Vinisha = a years
Age of anisha = x years, age of vinisha = y yeas
Then (x+y)/2 = 11, gives x + y = 22
Then (a+x)/10 = 11, gives a + x = 110
Then (a+y)/10 = 12, gives a + y = 120
Now there are 3 equations, solve and find all values, a = 104, x = 16, y = 6
Then average = (104+16+6)/11Incorrect
Explanation:
Let total age of 9 people except Anisha and Vinisha = a years
Age of anisha = x years, age of vinisha = y yeas
Then (x+y)/2 = 11, gives x + y = 22
Then (a+x)/10 = 11, gives a + x = 110
Then (a+y)/10 = 12, gives a + y = 120
Now there are 3 equations, solve and find all values, a = 104, x = 16, y = 6
Then average = (104+16+6)/11 
Question 5 of 10
5. Question
1 pointsCategory: Quantitative AptitudeIn a class of 70 students, ratio of number of boys to girls is 4 : 3. Among the girls ratio of number of girls coming by bus and not coming by bus is 1 : 4. If the ratio of total students coming by bus and not coming by bus in class is 8 : 27, find the ratio of number of boys coming by bus and not coming by bus.
Correct
Explanation:
Boys in class = (4/7)*70 = 40, so girls = 7040=30
Among girls, spots persons = (1/5)*30 = 6, so nonsports persons = 306 = 24
Total sports person in class = 8/(8+27) * 70 = 16, so nonsports
persons = 7016 = 54
So among boys sports person = 166=10, nonsports = 5424 = 30
So ratio = 10: 30 = 1: 3Incorrect
Explanation:
Boys in class = (4/7)*70 = 40, so girls = 7040=30
Among girls, spots persons = (1/5)*30 = 6, so nonsports persons = 306 = 24
Total sports person in class = 8/(8+27) * 70 = 16, so nonsports
persons = 7016 = 54
So among boys sports person = 166=10, nonsports = 5424 = 30
So ratio = 10: 30 = 1: 3 
Question 6 of 10
6. Question
1 pointsCategory: Quantitative AptitudeDirections (610): In the following questions, two equations numbered I and II are given. You have to solve both the equations and give
answer –9x^{2} – 9x + 2 = 0, 3y^{2} + 11y – 4 = 0
Correct
Explanation:
9x^{2} – 9x + 2 = 0
9x^{2} – 3x – 6x + 2 = 0
Gives x = 1/3, 2/3
3y^{2} + 11y – 4 = 0
3y^{2} + 12y – y – 4 = 0
Gives y = 4, 1/3
Put on number line
4…… 1/3 ….. 2/3Incorrect
Explanation:
9x^{2} – 9x + 2 = 0
9x^{2} – 3x – 6x + 2 = 0
Gives x = 1/3, 2/3
3y^{2} + 11y – 4 = 0
3y^{2} + 12y – y – 4 = 0
Gives y = 4, 1/3
Put on number line
4…… 1/3 ….. 2/3 
Question 7 of 10
7. Question
1 pointsCategory: Quantitative AptitudeDirections (610): In the following questions, two equations numbered I and II are given. You have to solve both the equations and give
answer –x^{2} – 10x + 25 = 0, y^{2} – 9y + 20 = 0
Correct
Explanation:
x^{2} – 10x + 25 = 0
x^{2} – 5x – 5x + 25 = 0
Gives x = 5
y^{2} – 9y + 20 = 0
y^{2} – 5y – 4y + 20 = 0
Gives y = 4,5
Put on number line
4……. 5Incorrect
Explanation:
x^{2} – 10x + 25 = 0
x^{2} – 5x – 5x + 25 = 0
Gives x = 5
y^{2} – 9y + 20 = 0
y^{2} – 5y – 4y + 20 = 0
Gives y = 4,5
Put on number line
4……. 5 
Question 8 of 10
8. Question
1 pointsCategory: Quantitative AptitudeDirections (610): In the following questions, two equations numbered I and II are given. You have to solve both the equations and give
answer –6x^{2} + 11x + 3 = 0, 20y^{2} + 9y + 1 = 0
Correct
Explanation:
6x^{2} + 11x + 3 = 0
6x^{2} + 2x + 9x + 3 = 0
Gives x = 1/3, 3/2
20y^{2} + 9y + 1 = 0
20y^{2} + 4y + 5y + 1 = 0
Gives y = 1/4, 1/5
Put on number line
3/2… 1/3……. 1/4…… 1/5Incorrect
Explanation:
6x^{2} + 11x + 3 = 0
6x^{2} + 2x + 9x + 3 = 0
Gives x = 1/3, 3/2
20y^{2} + 9y + 1 = 0
20y^{2} + 4y + 5y + 1 = 0
Gives y = 1/4, 1/5
Put on number line
3/2… 1/3……. 1/4…… 1/5 
Question 9 of 10
9. Question
1 pointsCategory: Quantitative AptitudeDirections (610): In the following questions, two equations numbered I and II are given. You have to solve both the equations and give
answer –30x^{2} + 17x ¬+ 2 = 0, 3y^{2} + 8y + 4 = 0
Correct
Explanation:
30x^{2} + 17x ¬+ 2 = 0
30x^{2} + 5x ¬+ 12x + 2 = 0
Gives x = 1/6, 2/5
3y^{2} + 8y + 4 = 0
3y^{2} + 6y + 2y + 4 = 0
Gives y = 2, 2/3
Put on number line
2…………. 2/3………. 2/5……… 1/6Incorrect
Explanation:
30x^{2} + 17x ¬+ 2 = 0
30x^{2} + 5x ¬+ 12x + 2 = 0
Gives x = 1/6, 2/5
3y^{2} + 8y + 4 = 0
3y^{2} + 6y + 2y + 4 = 0
Gives y = 2, 2/3
Put on number line
2…………. 2/3………. 2/5……… 1/6 
Question 10 of 10
10. Question
1 pointsCategory: Quantitative AptitudeDirections (610): In the following questions, two equations numbered I and II are given. You have to solve both the equations and give
answer –4x^{2} – 13x + 10 = 0, 2y^{2} – 15y + 22 = 0
Correct
Explanation:
4x^{2} – 13x + 10 = 0
4x^{2} – 8x – 5x + 10 = 0
Gives x = 2, 5/4
2y^{2} – 15y + 22 = 0
2y^{2} – 4y – 11y + 22 = 0
Gives y= 11/2 , 2
Put on number line
5/4……….. 2………. 11/2Incorrect
Explanation:
4x^{2} – 13x + 10 = 0
4x^{2} – 8x – 5x + 10 = 0
Gives x = 2, 5/4
2y^{2} – 15y + 22 = 0
2y^{2} – 4y – 11y + 22 = 0
Gives y= 11/2 , 2
Put on number line
5/4……….. 2………. 11/2
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