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Question 1 of 10
1. Question
1 pointsCategory: Quantitative AptitudeDirections (1 – 5): The pie chart below shows the year wise sales of mobile phones made by companies P and Q.
The table shows the ratio of sales of mobile phones made by companies P and Q resp. during the given period:YearRatio P Q 2006 3 7 2007 7 18 2008 2 3 2009 4 1 2010 11 9 2011 7 3 The total sales of mobile phones made by company Q during the years 2009 and 2010 together is?
Correct
Explanation:
(1/5)*(18/100)*3,25,000 + (9/20)*(20/100)*3,25,000
11700 + 29250Incorrect
Explanation:
(1/5)*(18/100)*3,25,000 + (9/20)*(20/100)*3,25,000
11700 + 29250 
Question 2 of 10
2. Question
1 pointsCategory: Quantitative AptitudeDirections (1 – 5): The pie chart below shows the year wise sales of mobile phones made by companies P and Q.
The table shows the ratio of sales of mobile phones made by companies P and Q resp. during the given period:YearRatio P Q 2006 3 7 2007 7 18 2008 2 3 2009 4 1 2010 11 9 2011 7 3 For company P the sale in 2007 is what percentage of the sales in 2008?
Correct
Explanation:
For P
Sales in 2007 = (7/25)*(10/100)* 3,25,000
Sales in 2008 = (2/5)*(16/100)* 3,25,000
Required% = Sales in 2007/ Sales in 2008 * 100Incorrect
Explanation:
For P
Sales in 2007 = (7/25)*(10/100)* 3,25,000
Sales in 2008 = (2/5)*(16/100)* 3,25,000
Required% = Sales in 2007/ Sales in 2008 * 100 
Question 3 of 10
3. Question
1 pointsCategory: Quantitative AptitudeDirections (1 – 5): The pie chart below shows the year wise sales of mobile phones made by companies P and Q.
The table shows the ratio of sales of mobile phones made by companies P and Q resp. during the given period:YearRatio P Q 2006 3 7 2007 7 18 2008 2 3 2009 4 1 2010 11 9 2011 7 3 If 40% of the sales by ‘P’ in the years 2007, and 2010 are sold at discount, how many mobile phones are sold by P without discount in these 2 years together?
Correct
Explanation:
In 2007 + 2010 + 2011 = [(7/25)*(10/100) + (11/20)*(20/100)] * 325000 = (69/500)*325000
So without discount = (60/100)* (69/500)*325000Incorrect
Explanation:
In 2007 + 2010 + 2011 = [(7/25)*(10/100) + (11/20)*(20/100)] * 325000 = (69/500)*325000
So without discount = (60/100)* (69/500)*325000 
Question 4 of 10
4. Question
1 pointsCategory: Quantitative AptitudeDirections (1 – 5): The pie chart below shows the year wise sales of mobile phones made by companies P and Q.
The table shows the ratio of sales of mobile phones made by companies P and Q resp. during the given period:YearRatio P Q 2006 3 7 2007 7 18 2008 2 3 2009 4 1 2010 11 9 2011 7 3 If company Q gets a profit of Rs 240 on each mobile phone sold during 2008, then find its profit in the year 2008?
Correct
Explanation:
By Q, sold in 2008 = (3/5)*(16/100)*325000 = 3120
So profit = 3120*240 = 748,800 = 74.88 lakhsIncorrect
Explanation:
By Q, sold in 2008 = (3/5)*(16/100)*325000 = 3120
So profit = 3120*240 = 748,800 = 74.88 lakhs 
Question 5 of 10
5. Question
1 pointsCategory: Quantitative AptitudeDirections (1 – 5): The pie chart below shows the year wise sales of mobile phones made by companies P and Q.
The table shows the ratio of sales of mobile phones made by companies P and Q resp. during the given period:YearRatio P Q 2006 3 7 2007 7 18 2008 2 3 2009 4 1 2010 11 9 2011 7 3 For company Q the ratio of the sales in the year 2006 to that in 2011 is?
Correct
Explanation:
(7/10)*(11/100)*3,25,000 : (3/10)*(25/100)*3,25,000
77 : 75Incorrect
Explanation:
(7/10)*(11/100)*3,25,000 : (3/10)*(25/100)*3,25,000
77 : 75 
Question 6 of 10
6. Question
1 pointsCategory: Quantitative AptitudeMixture A contains 75% milk and mixture B contains 90% milk. A milkman takes some mixture from mixture A and mixes it with four times the amount of mixture from mixture B. What is the percentage of milk in the new mixture?
Correct
Explanation:
Let x from A, 4x form B
So milk from A = (75/100)*x, milk from B = (90/100)*4x
So milk in new mixture is (75x/100) + (360x/100) = 4.35x
Total mixture in third is x+4x = 5x
So % of milk is (4.35x/5x)*100Incorrect
Explanation:
Let x from A, 4x form B
So milk from A = (75/100)*x, milk from B = (90/100)*4x
So milk in new mixture is (75x/100) + (360x/100) = 4.35x
Total mixture in third is x+4x = 5x
So % of milk is (4.35x/5x)*100 
Question 7 of 10
7. Question
1 pointsCategory: Quantitative Aptitude5 men started the work and worked for 4 days. Now they are replaced by 12 women where 1 man can do as much work as 2 women do in 1 day. If the works gets completed in a total of 10 days, in how many days 7 women can complete the entire work?
Correct
Explanation:
Given 1 m = 2w, so 12 women = 6 men
Let 1 man do work in x days, so 1 man’s 1 day work = 1/x, so 5 men 1 day’s work = (1/x)*5
And since they did work for 4 days, so there work becomes (5/x)* 4 = 20/x
Now total work completed in 10 days, this mean that 12 women or 6 men did in (104) = 6 days
So similarly as above, 1 man’s 1 day work = 1/x, so 6 men 1 day’s work = (1/x)*6
And since they did work for 6 days, so there work becomes (6/x)* 6 = 36/x
So (20/x) + (36/x) = 1
Solve, x = 56
1 man do work in 56 days, so 1 woman in 2*56 days, and 7 women in (2*56)/7Incorrect
Explanation:
Given 1 m = 2w, so 12 women = 6 men
Let 1 man do work in x days, so 1 man’s 1 day work = 1/x, so 5 men 1 day’s work = (1/x)*5
And since they did work for 4 days, so there work becomes (5/x)* 4 = 20/x
Now total work completed in 10 days, this mean that 12 women or 6 men did in (104) = 6 days
So similarly as above, 1 man’s 1 day work = 1/x, so 6 men 1 day’s work = (1/x)*6
And since they did work for 6 days, so there work becomes (6/x)* 6 = 36/x
So (20/x) + (36/x) = 1
Solve, x = 56
1 man do work in 56 days, so 1 woman in 2*56 days, and 7 women in (2*56)/7 
Question 8 of 10
8. Question
1 pointsCategory: Quantitative AptitudeThe incomes of 2 persons A and B is in the ratio 4 : 5. Their expenditures are in the ratio 2 : 3 respectively. If both save Rs 50,000, then what is B’s income?
Correct
Explanation:
Incomes 4x and 5x
Expenditures 2y and 3y
So 4x – 2y = 50000
And 5x – 3y = 50000
Solve both equations, x = 25,000
So B’s income = 5*25000Incorrect
Explanation:
Incomes 4x and 5x
Expenditures 2y and 3y
So 4x – 2y = 50000
And 5x – 3y = 50000
Solve both equations, x = 25,000
So B’s income = 5*25000 
Question 9 of 10
9. Question
1 pointsCategory: Quantitative AptitudeThe radius of sphere is 16 cm and cost of painting its surface Rs 50 per square cm. Suppose the radius of sphere and its cost are increased by 5% and 10% respectively. What will be the percentage increase in the total cost of painting per square cm?
Correct
Explanation:
Surface area of square is 4ᴨr^{2}
Radius increases by 5%, so by successive increase formula
surface area changes by 5+5+((5*5)/100) = 10.25%
Total cost of painting = Surface area of sphere * Cost
So again by successive increase formula
change in cost of painting will be = 10.25+10+((10.25*10)/100)Incorrect
Explanation:
Surface area of square is 4ᴨr^{2}
Radius increases by 5%, so by successive increase formula
surface area changes by 5+5+((5*5)/100) = 10.25%
Total cost of painting = Surface area of sphere * Cost
So again by successive increase formula
change in cost of painting will be = 10.25+10+((10.25*10)/100) 
Question 10 of 10
10. Question
1 pointsCategory: Quantitative Aptitude2 pipes A and B are such that A fills the bucket in 8 hrs and B empties the bucket in 4 hrs. If pipe A is opened first and after 2 hours B is also opened, in how much total time (in hours) will the bucket get empty?
Correct
Explanation:
In 2 hrs, bucket filled by A is 2 * 1/8 = 1/4
After 2 hrs, bucket filled by A + B in 1 hr is 1/8 1/4 = 1/8
This means after 2 hrs, 1/4th is full, now in next 1 hr 1/8 out of 1/4th already filled gets empty, so now 1/4 – 1/8 = 1/8 bucket is filled, in next 1 hr (A+B) empties this 1/8 too
So total 2 + 1 + 1Incorrect
Explanation:
In 2 hrs, bucket filled by A is 2 * 1/8 = 1/4
After 2 hrs, bucket filled by A + B in 1 hr is 1/8 1/4 = 1/8
This means after 2 hrs, 1/4th is full, now in next 1 hr 1/8 out of 1/4th already filled gets empty, so now 1/4 – 1/8 = 1/8 bucket is filled, in next 1 hr (A+B) empties this 1/8 too
So total 2 + 1 + 1
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