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Question 1 of 15
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 15
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 15
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 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 4 of 15
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 40% of the sales by ‘P’ in the years 2007, and 2010 are sold by P without at discount, how many mobile phones are sold 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 5 of 15
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 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 6 of 15
6. Question
1 pointsCategory: Quantitative AptitudeA and B can complete a piece of work in 20 days and 30 days respectively. They both start the work. After working for some days, A leaves. After 2 more days A joins again and now B leaves. If the work is completed in a total of 18 days, then B worked for how many days?
Correct
Explanation:
Let A left after x days
So A worked for (182) = 16 days and B worked for (x+2) days
(1/20)*16 + (1/30)*(x+2) = 1
Solve, x = 4 days, B worked for 4+2 daysIncorrect
Explanation:
Let A left after x days
So A worked for (182) = 16 days and B worked for (x+2) days
(1/20)*16 + (1/30)*(x+2) = 1
Solve, x = 4 days, B worked for 4+2 days 
Question 7 of 15
7. Question
1 pointsCategory: Quantitative AptitudeA and B started a business with Rs 2000 and Rs 2500 respectively. After 4 months C also joined with Rs 3000. After further 4 months, A and B withdrew Rs 500 each. If at the end of year C received Rs 7200 as his share from the total profit, then what is the total profit?
Correct
Explanation:
2000*8 + 1500*4 : 2500*8 + 2000*4 : 3000*8
11 : 14 : 12
So (12/37)*x = 7200Incorrect
Explanation:
2000*8 + 1500*4 : 2500*8 + 2000*4 : 3000*8
11 : 14 : 12
So (12/37)*x = 7200 
Question 8 of 15
8. Question
1 pointsCategory: Quantitative AptitudeA bag contains 4 blue and 5 red balls. 5 balls are chosen at random. What is the probability that there are more blue balls then red?
Correct
Explanation:
Case 1: 4 blue 1 red
^{4}C_{4}*^{5}C_{1}/^{9}C_{5}
Case 2: 3 blue 2 red
^{4}C_{3}*^{5}C_{2}/^{9}C_{5}
Add the casesIncorrect
Explanation:
Case 1: 4 blue 1 red
^{4}C_{4}*^{5}C_{1}/^{9}C_{5}
Case 2: 3 blue 2 red
^{4}C_{3}*^{5}C_{2}/^{9}C_{5}
Add the cases 
Question 9 of 15
9. Question
1 pointsCategory: Quantitative AptitudeThe speed of the stream is 5 km/hr and speed of the boat in still water is 9 km/hr. if a boat takes 55 hrs to travel downstream from point A to B and coming back to point C midway between A and B, the what is the total distance travelled by the boat?
Correct
Explanation:
Downstream speed = 9+5 = 14
Upstream speed = 95 = 4
Let the distance from A to B is x km
downstream distance = x, upstream distance is half of x because boat goes half way upstream, then
x/14 + (x/2)/4 = 55
solve, x = 280
so boat covered 280 + 280/2Incorrect
Explanation:
Downstream speed = 9+5 = 14
Upstream speed = 95 = 4
Let the distance from A to B is x km
downstream distance = x, upstream distance is half of x because boat goes half way upstream, then
x/14 + (x/2)/4 = 55
solve, x = 280
so boat covered 280 + 280/2 
Question 10 of 15
10. Question
1 pointsCategory: Quantitative AptitudeTwo cylindrical vessels of radius 7 cm and 14 cm and heights 10 cm and 12 cm respectively are filled with water and milk respectively. From both the vessels half of the total quantity is taken and mixed in a third vessel, what is the approximate percentage of water in the third vessel?
Correct
Explanation:
Water in 1st vessel = (22/7)*7*7*10 = 1540 cm^{3}
Milk in 2nd vessel = (22/7)*14*14*12 = 7392 cm^{3}
In third vessel, water taken is 1540/2 = 770 and milk taken is 7392/2 = 3696
So total is 770+3696 = 4516
So percentage of water = (770/4466)*100Incorrect
Explanation:
Water in 1st vessel = (22/7)*7*7*10 = 1540 cm^{3}
Milk in 2nd vessel = (22/7)*14*14*12 = 7392 cm^{3}
In third vessel, water taken is 1540/2 = 770 and milk taken is 7392/2 = 3696
So total is 770+3696 = 4516
So percentage of water = (770/4466)*100 
Question 11 of 15
11. Question
1 pointsCategory: Quantitative AptitudeA vessel is full of milk. A milkman takes out 5 l of milk and pours 10 l of water. This procedure is performed 1 more time. If after this the ratio of milk to water in the final solution is 2 : 3, what was the original quantity of milk in the vessel?
Correct
Explanation:
Suppose x l of milk
Now after 1st step:
Milk = x5, and water = 10 so ratio of milk to water is (x5) : 10
After step2:
Milk = (x5) – [(x5)/(x5+10)] * 5 and water = 10 – [10/(x5+10)] * 5
And 10 l water added too, so water is 10 – [10/(x5+10)] * 5 + 10 = 20 – [10/(x5+10)] * 5
So milk/water is now 6/5
Solve {(x5) – [(x5)/(x5+10)] * 5 } / {20 – [10/(x5+10)] * 5} = 2/3
[(x5)(x5) – (5x25)] / [20(x+5) – 50] = 2/3
Solve, 3x^{2} – 55x – 100 = 0
3x^{2} – 60x + 5x – 100 = 0
Solve, x = 20Incorrect
Explanation:
Suppose x l of milk
Now after 1st step:
Milk = x5, and water = 10 so ratio of milk to water is (x5) : 10
After step2:
Milk = (x5) – [(x5)/(x5+10)] * 5 and water = 10 – [10/(x5+10)] * 5
And 10 l water added too, so water is 10 – [10/(x5+10)] * 5 + 10 = 20 – [10/(x5+10)] * 5
So milk/water is now 6/5
Solve {(x5) – [(x5)/(x5+10)] * 5 } / {20 – [10/(x5+10)] * 5} = 2/3
[(x5)(x5) – (5x25)] / [20(x+5) – 50] = 2/3
Solve, 3x^{2} – 55x – 100 = 0
3x^{2} – 60x + 5x – 100 = 0
Solve, x = 20 
Question 12 of 15
12. Question
1 pointsCategory: Quantitative AptitudeThe outer diameter of cylinder is 1.6 cm and its thickness it 0.1 cm. If the height of cylinder is 40 cm, then how much water can the cylinder hold?
Correct
Explanation:
Radius of outer = 0.8 cm, so of inner = 0.8 – 0.1 = 0.7 cm
So water it can hold is (22/7) * 0.7 * 0.7 * 40Incorrect
Explanation:
Radius of outer = 0.8 cm, so of inner = 0.8 – 0.1 = 0.7 cm
So water it can hold is (22/7) * 0.7 * 0.7 * 40 
Question 13 of 15
13. Question
1 pointsCategory: Quantitative Aptitude2 persons P and Q who start at same time from point A reach point B at 2 PM and 4 PM respectively. The average speed of P is 15 km/hr and that of Q is 10 km/hr. What should be there average speed such that both reach at 1 PM?
Correct
Explanation:
Let distance from A to B is ‘d’ and time taken to reach at 1 PM is ‘t’
Then with 2 PM, time becomes ‘t+1’, so d/(t+1) = 15
With 4 PM, time becomes ‘t+3’, so d/(t+3) = 10
Solve the equations, t = 3 hrs, d = 60 km
So speed to reach at 3 PM = 60/3 = 20 km/hrIncorrect
Explanation:
Let distance from A to B is ‘d’ and time taken to reach at 1 PM is ‘t’
Then with 2 PM, time becomes ‘t+1’, so d/(t+1) = 15
With 4 PM, time becomes ‘t+3’, so d/(t+3) = 10
Solve the equations, t = 3 hrs, d = 60 km
So speed to reach at 3 PM = 60/3 = 20 km/hr 
Question 14 of 15
14. Question
1 pointsCategory: Quantitative AptitudeAge of Shweta is 4 years more than that of Raghav and also Shweta’s age is 8 years less than Reena. After 7 years ratio of ages of Raghav to Reena’s will be 7 : 13. What is the present age of Shweta?
Correct
Explanation:
Shweta = A, Raghav = B, Reena = C
A = B+4
A = C8
B+7 / C+ 7 = 7/13
Or (A4)+7 / (A+8)+7 = 7/13
A+3 / A+15 = 7/13
A = 11Incorrect
Explanation:
Shweta = A, Raghav = B, Reena = C
A = B+4
A = C8
B+7 / C+ 7 = 7/13
Or (A4)+7 / (A+8)+7 = 7/13
A+3 / A+15 = 7/13
A = 11 
Question 15 of 15
15. Question
1 pointsCategory: Quantitative AptitudeA cone of radius 8 cm and height 6 cm is mounted on a cylinder of radius 8 cm and height 12 cm. Find the total surface area of the figure thus formed.
Correct
Explanation:
Slant height of cone, l = √8^{2} + 6^{2} = 10 cm
Total surface area of final figure = curved surface area of cone + curved surface area of cylinder + area of base
= ᴨrl + 2ᴨrh + ᴨr^{2}
= ᴨr (l + 2h + r)
= (22/7) * 8 (10 + 2*12 +8)Incorrect
Explanation:
Slant height of cone, l = √8^{2} + 6^{2} = 10 cm
Total surface area of final figure = curved surface area of cone + curved surface area of cylinder + area of base
= ᴨrl + 2ᴨrh + ᴨr^{2}
= ᴨr (l + 2h + r)
= (22/7) * 8 (10 + 2*12 +8)
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