Aptitude Test 9
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Question 1 of 15
1. Question
1 pointsCorrectAnswer : A
Solution : let the marked price be Rs 100
then ,net S.P=95% of 90% of 80% of Rs 100
=Rs(95/100*90/100*80/100*100)=Rs68.40
IncorrectAnswer : A
Solution : let the marked price be Rs 100
then ,net S.P=95% of 90% of 80% of Rs 100
=Rs(95/100*90/100*80/100*100)=Rs68.40

Question 2 of 15
2. Question
1 points.
Shobha’s mathematics test had 75 problems i.e. 10 arithmetic, 30 algebra and 35 geometry problems. Although she answered 70% of the arithmetic ,40% of the algebra, and 60% of the geometry problems correctly. she did not pass the test because she got less than 60% of the problems right. How many more questions she would have to answer correctly to earn 60% of the passing grade?
CorrectAnswer : B
Solution : Number of questions attempted correctly=(70% of 10 + 40% of 30 + 60% 0f 35)
=7 + 12+21= 45
Questions to be answered correctly for 60% grade=60% of 75 = 45
Therefore required number of questions= (4540) = 5.
IncorrectAnswer : B
Solution : Number of questions attempted correctly=(70% of 10 + 40% of 30 + 60% 0f 35)
=7 + 12+21= 45
Questions to be answered correctly for 60% grade=60% of 75 = 45
Therefore required number of questions= (4540) = 5.

Question 3 of 15
3. Question
1 points.
There are two sections A and B of a class consisting of 36 and 44 students respectively. If the average weight of section A is 40kg and that of section B is 35kg, find the average weight of the whole class?
CorrectAnswer : C
Solution : total weight of(36+44) students=(36*40+44*35)kg =2980kg.
Therefore weight of the total class=(2980/80)kg =37.25kg.
IncorrectAnswer : C
Solution : total weight of(36+44) students=(36*40+44*35)kg =2980kg.
Therefore weight of the total class=(2980/80)kg =37.25kg.

Question 4 of 15
4. Question
1 points.
A third of Arun’s marks in mathematics exceeds a half of his marks in English by 80.if he got 240 marks In two subjects together how many marks did he got in English?
CorrectAnswer : D
Solution : Let Arun’s marks in mathematics and english be x and y
Then 1/3x1/2y=30
2x3y=180……>(1)
x+y=240…….>(2)
solving (1) and (2)
x=180
and y=60
IncorrectAnswer : D
Solution : Let Arun’s marks in mathematics and english be x and y
Then 1/3x1/2y=30
2x3y=180……>(1)
x+y=240…….>(2)
solving (1) and (2)
x=180
and y=60

Question 5 of 15
5. Question
1 points.
A dog takes 4 leaps for every 5 leaps of a hare but 3 leaps of a dog are equal to 4 leaps of the hare. Compare their speeds.
CorrectAnswer : A
Solution : Let the distance covered in 1 leap of the dog be x and that covered in 1 leap of the hare by y.
Then, 3x = 4y => x = 4/3 y => 4x = 16/3 y.
\ Ratio of speeds of dog and hare = Ratio of distances covered by them in the same time
= 4x: 5y = 16/3 y: 5y =16/3: 5 = 16:15
IncorrectAnswer : A
Solution : Let the distance covered in 1 leap of the dog be x and that covered in 1 leap of the hare by y.
Then, 3x = 4y => x = 4/3 y => 4x = 16/3 y.
\ Ratio of speeds of dog and hare = Ratio of distances covered by them in the same time
= 4x: 5y = 16/3 y: 5y =16/3: 5 = 16:15

Question 6 of 15
6. Question
1 points.
Two trains 100 meter’s and 120 meter’s long are running in the same direction with speeds of 72 km/hr ,In how much time will the first train cross the second?
CorrectAnswer : B
Solution : Relative speed of the trains = (72 – 54) km/hr = 18 km/hr
= (18 * 5/18) m/sec = 5 m/sec.
Time taken by the trains to cross each other
= Time taken to cover (100 + 120) m at 5 m /sec = (220/5) sec = 44 sec.
IncorrectAnswer : B
Solution : Relative speed of the trains = (72 – 54) km/hr = 18 km/hr
= (18 * 5/18) m/sec = 5 m/sec.
Time taken by the trains to cross each other
= Time taken to cover (100 + 120) m at 5 m /sec = (220/5) sec = 44 sec.

Question 7 of 15
7. Question
1 points.
A room is half as long again as it is broad. The cost of carpeting the at Rs. 5 per sq. m is Rs. 270 and the cost of papering the four walls at Rs. 10 per m^{2} is Rs. 1720. If a door and 2 windows occupy 8 sq. m, find the dimensions of the room.
CorrectAnswer : C
Solution : Let breadth = x metres, length = 3x metres, height = H metres.
Area of the floor=(Total cost of carpeting)/(Rate/m^{2})=(270/5)m^{2}=54m^{2}.
x* (3x/2) = 54 <=> x^2 = (54*2/3) = 36 <=> x = 6.
So, breadth = 6 m and length =(3/2)*6 = 9 m.
Now, papered area = (1720/10)m^{2} = 172 m^{2}.
Area of 1 door and 2 windows = 8 m^{2}.
Total area of 4 walls = (172 + 8) m^{2} = 180 m^{2}
2*(9+ 6)* H = 180 <=> H = 180/30 = 6 m.
IncorrectAnswer : C
Solution : Let breadth = x metres, length = 3x metres, height = H metres.
Area of the floor=(Total cost of carpeting)/(Rate/m^{2})=(270/5)m^{2}=54m^{2}.
x* (3x/2) = 54 <=> x^2 = (54*2/3) = 36 <=> x = 6.
So, breadth = 6 m and length =(3/2)*6 = 9 m.
Now, papered area = (1720/10)m^{2} = 172 m^{2}.
Area of 1 door and 2 windows = 8 m^{2}.
Total area of 4 walls = (172 + 8) m^{2} = 180 m^{2}
2*(9+ 6)* H = 180 <=> H = 180/30 = 6 m.

Question 8 of 15
8. Question
1 points.
How many spherical bullets can be made out of a lead cylinder 28cm high and with radius 6 cm, each bullet being 1.5 cm in diameter?
CorrectAnswer : D
Solution : Volume of cylinder = (∏ x 6 x 6 x 28) cm^{3} = ( 9∏/16) cm^{3.}
^{ }Number of bullet = Volume of cylinder = [(36 x 28)∏ x 16] /9∏ = 1792.
Volume of each bullet
IncorrectAnswer : D
Solution : Volume of cylinder = (∏ x 6 x 6 x 28) cm^{3} = ( 9∏/16) cm^{3.}
^{ }Number of bullet = Volume of cylinder = [(36 x 28)∏ x 16] /9∏ = 1792.
Volume of each bullet

Question 9 of 15
9. Question
1 points.
A thief is spotted by a policeman from a distance of 100 metres. When the policeman starts the chase, the thief also starts running. If the speed of the thief be 8km/hr and that of the policeman 10 km/hr, how far the thief will have run before he is overtaken?
CorrectAnswer : B
Solution : Relative speed of the policeman = (108) km/hr =2 km/hr.
Time taken by police man to cover 100m 100 x 1 hr = 1 hr.
1000 2 20
In 1/20 hrs. The thief covers a distance of 8 x 1/20 km = 2/5 km = 400 m
Relative speed of the policeman = (108) km/hr =2 km/hr.
Time taken by police man to cover 100m 100 x 1 hr = 1 hr.
1000 2 20
In 1/20 hrs. The thief covers a distance of 8 x 1/20 km = 2/5 km = 400 m
IncorrectAnswer : B
Solution : Relative speed of the policeman = (108) km/hr =2 km/hr.
Time taken by police man to cover 100m 100 x 1 hr = 1 hr.
1000 2 20
In 1/20 hrs. The thief covers a distance of 8 x 1/20 km = 2/5 km = 400 m
Relative speed of the policeman = (108) km/hr =2 km/hr.
Time taken by police man to cover 100m 100 x 1 hr = 1 hr.
1000 2 20
In 1/20 hrs. The thief covers a distance of 8 x 1/20 km = 2/5 km = 400 m

Question 10 of 15
10. Question
1 points.
If the manufacturer gains 10%,the wholesale dealer 15% and the retailer 25% ,then find the cost of production of a ,the retail price of which is Rs.1265?
CorrectAnswer : D
Solution : Let the cost of production of the table be Rs x
The, 125% of 115% of 110% of x=1265
125/100*115/100*110/100*x=1265=>253/160*x=>1265=>x=(1265*160/253)=Rs.800
IncorrectAnswer : D
Solution : Let the cost of production of the table be Rs x
The, 125% of 115% of 110% of x=1265
125/100*115/100*110/100*x=1265=>253/160*x=>1265=>x=(1265*160/253)=Rs.800

Question 11 of 15
11. Question
1 points.
Mr.Jones gave 40% of the money he had to his wife. He also gave 20% of the remaining amount to his 3 sons. Half of the amount now left was spent on miscellaneous items and the remaining amount of Rs.12000 was deposited in the bank. How much money did Mr.jones have initially?
CorrectAnswer : C
Solution : Let the initial amount with Mr.Jones be Rs.x
Then,(1/2)[100(3*20)]% of x=12000
ó (1/2)*(40/100)*(60/100)*x=12000
óx=((12000*25)/3)=100000
IncorrectAnswer : C
Solution : Let the initial amount with Mr.Jones be Rs.x
Then,(1/2)[100(3*20)]% of x=12000
ó (1/2)*(40/100)*(60/100)*x=12000
óx=((12000*25)/3)=100000

Question 12 of 15
12. Question
1 points.
A tin of oil was 4/5full.when 6 bottles of oil were taken out and four bottles of oil were poured into it, it was ¾ full. how many bottles of oil can the tin contain?
CorrectAnswer : C
Solution : Suppose x bottles can fill the tin completely
Then4/5x3/4x=64
X/20=2
X=40
Therefore required no of bottles =40
IncorrectAnswer : C
Solution : Suppose x bottles can fill the tin completely
Then4/5x3/4x=64
X/20=2
X=40
Therefore required no of bottles =40

Question 13 of 15
13. Question
1 points.
Excluding stoppages, the speed of the bus is 54kmph and including stoppages, it is 45kmph.for how many min does the bus stop per hr.
CorrectAnswer : A
Solution : Due to stoppages,it covers 9km less.
time taken to cover 9 km is [9/54 *60] min = 10min
IncorrectAnswer : A
Solution : Due to stoppages,it covers 9km less.
time taken to cover 9 km is [9/54 *60] min = 10min

Question 14 of 15
14. Question
1 points.
Find the smallest number which when divided by 6, 10 and 15 respectively leaves 5 as Remainder in each case?
CorrectAnswer : A
Solution : L.C.M of 6,10,15 — > 30
Required number =30 + 5 = 35
IncorrectAnswer : A
Solution : L.C.M of 6,10,15 — > 30
Required number =30 + 5 = 35

Question 15 of 15
15. Question
1 points.
If 1/3rd of a number subtracted from. Of that number, then the difference is 10more Than 1/7th of the same number. How much is that number?
CorrectAnswer : A
Solution : Let the number = x
: . x/2 – x/3 = x/7+ 10
— > 3x2x/6 = x/7+10
— > x/6 – x/7 = 10
— >7x6x/42 = 10 — > x = 420
IncorrectAnswer : A
Solution : Let the number = x
: . x/2 – x/3 = x/7+ 10
— > 3x2x/6 = x/7+10
— > x/6 – x/7 = 10
— >7x6x/42 = 10 — > x = 420
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