Aptitude Test 9
Questions Result & StatisticsQuizsummary
0 of 15 questions completed
Questions:
 1
 2
 3
 4
 5
 6
 7
 8
 9
 10
 11
 12
 13
 14
 15
Information
Aptitude
 Total number of questions : 15
 Time allotted : 25 minutes.
 Each question carry 1 mark, no negative marks.
 Click the “Finish quiz” button given in bottom of this page to submit your answer.
 Test will be submitted automatically if the time expired.
 Don’t refresh the page.
You have already completed the quiz before. Hence you can not start it again.
Quiz is loading...
You must sign in or sign up to start the quiz.
You have to finish following quiz, to start this quiz:
Results
Marks : 0 of 15 questions answered correctly  
Total number of questions  :  15 
Number of currect answered  :  0 
Your time  : 

Time has elapsed
You have reached 0 of 0 points, (0)
Categories
 Not categorized 0%

Thank you for submitting online aptitude test!
 1
 2
 3
 4
 5
 6
 7
 8
 9
 10
 11
 12
 13
 14
 15
 Answered
 Review
 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