Aptitude Test 7
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Question 1 of 15
1. Question
1 points.
One side of a rectangular field is 15 m and one of its diagonals is 17 m. Find the area of the field.
CorrectAnswer : A
Solution : Other side = ((17) ^{2}– (15)^{2)(1/2}) = (289 225)^{(1/2)} = (64)^{(1/2)} = 8 m.
Area = (15 x 8) m^{2} = 120 m^{2}.
IncorrectAnswer : A
Solution : Other side = ((17) ^{2}– (15)^{2)(1/2}) = (289 225)^{(1/2)} = (64)^{(1/2)} = 8 m.
Area = (15 x 8) m^{2} = 120 m^{2}.

Question 2 of 15
2. Question
1 points.
A train 220 m long is running with a speed of 59 kmph.. In what will it pass a man who is running at 7 kmph in the direction opposite to that in which the train is going?
CorrectAnswer : C
Solution : Speed of the train relative to man = (59 + 7) kmph
= 66 *5/18 m/sec = 55/3 m/sec.
Time taken by the train to cross the man = Time taken by it to cover 220 m
at (55/3) m / sec = (220 *3/55) sec = 12 sec
IncorrectAnswer : C
Solution : Speed of the train relative to man = (59 + 7) kmph
= 66 *5/18 m/sec = 55/3 m/sec.
Time taken by the train to cross the man = Time taken by it to cover 220 m
at (55/3) m / sec = (220 *3/55) sec = 12 sec

Question 3 of 15
3. Question
1 points.
45 men can complete a work in 16 days. Six days after they started working, 30 more men joined them. How many days will they now take to complete the remaining work?
CorrectAnswer : B
Solution : (45 x 16) men can complete the work in 1 day.
1 man’s 1 day’s work = 1/720
45 men’s 6 days’ work =(1/16*6)=3/8
Remaining work =(13/8)=5/8
75 men’s 1 day’s work = 75/720=5/48
Now, 5/48 work is done by them in 1 day.
5/8 work is done by them in (48/5 x 5/8)=6 days
IncorrectAnswer : B
Solution : (45 x 16) men can complete the work in 1 day.
1 man’s 1 day’s work = 1/720
45 men’s 6 days’ work =(1/16*6)=3/8
Remaining work =(13/8)=5/8
75 men’s 1 day’s work = 75/720=5/48
Now, 5/48 work is done by them in 1 day.
5/8 work is done by them in (48/5 x 5/8)=6 days

Question 4 of 15
4. Question
1 points.
The sum of two numbers is 184. If onethird of the one exceeds oneseventh of the other by 8, find the smaller number
CorrectAnswer : A
Solution : Let the numbers be x and (184 – x). Then,
(X/3) – ((184 – x)/7) = 8 => 7x – 3(184 – x) = 168 => 10x = 720 => x = 72.
So, the numbers are 72 and 112. Hence, smaller number = 72.
IncorrectAnswer : A
Solution : Let the numbers be x and (184 – x). Then,
(X/3) – ((184 – x)/7) = 8 => 7x – 3(184 – x) = 168 => 10x = 720 => x = 72.
So, the numbers are 72 and 112. Hence, smaller number = 72.

Question 5 of 15
5. Question
1 points.
Two pens and three pencils cost Rs 86. Four Pens and a pencil cost Rs. 112. Find the cost of a pen and that of a pencil.
CorrectAnswer : B
Solution : Let the cost of a pen and a pencil be Rs. X and Rs. Y respectively.
Then, 2x + 3y = 86 ….(i) and 4x + y =112.
Solving (i) and (ii), we get: x = 25 and y = 12.
Cost of a pen =Rs. 25 and the cost of a pencil =Rs. 12
IncorrectAnswer : B
Solution : Let the cost of a pen and a pencil be Rs. X and Rs. Y respectively.
Then, 2x + 3y = 86 ….(i) and 4x + y =112.
Solving (i) and (ii), we get: x = 25 and y = 12.
Cost of a pen =Rs. 25 and the cost of a pencil =Rs. 12

Question 6 of 15
6. Question
1 pointsCorrectAnswer : D
Solution : The given equations are:
2x+3y+z=55 …(i); x + z – y=4 …(ii); y x + z =12 …(iii)
Subtracting (ii) from (i), we get: x+4y=51 …(iv)
Subtracting (iii) from (i), we get: 3x+2y=43 …(v)
Multiplying (v) by 2 and subtracting (iv) from it, we get: 5x=35 or x=7.
Putting x=7 in (iv), we get: 4y=44 or y=11.
Putting x=7,y=11 in (i), we get: z=8.
IncorrectAnswer : D
Solution : The given equations are:
2x+3y+z=55 …(i); x + z – y=4 …(ii); y x + z =12 …(iii)
Subtracting (ii) from (i), we get: x+4y=51 …(iv)
Subtracting (iii) from (i), we get: 3x+2y=43 …(v)
Multiplying (v) by 2 and subtracting (iv) from it, we get: 5x=35 or x=7.
Putting x=7 in (iv), we get: 4y=44 or y=11.
Putting x=7,y=11 in (i), we get: z=8.

Question 7 of 15
7. Question
1 points.
The average weight of 10 oarsmen in a boat is increased by 1.8 kg when one of the crew, who weighs 53 kg is replaced by a new man. Find the weight of the new man.
CorrectAnswer : C
Solution : Total weight increased =(1.8 x 10) kg =18 kg.
:. Weight of the new man =(53 + 18) kg =71 kg.
IncorrectAnswer : C
Solution : Total weight increased =(1.8 x 10) kg =18 kg.
:. Weight of the new man =(53 + 18) kg =71 kg.

Question 8 of 15
8. Question
1 pointsCorrectAnswer : A
Solution : Here s={h,t} and e={h}.
P(E)=n(E)/n(S)=1/2
IncorrectAnswer : A
Solution : Here s={h,t} and e={h}.
P(E)=n(E)/n(S)=1/2

Question 9 of 15
9. Question
1 pointsCorrectAnswer : B
Solution : Let their edges be a and b. Then,
a^{3}/b^{3} = 1/27 (or) (a/b)^{3 }= (1/3)^{3 } (or) (a/b) = (1/3).
Ratio of their surface area = 6a^{2}/6b^{2 }= a^{2}/b^{2 }= (a/b)^{2} = 1/9, i.e. 1:9.
IncorrectAnswer : B
Solution : Let their edges be a and b. Then,
a^{3}/b^{3} = 1/27 (or) (a/b)^{3 }= (1/3)^{3 } (or) (a/b) = (1/3).
Ratio of their surface area = 6a^{2}/6b^{2 }= a^{2}/b^{2 }= (a/b)^{2} = 1/9, i.e. 1:9.

Question 10 of 15
10. Question
1 points.
A train running at 54 kmph takes 20 seconds to pass a platform. Next it takes.12 sec to pass a man walking at 6 kmph in the same direction in which the train is going . Find the length of the train and the length of the platform
CorrectAnswer : C
Solution : Let the length of train be x metres and length of platform be y metres.
Speed of the train relative to man = (54 – 6) kmph = 48 kmph
= 48*(5/18) m/sec = 40/3 m/sec.
In passing a man, the train covers its own length with relative speed.
Length of train = (Relative speed * Time) = ( 40/3)*12 m = 160 m.
Also, speed of the train = 54 *(5/18)m / sec = 15 m / sec.
(x+y)/15 = 20 <=> x + y = 300 <=> Y = (300 – 160) m = 140 m.
IncorrectAnswer : C
Solution : Let the length of train be x metres and length of platform be y metres.
Speed of the train relative to man = (54 – 6) kmph = 48 kmph
= 48*(5/18) m/sec = 40/3 m/sec.
In passing a man, the train covers its own length with relative speed.
Length of train = (Relative speed * Time) = ( 40/3)*12 m = 160 m.
Also, speed of the train = 54 *(5/18)m / sec = 15 m / sec.
(x+y)/15 = 20 <=> x + y = 300 <=> Y = (300 – 160) m = 140 m.

Question 11 of 15
11. Question
1 points.
A cyclist covers a distance of 750 m in 2 min 30 sec. What is the speed in km/hr. of the cyclist?
CorrectAnswer : D
Solution : Speed = {750/150} m/sec =5 m/sec = { 5 * 18 } km/hr =18km/hr
IncorrectAnswer : D
Solution : Speed = {750/150} m/sec =5 m/sec = { 5 * 18 } km/hr =18km/hr

Question 12 of 15
12. Question
1 points.
A can do a certain job in 12 days. B is 60% more efficient than A. How many days does B alone take to do the same job?
CorrectAnswer : B
Solution : Ratio of times taken by A and B = 160 : 100 = 8 : 5.
Suppose B alone takes x days to do the job.
Then, 8 : 5 :: 12 : x = 8x = 5 x 12 =x = 7 1/2 days
IncorrectAnswer : B
Solution : Ratio of times taken by A and B = 160 : 100 = 8 : 5.
Suppose B alone takes x days to do the job.
Then, 8 : 5 :: 12 : x = 8x = 5 x 12 =x = 7 1/2 days

Question 13 of 15
13. Question
1 points.
A retailer buys 40 pens at the market price of 36 pens from a wholesaler, if he sells these pens giving a discount of 1%, what is the profit %?
CorrectAnswer : D
Solution : let the market price of each pen be Rs 1
then,C.P of 40 pens = Rs 36 S.P of 40 pens =99% of Rs 40=Rs 39.60
profit %=((3.60*100)/36) %=10%
IncorrectAnswer : D
Solution : let the market price of each pen be Rs 1
then,C.P of 40 pens = Rs 36 S.P of 40 pens =99% of Rs 40=Rs 39.60
profit %=((3.60*100)/36) %=10%

Question 14 of 15
14. Question
1 points.
In an examination , 80% of the students passed in English , 85% in Mathematics and 75% in both English and Mathematics. If 40 students failed in both the subjects , find the total number of students.
CorrectAnswer : A
Solution : Let the total number of students be x .
Let A and B represent the sets of students who passed in English and Mathematics respectively
Then, number of students passed in one or both the subjects
= n (AÈB) =n (A)+n(B) n(AÇB)=80% of x + 85% of x –75% of x
=[(80/100)x+(85/100)x(75/100)x]=(90/100)x=(9/10)x
Students who failed in both the subjects = [x(9x/10)]=x/10.
So, x/10=40 of x=400.
Hence, total number of students = 400
IncorrectAnswer : A
Solution : Let the total number of students be x .
Let A and B represent the sets of students who passed in English and Mathematics respectively
Then, number of students passed in one or both the subjects
= n (AÈB) =n (A)+n(B) n(AÇB)=80% of x + 85% of x –75% of x
=[(80/100)x+(85/100)x(75/100)x]=(90/100)x=(9/10)x
Students who failed in both the subjects = [x(9x/10)]=x/10.
So, x/10=40 of x=400.
Hence, total number of students = 400

Question 15 of 15
15. Question
1 points.
The average age of a class of 39 students is 15 years. If the age of the teacher be included, then the average increases by3 months. Find the age of the teacher.
CorrectAnswer : C
Solution : Total age of 39 persons = (39 x 15) years = 585 years.
Average age of 40 persons= 15 yrs 3 months = 61/4 years.
Total age of 40 persons = (_(61/4 )x 40) years= 610 years.
:. Age of the teacher = (610 – 585) years=25 years.
IncorrectAnswer : C
Solution : Total age of 39 persons = (39 x 15) years = 585 years.
Average age of 40 persons= 15 yrs 3 months = 61/4 years.
Total age of 40 persons = (_(61/4 )x 40) years= 610 years.
:. Age of the teacher = (610 – 585) years=25 years.