Aptitude Test 10
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Aptitude
- Total number of questions : 15
- Time allotted : 25 minutes.
- Each question carry 1 mark, no negative marks.
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- Question 1 of 15
1. Question
1 points.
In an election between two candidates, 75% of the voters cast their votes, out of which 2% of the votes were declared invalid. A candidate got 9261 votes which were 75% of the total valid votes. Find the total number of votes enrolled in that election.
CorrectAnswer : A
Solution : Let the number of votes enrolled be x. Then,
Number of votes cast =75% of x. Valid votes = 98% of (75% of x).
75% of (98% of (75%of x)) =9261.
[(75/100)*(98/100)*(75/100)*x] =9261.
X = [(9261*100*100*100)/(75*98*75)] =16800
IncorrectAnswer : A
Solution : Let the number of votes enrolled be x. Then,
Number of votes cast =75% of x. Valid votes = 98% of (75% of x).
75% of (98% of (75%of x)) =9261.
[(75/100)*(98/100)*(75/100)*x] =9261.
X = [(9261*100*100*100)/(75*98*75)] =16800
- Question 2 of 15
2. Question
1 points.
An employer pays Rs.20 for each day a worker works and for feits Rs.3 for each day is ideal at the end of sixty days a worker gets Rs.280 . for how many days did the worker remain ideal?
CorrectAnswer : B
Solution : Suppose a worker remained ideal for x days then he worked for 60-x days
20*(60-x)-3x=280
1200-23x=280
23x=920
x=40
IncorrectAnswer : B
Solution : Suppose a worker remained ideal for x days then he worked for 60-x days
20*(60-x)-3x=280
1200-23x=280
23x=920
x=40
- Question 3 of 15
3. Question
1 points.
in a certain office 1/3 of the workers are women ½ of the women are married and 1/3 of the married women have children if ¾ of the men are married and 2/3 of the married men have children what part of workers are without children?
CorrectAnswer : D
Solution : Let the total no of workers be x
No of women =x/3
No of men =x-(x/3)=2x/3
No of women having children =1/3 of ½ ofx/3=x/18
No of men having children=2/3 of ¾ of2x/3=x/3
No of workers having children = x/8 +x/3=7x/18
Workers having no children=x-7x/18=11x/18=11/18 of all workers
IncorrectAnswer : D
Solution : Let the total no of workers be x
No of women =x/3
No of men =x-(x/3)=2x/3
No of women having children =1/3 of ½ ofx/3=x/18
No of men having children=2/3 of ¾ of2x/3=x/3
No of workers having children = x/8 +x/3=7x/18
Workers having no children=x-7x/18=11x/18=11/18 of all workers
- Question 4 of 15
4. Question
1 points.
A Cone and a sphere have equal radii and equal volumes. Find the ratio of the sphere of the diameter of the sphere to the height of the cone
CorrectAnswer : C
Solution : Let radius of each be R and height of the cone be H.
Then, (4/3) ∏ R3 = (1/3) ∏ R2H (or) R/H = ¼ (or) 2R/H = 2/4 =1/2
\Required ratio = 1:2.
IncorrectAnswer : C
Solution : Let radius of each be R and height of the cone be H.
Then, (4/3) ∏ R3 = (1/3) ∏ R2H (or) R/H = ¼ (or) 2R/H = 2/4 =1/2
\Required ratio = 1:2.
- Question 5 of 15
5. Question
1 points.
Three solid cubes of sides 1 cm, 6 cm and 8 cm are melted to form a new Cube. Find the surface area of the cube so formed.
CorrectAnswer : A
Solution : Volume of new cube = (13 + 63 + 83) cm+ = 729 cm3.
Edge of new cube = 3Ö729 cm = 9 cm.
\ Surface area of the new cube = (6 x 9 x 9) cm2 = 486 cm2.
IncorrectAnswer : A
Solution : Volume of new cube = (13 + 63 + 83) cm+ = 729 cm3.
Edge of new cube = 3Ö729 cm = 9 cm.
\ Surface area of the new cube = (6 x 9 x 9) cm2 = 486 cm2.
- Question 6 of 15
6. Question
1 pointsCorrectAnswer : B
Solution : a/2 = b/3 = c/5 = k
— > a = 2k, b = 3k, c = 5k
: . a+b+c/c = 2k+3k+5k/5k = 10k/5k = 2
IncorrectAnswer : B
Solution : a/2 = b/3 = c/5 = k
— > a = 2k, b = 3k, c = 5k
: . a+b+c/c = 2k+3k+5k/5k = 10k/5k = 2
- Question 7 of 15
7. Question
1 points.
Of the three numbers, second is twice the first and also thrice the third. If the average of the three numbers is 44, the largest number is
CorrectAnswer : C
Solution : Let the 3rd number be x, Then, second number = 3x
: . First number = 3x/2
: . x+3x + 3x/2 = (44*3)
— > 11x/2 = 132 — > x = 24
Largest number — > 3x
— > 3*24 = 72
IncorrectAnswer : C
Solution : Let the 3rd number be x, Then, second number = 3x
: . First number = 3x/2
: . x+3x + 3x/2 = (44*3)
— > 11x/2 = 132 — > x = 24
Largest number — > 3x
— > 3*24 = 72
- Question 8 of 15
8. Question
1 points.
Three numbers are in the ratio 3:4:5. The sum of the largest and the smallest equals the Sum of the third and 52. The smallest number is
CorrectAnswer : A
Solution : Let the numbers be x, y, z
: . x/3 = y/4 = z/5 = k
: . x = 3k, y = 4k, z = 5k
: . 3k + 5k= 4k + 52
— > k = 13
: . Smallest number = 39
IncorrectAnswer : A
Solution : Let the numbers be x, y, z
: . x/3 = y/4 = z/5 = k
: . x = 3k, y = 4k, z = 5k
: . 3k + 5k= 4k + 52
— > k = 13
: . Smallest number = 39
- Question 9 of 15
9. Question
1 points.
Vikas can cover a distance in 1hr 24min by covering 2/3 of the distance at 4 kmph and the rest at 5kmph.the total distance is?
CorrectAnswer : C
Solution : Let total distance be S
Total time=1hr24min
A to T: speed=4kmph
Distance=2/3S
T to S :: speed=5km
Distance=1-2/3S=1/3S
21/15 hr=2/3 S/4 + 1/3s /5
84=14/3S*3
S=84*3/14*3
= 6km
IncorrectAnswer : C
Solution : Let total distance be S
Total time=1hr24min
A to T: speed=4kmph
Distance=2/3S
T to S :: speed=5km
Distance=1-2/3S=1/3S
21/15 hr=2/3 S/4 + 1/3s /5
84=14/3S*3
S=84*3/14*3
= 6km
- Question 10 of 15
10. Question
1 points.
Two trains start from A & B and travel towards each other at speed of 50kmph and 60kmph resp. At the time of the meeting the second train has traveled 120km more than the first. The distance between them is
CorrectAnswer : D
Solution : Let the distance traveled by the first train be x km
then distance covered by the second train is x + 120km
x/50 = x+120 / 60
x= 600
so the distance between A & B is x + x + 120 = 1320 km
IncorrectAnswer : D
Solution : Let the distance traveled by the first train be x km
then distance covered by the second train is x + 120km
x/50 = x+120 / 60
x= 600
so the distance between A & B is x + x + 120 = 1320 km
- Question 11 of 15
11. Question
1 points.
The average ages of three persons is 27 years. Their ages are in the proportion of 1:3:5.What is the age in years of the youngest one among them.
CorrectAnswer : B
Solution : Let the age of three persons be x, 3x and 5x
— > 9x/3 = 27 — > x = 9
IncorrectAnswer : B
Solution : Let the age of three persons be x, 3x and 5x
— > 9x/3 = 27 — > x = 9
- Question 12 of 15
12. Question
1 pointsCorrectAnswer : D
Solution : Let the numbers be 2n, 2n+2 and 2n+4
2n + (2n+2) + (2n+4) = 48
6n = 48-6 = 42, n = 7
Hence the numbers are — > 14, 16 and 18
The least number is 14.
IncorrectAnswer : D
Solution : Let the numbers be 2n, 2n+2 and 2n+4
2n + (2n+2) + (2n+4) = 48
6n = 48-6 = 42, n = 7
Hence the numbers are — > 14, 16 and 18
The least number is 14.
- Question 13 of 15
13. Question
1 points.
13 sheeps and 9 pigs were bought for Rs. 1291.85.If the average price of a sheep be Rs. 74. What is the average price of a pig.
CorrectAnswer : C
Solution : Average price of a sheep = Rs. 74
: . Total price of 13 sheeps
= (74*13) = Rs. 962
But, total price of 13 sheeps and 9 pigs
= Rs. 1291.85
Total price of 9 pigs
= Rs. (1291.85-962) = Rs. 329.85
Hence, average price of a pig
= (329.85/9) = Rs. 36.65
IncorrectAnswer : C
Solution : Average price of a sheep = Rs. 74
: . Total price of 13 sheeps
= (74*13) = Rs. 962
But, total price of 13 sheeps and 9 pigs
= Rs. 1291.85
Total price of 9 pigs
= Rs. (1291.85-962) = Rs. 329.85
Hence, average price of a pig
= (329.85/9) = Rs. 36.65
- Question 14 of 15
14. Question
1 points.
A batsman in his 18th innings makes a score of 150 runs and thereby increasing his Average by 6. Find his average after 18th innings.
CorrectAnswer : B
Solution : Let the average for 17 innings is x runs
Total runs in 17 innings = 17x
Total runs in 18 innings = 17x + 150
Average of 18 innings = 17x + 150/18
: . 17x + 150/18 = x + 6 — > x = 42
Thus, average after 18 innings = 42
IncorrectAnswer : B
Solution : Let the average for 17 innings is x runs
Total runs in 17 innings = 17x
Total runs in 18 innings = 17x + 150
Average of 18 innings = 17x + 150/18
: . 17x + 150/18 = x + 6 — > x = 42
Thus, average after 18 innings = 42
- Question 15 of 15
15. Question
1 points.
P is able to do a piece of work in 15 days and Q can do the same work in 20 days. If they can work together for 4 days, what is the fraction of work left?
CorrectAnswer : B
Solution : Amount of work P can do in 1 day = 1/15
Amount of work Q can do in 1 day = 1/20
Amount of work P and Q can do in 1 day = 1/15 + 1/20 = 7/60
Amount of work P and Q can together do in 4 days = 4 × (7/60) = 7/15
Fraction of work left = 1 – 7/15= 8/15
IncorrectAnswer : B
Solution : Amount of work P can do in 1 day = 1/15
Amount of work Q can do in 1 day = 1/20
Amount of work P and Q can do in 1 day = 1/15 + 1/20 = 7/60
Amount of work P and Q can together do in 4 days = 4 × (7/60) = 7/15
Fraction of work left = 1 – 7/15= 8/15
