Aptitude Test 6
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 Total number of questions : 15
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 Question 1 of 15
1. Question
1 pointsCorrectAnswer : A
Solution : Smallest number of 6 digits is 100000.
On dividing 100000 by 111, we get 100 as remainder.
Number to be added = (111 – 100) – 11.
Hence, required number = 100011
IncorrectAnswer : A
Solution : Smallest number of 6 digits is 100000.
On dividing 100000 by 111, we get 100 as remainder.
Number to be added = (111 – 100) – 11.
Hence, required number = 100011
 Question 2 of 15
2. Question
1 points.
A crate of mangoes contains one bruised mango for every thirty mango in the crate. If three out of every four bruised mango are considerably unsaleable and there are 12 unsaleable mangoes in the crate then how many mango are there in the crate?
CorrectAnswer : B
Solution : Let the total no of mangoes in the crate be x
Then the no of bruised mango = 1/30 x
Let the no of unsalable mangoes =3/4 (1/30 x)
1/40 x =12
x=480
IncorrectAnswer : B
Solution : Let the total no of mangoes in the crate be x
Then the no of bruised mango = 1/30 x
Let the no of unsalable mangoes =3/4 (1/30 x)
1/40 x =12
x=480
 Question 3 of 15
3. Question
1 points.
In a caravan, in addition to 50 hens there are 45 goats and 8 camels with some keepers. If the total number of feet be 224 more than the number of heads, find the number of keepers
CorrectAnswer : D
Solution : Let the number of keepers be x then,
Total number of heads =(50 + 45 + 8 + x)= (103 + x).
Total number of feet = (45 + 8) x 4 + (50 + x) x 2 =(312 +2x).
(312 + 2x)(103 + x) =224ó x =15.
Hence, number of keepers =15.
IncorrectAnswer : D
Solution : Let the number of keepers be x then,
Total number of heads =(50 + 45 + 8 + x)= (103 + x).
Total number of feet = (45 + 8) x 4 + (50 + x) x 2 =(312 +2x).
(312 + 2x)(103 + x) =224ó x =15.
Hence, number of keepers =15.
 Question 4 of 15
4. Question
1 points.
The ages of two persons differ by 16 years. If 6 years ago, the elder one be 3 times as old as the younger one, find their present ages.
CorrectAnswer : B
Solution : Let the age of the younger person be x years.
Then, age of the elder person = (x + 16) years.
3 (x – 6) = (x + 16 – 6) 3x 18 = x + 10 2x = 28 ó x = 14.
Hence, their present ages are 14 years and 30 years.
IncorrectAnswer : B
Solution : Let the age of the younger person be x years.
Then, age of the elder person = (x + 16) years.
3 (x – 6) = (x + 16 – 6) 3x 18 = x + 10 2x = 28 ó x = 14.
Hence, their present ages are 14 years and 30 years.
 Question 5 of 15
5. Question
1 points.
If A`s salary is 20% less then B`s salary , by how much percent is B`s salary more than A`s ?
CorrectAnswer : A
Solution : Required percentage = [(20*100)/(10020)]%=25%.
IncorrectAnswer : A
Solution : Required percentage = [(20*100)/(10020)]%=25%.
 Question 6 of 15
6. Question
1 points.
A is twice as good a workman as B and together they finish a piece in 18 days. In how many days will A alone finish the work?
CorrectAnswer : D
Solution : (A’s 1 day’s work):(B’s 1 days work) = 2 : 1.
(A + B)’s 1 day’s work = 1/18
Divide 1/18 in the ratio 2 : 1.
A’s 1 day’s work =(1/18*2/3)=1/27
Hence, A alone can finish the work in 27 days.
IncorrectAnswer : D
Solution : (A’s 1 day’s work):(B’s 1 days work) = 2 : 1.
(A + B)’s 1 day’s work = 1/18
Divide 1/18 in the ratio 2 : 1.
A’s 1 day’s work =(1/18*2/3)=1/27
Hence, A alone can finish the work in 27 days.
 Question 7 of 15
7. Question
1 points.
If a man walks at the rate of 5 kmph, he misses a train by 7 minutes. However, if he walks at the rate of 6 kmph, he reaches the station 5 minutes before the arrival of the train. Find the distance covered by him to reach the station.
CorrectAnswer : C
Solution : Let the required distance be x km
Difference in the time taken at two speeds=1 min =1/2 hr
Hence x/5x/6=1/5<=>6x5x=6
x=6
Hence, the required distance is 6 km
IncorrectAnswer : C
Solution : Let the required distance be x km
Difference in the time taken at two speeds=1 min =1/2 hr
Hence x/5x/6=1/5<=>6x5x=6
x=6
Hence, the required distance is 6 km
 Question 8 of 15
8. Question
1 points.
A man is standing on a railway bridge which is 180 m long. He finds that a train crosses the bridge in 20 seconds but himself in 8 seconds. Find the length of the train and its speed?
CorrectAnswer : C
Solution : Let the length of the train be x metres,
Then, the train covers x metres in 8 seconds and (x + 180) metres in 20 sec
x/8=(x+180)/20 ó 20x = 8 (x + 180) <=> x = 120.
Length of the train = 120 m.
Speed of the train = (120/8) m / sec = m / sec = (15 *18/5) kmph = 54 km
IncorrectAnswer : C
Solution : Let the length of the train be x metres,
Then, the train covers x metres in 8 seconds and (x + 180) metres in 20 sec
x/8=(x+180)/20 ó 20x = 8 (x + 180) <=> x = 120.
Length of the train = 120 m.
Speed of the train = (120/8) m / sec = m / sec = (15 *18/5) kmph = 54 km
 Question 9 of 15
9. Question
1 points.
A man can row upstream at 7 kmph and downstream at 10 kmph.Find man’s rate in still water and the rate of current
CorrectAnswer : A
Solution : Rate in still water=1/2(10+7)km/hr=8.5 km/hr.
Rate of current=1/2(107)km/hr=1.5 km/hr
IncorrectAnswer : A
Solution : Rate in still water=1/2(10+7)km/hr=8.5 km/hr.
Rate of current=1/2(107)km/hr=1.5 km/hr
 Question 10 of 15
10. Question
1 pointsCorrectAnswer : B
Solution : Area of the square = (1/2)* (diagonal)^{ 2} = [(1/2)*3.8*3.8 ]m^{2} = 7.22 m^{2}
IncorrectAnswer : B
Solution : Area of the square = (1/2)* (diagonal)^{ 2} = [(1/2)*3.8*3.8 ]m^{2} = 7.22 m^{2}
 Question 11 of 15
11. Question
1 points.
The difference between two parallel sides of a trapezium is 4 cm. perpendicular distance between them is 19 cm. If the area of the trapezium is 475 find the lengths of the parallel sides.
CorrectAnswer : A
Solution : Let the two parallel sides of the trapezium be a em and b em.
Then, a – b = 4
And, (1/2) x (a + b) x 19 = 475 ó (a + b) =((475 x 2)/19) ó a + b = 50
Solving (i) and (ii), we get: a = 27, b = 23.
So, the two parallel sides are 27 cm and 23 cm
IncorrectAnswer : A
Solution : Let the two parallel sides of the trapezium be a em and b em.
Then, a – b = 4
And, (1/2) x (a + b) x 19 = 475 ó (a + b) =((475 x 2)/19) ó a + b = 50
Solving (i) and (ii), we get: a = 27, b = 23.
So, the two parallel sides are 27 cm and 23 cm
 Question 12 of 15
12. Question
1 pointsCorrectAnswer : D
Solution : Required number of ways = ^{15}c_{11 }= ^{15}c_{(1511)} = ^{11}c_{4}
= 15x14x13x12/4x3x2x1 = 1365.
IncorrectAnswer : D
Solution : Required number of ways = ^{15}c_{11 }= ^{15}c_{(1511)} = ^{11}c_{4}
= 15x14x13x12/4x3x2x1 = 1365.
 Question 13 of 15
13. Question
1 points.
A Cone, a hemisphere and a cylinder stand on equal bases and have the same height. Find ratio of their volumes.
CorrectAnswer : C
Solution : Let R be the radius of each
Height of the hemisphere = Its radius = R.
Height of each = R.
Ratio of volumes = (1/3)∏ R^{2 }x R : (2/3)∏R^{3} : ∏ R^{2 }x R = 1:2:3
IncorrectAnswer : C
Solution : Let R be the radius of each
Height of the hemisphere = Its radius = R.
Height of each = R.
Ratio of volumes = (1/3)∏ R^{2 }x R : (2/3)∏R^{3} : ∏ R^{2 }x R = 1:2:3
 Question 14 of 15
14. Question
1 points.
A man can row 7 ½ kmph in still water.if in a river running at 1.5 km/hr an hour,it takes him 50 minutes to row to a place and back,how far off is the place?
CorrectAnswer : D
Solution : Speed downstream =(7.5+1.5)km/hr=9 km/hr;
Speed upstream=(7.51.5)kmph=6kmph.
Let the required distance be x km.then,
x/9+x/6=50/60.
2x+3x=(5/6*18)
5x=15
x=3.
Hence,the required distance is 3km.
IncorrectAnswer : D
Solution : Speed downstream =(7.5+1.5)km/hr=9 km/hr;
Speed upstream=(7.51.5)kmph=6kmph.
Let the required distance be x km.then,
x/9+x/6=50/60.
2x+3x=(5/6*18)
5x=15
x=3.
Hence,the required distance is 3km.
 Question 15 of 15
15. Question
1 pointsCorrectAnswer : A
Solution : Let principal = P. Then, S.l. = P and T = 16 yrs.
Rate = (100 x P)/(P*16)% = 6 ¼ % p.a
IncorrectAnswer : A
Solution : Let principal = P. Then, S.l. = P and T = 16 yrs.
Rate = (100 x P)/(P*16)% = 6 ¼ % p.a