Aptitude Test 3
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 Total number of questions : 15
 Time allotted : 25 minutes.
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Question 1 of 15
1. Question
1 points.
10 cats caught 20 rats in 10 seconds. How many cats are required to catch 100 rats in 50 seconds?
CorrectAnswer: Option C
IncorrectAnswer: Option C

Question 2 of 15
2. Question
1 points.
Average age of students of an adult school is 40 years. 120 new students whose average age is 32 years joined the school. As a result the average age is decreased by 4 years. Find the number of students of the school after joining of the new students:
CorrectAnswer: Option A
Solution: Let the original no. of students be x
A.T.S. 40x + 120 × 32 = (x + 120)36 ⇒ x = 120
∴ Required no. of students after joining the new student
= x + 120 = 240
IncorrectAnswer: Option A
Solution: Let the original no. of students be x
A.T.S. 40x + 120 × 32 = (x + 120)36 ⇒ x = 120
∴ Required no. of students after joining the new student
= x + 120 = 240

Question 3 of 15
3. Question
1 points.
A businessman purchased 35 kg dal of Rs 525 and sells it at the rate of Rs 18 per kg. Then the rate of profit or loss is:
CorrectAnswer: Option D
Solution: C.P. of 1 kg dal = 525/35 = Rs. 15 ; S.P. = Rs. 18
Profit = S.P. – C.P. = 18 – 15 = Rs 3, P% = (3/15)x100 = 20%
IncorrectAnswer: Option D
Solution: C.P. of 1 kg dal = 525/35 = Rs. 15 ; S.P. = Rs. 18
Profit = S.P. – C.P. = 18 – 15 = Rs 3, P% = (3/15)x100 = 20%

Question 4 of 15
4. Question
1 points.
Arjun and Sajal are friends . Each has some money. If Arjun gives Rs. 30 to Sajal, the Sajal will have twice the money left with Arjun. But, if Sajal gives Rs. 10 to Arjun, the Arjun will have thrice as much as is left with Sajal. How much money does each have?
CorrectAnswer: Option B
Solution: Suppose Arjun has Rs. x and Sajal has Rs. y. Then,
2(x30) = y+30 ==> 2x–y =90 …(i)
and x +10 =3(y10) => x3y = – 40 …(ii)
Solving (i) and (ii), we get x =62 and y =34.
Arjun has Rs. 62 and Sajal has Rs. 34.
IncorrectAnswer: Option B
Solution: Suppose Arjun has Rs. x and Sajal has Rs. y. Then,
2(x30) = y+30 ==> 2x–y =90 …(i)
and x +10 =3(y10) => x3y = – 40 …(ii)
Solving (i) and (ii), we get x =62 and y =34.
Arjun has Rs. 62 and Sajal has Rs. 34.

Question 5 of 15
5. Question
1 points.
Village A has a population of 68000, which is decreasing at the rate of 1200 per year. Village B has a population of 42000, which is increasing at the rate of 800 per year. In how many years will the population of the two villages be equal?
CorrectAnswer: Option C
Solution: Let the population of two villages be equal after x years
Then, 68000 – 1200x = 42000 + 800x
2000x = 26000
x=13 Years
IncorrectAnswer: Option C
Solution: Let the population of two villages be equal after x years
Then, 68000 – 1200x = 42000 + 800x
2000x = 26000
x=13 Years

Question 6 of 15
6. Question
1 points.
The average weight of a,b,c is 45 kg. The average weight of a & b is 40 kg & that of b,c is 43 kg. Find the weight of b.
CorrectAnswer: Option A
Solution: Let a,b,c represent their individual weights.
Then, a+b+c=(45*3)kg = 135 kg
a+b = (40*2)kg = 80 kg & b+c = (43*2)kg = 86 kg
b=(a+b)+(b+c)(a+b+c)
=(80+86135)kg =31 kg
IncorrectAnswer: Option A
Solution: Let a,b,c represent their individual weights.
Then, a+b+c=(45*3)kg = 135 kg
a+b = (40*2)kg = 80 kg & b+c = (43*2)kg = 86 kg
b=(a+b)+(b+c)(a+b+c)
=(80+86135)kg =31 kg

Question 7 of 15
7. Question
1 pointsCorrectAnswer: Option C
Solution: This is a simple alternating subtraction series, with a pattern 2, 5, 2, 5 ….
IncorrectAnswer: Option C
Solution: This is a simple alternating subtraction series, with a pattern 2, 5, 2, 5 ….

Question 8 of 15
8. Question
1 points.
The product of the ages of Ankit and Nikita is 240. If twice the age of Nikita is more than Ankit’s age by 4 years, what is Nikita’s age?
CorrectAnswer: Option D
Solution: Let Ankit’s age be x years. Then, Nikita’s age = 240/x years.
∴ 2 x (240 /x ) – x = 4 ==> 480 – x^{2} = 4x ==> x^{2} + 4x – 480 = 0
==> ( x + 24) (x20) = 0 ==> x = 20.
Hence, Nikita’s age = (240/20) years = 12 years
IncorrectAnswer: Option D
Solution: Let Ankit’s age be x years. Then, Nikita’s age = 240/x years.
∴ 2 x (240 /x ) – x = 4 ==> 480 – x^{2} = 4x ==> x^{2} + 4x – 480 = 0
==> ( x + 24) (x20) = 0 ==> x = 20.
Hence, Nikita’s age = (240/20) years = 12 years

Question 9 of 15
9. Question
1 pointsCorrectAnswer: Option A
Solution: Each number in the series is the preceding number multiplied by 2 and then increased by 1
IncorrectAnswer: Option A
Solution: Each number in the series is the preceding number multiplied by 2 and then increased by 1

Question 10 of 15
10. Question
1 points.
a and b together can complete a piece of work in 4 days. If a alone can complete the same work in 12 days, in how many days can b alone complete that work?
CorrectAnswer: Option B
Solution: (a + b)’s 1 day’s work = (1/4).
a ‘s 1 day’s work = (1/12).
b‘s 1 day’s work =((1/4)(1/12))=(1/6)
hence, b alone can complete the work in 6 days.
IncorrectAnswer: Option B
Solution: (a + b)’s 1 day’s work = (1/4).
a ‘s 1 day’s work = (1/12).
b‘s 1 day’s work =((1/4)(1/12))=(1/6)
hence, b alone can complete the work in 6 days.

Question 11 of 15
11. Question
1 points.
A train is moving at a speed of 132 km/hr. If the length of the train is 110 m, how long will it take to cross a railway platform 165 m long?
CorrectAnswer: Option C
Solution: Speed of train = 132 *(5/18) m/sec = 110/3 m/sec.
Distance covered in passing the platform = (110 + 165) m = 275 m.
Time taken =275 *(3/110) sec = 15/2 sec = 7 ½ sec
IncorrectAnswer: Option C
Solution: Speed of train = 132 *(5/18) m/sec = 110/3 m/sec.
Distance covered in passing the platform = (110 + 165) m = 275 m.
Time taken =275 *(3/110) sec = 15/2 sec = 7 ½ sec

Question 12 of 15
12. Question
1 points.
A man can row 18 km/hr in still water. It takes him thrice as long to row up as to row down the river. Find the rate of stream.
CorrectAnswer: Option A
Solution: Let man’s rate upstream be x km/h. Then, His rate downstream = 3x km/h.
So, 2x = 18 or x=9.
Rate upstream=9 km/hr, Rate downstream = 27 km/hr.
Hence, rate of stream = 1/2(279) km/hr = 9 km/hr.
IncorrectAnswer: Option A
Solution: Let man’s rate upstream be x km/h. Then, His rate downstream = 3x km/h.
So, 2x = 18 or x=9.
Rate upstream=9 km/hr, Rate downstream = 27 km/hr.
Hence, rate of stream = 1/2(279) km/hr = 9 km/hr.

Question 13 of 15
13. Question
1 points.
Find the cost of carpeting a room 13 m long and 9 m broad with a carpet 75 cm wide at the rate of Rs. 12.40 per square meter.
CorrectAnswer: Option B
Solution: Area of the carpet = area of the room = (13 * 9) m^{2} = 117 m^{2}.
Length of the carpet = (area/width) = 117 *(4/3) m = 156 m.
Therefore cost of carpeting = Rs. (156 * 12.40) = Rs. 1934.40
IncorrectAnswer: Option B
Solution: Area of the carpet = area of the room = (13 * 9) m^{2} = 117 m^{2}.
Length of the carpet = (area/width) = 117 *(4/3) m = 156 m.
Therefore cost of carpeting = Rs. (156 * 12.40) = Rs. 1934.40

Question 14 of 15
14. Question
1 pointsCorrectAnswer: Option A
Solution: Go on multiplying 2 and adding 1 to get the next number. So, 39 is wrong.
IncorrectAnswer: Option A
Solution: Go on multiplying 2 and adding 1 to get the next number. So, 39 is wrong.

Question 15 of 15
15. Question
1 points.
Water flows into a tank 200 m x 160 m through a rectangular pipe of 1.5 m x 1.25 m @ 20 km/hr . In what time (in minutes) will the water rise by 2 meter?
CorrectAnswer: Option C
Solution: Volume required in the tank = (200 x 150 x 2) m^{3} = 60000 m^{3}.
Length of water column flown in1 min = (20*1000)/60 m = 1000/3 m
Volume flown per minute = 1.5 * 1.25 * (1000/3) m^{3} = 625 m^{3}.
∴ Required time = (60000/625) minutes = 96 minutes.
IncorrectAnswer: Option C
Solution: Volume required in the tank = (200 x 150 x 2) m^{3} = 60000 m^{3}.
Length of water column flown in1 min = (20*1000)/60 m = 1000/3 m
Volume flown per minute = 1.5 * 1.25 * (1000/3) m^{3} = 625 m^{3}.
∴ Required time = (60000/625) minutes = 96 minutes.