 Number System Quiz1
 Number System Quiz2
Engineering Aptitude
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Engineering Aptitude quiz2
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Question 1 of 10
1. Question
1 pointsWhat is the unit digit in(7^{95} – 3^{58})? Correct
Unit digit in 7^{95} = Unit digit in [(7^{4})^{23} x 7^{3}]
= Unit digit in [(Unit digit in(2401))^{23} x (343)]
= Unit digit in (1^{23} x 343)
= Unit digit in (343)
= 3Unit digit in 3^{58} = Unit digit in [(3^{4})^{14} x 3^{2}]
= Unit digit in [Unit digit in (81)^{14} x 3^{2}]
= Unit digit in [(1)^{14} x 3^{2}]
= Unit digit in (1 x 9)
= Unit digit in (9)
= 9Unit digit in (7^{95} – 3^{58}) = Unit digit in (343 – 9) = Unit digit in (334) = 4.
So, Option B is the answer.
Incorrect
Unit digit in 7^{95} = Unit digit in [(7^{4})^{23} x 7^{3}]
= Unit digit in [(Unit digit in(2401))^{23} x (343)]
= Unit digit in (1^{23} x 343)
= Unit digit in (343)
= 3Unit digit in 3^{58} = Unit digit in [(3^{4})^{14} x 3^{2}]
= Unit digit in [Unit digit in (81)^{14} x 3^{2}]
= Unit digit in [(1)^{14} x 3^{2}]
= Unit digit in (1 x 9)
= Unit digit in (9)
= 9Unit digit in (7^{95} – 3^{58}) = Unit digit in (343 – 9) = Unit digit in (334) = 4.
So, Option B is the answer.

Question 2 of 10
2. Question
1 points(x^{n} – a^{n}) is completely divisible by (x – a), when Correct
For every natural number n, (x^{n} – a^{n}) is completely divisible by (x – a).
Incorrect
For every natural number n, (x^{n} – a^{n}) is completely divisible by (x – a).

Question 3 of 10
3. Question
1 pointsWhat will be remainder when 17^{200} is divided by 18 ? Correct
When n is even. (x^{n} – a^{n}) is completely divisibly by (x + a)
(17^{200} – 1^{200}) is completely divisible by (17 + 1), i.e., 18.
(17^{200} – 1) is completely divisible by 18.
On dividing 17^{200} by 18, we get 1 as remainder.
Incorrect
When n is even. (x^{n} – a^{n}) is completely divisibly by (x + a)
(17^{200} – 1^{200}) is completely divisible by (17 + 1), i.e., 18.
(17^{200} – 1) is completely divisible by 18.
On dividing 17^{200} by 18, we get 1 as remainder.

Question 4 of 10
4. Question
1 points(1^{2} + 2^{2} + 3^{2} + … + 10^{2}) = ?
Correct
We know that (1^{2} + 2^{2} + 3^{2} + … + n^{2}) = 1 n(n + 1)(2n + 1) 6 Putting n = 10, required sum = 1 x 10 x 11 x 21 = 385 6 Incorrect
We know that (1^{2} + 2^{2} + 3^{2} + … + n^{2}) = 1 n(n + 1)(2n + 1) 6 Putting n = 10, required sum = 1 x 10 x 11 x 21 = 385 6 
Question 5 of 10
5. Question
1 pointsWhen a number is divided by 13, the remainder is 11. When the same number is divided by 17, then remainder is 9. What is the number ? Correct
x = 13p + 11 and x = 17q + 9
13p + 11 = 17q + 9
17q – 13p = 2
q = 2 + 13p 17 The least value of p for which q = 2 + 13p is a whole number is p = 26 17 x = (13 x 26 + 11)
= (338 + 11)
= 349
Incorrect
x = 13p + 11 and x = 17q + 9
13p + 11 = 17q + 9
17q – 13p = 2
q = 2 + 13p 17 The least value of p for which q = 2 + 13p is a whole number is p = 26 17 x = (13 x 26 + 11)
= (338 + 11)
= 349

Question 6 of 10
6. Question
1 pointsA, B and C start at the same time in the same direction to run around a circular stadium. A completes a round in 252 seconds, B in 308 seconds and c in 198 seconds, all starting at the same point. After what time will they again at the starting point ?
Correct
L.C.M. of 252, 308 and 198 = 2772.
So, A, B and C will again meet at the starting point in 2772 sec. i.e., 46 min. 12 sec.
Incorrect
L.C.M. of 252, 308 and 198 = 2772.
So, A, B and C will again meet at the starting point in 2772 sec. i.e., 46 min. 12 sec.

Question 7 of 10
7. Question
1 pointsFind the number of zeros in 75!
Correct
Incorrect

Question 8 of 10
8. Question
1 pointsWhat is the number of trailing zeroes in 1123!?
Correct
[1123/5] = 224
[224/5] = 44
[44/5] = 8
[8/5]=1. It is less than 5, so we stop here.
The answer is = 224 + 44 + 8 + 1 =277
Incorrect
[1123/5] = 224
[224/5] = 44
[44/5] = 8
[8/5]=1. It is less than 5, so we stop here.
The answer is = 224 + 44 + 8 + 1 =277

Question 9 of 10
9. Question
1 pointsWhich of the following shows the prime factorization of 84?
Correct
Explanation
Step 1:
We can breakdown 84 into its factors as shown below.
84 = 2 × 42;
42 = 2 × 21;
21 = 3 × 7;
Step 2:
So the prime factorization of 84 is
= 2 × 2 × 3 × 7 = 2^{2} × 3 × 7
Incorrect
Explanation
Step 1:
We can breakdown 84 into its factors as shown below.
84 = 2 × 42;
42 = 2 × 21;
21 = 3 × 7;
Step 2:
So the prime factorization of 84 is
= 2 × 2 × 3 × 7 = 2^{2} × 3 × 7

Question 10 of 10
10. Question
1 pointsSix bells commence tolling together and toll at intervals of 2, 4, 6, 8 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together ?
Correct
L.C.M. of 2, 4, 6, 8, 10, 12 is 120.
So, the bells will toll together after every 120 seconds(2 minutes).
In 30 minutes, they will toll together 30 + 1 = 16 times. 2 Incorrect
L.C.M. of 2, 4, 6, 8, 10, 12 is 120.
So, the bells will toll together after every 120 seconds(2 minutes).
In 30 minutes, they will toll together 30 + 1 = 16 times. 2
 Time and Work Quiz3
Engineering Aptitude
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Time and Work Quiz1
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Question 1 of 10
1. Question
1 pointsA can do a work in 15 days and B in 20 days. If they work on it together for 4 days, then the fraction of the work that is left is :
Correct
A’s 1 day’s work = 1 ; 15 B’s 1 day’s work = 1 ; 20 (A + B)’s 1 day’s work = 1 + 1 = 7 . 15 20 60 (A + B)’s 4 day’s work = 7 x 4 = 7 . 60 15 Therefore, Remaining work = 1 – 7 = 8 . 15 15 Incorrect
A’s 1 day’s work = 1 ; 15 B’s 1 day’s work = 1 ; 20 (A + B)’s 1 day’s work = 1 + 1 = 7 . 15 20 60 (A + B)’s 4 day’s work = 7 x 4 = 7 . 60 15 Therefore, Remaining work = 1 – 7 = 8 . 15 15 
Question 2 of 10
2. Question
1 pointsA, B and C can do a piece of work in 20, 30 and 60 days respectively. In how many days can A do the work if he is assisted by B and C on every third day?
Correct
A’s 2 day’s work = 1 x 2 = 1 . 20 10 (A + B + C)’s 1 day’s work = 1 + 1 + 1 = 6 = 1 . 20 30 60 60 10 Work done in 3 days = 1 + 1 = 1 . 10 10 5 Now, 1 work is done in 3 days. 5 Whole work will be done in (3 x 5) = 15 days.
Incorrect
A’s 2 day’s work = 1 x 2 = 1 . 20 10 (A + B + C)’s 1 day’s work = 1 + 1 + 1 = 6 = 1 . 20 30 60 60 10 Work done in 3 days = 1 + 1 = 1 . 10 10 5 Now, 1 work is done in 3 days. 5 Whole work will be done in (3 x 5) = 15 days.

Question 3 of 10
3. Question
1 pointsA alone can do a piece of work in 6 days and B alone in 8 days. A and B undertook to do it for Rs. 3200. With the help of C, they completed the work in 3 days. How much is to be paid to C?
Correct
C’s 1 day’s work = 1 – 1 + 1 = 1 – 7 = 1 . 3 6 8 3 24 24 A’s wages : B’s wages : C’s wages = 1 : 1 : 1 = 4 : 3 : 1. 6 8 24 C’s share (for 3 days) = Rs. 3 x 1 x 3200 = Rs. 400. 24 Incorrect
C’s 1 day’s work = 1 – 1 + 1 = 1 – 7 = 1 . 3 6 8 3 24 24 A’s wages : B’s wages : C’s wages = 1 : 1 : 1 = 4 : 3 : 1. 6 8 24 C’s share (for 3 days) = Rs. 3 x 1 x 3200 = Rs. 400. 24 
Question 4 of 10
4. Question
1 pointsA can do a piece of work in 4 hours; B and C together can do it in 3 hours, while A and C together can do it in 2 hours. How long will B alone take to do it? Correct
A’s 1 hour’s work = 1 ; 4 (B + C)’s 1 hour’s work = 1 ; 3 (A + C)’s 1 hour’s work = 1 . 2 (A + B + C)’s 1 hour’s work = 1 + 1 = 7 . 4 3 12 B’s 1 hour’s work = 7 – 1 = 1 . 12 2 12 B alone will take 12 hours to do the work.
Incorrect
A’s 1 hour’s work = 1 ; 4 (B + C)’s 1 hour’s work = 1 ; 3 (A + C)’s 1 hour’s work = 1 . 2 (A + B + C)’s 1 hour’s work = 1 + 1 = 7 . 4 3 12 B’s 1 hour’s work = 7 – 1 = 1 . 12 2 12 B alone will take 12 hours to do the work.

Question 5 of 10
5. Question
1 pointsIf 6 men and 8 boys can do a piece of work in 10 days while 26 men and 48 boys can do the same in 2 days, the time taken by 15 men and 20 boys in doing the same type of work will be:
Correct
Let 1 man’s 1 day’s work = x and 1 boy’s 1 day’s work = y.
Then, 6x + 8y = 1 and 26x + 48y = 1 . 10 2 Solving these two equations, we get : x = 1 and y = 1 . 100 200 (15 men + 20 boy)’s 1 day’s work = 15 + 20 = 1 . 100 200 4 15 men and 20 boys can do the work in 4 days.
Incorrect
Let 1 man’s 1 day’s work = x and 1 boy’s 1 day’s work = y.
Then, 6x + 8y = 1 and 26x + 48y = 1 . 10 2 Solving these two equations, we get : x = 1 and y = 1 . 100 200 (15 men + 20 boy)’s 1 day’s work = 15 + 20 = 1 . 100 200 4 15 men and 20 boys can do the work in 4 days.

Question 6 of 10
6. Question
1 pointsA can finish a work in 18 days and B can do the same work in 15 days. B worked for 10 days and left the job. In how many days, A alone can finish the remaining work?
Correct
B’s 10 day’s work = 1 x 10 = 2 . 15 3 Remaining work = 1 – 2 = 1 . 3 3 Now, 1 work is done by A in 1 day. 18 1 work is done by A in 18 x 1 = 6 days. Incorrect
B’s 10 day’s work = 1 x 10 = 2 . 15 3 Remaining work = 1 – 2 = 1 . 3 3 Now, 1 work is done by A in 1 day. 18 1 work is done by A in 18 x 1 = 6 days. 
Question 7 of 10
7. Question
1 pointsIn the beginning, Ram works at a rate such that he can finish a piece of work in 24 hrs, but he only works at this rate for 16 hrs. After that, he works at a rate such that he can do the whole work in 18 hrs. If Ram is to finish this work at a stretch, how many hours will he take to finish this work?
Correct
Ram’s 16 hr work = 16/24 = 2/3. Remaining work = 1 – 2/3 = 1/3.
Using work and time formula: This will be completed in 1/3 × 18 i.e. 6 hrs.
So, total time taken to complete work = 16 + 6= 22 hrs.Incorrect
Ram’s 16 hr work = 16/24 = 2/3. Remaining work = 1 – 2/3 = 1/3.
Using work and time formula: This will be completed in 1/3 × 18 i.e. 6 hrs.
So, total time taken to complete work = 16 + 6= 22 hrs. 
Question 8 of 10
8. Question
1 pointsA can do a piece of work in 10 days, and B can do the same work in 20 days. With the help of C, they finished the work in 4 days. C can do the work in how many days, working alone?
Correct
Their combined 4 day work = 4(1/10 + 1/15) = 12/20 = 3/5.
Remaining work = 1 – 3/5 = ⅖.
This means C did 2/5 work in 4 days, hence he can finish the complete work in 5/2 × 4 = 10 days.Incorrect
Their combined 4 day work = 4(1/10 + 1/15) = 12/20 = 3/5.
Remaining work = 1 – 3/5 = ⅖.
This means C did 2/5 work in 4 days, hence he can finish the complete work in 5/2 × 4 = 10 days. 
Question 9 of 10
9. Question
1 pointsA can do a piece of work in 12 days. B can do this work in 16 days. A started the work alone. After how many days should B join him, so that the work is finished in 9 days?
Correct
A’s work in 9 days = 9/12 = 3/4. Remaining work = 1/4.
This work was done by B in 1/4 × 16 = 4 days.
∴ B would have joined A after 9 – 4 = 5 days.Q.4.A and B can do a piece of work in 4 days,Incorrect
A’s work in 9 days = 9/12 = 3/4. Remaining work = 1/4.
This work was done by B in 1/4 × 16 = 4 days.
∴ B would have joined A after 9 – 4 = 5 days.Q.4.A and B can do a piece of work in 4 days, 
Question 10 of 10
10. Question
1 pointsA, B, C, and D can do a piece of work in 20 days. If A and B can do it together in 50 days, and C alone in 60 days, find the time in which D alone can do it.
Correct
D alone will take 1/20 – 1/50 – 1/60 = 4/300 = 1/75
⇒ 75 days to complete the work.Incorrect
D alone will take 1/20 – 1/50 – 1/60 = 4/300 = 1/75
⇒ 75 days to complete the work.
 Time, Speed and distance Quiz4
Engineering Aptitude
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time speed and distance Quiz4
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Question 1 of 10
1. Question
1 pointsA person crosses a 600 m long street in 5 minutes. What is his speed in km per hour?
Correct
Speed = 600 m/sec. 5 x 60 = 2 m/sec.
= 2 x 18 km/hr 5 = 7.2 km/hr.
Incorrect
Speed = 600 m/sec. 5 x 60 = 2 m/sec.
= 2 x 18 km/hr 5 = 7.2 km/hr.

Question 2 of 10
2. Question
1 pointsA man goes to Mumbai from Pune at a speed of 4 km/hr and returns to Pune at speed of 6km/hr. What is his average speed of the entire journey?
Correct
Incorrect

Question 3 of 10
3. Question
1 pointsIf a person walks at 14 km/hr instead of 10 km/hr, he would have walked 20 km more. The actual distance travelled by him is:
Correct
Let the actual distance travelled be x km.
Then, x = x + 20 10 14 14x = 10x + 200
4x = 200
x = 50 km.
Incorrect
Let the actual distance travelled be x km.
Then, x = x + 20 10 14 14x = 10x + 200
4x = 200
x = 50 km.

Question 4 of 10
4. Question
1 pointsExcluding stoppages, the speed of a bus is 54 kmph and including stoppages, it is 45 kmph. For how many minutes does the bus stop per hour?
Correct
Due to stoppages, it covers 9 km less.
Time taken to cover 9 km = 9 x 60 min = 10 min. 54 Incorrect
Due to stoppages, it covers 9 km less.
Time taken to cover 9 km = 9 x 60 min = 10 min. 54 
Question 5 of 10
5. Question
1 pointsIn a flight of 600 km, an aircraft was slowed down due to bad weather. Its average speed for the trip was reduced by 200 km/hr and the time of flight increased by 30 minutes. The duration of the flight is:
Correct
Let the duration of the flight be x hours.
Then, 600 – 600 = 200 x x + (1/2) 600 – 1200 = 200 x 2x + 1 x(2x + 1) = 3
2x^{2} + x – 3 = 0
(2x + 3)(x – 1) = 0
x = 1 h
Incorrect
Let the duration of the flight be x hours.
Then, 600 – 600 = 200 x x + (1/2) 600 – 1200 = 200 x 2x + 1 x(2x + 1) = 3
2x^{2} + x – 3 = 0
(2x + 3)(x – 1) = 0
x = 1 h

Question 6 of 10
6. Question
1 pointsA man completes a journey in 10 hours. He travels first half of the journey at the rate of 21 km/hr and the second half at the rate of 24 km/hr. Find the total journey in km.
Correct
1/2)x + (1/2)x = 10 21 24 x + x = 20 21 24 15x = 168 x 20
x = 168 x 20 = 224 km. 15 Incorrect
1/2)x + (1/2)x = 10 21 24 x + x = 20 21 24 15x = 168 x 20
x = 168 x 20 = 224 km. 15 
Question 7 of 10
7. Question
1 pointsA farmer travelled a distance of 61 km in 9 hours. He travelled partly on foot @ 4 km/hr and partly on bicycle @ 9 km/hr. The distance travelled on foot is:
Correct
Let the distance travelled on foot be x km.
Then, distance travelled on bicycle = (61 –x) km.
So, x + (61 –x) = 9 4 9 9x + 4(61 –x) = 9 x 36
5x = 80
x = 16 km.
Incorrect
Let the distance travelled on foot be x km.
Then, distance travelled on bicycle = (61 –x) km.
So, x + (61 –x) = 9 4 9 9x + 4(61 –x) = 9 x 36
5x = 80
x = 16 km.

Question 8 of 10
8. Question
1 pointsA boat can travel with a speed of 13 km/hr in still water. If the speed of the stream is 4 km/hr, find the time taken by the boat to go 68 km downstream.
Correct
Speed downstream = (13 + 4) km/hr = 17 km/hr.
Time taken to travel 68 km downstream = 68 hrs = 4 hrs. 17 Incorrect
Speed downstream = (13 + 4) km/hr = 17 km/hr.
Time taken to travel 68 km downstream = 68 hrs = 4 hrs. 17 
Question 9 of 10
9. Question
1 pointsA boat running upstream takes 8 hours 48 minutes to cover a certain distance, while it takes 4 hours to cover the same distance running downstream. What is the ratio between the speed of the boat and speed of the water current respectively? Correct
Let the man’s rate upstream be x kmph and that downstream be y kmph.
Then, distance covered upstream in 8 hrs 48 min = Distance covered downstream in 4 hrs.
x x 8 4 = (y x 4) 5 44 x =4y 5 y = 11 x. 5 Required ratio = y + x : y – x 2 2 = 16x x 1 : 6x x 1 5 2 5 2 = 8 : 3 5 5 = 8 : 3.
Incorrect
Let the man’s rate upstream be x kmph and that downstream be y kmph.
Then, distance covered upstream in 8 hrs 48 min = Distance covered downstream in 4 hrs.
x x 8 4 = (y x 4) 5 44 x =4y 5 y = 11 x. 5 Required ratio = y + x : y – x 2 2 = 16x x 1 : 6x x 1 5 2 5 2 = 8 : 3 5 5 = 8 : 3.

Question 10 of 10
10. Question
1 pointsA motorboat, whose speed in 15 km/hr in still water goes 30 km downstream and comes back in a total of 4 hours 30 minutes. The speed of the stream (in km/hr) is:
Correct
Let the speed of the stream be x km/hr. Then,
Speed downstream = (15 + x) km/hr,
Speed upstream = (15 – x) km/hr.
30 + 30 = 4 1 (15 + x) (15 – x) 2 900 = 9 225 – x^{2} 2 9x^{2} = 225
x^{2} = 25
x = 5 km/hr.
Incorrect
Let the speed of the stream be x km/hr. Then,
Speed downstream = (15 + x) km/hr,
Speed upstream = (15 – x) km/hr.
30 + 30 = 4 1 (15 + x) (15 – x) 2 900 = 9 225 – x^{2} 2 9x^{2} = 225
x^{2} = 25
x = 5 km/hr.
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 Number System Quiz1
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