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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