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Question 1 of 5
1. Question
When a ball bounces, it rise to 2/3 of the height from which it fell. If the ball is dropped from a height of 36 m, then how high will it rise at the third bounce?
Correct
Ans: a
After first bounce, height of ball = (2/3)1 x36
And after third bounce, height of ball = (2/3)3 x36
= (8/27)x36 =[ (8/4)/3] = 32/3m = =10(2/3)m
Hence, the required height at third bounce is 10(2/3)m.
Incorrect
Ans: a
After first bounce, height of ball = (2/3)1 x36
And after third bounce, height of ball = (2/3)3 x36
= (8/27)x36 =[ (8/4)/3] = 32/3m = =10(2/3)m
Hence, the required height at third bounce is 10(2/3)m.
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Question 2 of 5
2. Question
What is the remainder when (1723 + 2323+2923) is divided by 23?
Correct
Ans: a
We have, [(1723 +2323+2923)/23]
=[ (23×1 -6)23 +(23×1+0)23+(23×1+6)23]/23
Remainder = (-6)23 +0+(6)23 =0
Incorrect
Ans: a
We have, [(1723 +2323+2923)/23]
=[ (23×1 -6)23 +(23×1+0)23+(23×1+6)23]/23
Remainder = (-6)23 +0+(6)23 =0
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Question 3 of 5
3. Question
The difference of two consecutive cubes
Correct
Ans: b
The difference of two consecutive cubes is never divisible by 2.
Illustration 1. Let the two consecutive numbers be 4 and 5
∴ (5)3-(4)3 = 125 -64 =61
Illustration 2. Let the two consecutive numbers be 9 and 10
∴ (10)3-(9)3 = 1000-729 = 271
Incorrect
Ans: b
The difference of two consecutive cubes is never divisible by 2.
Illustration 1. Let the two consecutive numbers be 4 and 5
∴ (5)3-(4)3 = 125 -64 =61
Illustration 2. Let the two consecutive numbers be 9 and 10
∴ (10)3-(9)3 = 1000-729 = 271
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Question 4 of 5
4. Question
Product of four consecutive numbers plus one is
Correct
Ans: c
Product of four consecutive numbers plus one is always a square.
Illustration 1. Let four consecutive numbers be 3, 4, 5 and 6
∴(3x 4x5x 6)+ 1 = 361= (19)2
Illustration 2. Let four consecutive numbers be 9, 10, 11 and 12
∴(9x 10 x 11x 12)+ 1=11881= (109)2
Incorrect
Ans: c
Product of four consecutive numbers plus one is always a square.
Illustration 1. Let four consecutive numbers be 3, 4, 5 and 6
∴(3x 4x5x 6)+ 1 = 361= (19)2
Illustration 2. Let four consecutive numbers be 9, 10, 11 and 12
∴(9x 10 x 11x 12)+ 1=11881= (109)2
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Question 5 of 5
5. Question
What is the remainder when 41000 is divided by 7?
Correct
Ans: c
We know that, (41/7) gives remainder 4; (42/7) gives remainder 2, (43/7)
gives remainder 1; (44/7) gives remainder 4
NOW, 44 gives us the same remainder as 41, so cyclicity is of 3. So,
any power of 3 or multiple of 3 will give the remainder 1.
So, 4999 will give remainder 1 and 41000 will give the next remainder
in the cycle which is 4
Incorrect
Ans: c
We know that, (41/7) gives remainder 4; (42/7) gives remainder 2, (43/7)
gives remainder 1; (44/7) gives remainder 4
NOW, 44 gives us the same remainder as 41, so cyclicity is of 3. So,
any power of 3 or multiple of 3 will give the remainder 1.
So, 4999 will give remainder 1 and 41000 will give the next remainder
in the cycle which is 4
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