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1) When divided equal part of 10 meter each, a piece of 5 meter is left.

2) When divided equal of 6 meter each, a piece of 1 meter is left.

What is the value of length n<100 meter of wire?

(1) When divided equal part of 10 meter each, a piece of 5 meter is left --> \(n=10q+5\): 5, 15, 25, ..., 95. Not sufficient.

(2) When divided equal of 6 meter each, a piece of 1 meter is left --> \(n=6p+1\): 1, 7, 13, 19, 25, ..., 97. Not sufficient.

(1)+(2) General formula of \(n\) based on \(n=10q+5\) and \(n=6p+1\) would be \(n=30m+25\) --> \(n\) can be: 25, 55 or 85. Not sufficient.

Answer: E.

To elaborate more. How to derive general formula of \(n\) based on \(n=10q+5\) and \(n=6p+1\): divisor will be the least common multiple of above two divisors 6 and 10, hence 30. Remainder will be the first common integer in above two patterns, hence 25. So, to satisfy both conditions, \(n\) must be of a type \(n=30m+25\): 25, 55 or 85.

1) When divided equal part of 10 meter each, a piece of 5 meter is left.

2) When divided equal of 6 meter each, a piece of 1 meter is left.

What is the value of length n<100 meter of wire?

(1) When divided equal part of 10 meter each, a piece of 5 meter is left --> \(n=10q+5\): 5, 15, 25, ..., 95. Not sufficient.

(2) When divided equal of 6 meter each, a piece of 1 meter is left --> \(n=6p+1\): 1, 7, 13, 19, 25, ..., 97. Not sufficient.

(1)+(2) General formula of \(n\) based on \(n=10q+5\) and \(n=6p+1\) would be \(n=30m+25\) --> \(n\) can be: 25, 55 or 85. Not sufficient.

Answer: E.

To elaborate more. How to derive general formula of \(n\) based on \(n=10q+5\) and \(n=6p+1\): divisor will be the least common multiple of above two divisors 6 and 10, hence 30. Remainder will be the first common integer in above two patterns, hence 25. So, to satisfy both conditions, \(n\) must be of a type \(n=30m+25\): 25, 55 or 85.

To elaborate more. How to derive general formula of \(n\) based on \(n=10q+5\) and \(n=6p+1\): divisor will be the least common multiple of above two divisors 6 and 10, hence 30. Remainder will be the first common integer in above two patterns, hence 25. So, to satisfy both conditions, \(n\) must be of a type \(n=30m+25\): 25, 55 or 85.

Thanks very much. Above would have been very difficult to figure out in the real exam.

Re: What is the value of length n<100 meter of wire? [#permalink]

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11 Aug 2013, 09:36

N<100

What is N?

(1).

N =10A +5 ..... N can be 5,15,25,35 and so on

INSUFFICIENT

(2).

N= 6B + 1 .... N can be 1,7,13,19,25 and so on

INSUFFICIENT

Combining (1).& (2).

We get N = 30X + 25

N can be 25,55,85

Hence INSUFFICIENT

(E) it is !!
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Re: What is the value of length n<100 meter of wire? [#permalink]

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30 Jul 2014, 03:14

Actually, this problem you can solve at most in 10 seconds:)

The main point here is that all numbers with exact remainder form arithmetic progression with difference=divisor.

For example, all x such that "when x is divided by 5 the remainder is 1" form arithmetic progression with first element 1 and difference 5: 1, 6, 11, 16, 21.....

If 50<x<100 for example, I can definitely say that there are several such x, because the distance between all such numbers is 5.

To solve this problem you need just to check if the divisor=(distance between numbers) large enough to have only 1 number inside interval.

So, I need to find exact number less than 100. (1) The difference=divisor=10 is quite small for 100. Insufficient. (2) The difference=divisor=6 is quite small for 100. Insufficient.

(1)+(2) The new difference=least common multiple of 10 and 6=30 is small for 100. Insufficient.

The correct answer is E

You don't really need here to write formula for x and first several values for each statement. _________________

I'm happy, if I make math for you slightly clearer And yes, I like kudos:)

Re: What is the value of length n<100 meter of wire? [#permalink]

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10 Mar 2015, 11:59

smyarga wrote:

Actually, this problem you can solve at most in 10 seconds:)

The main point here is that all numbers with exact remainder form arithmetic progression with difference=divisor.

For example, all x such that "when x is divided by 5 the remainder is 1" form arithmetic progression with first element 1 and difference 5: 1, 6, 11, 16, 21.....

If 50<x<100 for example, I can definitely say that there are several such x, because the distance between all such numbers is 5.

To solve this problem you need just to check if the divisor=(distance between numbers) large enough to have only 1 number inside interval.

So, I need to find exact number less than 100. (1) The difference=divisor=10 is quite small for 100. Insufficient. (2) The difference=divisor=6 is quite small for 100. Insufficient.

(1)+(2) The new difference=least common multiple of 10 and 6=30 is small for 100. Insufficient.

The correct answer is E

You don't really need here to write formula for x and first several values for each statement.

For (1)+(2), we need to know that the first number is 25. Only then we could say that this is insufficient. If the first number was >70, (1)+(2), could've been sufficient.

Re: What is the value of length n<100 meter of wire? [#permalink]

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25 Sep 2017, 23:35

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