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.

For more about this concept see:
manhattan-remainder-problem-93752.html#p721341
when-positive-integer-n-is-divided-by-5-the-remainder-is-90442.html#p722552
when-the-positive-integer-a-is-divided-by-5-and-125591.html#p1028654

Hope it helps.
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clear explanation

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.

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 !!

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.

How about this?

n < 100 ----- (1)

S1- When divided equal part of 10 meter each, a piece of 5 meter is left.

implies n = 10k + 5, where k is some integer

10k + 5 < 100 (from 1)

solving k < 9.5 implies no unique solution for n as it can have multiple values. Therefore insufficient.

S2- When divided equal part of 6 meter each, a piece of 1 meter is left.

solving as above we get k < 16.5 which again means it is insufficient.

Even combining S1 and S2 we don't reach a definite solution. Hence answer is E.

n=10q+5
n=6p+1
→10q-6p=-4
least values of q and p are 2 and 4 respectively
substituting, least value of n=25
lcm of divisors 10 and 6=30
so 3 possible <100 meter values of n: 25, 55, 85
neither 1 nor 2 sufficient
E