either 7 or 77

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either 7 or 77

by sanju09 » Thu Apr 09, 2009 3:38 am
If each term in the sum a1 + a2 + a3 + ... + an is either 7 or 77 and the sum is equal to 350, which of the following could equal to n?
A. 38
B. 39
C. 40
D. 41
E. 42
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by bluementor » Thu Apr 09, 2009 3:53 am
7x + 77y = 350, where x and y are nonzero integers.
x + 11y = 50 …eqn 1

x + y = n …eqn 2

from equation eqn 1, y cannot be greater than 4. So by plugging values for y:

y=4, then x = 2 ->n=6 (not available in answer choices)
y =3, then x = 17 -> n=20 (not available in answer choices)
y = 2, then x = 28 -> n=30 (not available in answer choices)
y = 1, then x = 39 -> n=40 (answer choice C).

C

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by Thonk02 » Thu Apr 09, 2009 9:33 am
How does X + Y = N?

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by kapsii » Thu Apr 09, 2009 10:13 am
Thonk02, BM has split N into X number of 7s and Y number of 77s. Thus he formed the two equations.

I try not to form too many equations, it takes up time and need a lot more attention, so what I did is, for the sum to be 350 and number of terms to be between 38 & 42 (inclusive), the series has to contain more 7s than 77s.
Next, I took the option which would be easier to multiply (i.e. 40) and multiplied it by 7.
Then I subtracted the answer from 350 (i.e. 350 - 280) which gave me 70. So, immediately I deduced I must replace one of my 7 with 77. Which still left me with 40 as the answer, I got lucky!

had I for some reason chosen 39 to be multiplied with 7, I would have got the difference as 77, which means I either have to add 11 7s or just one 77. adding one 77 takes the number of terms to 40.

It sounds like a long method, but it really saves time for me.
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by GID09 » Thu Apr 09, 2009 1:35 pm
If only 7 & 77 are there in the series, don't we require 10x(multiples of 10) 7s to get a number that ends with 0. Answer is straight C, am I missing something here??

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by kapsii » Thu Apr 09, 2009 10:04 pm
Good observation GID09, come what may, the number of terms would end with 0,
50 -> 50 7's and 0 77's
40 -> 39 7s and 1 77.
30 -> 28 7s and 2 77's
20 -> 17 7's and 3 77's
10 -> 6 7's and 4 77's

If we could think like this in the exam and back ourselves, it would be great.
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by sanju09 » Fri Apr 10, 2009 3:05 am
If x is the frequency of 7 in the data then (n - x) is the frequency of 77 in the data here; making us have

7 x + 77 (n - x) = 350

or x = (11 n - 50)/10 and x has to be a non-negative integer here.

Take a careful look to realize that n cannot be anything other than some multiple of 10 here. 40 is the only multiple of 10 in the given choices.
The mind is everything. What you think you become. -Lord Buddha



Sanjeev K Saxena
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