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by abhasjha » Thu Oct 16, 2014 10:39 pm
here an easy way to solve questions of these type

7A + 77 B = 350

7A =350- 77B

now by hit and trial method you need to subsitute the values of B in such a manner that the values of A is also an integer . (A little bit of google search shows these are diophantine equation ) . We are basically interested in simultaneous integral solution for - 7A= 350- 77B .

I test values let B = 0 in the equation - 7A= 350- 77B
7A = 350-0
7A = 350. A= 50 .BUT 50 is not given in answer choice . So i put B =1 in the equation - 7A= 350- 77B
7A = 350-77= 273
A=273/7 = 39 . we get again a integer value for A . and we have 39 A and 1 B . which means there are 39 (7's ) and one 77 in the series . Total number of term in the series = 40 .

Answer is C.

now try to solve two such questions one from OG and one from Gprep . here they go:
Q1.
Joanna bought only $0.15 stamps and $0.29 stamps. How many $0.15 stamps did she buy?

(1) She bought $4.40 worth of stamps.
(2) She bought an equal number of $0.15 stamps and $0.29 stamps

OA- A

Q2. Martha bought several pencils. If each pencil was either a 23 cent pencil or a 21 pencil, how many 23 cent pencils did martha buy?

(1) Martha bought a total of 6 pencils

(2) The total value of the pencils Martha bought was 130 cents.

OA- B
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by Mathsbuddy » Thu Oct 16, 2014 11:50 pm
A quicker way is to observe that the last digit in the sum of the series must be a multiple of 7 and n.
As 350 ends in a zero, then n must be a multiple of 10.
ANSWER = C
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by GMATGuruNY » Fri Oct 17, 2014 1:21 am
If each term in the sum a1+a1+a3+a4+...+aN is either 7 or 77 and the sum equals 350, which of the following could be equal to n?

1) 38
2) 39
3) 40
4) 41
5) 42
350/7 = 50.
So if each term were 7, there would be 50 terms.
The answer choices are all a little less than 50.
Implication:
Most -- but not all -- of the terms will be 7.

Case 1: 1 term = 77
Amount remaining = 350-77 = 273.
273/7 = 39
This works!
Thus, 1 term = 77, while 39 terms = 7, for a total of 40 terms.

The correct answer is C.

Alternate approach:

Let x = the number of 7's and y = the number of 77's.
Since the sum is 350, we get:
7x + 77y = 350
x + 11y = 50.

The answer choices imply that the total number of terms -- x+y -- is between 38 and 42, inclusive.
Thus:
38 ≤ x+y ≤ 42.
Substituting x = 50-11y into 38 ≤ x+y ≤ 42, we get:
38 ≤ (50-11y) +y ≤ 42
38 ≤ 50 - 10y ≤ 42
-12 ≤ - 10y ≤ -8
12 ≥ 10y ≥ 8.

Only one integer value satisfies the resulting inequality:
y = 1.
Since y=1 and x=50-11y, we get:
x = 50 - 11*1 = 39.
Thus, the total number of terms = x+y = 39+1 = 40.
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by Brent@GMATPrepNow » Fri Oct 17, 2014 6:11 am
If each term in the sum a1 + a2 + ... + an is either 7 or 77, and the sum equals 350, which of the following could be equal to n?

A)37
B)39
C)40
D)41
E)42
Notice that 77 does not divide into 350 many times.
In fact, there can be, at most, four 77's in the sum
So, there are only 5 cases to consider (zero 77's, one 77, two 77's, three 77's and four 77's)
It shouldn't take long to check the cases.

case 1: zero 77's in the sum
If every term is 7, the total number of terms is 50.
50 is not one of the answer choices, so move on.

case 2: one 77 in the sum
350 - 77 = 273
273/7 = 39
So, there could be thirty-nine 7's and one 77 in the sum, for a total of 40 terms.

This matches one of the answer choices, so the correct answer is C

Cheers,
Brent
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by Brent@GMATPrepNow » Fri Oct 17, 2014 6:11 am
If each term in sum a1+a2+ ... +an is either 7 or 77 and the sum equals 350, which of the following could be equal to n?

A)37
B)39
C)40
D)41
E)42
Another possible approach it to look for a pattern:

Since both 7 and 77 have 7 as their units digit, we know that if we take any two terms, their sum will have a units digit of 4 (e.g., 7 + 7 = 14, 7 + 77 = 84, 77 + 77 = 154)

Similarly, if we take any three terms, their sum will have a units digit of 1. (e.g., 7 + 7 + 7 = 21, 7 + 7 + 77 = 91, etc.)

Now let's look for a pattern.

The sum of any 1 term will have units digit 7
The sum of any 2 terms will have units digit 4
The sum of any 3 terms will have units digit 1
The sum of any 4 terms will have units digit 8
The sum of any 5 terms will have units digit 5
The sum of any 6 terms will have units digit 2
The sum of any 7 terms will have units digit 9
The sum of any 8 terms will have units digit 6
The sum of any 9 terms will have units digit 3
The sum of any 10 terms will have units digit 0
The sum of any 11 terms will have units digit 7 (at this point, the pattern repeats)

From this, we can conclude that the sum of any 20 terms will have units digit 0
And the sum of any 30 terms will have units digit 0, and so on.

We are told the sum of the terms is 350 (units digit 0), so the number of terms must be 10 or 20 or 30 or . . .

Since C is a multiple of 10, this must be the correct answer.

Cheers,
Brent
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by Mathsbuddy » Fri Oct 17, 2014 7:16 am
I can see that Brent's answer is a nicely detailed version of what I meant earlier.
You can skip all the pattern seeking though, because being 350, the number of terms MUST BE a multiple of 10. And 10 terms x 7 (last digit) will ALWAYS give an answer ending with ZERO.
Only one given answer (C) matches this.
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