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Martha bought several pencils. If each pencil was either a 23-cent pencil or a 21-cent pencil, ho wmany 23-cent pencils did Marta buy?

(1) Martha bought a total of 6 pencils.

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

Please explain.

I chose B. 23x+21y=130. Was sure I am going to try x to be max of 5. So with little working for x from 1 to 5, found that x to be 2 and y to be 4. Hence B

Martha bought several pencils. If each pencil was either a 23-cent pencil or a 21-cent pencil, ho wmany 23-cent pencils did Marta buy?

(1) Martha bought a total of 6 pencils.

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

Please explain.

I think it is C
Here is my reasoning
1)alone insuff we can't figure out anything but the quant of pencils!
2)same here
1&2 together
Let x be 23 cents pencils and y be 21 cents pencils so X+Y=6
From 2 statement 23X+21Y=130
Hence we have
X+Y=6
23X+21Y=130
X=6-Y substitute 23*(6-Y)+21Y=130--->138-23Y+21Y=130--->2Y=8--->Y=4,X=2

(2) 23x + 21y = 130 --> only one combination can lead to this: x = 2, y = 4

Ans B

It's pretty easy to manipulate. You can see that x cannot be 5-10 as that will result in y being fractional. You can then quickly work out x for values of 4, then 3, then 2... each time you do that, subtract it off from 130 and see if it's divisible by21. Sounds like a lot of work, but it's very easy and fast math really...

Martha bought several pencils. If each pencil was either a 23-cent pencil or a 21-cent pencil, ho wmany 23-cent pencils did Marta buy?

(1) Martha bought a total of 6 pencils.

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

Please explain.

By no means would I have picked B on the real test day. Its got no straight forward step-by-step approach In my mind it was a clear C. But boy was I wrong! I think the BAD strategy is going to be on my check list as well.
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