pencils - Trap Q

the attachments in this post are quite large. I am reattaching as word documents. Also, apologies for incorrectly putting this into the PS Forum

23 cent pencils & 21 cent pencils

23 cent pencils ?

Statement I

total number of pencils 6
we dont know the exact amount.

insufficient.

Statement II

total amount 130

21x + 23y = 130


remember x and y has to be integers and after multiplying the unit digits should add to 0.

This can only be done one way

when x=4 and y=2

21*4 = 84
23*2 = 46

84+46=130

Sufficient.

Hence B is the answer.

all the best.

LSB wrote:the attachments in this post are quite large. I am reattaching as word documents. Also, apologies for incorrectly putting this into the PS Forum

statement 1 . insufficient

statement 2 . since 130 is not a factor of either 23 or 21... she would have definitely atleast 1 of each pencil.

now we know 130/23 <n< 130/21

n must be a integer so its 6.

assume she bought X pencils for 23 cents..

she wud have bought 6-X for 21

therefore

23x + 21 ( 6 – x)= 130

hence X = 2.

hence B

Hope that helps

Makes sense. I guess you can convert the equation into having one variable. I got the right answer (it was a sledgehammer approach though)

Was just unsure how we can have only one equation that solves for two variables. Thanks guys.

I think the key thing to realise here is that this question is also testing your knowledge of digits ... as parallel_chase pointed out.

Essentially, with statement 2 we have to figure out how many 23's and how many 21's we can fit into 130.

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