Martha bought several pencils. If each pencil was either a

Actually the second statement also gives us two equations. Albeit, one is not explicit.

The equation that be derived clearly from the second statement is:
23x + 21y = 130

Now, how do we have another equation here?

As IanStewart has also explained,

If the pencils cost either 21c or 23c, and the total value of the pencils was 130 cents, the total pencils had to be six. Had to be.

Here's why:

1. What's the minimum # of pencils she must have bought?
I'm thinking in terms of the expensive pencils.
I'll start with a relaxed number.
e.g. 23 x 3 = 69. Less than 130. Too few.
23 x 4 = 92. Less than 130. Too few.
23 x 5 = 115. Less than 130. Too few.
23 x 6 > 130. Wait. That's too many.
Even if all the pencils were the expensive ones and she bought five pencils, her total would have been less than 130. So, she would have bought more than 5 pencils to get a total of 130 cents.


2. What's the maximum # of pencils she must have bought?
I'm thinking in terms of the cheaper pencils now.
21 x 10 = 210. Greater than 130. Too many.
21 x 7 = 140 + 7. Greater than 130. Too many.
21 x 6 = 126. Ok, that's too few.
Even if all the pencils were the cheaper ones and she bought seven pencils, her total would have been greater than 130. So, she would have bought fewer than 7 pencils to get a total of 130 cents.

Combining these two, we know for sure that she bought six pencils total.

So, we have our second equation now:
x + y = 6

This can be solved using the system of coordinate geometry. We know when an equation has a negative slope then it is called a decreasing line, in terms of Co-ord Geometry. The particular property of a decreasing line is this: for any satisfying value of (x,y), the next value of x, y will take the opposite coordinates of (x,y):

What I meant is this : We know that 23x + 21 y= 130

Now, since the constraint here is nos. of pencils has to be an integer, the nearest value will be 23 x 2 + 21 x 4 . How do I know because the unit digit will become zero and hence that will be my first iteration. other times what I do is take any value of x or y to be zero, if the constant is divisible by any of coefficient of x or y,

So we have the initial value x=2 and y =4/ We need to check if there are any other values within the range,
So the property dictates this , Since it is a decreasing line, if x increases, y has to decrease. or vice versa to satisfy the equation.

Now the next value of x will change by the coefficient of y,
So what was the coefficient of y?
It was 21. !

So x will take the value of x =2 +21 =23

Now when x =23, y will take the coefficient of x, which is 23 but in a decreasing way because you increased x. So y will be 4-23= -19, which is not possible as no of pencils cannot be negative

Similarly, if you increase y by the coefficient of x, y becomes = 4+ 23 =27, so x will have to decrease by the coefficient of y which is 21. So x will become, 2-21= -19, which is not possible again,

So both are not possible and hence statement 2 has a unique solution. Hence B.

Solve these kinds of questions with this logic, and mostly you may not fail.
Hope it helps

The way I thought this problem was the following.

(1) is clearly insufficient bc we only know the sum of the quantities but not the money spent.

(2) 23x+21y=130
I realized this can be solve by trying different numbers since we know x and y are integers since they are pencils and we also know they must be positive. So we can use info from (1) to help us to solve just using (2) without actually use (1). So we have a clue that x+y=6 so instead of trying a lot of numbers in (2) to get 130 we just try the pair of numbers that sum 6. (0 and 6) (1 and 5) (2 and 4) ( 3 and 3)

x=0, y =6, 23x+21y=130 --> 126=130, NO
x=6, y=0, -- >138=130, NO
x=y=3, 69+63= 132 -->132=130 NO
x=1, y=5, 23+105= 128, -->128=130 NO
x=5, Y=1, 115+21=136. --> 136=130 NO
x=2, Y=4; 46+84= 130. --> 130=130 = YES
x=4, Y=2, 92+42=134. --> 134=130 NO

so the only possible combination is x=2 and Y=4
so (2) is sufficient by itself, even though we use (1) as a clue, it is not necessary to solve the problem

IMO B

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