Question 1 (nov. 25)

lets have the x1 as the measurement between x & 3

as teh the whole region is square , we have

3+3+3 = x+ x1 + 3 --------> x+x1 = 6 =>eq 1

as shaded area is equal to unshaded area,

shaded area = 9+9+3X1 + 2(3X)
unshaded area = 9+2(3X1) + 3X

equating these two equations gives X-X1 = -3 +> eq 2

solving eq 1 & 2

we get X = 3/2 => 1.5 ==> choice C


substituing the value X = 1.5 & X1 = 4.5 , solves right for shaded equal to unshaded ares, whereas OA answer 2 & 4 doesnt.

can someone exlain/confirm plz. TIA

bacali wrote:The figure above represents a square garden that is divided into 9 rectangular regions with indicated dimensions in meters. The shaded regions are planted with peas, and the unshaded regions are planted with tomatoes. If the sum of the areas of the regions planted with peas is equal to the sum of the areas of the regions planted with tomatoes, what is the value of x?


A. 0.5
B. 1
C. 1.5
D. 2
E. 2.5

We know that the whole region is a square. Therefore, the base is also 9. Instead of calling the unknown length x1, let's call it (6-x).

So, we know that:

Shaded = Unshaded

9 + 9 + 3x + 3x + 3(6-x) = 9 + 3x + 3(6-x) + 3(6-x)

18 + 6x + 18 - 3x = 9 + 3x + 18 - 3x + 18 - 3x
36 + 3x = 45 - 3x
6x = 9
x = 1.5

I concur that (C) is the right answer.

Note that this would have also been a great question for backsolving. Start by plugging in either (B) or (D), then decide if you need a bigger value for x or a smaller one, then plug in one more choice.

If we had started with x=2, we'd get 6-x = 4, so:

Shaded = Unshaded

9 + 9 + 6 + 6 + 12 = 9 + 12 + 12 + 6
42 = 39

Plugging in x=2, the unshaded regions were too big, so we need to decrease the value of x to help equalize. Eliminate (D) and (E), try:
(B) x = 1; 6-x = 5

9 + 9 + 3 + 3 + 15 = 9 + 15 + 15 + 3
39 = 42

Now we've gone too far in the other direction: eliminate (A) and (B), choose (C), the only option remaining.

each area in the middle row mirrors the one underneath it in the bottom row. So the total shaded area from the middle and bottom rows is always the same as the unshaded one in the same rows, regardless of the value of x. In essence, we can ignore the bottom and middle rows completely.

This means the shaded area in the top row needs to equal the unshaded one in this row. When we start calculating each area we will be multiplying by the same height: 3. So we can even ignore the height.

The only thing that we need to make sure is that the unshaded width equals the shaded one:

x + 3 = 6 - x

xyzabc123 wrote:each area in the middle row mirrors the one underneath it in the bottom row. So the total shaded area from the middle and bottom rows is always the same as the unshaded one in the same rows, regardless of the value of x. In essence, we can ignore the bottom and middle rows completely.

This means the shaded area in the top row needs to equal the unshaded one in this row. When we start calculating each area we will be multiplying by the same height: 3. So we can even ignore the height.

The only thing that we need to make sure is that the unshaded width equals the shaded one:

x + 3 = 6 - x

thanks a lot ....

Yeah, this is from a PBT. The OA is D, which is wrong. IMO, it's C