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Every square is a rectangle however every rectangle need not be a square.
Considering the same, the answer comes up correct = 20 = Answer = B
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it doesn't matter here if its a square o rectanlge value gets replaced later. take it as s^2 or a*b
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it doesn't matter here if its a square o rectanlge value gets replaced later. take it as s^2 or a*b
Agreed, it gets cancelled, sorry for the trouble :oops:
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Fantabulous

Did you consider a rectangle as a square? Because if it is a rectangle then the sides may or may not be equal. Please clarify.

I have assumed that the rectangle is a square since square is also a type of rectangle (but I see I missed writing it).
It is a PS question so it will have a unique answer. This means a - b will be the same for every such group of rectangles. So we can easily take one such group such that the rectangles are squares to make our calculations easier.
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Could you please let me know why you do not distribute the s^2 through the parenthesis? I asked this in my own topic but it has since been locked and no further information was provided to me except to link to this thread.

\(s^2(a+b)(a−b)=300\) ...........(II)

Why does this not simplify to:
\(15∗s^2a−s^2b)\)

veritas-prep-the-sides-of-rectangle-x-are-each-multiplied-by-a-to-for-196927.html
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Could you please let me know why you do not distribute the s^2 through the parenthesis? I asked this in my own topic but it has since been locked and no further information was provided to me except to link to this thread.

\(s^2(a+b)(a−b)=300\) ...........(II)

Why does this not simplify to:
\(15∗s^2a−s^2b)\)

veritas-prep-the-sides-of-rectangle-x-are-each-multiplied-by-a-to-for-196927.html

\((a+b)*(a-b) = a^2 - b^2\)
You can simplify this to \(s^2a^2 - s^2b^2 = 300\) but how will that take you to the answer? You need the value of \((a - b)\) so you need to get rid of only \(s^2(a+b)\).
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Let rectangle X length=l width=w => Area = l*w
So now: rectangle Y length=a*l width=a*w => Area =a^2* l*w
rectangle Z length=b*l width=b*w => Area =b^2* l*w

given: a* Area of X = 10 = a*l*w
b* Area of X = 5 = b*l*w

so we can write a*l*w+b*l*w=10+5
=> l*w(a+b)=15

Also given => Area of Y - area of Z = 300 = a^2*l*w -b^2*l*w

so lw(a^2-b^2)=300
=> l*w(a+b)(a-b)=300
=> 15 (a-b)=300
=> (a-b)= 20


Answer: B
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VeritasPrepKarishma
goodyear2013
The sides of rectangle X are each multiplied by a to form rectangle Y and by b to form rectangle Z. a times the area of X is 10, and b times the area of X is 5. If the difference in area between Y and Z is 300, what is a - b?
5
20
30
50
60

Set sides of rectangle X are L and W
Area of X: = LW Area of Y = (aL)(aW) = a^2 * LW Area of Z = (bL)(bW) = b^2 * LW a times area X = aLW = 10 b times area X = bLW = 5
Which means that difference between Y and Z = aaLW – bbLW = 300.
And since you're looking to solve for a - b, you can try to get a and b alone by factoring out the common LW terms:
LW(a^2 - b^2) = 300
Which gives you the Difference of Squares setup that allows you to get (a - b) alone:
LW(a + b)(a - b) = 300
Then you should see that if you distribute LW across the first set of parentheses, you can get aLW and bLW, for which you have actual values:
(aLW + bLW)(a - b) = 300
(10 + 5)(a - b) = 300 15(a - b) = 300 (a - b) = 20

Hi, I want to know if we have the simpler solution, please

Assume the side of rectangle X is s.

X (side s, area s^2)
Y (side as, area (as)^2)
Z (side bs, area (bs)^2)

Given: \(as^2 = 10\), \(bs^2 = 5\) ..........(I)
\((as)^2 - (bs)^2 = 300\)
\(s^2 (a+b)(a-b) = 300\) ...........(II)

We need to find (a-b).
From (I), \(as^2 + bs^2 = 15 = s^2(a+b)\) (you do this because you need to get rid of s^2 and (a+b) in equation (II) above)
Substitute this in (II) to get \(15*(a-b) = 300\)
(a-b) = 20

Another way of doing this..

Let the original sides be x and y ---- area = \(xy\)

For Rectangle X -- ax and ay ---- area = \(a^2xy\)

For Rectangle Y -- bx an by ---- area = \(b^2xy\)

Now, it's given that

\(axy = 10\)
Thus
\(a = \frac{10}{(xy)}\)

Similarly we can write for Z as

\(bxy = 5\)
Thus
\(b = \frac{5}{(xy)}\)

It can clearly be seen that

\(a = 2b\)

Substituting "axy" in the original area equation..we can see that the area of X can be written as = 10a
Similarly for B it can be written as = 5b

Difference in areas

\(10a - 5b = 300\)

Using value of a as 2b, we get

b = 20
a = 40
Thus
a-b = 20
(B)
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Area X = l*w
l*w*a = 10
l*w*b = 5
Dividing the first by the second, we know that a/b = 2, so a=2b

Difference in area: l*w*a^2 - l*w*b^2 = 300
Substituting the above: 10*a - 5*b = 300
2a - b = 60
Since we know a=2b from above, 3b = 60
b = 20, so a = 40
40 - 20 = 20
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Let, each side of X =1, area = 1
So, each side of Y = a *1=a, and area=a^2
each side of Z = b*1= b, and area =b^2
As per question, a^2-b^2 =300
=> (a+b) (a-b) =300 => (10+5) (a-b)=300
So, (a-b)= 20.
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This is a high-quality question.
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In this question, do we just assume that the rectangle is square with equal sides?
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YZLIM1994
In this question, do we just assume that the rectangle is square with equal sides?
We don't necessarily need to assume that.

X: m n
Y: am an
Z: bm bn

amn = 10
bmn = 5

Adding the two above, mn(a+b) = 15

a^2*mn - b^2*mn = 300

a - b = ?

mn(a^2 - b^2) = 300

mn(a + b)(a - b) = 300

Substituting (a+b) = 15 into the above equation:

mn(a + b)(a - b) = 300

15*(a-b) = 300
So, a - b = 20
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