A rectangular solid is changed such that the width and lengt

Ok, so here is what I know:
~Old Volume = New Volume
~(L+1)(W+1)(H-9) = (L*W*H)
~(H-9) = 4w
~width, length of original rectangular are equal
So, from that I get:
(L+1)(W+1)(4w)=(W*W*H)
But in the book, the equation differs from mine in two ways.
For starters, mine is L+1)(W+1)(4w)=(W*W*H) while theirs is (W+1)(W+1)(4w)=(W*W*H)

If L=W then substitute L with W in the equation and obtain
\((W+1)(W+1)(4W)=(W*W*H)\)- the one in the book

But here is my (apparently incorrect) reasoning.
L=W in the old rectangle, so why plug "W" into the new rectangle volume?
if (H-9)=4w, then why do I plug 4w into the new rectangle volume and h=4w-9 into the old rectangle formula? It seems unnecessary to have to plug in 4w for (h-9) then derive h=4w-9 and plug in on the other side.

You have
\((W+1)(W+1)(H-9)=(W*W*H)\) and in order to solve it you need to express H in terms of W, so from \(H-9=4w\) you get \(H=4W+9\) and \(H-9=4W\)
\((W+1)(W+1)(4W)=(W*W*(4W+9))\)

The above equation is in W and H, so have to express all the variables in one term (W) in order to solve it. You have to plug in those values at both sides in order to solve the equation.

Hope it's clear, let me know

Zarrolou, I followed your approach and I got the final quadratic equation as \(w^2-4w=0\)---->w=4. Then we get h=4w+9=25. Also l=w=4. Volume is 4*4*25=400. Your approach is correct and I don't think there is any other possibility.

Hi Karishma/Bunnel,

Any more such questions?

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Hope this helps.

Its hard to understand the increase & decrease thing in provided explanation. Please help me ..

Here's how i solved it

Rectangle 1: length = l, width = w and height = h. Now l=w.
so Volume V1 = l(squared)h

Rectangle 2: length = l+1, width = w+1 or l+1 and height = h-9.
So, volume V2 = (l+1)squared(h-9)

Now new height = 4 times previous width => h-9 = 4w or (h-9) = 4l => h = 4l + 9

Substituting the value of h:-
Volume V1 = l(squared)(4l+9) = 4l(cube) + 9 l(squared)

and Volume V2 = (l+1)squared(4l) = 4lcube + 8lsquared = 4l

Equating V1 and V2 we get: l = 0, 4. since length isnt 0, l=4 => w=4 and h = 16+9 = 25

therefore volume V = lwh = 4x4x25 = 400
Answer E

The increase/decrease method has fewer calculations and steps but its certainly a little trickier...
Try to make diagrams at each step to understand what is going on.

Hi VeritasPrepKarishma,
I tried to solve the question by increase=decrease method. I took increase first and equated the same to decrease keeping the length and breadth same. Equation was:
1*b*h + l*1*h = (l+1)(b+1)*9 -> Keeping the length and breadth same. As l=b,
-> 2lh = (l+1)² *9 ; h-9= 4l -> h = 9+4l
-> 2l*(4l+9) = (l²+2l+1)9
-> 8l²+18l = 9l²+18l+9
As you can see , there is not definite solution with this equation. Can you please point out the flaw in the method?

Shouldn't Increase = 1*(l+1)*(h-9) + 1*w*(h-9) = (h-9)*(l+w+1) be Increase = 1*(l+1)*(h-9) + 1*(w+1)*(h-9) = (h-9)*(l+w+2) since the width increased by 1 too?

Hello,

I really liked the increase/decrease method.

Could you please suggest how you derived this equation:

Increase = 1*(l+1)*(h-9) + 1*w*(h-9) = (h-9)*(l+w+1)

thank you.

So in step 1, you chop off a block to decrease the area. The area of that block is l*w*9

Then in step 2, you add a block to increase the length. The area of this block is 1*w*(h-9)

Then you add another block to increase the width. The area of this block is 1*(l + 1)*(h - 9)

The decrease = Both increases
l*w*9 = 1*w*(h-9) + 1*(l+1)*(h - 9)

Ok, so here is what I know:
~Old Volume = New Volume
~(L+1)(W+1)(H-9) = (L*W*H)
~(H-9) = 4w
~width, length of original rectangular are equal
So, from that I get:
(L+1)(W+1)(4w)=(W*W*H)
But in the book, the equation differs from mine in two ways.
For starters, mine is L+1)(W+1)(4w)=(W*W*H) while theirs is (W+1)(W+1)(4w)=(W*W*H)

If L=W then substitute L with W in the equation and obtain
- the one in the book

But here is my (apparently incorrect) reasoning.
L=W in the old rectangle, so why plug "W" into the new rectangle volume?
if (H-9)=4w, then why do I plug 4w into the new rectangle volume and h=4w-9 into the old rectangle formula? It seems unnecessary to have to plug in 4w for (h-9) then derive h=4w-9 and plug in on the other side.

You have
and in order to solve it you need to express H in terms of W, so from you get and


The above equation is in W and H, so have to express all the variables in one term (W) in order to solve it. You have to plug in those values at both sides in order to solve the equation.
Hope it's clear, let me know





Can i considered this correct?
Any Experts! Please tell me!

Posted from my mobile device

Straight forward method:

Length = Width
So
\(L*L*H = (L+1)^2 (H-9)\) (Volume stays the same)
H-9 = 4L so H = 4L + 9

\(L*L*(4L + 9) = (L+1)^2 4L\)
\(L(4L + 9) = 4(L+1)^2\)
\(L = 4\)

\(H = 4L + 9 = 25\)

\(Volume = L^2*H = 4^2 * 25 = 400\)

Check this video for discussion on cubes and cuboids: https://youtu.be/kqVgiZRd6yM