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If w widgets cost c cents, how many widgets can you get for d dollars?

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Bunuel
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If w widgets cost c cents, how many widgets can you get for d dollars? [#permalink] New post   04 Sep 2018, 23:10
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If w widgets cost c cents, how many widgets can you get for d dollars?


(A) \(\frac{100dw}{c}\)

(B) \(\frac{dw}{100c}\)

(C) \(100cdw\)

(D) \(\frac{dw}{c}\)

(E) \(cdw\)
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Re: If w widgets cost c cents, how many widgets can you get for d dollars? [#permalink] New post   15 Sep 2018, 08:51
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Bunuel wrote:
If w widgets cost c cents, how many widgets can you get for d dollars?


(A) \(\frac{100dw}{c}\)

(B) \(\frac{dw}{100c}\)

(C) \(100cdw\)

(D) \(\frac{dw}{c}\)

(E) \(cdw\)


1 Dollar = 100 Cents

So, d Dollar = 100 d Cents

For c cents w widgets can be bought

So, For 1 cents \(\frac{w}{c}\) widgets can be bought

Thus, for 100 d Cents \(\frac{100dw}{c}\) widgets can be bought, Answer must be (A)
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Re: If w widgets cost c cents, how many widgets can you get for d dollars? [#permalink] New post   05 Sep 2018, 01:30
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For c cents, w widgets can be bought

......1 cent.., \(\frac{w}{c,}\) widgets can be bought

.......100d cents...., \(\frac{w}{c}\)*100d widgets can be bought

= \(\frac{100dw}{c}\)

Answer: A.
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Re: If w widgets cost c cents, how many widgets can you get for d dollars? [#permalink] New post   03 Sep 2019, 22:04
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Abhishek009 wrote:
Bunuel wrote:
If w widgets cost c cents, how many widgets can you get for d dollars?


(A) \(\frac{100dw}{c}\)

(B) \(\frac{dw}{100c}\)

(C) \(100cdw\)

(D) \(\frac{dw}{c}\)

(E) \(cdw\)


1 Dollar = 100 Cents

So, d Dollar = 100 d Cents

For c cents w widgets can be bought

So, For 1 cents \(\frac{w}{c}\) widgets can be bought

Thus, for 100 d Cents \(\frac{100dw}{c}\) widgets can be bought, Answer must be (A)


I dont get the multiplication by hundred
Does this mean the problem is based on the fact that you write C's as a non comma? Like 2 cents = 2 and not 0,02?
I think that was my mistake

because I wrote it like W / C and said if C = 0,05 that means I can buy 20 widgets for one Dollar so D dollar give me D*20 widges
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Re: If w widgets cost c cents, how many widgets can you get for d dollars? [#permalink] New post   03 Sep 2019, 22:16
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Bunuel wrote:
If w widgets cost c cents, how many widgets can you get for d dollars?


(A) \(\frac{100dw}{c}\)

(B) \(\frac{dw}{100c}\)

(C) \(100cdw\)

(D) \(\frac{dw}{c}\)

(E) \(cdw\)



We know

1 dollar = 100 cents

Simply use the unitary method:

c cents buys us w widgets

Then 1 cent buys us w/c widgets

Then, d dollars (= 100d cents) buys us (w/c)*100d widgets = 100dw/c widgets.
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Re: If w widgets cost c cents, how many widgets can you get for d dollars? [#permalink] New post   01 Oct 2020, 04:14
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'w' widgets = 'c' cents

1 cent: \(\frac{w}{c}\) widgets

1 dollar = 100 cents, hence multiplying both the sides by '100'

=> 100 cents = \(\frac{w}{c}\) * 100

=> 1 dollar = \(\frac{100w }{c}\)

=> 'd dollars: \(\frac{100dw }{c}\)

Answer A
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Re: If w widgets cost c cents, how many widgets can you get for d dollars? [#permalink] New post   05 Sep 2018, 01:00
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Bunuel wrote:
If w widgets cost c cents, how many widgets can you get for d dollars?


(A) \(\frac{100dw}{c}\)

(B) \(\frac{dw}{100c}\)

(C) \(100cdw\)

(D) \(\frac{dw}{c}\)

(E) \(cdw\)


Unit cost of widgets=\(\frac{c}{w}\) cents
Or, \(\frac{c}{100w}\) dollars are required to purchase 1 widget
Or, $1 is required to purchase \(\frac{100w}{c}\)widgets
hence $d is required to purchase \(\frac{100dw}{c}\)

Ans. (A)
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If w widgets cost c cents, how many widgets can you get for d dollars? [#permalink] New post   15 Sep 2018, 08:40
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One of my favorite approaches for questions with all variables is plug-in method.

Let w=100 pcs, c=100 cents, d = $1 (actually d=c, so for 'd' we can buy the same number of 100 pcs - now we have to find an option that will give us the same 100 pcs.). Only option A gives us 100. Hence this is the right choice.

Now if you wonder how efficient this method is: 01:10 for the question above and I got the right answer, with my brain in a low level of energy consumption mode. Belive me, the saved brain stamina will be very useful for hard level questions later on the test. I highly recommend to use algebra only in cases in which it is really necessary.
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Re: If w widgets cost c cents, how many widgets can you get for d dollars? [#permalink] New post   01 Sep 2019, 08:41
can you explain this question in simple way ? a little confusing quantity or price and number of item
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Re: If w widgets cost c cents, how many widgets can you get for d dollars? [#permalink] New post   15 Dec 2022, 13:39
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Bunuel wrote:
If w widgets cost c cents, how many widgets can you get for d dollars?


(A) \(\frac{100dw}{c}\)

(B) \(\frac{dw}{100c}\)

(C) \(100cdw\)

(D) \(\frac{dw}{c}\)

(E) \(cdw\)


We can let w = 2, c = 10, and d = 4.

Thus, 2 widgets cost 10 cents, or each widget costs 10/2 = 5 cents.

So for 400 cents, 400/5 = 80 widgets can be purchased.

Plugging these values back into answer choice A we have:

(100 x 4 x 2)/10 = 800/10 = 80, so answer A is correct.
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Re: If w widgets cost c cents, how many widgets can you get for d dollars? [#permalink] New post   22 Oct 2023, 02:39
Bunuel could you attach a few links of this question type?

Thanks :)
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Re: If w widgets cost c cents, how many widgets can you get for d dollars? [#permalink]
22 Oct 2023, 02:39
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