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# A retailer orders u units of an item for which he pays d dol

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Joined: 09 Feb 2013
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A retailer orders u units of an item for which he pays d dol  [#permalink]

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Updated on: 12 Mar 2013, 01:55
1
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Difficulty:

15% (low)

Question Stats:

83% (01:58) correct 17% (02:36) wrong based on 163 sessions

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A retailer orders u units of an item for which he pays d dollars. However, he mistakenly receives u - 3 items. How much was he overcharged, in terms of d and u?

A. 3d - u
B. 3du - 9
C. (3d)/u
D. (u−d)/3
E. (du)/3

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Originally posted by emmak on 11 Mar 2013, 23:38.
Last edited by Bunuel on 12 Mar 2013, 01:55, edited 1 time in total.
Edited the question.
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Posts: 57255
Re: A retailer orders u units of an item for which he pays d dol  [#permalink]

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12 Mar 2013, 02:01
1
emmak wrote:
A retailer orders u units of an item for which he pays d dollars. However, he mistakenly receives u - 3 items. How much was he overcharged, in terms of d and u?

A. 3d - u
B. 3du - 9
C. (3d)/u
D. (u−d)/3
E. (du)/3

Say the retailer ordered u = 4 items for d = 4$(1$ per item). Thus:

He received 4 - 3 = 1 unit for 4$, which is 3$ more than the item actually costs.

Now, plug u = 4 and d = 4 into the answer choices to see which one yields 3. Only C fits: (3d)/u = (3*4)/4 = 3.

Hope it's clear.
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Re: A retailer orders u units of an item for which he pays d dol  [#permalink]

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14 Mar 2013, 01:55
Bunuel wrote:
emmak wrote:
A retailer orders u units of an item for which he pays d dollars. However, he mistakenly receives u - 3 items. How much was he overcharged, in terms of d and u?

A. 3d - u
B. 3du - 9
C. (3d)/u
D. (u−d)/3
E. (du)/3

Say the retailer ordered u = 4 items for d = 4$(1$ per item). Thus:

He received 4 - 3 = 1 unit for 4$, which is 3$ more than the item actually costs.

Now, plug u = 4 and d = 4 into the answer choices to see which one yields 3. Only C fits: (3d)/u = (3*4)/4 = 3.

Hope it's clear.

Amount per items actually recieved = d/u-3

Amount per items to be recieved = d/u

Overcharged amount = d/u-3 - d/u = 3d/u(u-3)
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Posts: 57255
Re: A retailer orders u units of an item for which he pays d dol  [#permalink]

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14 Mar 2013, 02:07
1
akshaychandna wrote:
Bunuel wrote:
emmak wrote:
A retailer orders u units of an item for which he pays d dollars. However, he mistakenly receives u - 3 items. How much was he overcharged, in terms of d and u?

A. 3d - u
B. 3du - 9
C. (3d)/u
D. (u−d)/3
E. (du)/3

Say the retailer ordered u = 4 items for d = 4$(1$ per item). Thus:

He received 4 - 3 = 1 unit for 4$, which is 3$ more than the item actually costs.

Now, plug u = 4 and d = 4 into the answer choices to see which one yields 3. Only C fits: (3d)/u = (3*4)/4 = 3.

Hope it's clear.

Amount per items actually recieved = d/u-3

Amount per items to be recieved = d/u

Overcharged amount = d/u-3 - d/u = 3d/u(u-3)

The retailer received u - 3 items for the price of u items .

The price of u items is d dollars (so, the price of an items d/u dollars);
The price of u - 3 items is d/u*(u -3) dollars;

The difference is d - d/u*(u -3) = (du - du +3d)/u = 3d/u.

Hope it helps.
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Re: A retailer orders u units of an item for which he pays d dol  [#permalink]

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14 Mar 2013, 03:17
emmak wrote:
A retailer orders u units of an item for which he pays d dollars. However, he mistakenly receives u - 3 items. How much was he overcharged, in terms of d and u?

A. 3d - u
B. 3du - 9
C. (3d)/u
D. (u−d)/3
E. (du)/3

Algebraic Solution
Amount overcharged = Amount paid by him - Amount he should have paid
Amount he should have paid=
original price of items*total items purchased
$$\frac{d}{u}*(u-3)$$
Amount paid by him was d dollars
Thus amount overcharged = $$d-d*\frac{(u-3)}{u}$$
On solving we get 3d/u
Thus C
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Re: A retailer orders u units of an item for which he pays d dol  [#permalink]

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02 Oct 2014, 02:05
1
Price ............. Quantity ............ Total

$$\frac{d}{u}$$ .................... u .................... d

$$\frac{d}{u}$$ ...................... (u-3) ............... $$d - \frac{3d}{u}$$ (Three pieces less received)

$$Overcharge = d - d + \frac{3d}{u} = \frac{3d}{u}$$

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Re: A retailer orders u units of an item for which he pays d dol  [#permalink]

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11 Jan 2018, 10:54
Bunuel

I am making a silly mistake! can't put my finger on it!

I plugged in 10 for u and 10 for d.

10u=10d so 1u= 1d
10-3=7
7u = 10d so 1u=10/7d

1d- 10-7d= -3/7

which is wrong! i cant figure out what am i doing wrong!

Thanks
VP
Joined: 09 Mar 2016
Posts: 1255
A retailer orders u units of an item for which he pays d dol  [#permalink]

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11 Jan 2018, 13:31

Amount per items actually recieved = d/u-3

Amount per items to be recieved = d/u

Overcharged amount = d/u-3 - d/u = 3d/u(u-3)[/quote]

The retailer received u - 3 items for the price of u items .

The price of u items is d dollars (so, the price of an items d/u dollars);
The price of u - 3 items is d/u*(u -3) dollars;

The difference is d - d/u*(u -3) = (du - du +3d)/u = 3d/u.

Hope it helps.[/quote]

Hi Bunuel can you explain how from this d - d/u*(u -3) you got this (du - du +3d)/u and this 3d/u ?

btw any nice link to know better how to simplify such algebraic expressions ? I always experience such technical issues with algebra

I know this time you gonna reply thanks!
Math Expert
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Posts: 57255
Re: A retailer orders u units of an item for which he pays d dol  [#permalink]

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11 Jan 2018, 20:39
srishti201996 wrote:
A retailer orders u units of an item for which he pays d dollars. However, he mistakenly receives u - 3 items. How much was he overcharged, in terms of d and u?

A. 3d - u
B. 3du - 9
C. (3d)/u
D. (u−d)/3
E. (du)/3

Bunuel

I am making a silly mistake! can't put my finger on it!

I plugged in 10 for u and 10 for d.

10u=10d so 1u= 1d
10-3=7
7u = 10d so 1u=10/7d

1d- 10-7d= -3/7

which is wrong! i cant figure out what am i doing wrong!

Thanks

The retailer ordered u = 10 items for d = 10$(1$ per item). Thus:

He received 10 - 3 = 7 unit for 10$, which is 3$ more than the items actually costs.

Now, plug u = 10 and d = 10 into the answer choices to see which one yields 3. Only C fits: (3d)/u = (3*10)/10 = 3.

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Posts: 57255
Re: A retailer orders u units of an item for which he pays d dol  [#permalink]

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11 Jan 2018, 20:42
dave13 wrote:

Amount per items actually recieved = d/u-3

Amount per items to be recieved = d/u

Overcharged amount = d/u-3 - d/u = 3d/u(u-3)

The retailer received u - 3 items for the price of u items .

The price of u items is d dollars (so, the price of an items d/u dollars);
The price of u - 3 items is d/u*(u -3) dollars;

The difference is d - d/u*(u -3) = (du - du +3d)/u = 3d/u.

Hope it helps.[/quote]

Hi Bunuel can you explain how from this d - d/u*(u -3) you got this (du - du +3d)/u and this 3d/u ?

btw any nice link to know better how to simplify such algebraic expressions ? I always experience such technical issues with algebra

I know this time you gonna reply thanks!
[/quote]

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Re: A retailer orders u units of an item for which he pays d dol  [#permalink]

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12 Jan 2018, 08:52
Bunuel wrote:
dave13 wrote:

Amount per items actually recieved = d/u-3

Amount per items to be recieved = d/u

Overcharged amount = d/u-3 - d/u = 3d/u(u-3)

The retailer received u - 3 items for the price of u items .

The price of u items is d dollars (so, the price of an items d/u dollars);
The price of u - 3 items is d/u*(u -3) dollars;

The difference is d - d/u*(u -3) = (du - du +3d)/u = 3d/u.

Hope it helps.

Hi Bunuel can you explain how from this d - d/u*(u -3) you got this (du - du +3d)/u and this 3d/u ?

btw any nice link to know better how to simplify such algebraic expressions ? :) I always experience such technical issues with algebra :)

I know this time you gonna reply :) thanks!
[/quote]

Hello Bunuel ! Hope you have an amazing day! :)

Thank you for the link, i reviewed the information provided in the website but I still do have doubts.

Ok So we have this: $$d$$ - $$\frac{d}{u}$$*$$(u -3)$$

First step: multiply $$\frac{d}{u}$$ by $$(u -3)$$ ---> -d + d3 <--- so here when we multiply $$\frac{d}{u}$$ by $$U$$ --- U cancel out

Now we have d- d + d3

So where am i wrong Bunuel ?

Many thanks for your methodical explanation !
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Joined: 09 Mar 2016
Posts: 1255
Re: A retailer orders u units of an item for which he pays d dol  [#permalink]

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12 Jan 2018, 09:03
dave13 wrote:
Bunuel wrote:
dave13 wrote:

Amount per items actually recieved = d/u-3

Amount per items to be recieved = d/u

Overcharged amount = d/u-3 - d/u = 3d/u(u-3)

The retailer received u - 3 items for the price of u items .

The price of u items is d dollars (so, the price of an items d/u dollars);
The price of u - 3 items is d/u*(u -3) dollars;

The difference is d - d/u*(u -3) = (du - du +3d)/u = 3d/u.

Hope it helps.

Hi Bunuel can you explain how from this d - d/u*(u -3) you got this (du - du +3d)/u and this 3d/u ?

btw any nice link to know better how to simplify such algebraic expressions ? I always experience such technical issues with algebra

I know this time you gonna reply thanks!

Hello Bunuel ! Hope you have an amazing day!

Thank you for the link, i reviewed the information provided in the website but I still do have doubts.

Ok So we have this: $$d$$ - $$\frac{d}{u}$$*$$(u -3)$$

First step: multiply $$\frac{d}{u}$$ by $$(u -3)$$ ---> -d + d3 <--- so here when we multiply $$\frac{d}{u}$$ by $$U$$ --- U cancel out

Now we have d- d + d3

UPDATE: I got i forgot to multiplt d/u 3!!! i multiplied jusd D thinking that U canceled out! I am getting smarter lol

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Re: A retailer orders u units of an item for which he pays d dol  [#permalink]

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10 Aug 2018, 18:46
emmak wrote:
A retailer orders u units of an item for which he pays d dollars. However, he mistakenly receives u - 3 items. How much was he overcharged, in terms of d and u?

A. 3d - u
B. 3du - 9
C. (3d)/u
D. (u−d)/3
E. (du)/3

Since the price per unit is d/u, the cost of u - 3 units is d/u x (u - 3) = (ud - 3d)/u, so the retailer was overcharged by:

d - (ud - 3d)/u

ud/u - (ud - 3d)/u

(ud - (ud - 3d))/u

(ud - ud + 3d)/u

(3d)/u

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Re: A retailer orders u units of an item for which he pays d dol  [#permalink]

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11 Aug 2018, 06:29
1
Top Contributor
emmak wrote:
A retailer orders u units of an item for which he pays d dollars. However, he mistakenly receives u - 3 items. How much was he overcharged, in terms of d and u?

A. 3d - u
B. 3du - 9
C. (3d)/u
D. (u−d)/3
E. (du)/3

A retailer orders u units of an item for which he pays d dollars.
So, EACH unit costs d/u dollars

However, he mistakenly receives u - 3 items.
In other words, he received 3 fewer units than he wanted
So, he paid for 3 units that he never received (i.e., he was overcharged for 3 units)

How much was he overcharged, in terms of d and u?
He was overcharged for 3 units, and EACH unit costs d/u dollars
TOTAL overcharge = (3)(d/u)
= 3d/u

Cheers,
Brent
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Re: A retailer orders u units of an item for which he pays d dol   [#permalink] 11 Aug 2018, 06:29
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