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

Question

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 · Answer

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.

Answer: C.

Hope it's clear.

akshaychandna · Answer

Please help me understand where am i going wrong ..

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)

Bunuel · Answer

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 .

Answer: C.

Hope it helps.

Dipankar6435 · Answer

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

PareshGmat · Answer

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}

Answer = C

srishti201996 · Answer

Bunuel

I am making a silly mistake! can't put my finger on it!
can you please tell me?

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

dave13 · Answer

Please help me understand where am i going wrong ..

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.

Answer: C.

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!

I always experience such technical issues with algebra

I know this time you gonna reply

thanks!

Bunuel · Answer

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.

Answer: C.

Bunuel · Answer

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 . Answer: C. 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! https://www.themathpage.com/ARITH/add-fr ... ions-1.htm

dave13 · Answer

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! https://www.themathpage.com/ARITH/add-fr ... ions-1.htm 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 !

dave13 · Answer

https://www.themathpage.com/ARITH/add-fr ... ions-1.htm 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

ScottTargetTestPrep · Answer

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

Answer: C

BrentGMATPrepNow · Answer

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

Answer: C

Cheers,
Brent

Logo · Answer

Can someone help me out? I do not understand why you take 3d/u. Wouldn't the amount overcharged be the amount overcharged/the price that should have been paid. For example, 
u=10
d=2
total= 20

3*2/20-(3*2)=
6/14

Archit3110 · Answer

let u be 10 units & d be $ 10 i.e 1 $ per unit cost

so u-3 ; 7 units received against 10 so he was overcharged by $3

option C ; gives same value (3d)/u

Basshead · Answer

Plug in numbers:

A retailer orders u units =  12 units 
Pays d dollars =  3 dollars 
Each unit =  0.25

He receives  u-3 units = 9 units

9/0.25  = 2.25

He was overcharged 0.75. We need to find an answer that equals 0.75.

C =  \frac{(3d)}{u} = \frac{3(3)}{12} = \frac{9}{12} = 0.75 .

Answer is C.

jcerdae · Answer

Could someone help me! cant understand what is wrong with my approach!

U= Units ordered
D= Dollars Payed
U-3= Units actually received

How much overcharged = Actually Paid(In monetary units) - Received (in monetary units)
where 
 Actually Paid= U*D
 Received= (U-3)*D

Then
 U*D-(U-3)*D = 2D-UD But that answer is not there!!

Help me please!

Abhishek009 · Answer

Price per unit is  \frac{d}{u}

Cost of u - 3 items is  \frac{d(u-3)}{u}

Thus, over charge is  d - \frac{d(u-3)}{u}

Or,  d - \frac{du-3d}{u}

Or,  \frac{3d}{u} , Answer is (C)

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

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