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Profit/loss Arithmatic

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Profit/loss Arithmatic

by rrobiinn » Mon Sep 10, 2012 9:49 pm
A reduction of 20% in the price of sugar enables a purchaser to obtain 4 kg more for $160. What is the reduced price per kg?

* I became depressed thinking that with easy questions like this and I can't solve, would bring poor score in Gmat :(
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by das.ashmita » Mon Sep 10, 2012 10:06 pm
My approach:

let the initial cost per kg be $'p' and amt be 'x'kg

given, px = (0.8p)(x+4) = 160

solving px = (0.8p)(x+4) we get, x=16

px = 160
=>p = 160/16 = 10

reduced price per kg = (0.8)p = $8
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by rrobiinn » Tue Sep 11, 2012 12:00 am
das.ashmita wrote:My approach:

let the initial cost per kg be $'p' and amt be 'x'kg

given, px = (0.8p)(x+4) = 160

solving px = (0.8p)(x+4) we get, x=16

px = 160
=>p = 160/16 = 10

reduced price per kg = (0.8)p = $8
How
px = (0.8p)(x+4)
and px = 160
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by KapilKapoor » Tue Sep 11, 2012 12:31 am
Hi rrobiinn,

What we know is, total material was purchased with 160$ both the time.
First time = p*x
Second time = (0.8p) * (x+4)
Redu Rate Total qty
received

=> px = (0.8p)(x+4)
and px = 160


Moreover you can see for the other method also as mentioned below:

let the rate = R $/Kg
How much will be the wt purchased out of 160$ = 160/R Kg
The reduced rate = 0.8R $/Kg
How much will be the wt purchased out of 160$ now = 160/(0.8R) Kg

According to the question:

The new Wt = 4 + Old Wt (If the material was purchased of the same 160$ only)
160/(0.8R) = 4 + 160/R

R = 10 and 0.8R = 8$ (Ans)

Hope this will clear your doubt.

Regards,
Kapil
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by neelgandham » Tue Sep 11, 2012 12:33 am
rrobiinn wrote:My approach:
How
px = (0.8p)(x+4)
and px = $160
let the initial cost per kg be $'p' and amt be 'x'kg
Case 1: Cost of purchase = Cost per kilogram * Number of kilograms purchased
= $p*x

Case 2: If the price of sugar drops by 20% then the cost per kilogram is p - (20/100)*p = $0.8p
Amount of sugar purchased = x+4
Cost of purchase = Cost per kilogram * Number of kilograms purchased
= $0.8p * (x+4)

But we know that the total cost of purchase is 160 in both the cases.
So, p*x = 0.8p * (x+4)
px = 0.8px + 3.2p
0.2px = 3.2p
x = 16 and p = $10

Reduced price = 0.8 * p = 0.8*10 = $8
Anil Gandham
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by das.ashmita » Tue Sep 11, 2012 12:38 am
Hi rrobinn

If we assume initial cost per kg be $'p' and amt be 'x'kg

A reduction of 20% in the price of sugar enables a purchaser to obtain 4 kg more for $160
From the above statement, we can say 2 things
1. before reduction : price of x kg was $160
In other words , px = 160

2. after reduction : price of x+4 kg was $160 & price / kg = (1-20/100)p = 0.8p
In other words , (0.8p)(x+4) = 160

Hope its clear :)
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by GMATGuruNY » Tue Sep 11, 2012 1:56 am
rrobiinn wrote:A reduction of 20% in the price of sugar enables a purchaser to obtain 4 kg more for $160. What is the reduced price per kg?

* I became depressed thinking that with easy questions like this and I can't solve, would bring poor score in Gmat :(
A reduction of 20% implies that the original price is almost certainly a multiple of 10.
Thus, we can GUESS AND CHECK the correct answer very quickly, no algebra required.

Let the original price per kg = 10.
Amount that can be purchased for 160 = 160/10 = 16.
Reduced price = 10 - .2(10) = 8.
New amount that can be purchased for 160 = 160/8 = 20.
Success! At the reduced price, 4 more kg can be purchased.
Thus, the reduced price = 8.

For a similar problem that I solved with this approach, check here:

https://www.beatthegmat.com/profit-loss-t115625.html
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