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BhavyaKamana
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GMAT 1: 800 Q51 V51
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Ok lets dumb this down for us. "A problem well defined is halfway through to the solution - Kidlins law"

Visualization
Now lets imagine this situation more clearly - in "real" terms
Lets assume the package used to be 10oz being sold at $100 and now they are selling 14oz for $100.

Translation
So lets compare the per oz rates
earlier - $100/10 = $10 per oz
now - $100/14 per oz (dont simplify till the end)

Approach
Lets just now lets calculate the change in price
(old-new)/old - check below image

Conclusion & Estimation
So the percentage come to around 200/7 - which we know its less than 30% and it is greater than 25% - thus the only workable answer for us 29%


Shu "TheGMATTutor"
[color=#302f2f]The GMAT concepts are relatively basic, but they require a deep understanding to tackle nuanced questions effectively. So lets dumb it down for us. This is not just a math and English test—it is a deductive test designed to assess your problem-solving skills[/color]
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Screenshot 2024-10-14 at 16.30.21.png
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BhavyaKamana
For a sales promotion, a dog food producer produces packages with 40 percent more dog food than the regular-size package and sells the larger package for the same price as the usual price of the regular-size package. Compared to the regular-size package at its usual price, approximately what percent less is the price per ounce of the larger package of dog food?

A. 25%
B. 29%
C. 37.5%
D. 40%
E. 60%

Attachment:
2024-01-27_13-51-20.png

Assume numbers that simplify calculations.

Say the regular package quantity is 100 ml so large package quantity is 140 ml.
Say usual price of regular package is $140 (not $100 because then price per unit will become a fraction) and that stays the same for the large package.

So price per unit for regular package is $1.4 and for the large package is $1.

So the price per unit for large package is less by approximately \(\frac{.4}{1.4} * 100 = \frac{2}{7} * 100 = 28.56%\) (percent fraction equivalent taught us that 1/7 = 14.28)

Percent fraction equivalents are discussed here: https://youtu.be/HxnsYI1Rws8

Answer (B)
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