Explanation

Let us assume that the car was manufactured for $100.

Price it was sold to dealer (@20% higher)= \($100 + $100 \times \frac{20}{100}\) = $120.

Now dealer sells it @ 15% higher = \($120 + $120 \times \frac{15}{100}\) = $138.

We can see by putting d = 100. option E is correct.
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the car cost the consumer, in dollars = d *1.15 *1.2 = 1.38d

great explanation sandy. this question doesnt seem like it should be in the hard category

I would typically solve this questioning the same way Sandy did. However, here is an algebraic approach:

A car manufacturer produced a car at a cost of d dollars and sold it to a dealer at a price 20 percent higher than the production cost.
Dealer price = (cost price) + (20% of cost price)
Dealer price = d + (20% of d)
Dealer price = 1d + 0.2d
Dealer price = 1.2d

If the dealer sold the car to a consumer for 15 percent more than the dealer paid for it, what did the car cost the consumer, in dollars?
Consumer price = (dealer price) + (15% of dealer price)
Consumer price = (1.2d) + (15% of 1.2d)
Consumer price = (1.2d) + (0.15)(1.2d)
Consumer price = (1.2d) + (0.18d)
Consumer price = 1.38d

Answer: E
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Is this hard type qstn. is from ETS?

This is from the power prep free software which usually is on the easier side.

On the test day are harder, if that is what you meant

How can I find ALL the QUANT QUESTIONS
from all the sources of OG i.e ETS only?
Plz help!!!!!!!

Here https://greprepclub.com/forum/search.ph ... earch_tags question bank. Pretty easy sir to find

The tag OG of course stands for Official Guide or big book or powerprep

You can also use the filter on top of every section, for instance Quatitative comparison

https://greprepclub.com/forum/quantitat ... stions-20/

use the filter on top, pretty intuitive