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Quick way to solve this -

As price of 1 Large can equals that of 5 medium cans, number of large cans cannot go beyond 40 for 200 medium cans.

Only option A satisfies this.
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I applied "Try Options" approach and fortunately the very first option appeared correct.

35 small cans=25 medium
Remaining 200-25=175 medium cans
Now 175 medium cans=35 large cans
which satisfies the question that equal no. of large and small cans were bought by the customer.
Hence A)
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I applied "Try Options" approach and fortunately the very first option appeared correct.

35 small cans=25 medium
Remaining 200-25=175 medium cans
Now 175 medium cans=35 large cans
which satisfies the question that equal no. of large and small cans were bought by the customer.
Hence A)
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Bunuel
A merchant sells three different sizes of canned tomatoes. A large can costs as much as 5 medium cans or 7 small cans. If a customer buys an equal number of small and large cans of tomatoes for the exact amount of money that would buy 200 medium cans, how many small cans will she buy?

A. 35
B. 45
C. 72
D. 199
E. 208


The max no of medium cans she can buy is 200/5= 40 so small cans has to be < medium as given in relation only option which goes well is A 35..
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eswarchethu135
L = 5M , L = 7S

5M = 7S => S = \(\frac{5M}{7}\)

200M = x(S) + y(L)

200M = x(\(\frac{5M}{7}\)) + y(5M)

200 = x(\(\frac{5}{7}\)) + y(5)

40 = \(\frac{x}{7}\) + y

X has to be an integer which is divisible by 7. Out of options, only A satisfies it.

OPTION: A

x=y is already given.
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I solved by defining the prices of each respective can.

Let Large Can = $35 (choose smart number, 7*5=35)

L=7s hence $35=7s
Small = $5

Medium = $7

200 medium cans @ $7 = $1400

Since number of small cans = number of large cans, lets substitute values from options.

Start with A.

S=35

Cost of 35 small cans = 35*5
= $175

Cost of 35 large cans = 35*35
= $1225

Total = 1225+175
= $1400

35 large + 35 small = 200 medium cans
A is correct.
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