John purchased large bottles of water for $2 each and small

Hi Bunuel,

There is small error in one of the calculations.
\(x=15\) and \(y=3\)
should be
\(x=15\) and \(y=2\)

Kindly correct me if i am wrong.

Typo edited. Thank you. +1.

Let x be the number of large bottles of water and y be the number of small bottles of water, from the question stem we get:
2*x+1.5*y=33, thus 1 is INSUFFICIENT.
Mowing to 2 condition:
(2*x+1.5*y)/(x+y) = 1.65 ------>>>> 2x+1.5y = 1.65x+1.65y, from here we easily get that 7x=3y, OR x = 3y/7 now we now x we can easily find the ratio of y in total of bottles: y/(y+3y/7) = 7/10 or 70%

Please correct me, if I went awry. Actually we do not need the solution, as it is data sufficiency so 2 is SUFFICIENT

Dealing with Statement (2): remember weighted average (also used when dealing with mixture problems).

If we have \(N_1\) numbers with average \(A_1\), and \(N_2\) numbers with average \(A_2\), the final average being A, then the differences between the final average and the initial averages are inversely proportional to the two numbers of numbers (assume \(A_1>A_2\)):

\((A_1-A)N_1=(A-A_2)N_2\) or \(\frac{A_1-A}{A-A_2}=\frac{N_2}{N_1}\).

This follows from the equality \(\frac{N_1A_1+N_2A_2}{N_1+N_2}=A.\)

In our case we know \(A, A_1,A_2\) so we can find the ratio \(\frac{N_2}{N_1}\) and then, obviously \(\frac{N_2}{N_1+N_2}\).

1. small bottles can be 2, and 15 large or 6 small and 12 large. 1 insufficient.
2. 2L+3S/L+S = 1.65
we are given proportions. thus, we can solve the question.
2L+1.5S=1.65L + 1.65S
0.35L=0.15S.
multiply by 100
35L=15S.
S/L = 35/15

S/S+L = 35/15+35 = 35/50 or 7/10. 70%

(1)Not suff.
(2) I will use weighted avg. method here
W1/W2=(A2-Avg.)/(Avg.-A2)
W1=No. of large bottles
W2=No. of large bottles
A1=2
A2=1.5
Avg.=1.65
W1/W2=3/7
%age of small bottles= 7/(3+7)------>7/10----70%
Ans B

We are given that John purchased large bottles of water for $2 each and small bottles of water for $1.50 each. We can let s = the number of small bottles purchased, and b = the number of larger bottles purchased. We need to determine what percentage of the bottles purchased were small bottles, i.e. the value of s/(s+b) x 100.

Statement One Alone:

John spent $33 on the bottles of water.

Using the information in statement one, we can create the following equation:

2b + 1.5s = 33

We can multiply the entire equation by 2 and we have:

4b + 3s = 66

4b = 66 - 3s

4b = 3(22 - s)

b = [3(22 - s)]/4

Since b must be an integer, 3(22 - s) must be a multiple of 4.

3(22 - s) is a multiple of 4 when s = 2, 6, 10, 14, or 18.

Since we have multiple values of s, we will also have multiple values of b, and thus we do not have enough information to answer the question.

Statement Two Alone:

The average price of bottles purchased was $1.65.

Using the information in statement two, we can create the following equation:

1.65 = (2b + 1.5s)/(b + s)

1.65(b + s) = 2b + 1.5s

165(b + s) = 200b + 150s

165b + 165s = 200b + 150s

15s = 35b

3s = 7b

(3/7)s = b

We can now determine the value of s/(s+b) x 100 by substituting (3/7)s for b:

s/[s+(3/7)s] x 100

s/[(10/7)s] x 100

1/(10/7) x 100

7/10 x 100 = 70

So the small bottles account for 70 percent of the bottles purchased. We have answered the question.

Answer: B

hey Bunuel VeritasKarishma could you please assist me with st 1.
when i take it as follows:

2x+1.5y=33
20x+15y=330

for this equation, only two valid solutions are there (as per the exchange rate method)
x=0 and y=22
now adding 15 and subtracting 20 to get another sol:
x=15 and y=2
i cant subtract or add further.

no other sol is valid as per me, hence i had marked A as sufficient. but when i convert the same into 4x+3y=66 then ofc many sols are valid. could pls help me with what I am doing wrong?

The equation needs to be in lowest terms to get all the possible solutions. When it is a multiple, you lose out on solutions.
Note what happens when you make it 40x + 15y = 660 and try to solve for x and y.

of course! you cover this in the blog as well. my mistake. thanks a lot for the clarification.

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