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Sub 505 (Easy)|   Fractions and Ratios|   Word Problems|                  
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OA - C

St.1 is insufficient - Blue are 1/3 and average of blue is 21000. Hence, blue is one out of three, but we do not know the ratio of white.

St.2 is insufficient - White are 2/3 and average of white is 24000. Hence, white is two out of three, but we do not know the ratio of blue.

St.1 and 2 combined - blue is 1 and white are two. So, average of 3 is 21000+24000+24000 = 69000/3 = 23000, Hence sufficient.
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But we still dont know what the total number of cars was. Shouldnt we have taken (1/3)*x and (2/3)*x
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vrinda6
But we still dont know what the total number of cars was. Shouldnt we have taken (1/3)*x and (2/3)*x

No. We do not need the total number of cars, because the fractions 1/3 and 2/3 already give the weights of the two groups. Using x would give the same result anyway:

Total value = x/3*21000 + 2x/3*24000
Average = (x/3*21000 + 2x/3*24000)/x = 23000

So x cancels out.
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What we know:
- Cars are blue (B) or white (W)
- Make X the price of B and Y the price of white
- We are looking to solve for x*B + y*W

1) 1/3=B, thus 2/3=W
1/3*$21,000=$7,000 --> B

$21k + 2/3 * Y=
Insufficient as we still don't know Y

2) 2/3=W, thus 1/3=B
2/3*$24,000=$16,000
1/3*X+$16k=

Insufficient as we still don't know X

Combined:

We know we can solve so we can stop here. See below for completeness:

Blue = $7k
White = $16k
Total = $23k
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