Answer: Option B

Please check the video for the step-by-step solution.

GMATinsight's Solution





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Solution:
Selling Price of the mixture is,
=> x * 0.85 = 17
=> x = $20

Given, price of B is $22,

Let, A be 4y and B be 7y in qunatity, then,

=> 4 * P + 7 * 22 = 20 (7+4)
=> 4P = 20 * 11 - 154
=> P = 66/4
=> P = 16.5

16.5 is the answer, which is option B

17 per kg incurred a loss of 15% so actual cost price is 20 per kg for mixture. Using allegations
4:7==2:(20-x)
40-2x=7
X=33/2=16.5

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CP of A = A
CP of B = 22/Kg
SP of mixture = 17/kg
Loss % = 15%

=> (CP - SP)/CP = 0.15
=> SP = 0.85CP
=> CP of mixture = 17/0.85 = 20/Kg

Now,
Ratio of mixtures = 4:7

By concept of Weighted Average

Average CP of mixture = [4(A) + 7(22)] / (4+7)

=> 20*11 = 4A + 7*22
=> 4A = 220 - 154
=> A = 16.5

Answer:B [\b]

[b]Posted from my mobile device
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Since A &B are mixed in the ratio 4:7, let's assume the weight of A as 4m and the weight of B as 7m.

Since the seller incurs a loss of 15% by selling the mixture at $17 i.e. 85% of the cost price he paid for the mixture.

So 85%($20) is $17.

Therefore original price of mixture is $20.

Since this price is the average price of the mixture i.e. (4m *n + 22 * 7m) / (4m + 7m) = 20 ; Assuming n as the CP per kg of A

=> 4m + 7 * 22 = 20 * 11

=> 4m = 66

=> m = 16.5

Therefore B

The ratio and mixture question.

A = the cost price per kg of A

A mixture of A and B is 4:7

The equation is 0.85*( 4*A + 7*22 ) = (7+4)*17

Find A:

4*A +154 = 220
4A = 66
A = 16.5

So I choose B

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IMO B

Given Data:
(i) Ratio of A & B in Mixture= 4:7
(ii) CP of B = $22/Kg
(iii) SP of Mixture = $17/kg & Loss % = 15%

Solution:
For Mixture, Profit/Loss % = (SP-CP)/CP x 100
=> -15 = (SP/CP-1) x 100
=> SP/CP= 0.85 => CP = SP/0.85 = 17/0.85 = 20 -----------(I)

Now, Let Cost Price of A = $x /Kg

Therefore, CP of Mixture =$ (4x + 7x22)/(4+7) per Kg -----------(II)

Equating (I) & (II),
(4x + 7x22)/(4+7) = 20
=> 4x +154 =220
=> x = 66/4 = 16.5
Cost Price of A = $ 16.5 /Kg

B. 16.5

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IMO ans is B i.e 16.5



Data from Question:-

1) Ratio of rice A & B in Mixture = 4:7

2) Selling price of rice Mixture = $17/Kg at loss of 15%

2) Cost Price of rice B = $22/Kg



CP= Cost Price, SP = Selling Price



Ans:-



Loss % = (Loss *100 )/CP = (CP-SP)/CP *100



=> From Input above find CP of Rice Mixture A & B

=> 15 = (CP-17)/CP *100 => 3CP=20CP-340

=> CP of Rice mixture A & B = $20/Kg



Let’s CP of A be $x/kg



So for rice mixture A & B



=> 4/11x+7/11*22 = 20

=>4/11x=6

=>x=66/4

=> x= $16.5/kg



Cost Price of Rice A is $16.5/kg

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The answer is A.

This question took 2 hours to solve..
I was having a hard time to get the meaning of “a loss of 15%”.

I think that this question is a wake up call for me. Because I’ve always solved the profit/loss question using the formula «\(x*(1± \frac{r}{100})\)» without proper understanding of the formula, this 2-hours-of-digging-a-hole situation happened ?.

So this is what I learnt from this question.

“I bought a lemon at $10 and want a profit of 50% from reselling the lemon. At what price I should sell the lemon?”

: She should sell it at $15. Because the profit she want is 50% from what she spent on the lemon. 50% of $10 is $5, using the formula \(10*\frac{50}{100}\)
Thus, To calculate selling price, we use\(10\)(the cost)\(+\)\((10*\frac{50}{100})\)(the profit), which is exactly same as \(10*(1+\frac{50}{100})\)

“I bought a lemon at $10 and incurred a loss of 40% because of miscalculation of selling price. Guess how much I sold the lemon”
: She sold the lemon at $6. Because she had \(10*\frac{40}{100}=$4\) loss. So she earned $0 and left with $-4 account.
To calculate selling price, \(10\)(the cost)\(+\)\((10*\frac{-40}{100})\)(the loss), which is exactly same as \(10*(1-\frac{40}{100})\).

Back to the question,

Rice of two different varieties A and B are mixed together in the ratio of 4:7. Upon selling the mixture at the price of $17 per kg, the seller incurred a loss of 15%. Cost price of B is $22 per kg. What is the cost price per kg of A?

-> since we want to know cost price per kg of A, let’s say that the total kg of the mixture is 1kg, and the cost price per kg of A is \(x\)
In the mixture, there is \(\frac{4}{11}kg\) of A, \(\frac{7}{11}kg\) of B.
Cost price of each is \($\frac{4x}{11}\) and \($\frac{7*22}{11}=\frac{154}{11}\). Thus, total cost price is \(\frac{4x+154}{11}\).
Sale price is \($17\).
$ loss amount is 15% of the cost; \(\frac{4x+154}{11}*\frac{15}{100}=\frac{12x+462}{220}\).
Therefore, \(cost-loss=sale price, \frac{4x+154}{11}-\frac{12x+462}{220}=17\).

When you solve the equation above, you’ll get \(x=15\)

Thus, the cost price per kg of A is $15.


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let 11 be the total amount of rice solved and cost of the individual rice be A and B
Then to make a 15 percent loss the eqn becomes

=>4A + 7B -11(17) / 4A +7B = 15/100
and B= 22 given

=>4A+ 11*14 -11(17) / 4A +11*14 = 15/100

Upon simplification we get 33/2 =16.5
THerefore IMO B