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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]

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

A. 15
B. 16.5
C. 17.75
D. 18
E. 19

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

A. 15
B. 16.5
C. 17.75
D. 18
E. 19


 

This question was provided by GMATWhiz
for the Heroes of Timers Competition

 

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
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