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Bunuel
In assembling a Bluetooth device, a factory uses one of two kinds of modules. One module costs $6.25 and the other one, that is cheaper, costs $2.50. The factory holds a $70 worth stock of 22 modules. How many of the modules in the stock are of the cheaper kind?

(A) 4
(B) 8
(C) 12
(D) 16
(E) 18
 
Let the number of costly and cheaper modules be x and y respectively
Given:
6.25x + 2.50y = 70
Also given:
x + y = 22
6.25x + 2.50(22 -x) = 70
6.25x + 55 - 2.50x = 70
3.75x = 15
x = 15/3.75 = 4
No of modules of cheaper kind = 4 + y = 22
y = 18. IMO E­
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6.25x + 2.5y = 70
and
x + y = 22

find y

y is 18

Option (E) is correct
Bunuel
In assembling a Bluetooth device, a factory uses one of two kinds of modules. One module costs $6.25 and the other one, that is cheaper, costs $2.50. The factory holds a $70 worth stock of 22 modules. How many of the modules in the stock are of the cheaper kind?

(A) 4
(B) 8
(C) 12
(D) 16
(E) 18
 
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x + y = 22
2.5 x + 6.25 y = 70
Solve these two equations, get x = 18 (E)
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You can also use weighted average principles. The average cost per module is $70/22, or just over $3 per module. That's way closer to $2.50 than to $6.25, so we must have many more of the cheap module, and only D or E are even remotely plausible. It's now easy to either test one answer choice to decide between those two choices, or to use weighted average principles to decide: an average of exactly $3.25 would divide the distance from 2.50 to 6.25 in a 1 to 4 ratio, which would mean exactly 4/5 of the modules were the cheap kind. But our actual average is even closer to $2.50 (since 70/22 ~ 3.18 < 3.25), so more than 4/5 must be cheap, and the only possible answer is 18.
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Bunuel
In assembling a Bluetooth device, a factory uses one of two kinds of modules. One module costs $6.25 and the other one, that is cheaper, costs $2.50. The factory holds a $70 worth stock of 22 modules. How many of the modules in the stock are of the cheaper kind?

(A) 4
(B) 8
(C) 12
(D) 16
(E) 18

I agree with all the above-mentioned solutions, But I feel this question doesn't require a pen and paper to solve it.

Here is my approach :

Modlue X : $6.25 , Module Y : $2.50, We need to know how many modules of Y are used

Step 1 : What is the average price of module ?? $70/22 = ~$3...
That means there are more n.o of Ys than Xs in the group. (Avg and ratio concept)

So Y should be greater than 11, C,D,E are potential choices. I think Y should be nearly double that of X.

Just go by options: So start from E (decreasing order for the above-mentioned reason)

18*2.5= 36+9 =45;
4*6.25 = 24+1 =25

Bang !! We got our answer.

Although this seems a bit tedious, With practice we can do mental calculations and get the anser. It's always better to do mental calculations.

IMO E
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