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In assembling a Bluetooth device, a factory uses one of two kinds of [#permalink]

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21 Feb 2014, 18:34

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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?

Re: In assembling a Bluetooth device, a factory uses one of two kinds of [#permalink]

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21 Feb 2014, 23:21

anujtsingh wrote:

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

It is easy go : consider x is count of costlier module. y is count of cheaper module.

6.25*x + 2.5*y = 70; x+y = 22;

by solving these two equation we get y = 18
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Re: In assembling a Bluetooth device, a factory uses one of two kinds of [#permalink]

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01 Nov 2015, 12:13

i believe it is a 700 lvl question just because it involves decimals. a+b=22 6.25a + 2.5b = 70 / multiply by 4 find b: 25a+10b = 280 a=22-b 25(22-b)+10b = 280 550-25b+10b = 280 550-280 = 15b 270 = 15b b = 270/15 b=18.

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

We can also solve the question using ONE VARIABLE.

Let x = the number of cheap modules So, 22 - x = the number of expensive modules

So, the VALUE of the cheap modules = 2.5x So, the VALUE of the cheap expensive = 6.25(22 - x)

The TOTAL value = 70 dollars, so we can write: 2.5x + 6.25(22 - x) = 70 Expand to get: 2.5x + 137.5 - 6.25x = 70 Simplify: -3.75x + 137.5 = 70 Rearrange to get: -3.75x = -67.5 Solve: x = 18

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

Shortlist the options by using approximation and then logic will get you to the answer.

Cost of one module is 6.25 and of the other is 2.50. The average cost of the total modules = 70/20 = approx 3.5. It is a bit less than 3.5 because actually the number of modules is 22, not 20. So the average is way closer to 2.50 and hence out of 22, many more modules are the cheaper variety. So only possible options are 16 and 18. Now note that with 16 modules of $2.50, you will get the total cost as 40 but if remaining 6 modules are of cost $6.25 each, you will not get a whole number for total cost. But the total cost is $70 - a whole number. So the number of $2.50 modules must be 18 so that the leftover 4 modules are of $6.25 which will give a whole number.

Could the values be multiplied by 10 to avoid working with decimals?

Thanks

Yes - 625, 250 and 7000 if you want to avoid decimals. It does make the numbers much larger and that I don't like but it is up to you.
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