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# In a certain box of of widgets there are only brass widgets and iron

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Math Expert
Joined: 02 Sep 2009
Posts: 55150
In a certain box of of widgets there are only brass widgets and iron  [#permalink]

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10 Sep 2017, 05:49
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Difficulty:

15% (low)

Question Stats:

89% (01:42) correct 11% (03:07) wrong based on 44 sessions

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In a certain box of of widgets there are only brass widgets and iron widgets. If 42 of the widgets are brass and 14 are iron, what is the total weight, in kilograms of the box?

(1) The weight of seven brass widgets is three times the weight of an iron widget.
(2) The total weight of a brass widget and five iron widgets is 38 kilograms.

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Re: In a certain box of of widgets there are only brass widgets and iron  [#permalink]

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10 Sep 2017, 07:14
Lets say that
b - weight of the brass widget
i - weight of iron widget

The question is 42 * b + 14 * i = ?

Let's express given statements with formulas:
1) 7 * b = 3 * i
2) b + 5 *i = 38
In both we can express one variable through another and substitute in the equation, but as the remaining variable is unknown we can't calculate the total weight.
Both are NS

Taking statements together, we can solve.
Multiply second statement by 7
7 * b + 35 * i = 7 * 38
Substitute by the expression from the first statement (7 * b = 3 * i)
3 * i + 35 * i = 7 * 38
38 * i = 7 * 38 => i = 7
b = 3 * i / 7 = > b = 3
Total weight will be 42 * 3 + 14 * 7 = 224

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Re: In a certain box of of widgets there are only brass widgets and iron  [#permalink]

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10 Sep 2017, 11:07
Zarevl wrote:
Lets say that
b - weight of the brass widget
i - weight of iron widget

The question is 42 * b + 14 * i = ?

Let's express given statements with formulas:
1) 7 * b = 3 * i
2) b + 5 *i = 38
In both we can express one variable through another and substitute in the equation, but as the remaining variable is unknown we can't calculate the total weight.
Both are NS

Taking statements together, we can solve.
Multiply second statement by 7
7 * b + 35 * i = 7 * 38
Substitute by the expression from the first statement (7 * b = 3 * i)
3 * i + 35 * i = 7 * 38
38 * i = 7 * 38 => i = 7
b = 3 * i / 7 = > b = 3
Total weight will be 42 * 3 + 14 * 7 = 224

Buddy, in statement 2, we indeed have enough information to get the value of iron and brass widget..
$$B+5I=38$$ --> $$B=38-5I$$, we can substitute in $$42B+14I$$ equation.

Am I correct?
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Re: In a certain box of of widgets there are only brass widgets and iron  [#permalink]

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11 Sep 2017, 00:29
Bunuel wrote:
In a certain box of of widgets there are only brass widgets and iron widgets. If 42 of the widgets are brass and 14 are iron, what is the total weight, in kilograms of the box?

(1) The weight of seven brass widgets is three times the weight of an iron widget.
(2) The total weight of a brass widget and five iron widgets is 38 kilograms.

S1:
Let I be weight of 1 Iron Widget.
Weight of 7 Brass = 3I
Weight of 42 Brass Widgets = 18I.
So total weight = 18I+14I. we dont what W is in kgs. So insufficient.

S2:
B+5I= 38
B= 38-5I;. Still not getting values of total widgets asked in the question. Insufficinet

Combining
42(38-5I)+14I=32I;
42(38-5I)=18I
7*(38-5I)=3I
7*38= 38 I;I=7;
Total Weight = 7*32= 224kgs;

ANs:C
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In a certain box of of widgets there are only brass widgets and iron  [#permalink]

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11 Sep 2017, 05:19
Buddy, in statement 2, we indeed have enough information to get the value of iron and brass widget..
$$B+5I=38$$ --> $$B=38-5I$$, we can substitute in $$42B+14I$$ equation.

Am I correct?[/quote]

Sorry, no
You expressed B in terms of I, but we still don't know the value of I

42 * B + 14 * I = ?
42 * (38 - 5 * I) + 14 * I = ?
42 * 38 - 42 * 5 * I + 14 * I
42*32 - 196 * I = ?
There are several possible values for I, so that this equation will give integer result.
In a certain box of of widgets there are only brass widgets and iron   [#permalink] 11 Sep 2017, 05:19
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