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# Each Machine of type A has 3 steel parts and 2 chrome parts.

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Each Machine of type A has 3 steel parts and 2 chrome parts.  [#permalink]

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14 Mar 2011, 14:31
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Question Stats:

77% (02:14) correct 23% (02:24) wrong based on 269 sessions

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Each Machine of type A has 3 steel parts and 2 chrome parts. Each machine of type B has 4 steel parts and 7 chrome parts. If a certain group of type A and type B machines has a total of 20 steel parts and 22 chrome parts, how many machines are in the group

A. 2
B. 3
C. 4
D. 6
E. 9
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14 Mar 2011, 14:57
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3
tonebeeze wrote:
Each Machine of type A has 3 steel parts and 2 chrome parts. Each machine of type B has 4 steel parts and 7 chrome parts. If a certain group of type A and type B machines has a total of 20 steel parts and 22 chrome parts, how many machines are in the group

a. 2
b. 3
c. 4
d. 6
e. 9

Given:
(i) $$3a+4b=20$$ and (ii) $$2a+7b=22$$, where $$a$$ and $$b$$ are # of type A and type B machines respectively.

Multiply (i) by 2 and (ii) by 3 and subtract: $$(6a+21b)-(6a+8b)=66-40$$ --> $$b=2$$ --> $$a=4$$ --> $$a+b=6$$.

Or you can notice that from (i) $$3a=20-4b$$: 20 minus multiple of 4 to be a positive multiple of 3 $$b$$ must be either 2 or 5 (which doesn't fit into the second equation) --> $$b=2$$ --> $$a=4$$ --> $$a+b=6$$.

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Joined: 11 Feb 2011
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14 Mar 2011, 21:20
Look at the below representation of the problem:
Steel Chrome total
A 3 2 20 >>no. of type A machines=20/5=4

B 4 7 22 >>no. of type B machines=22/11=2

So the answer is 6 i.e D.

Hope its clear .
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14 Mar 2011, 21:40
Let there be x # of machine A and y # of machine B

So we have :

3x + 4y = 20

4x + 7y = 22

=> y = 2 and x = (20 - 8)/3 = 4

So total = 6, answer is D.
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15 Mar 2011, 23:06
tonebeeze wrote:
Each Machine of type A has 3 steel parts and 2 chrome parts. Each machine of type B has 4 steel parts and 7 chrome parts. If a certain group of type A and type B machines has a total of 20 steel parts and 22 chrome parts, how many machines are in the group

a. 2
b. 3
c. 4
d. 6
e. 9

3a + 4b = 20 ......1

4a + 7b = 22.......2

solving (!) n (2)
b=2 a=4

thus total=6
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02 Jun 2012, 14:54
We have ratios, so we can use unknown multiplier

3x + 2x = 20 ---> x=2

4x + 7x = 22 ----> x= 4

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Re: Each Machine of type A has 3 steel parts and 2 chrome parts.  [#permalink]

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15 Dec 2018, 19:34
1
Hi All,

We're told that each Machine of type A has 3 steel parts and 2 chrome parts, each machine of type B has 4 steel parts and 7 chrome parts and a certain group of type A and type B machines has a total of 20 steel parts and 22 chrome parts. We're asked for the total number of machines. This question can be approached in a number of different ways; the easiest approach would likely be to focus on the 'multiples' involved and play around a little bit with the basic Arithmetic.

While it might seem a bit weird, we can actually ignore whether the parts are 'steel' or 'chrome.' Each Machine A has a total of 5 parts and each Machine B has a total of 11 parts. The 'group' has a total of 42 parts, so we need to add a multiple of 5 to a multiple of 11 and end up with 42. The number 42 is relatively small, so there's likely just one way to get to that total... You might recognize that 11(2) = 22... meaning that there would be 42 - 22 = 20 parts remaining. Since 20 is a multiple of 5, we know that there would be 4 type B machines to go along with the 2 Type A machines. That's the only way to get to 42 total parts, so 4+2 = 6 must be the answer.

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Re: Each Machine of type A has 3 steel parts and 2 chrome parts.  [#permalink]

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07 Jan 2019, 16:03
tonebeeze wrote:
Each Machine of type A has 3 steel parts and 2 chrome parts. Each machine of type B has 4 steel parts and 7 chrome parts. If a certain group of type A and type B machines has a total of 20 steel parts and 22 chrome parts, how many machines are in the group

A. 2
B. 3
C. 4
D. 6
E. 9

Let there be a type A machines and b type B machines:
Need: a+b

3a + 4b = 20
2a + 7b = 22

Solving those 2 equations: a=4, b=2
Therefore: 6
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Re: Each Machine of type A has 3 steel parts and 2 chrome parts.   [#permalink] 07 Jan 2019, 16:03
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