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

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

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

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

thus total=6

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.

Final Answer:

GMAT assassins aren't born, they're made,
Rich
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