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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.

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Source: GMAT Paper Tests

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

The OA is D.

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BTGmoderatorLU wrote:
Source: GMAT Paper Tests

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
(3 steel parts in each type A machine) + (4 steel parts in each type B machine) = 20 steel parts:
3A + 4B = 20

3A = a multiple of 3 less than 20.
Options for 3A:
3, 6, 9, 12, 15, 18

4B = a multiple of 4 less than 20.
Options for 4B:
4, 8, 12, 16

Only the two values in blue yield a sum of 20.
Thus:
3A = 12, implying that A=4
4B = 8, implying that B=2
A+B = 4+2 = 6

The correct answer is D.

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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: D

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

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Contact Rich at Rich.C@empowergmat.com

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BTGmoderatorLU wrote:
Source: GMAT Paper Tests

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

The OA is D.
We can let a = the number of type A machines and b = the number of type B machines; thus:

3a + 4b = 20 (This is the equation for the steel parts.)

and

2a + 7b = 22 (This is the equation for the chrome parts.)

Multiplying the first equation by -2 and the second by 3, we have:

-6a - 8b = -40

and

6a + 21b = 66

Adding the equations together, we are left with:

13b = 26

b = 2, so a is:

3a + 4 x 2 = 20

3a = 12

a = 4

So there are 6 machines.

Answer: D

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Scott Woodbury-Stewart
Founder and CEO
scott@targettestprep.com



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