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

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.

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

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

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

thus total=6

We have ratios, so we can use unknown multiplier

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

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

4+2=6 .........40 sec about

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:

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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

Alternate method:

We can form equations based on each component.

Steel:
Let x and y equal the number of machines of A and B respectively.

3x + 4y = 20

Test y=1 machine
3x+4(1)=20
3x = 16

this won't work since x is at least equal to 1 and is an integer (we are dealing with machines here)
Keep testing
3x + 4(2) = 20 --> 3x=12
3x + 4(3) = 20 --> 3x= 8
3x + 4(4) = 20 --> 3x= 4

The only mix that satisfies this equation is when y = 2 and x=4

We can test in the second equation but its unnecessary.

This seemed to me like an allegation/mixture problem but couldn't solve that way. Saw the solutions n realised that its a number property one. How do i differentiate between the two? Please help.

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EducationAisle Bunuel Please could you help me understand why the following method doesn't work?

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

adding the two we get

5a + 11b = 44

a = \(\frac{44-11b}{5}\)

Now "a" and "b" have to be integers using this restriction why can't we get the values of "a" and "b" from the above equation?

It should be: 5a + 11b = 42

Oops

Thanks for pointing that one out

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