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: \((6a+21b)-(6a+8b)=66-40\) --> \(b=2\) --> \(a=4\) --> \(a+b=6\).

Answer: D.

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) --> \(b=2\) --> \(a=4\) --> \(a+b=6\).

Answer: D.
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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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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
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