X is the set of positive integer multiples of 3, and Y is th

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[GMAT math practice question]

X is the set of positive integer multiples of 3, and Y is the set of positive integer multiples of 7. Define X+Y as {x+y|x ∈ X and y ∈ Y}. How many elements of X + Y are less than or equal to 21?

A. 4
B. 6
C. 8
D. 11
E. 13
Source: — Problem Solving |

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by Max@Math Revolution » Tue Aug 27, 2019 11:38 pm
=>

X = { 3, 6, 9, 12, 15, 18, 21, ... } and Y = { 7, 14, 21, ... }
The following pairs ( x, y ) satisfy x + y <= 21:
(3, 7), (3, 14), (6, 7), (6, 14), (9,7) and (12,7).
There are 6 such pairs, (x,y).

Therefore, B is the answer.
Answer: B

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by deloitte247 » Thu Aug 29, 2019 6:27 pm
Set X=> contains positive integers which are multiples of 3.
$$X=\left\{3,6,9,12,15,18,21,24,...,\ \inf inity\right\}$$
Set Y=> contains positive integers which are multiples of 7
$$Y=\left\{7,14,21,28,35,\ ...\ ,\ \inf inity\right\}$$
$$Define\ a\ new\ set\ of\ X+Y\ with\ constructor\ x+y\ where\ x\ is\ an\ element\ of\ set\ X\ and\ Y\ is\ an\ element\ of\ set\ Y\ which\ must\ be\ \le21.$$
$$X+Y=\left\{\left(7+3\right),\left(7+6\right),\left(7+9\right),\left(7+12\right),\left(14+6\right),\left(14+3\right)\right\}$$
$$X+Y=\left\{10,13,16,19,20,17\right\}$$
$$Total\ elements\ of\ X+Y=6\ elements\ \le21$$

Answer is option B.