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Re: GMAT Club World Cup 2022 (DAY 1): On a 2022 World Cup De Bruyne, [#permalink]
2
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Answer C
R+D+Mb+Me=X
R=X-22
D=X-27
Mb=X-20
Me=X-18

We got 4X-87=X which is X = 87/3 = 29
And we can get Messi =29-18 = 11
Thus, the answer is C
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Re: GMAT Club World Cup 2022 (DAY 1): On a 2022 World Cup De Bruyne, [#permalink]
1
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Let the number of goals scored by De Bruyne = p
Let the number of goals scored by Ronaldo = q
Let the number of goals scored by Mbappe = r
Let the number of goals scored by Messi= s

Ques s ?

Given
q + r +s = 27
p + r +s = 22
p + q +s = 20
p + q + r = 18

Adding the above equations we get

3(p+q+r+s) = 87

p + q + r + s = \(\frac{87}{3}\) = 29

s = 29 - 18

s = 11

Option C
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Re: GMAT Club World Cup 2022 (DAY 1): On a 2022 World Cup De Bruyne, [#permalink]
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Re: GMAT Club World Cup 2022 (DAY 1): On a 2022 World Cup De Bruyne, [#permalink]
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