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2 Answers

If there are K kids, they each eat 6K strawberries. There are 7-K adults, who each eat 10(7-K) strawberries.

Together they eat 68 strawberries, so 6K+10(7-K)=68, 6K+70-10K=68, 2=4K, so K=½!

So there’s something wrong with the question!

My guess is that they ate 58 strawberries, so 6K+10(7-K)=58, 6K+70-10K=58, 12=4K, K=3. There would be 3 kids.

ago by Top Rated User (775k points)
Let number of adults be x and number of kids be y.

Given,

x + y =7 => x = 7-y  --------------------(1)

and,

10x + 6y = 68 -----------------(2)

Substituting the value of eq(1) in eq(2) we get,

10(7-y) + 6y =68

=> 70 - 10y + 6y = 68

=> 70 -4y = 68

=> 4y = 2

=> y=1/2

The number of children cannot be a fraction. Please recheck your question.
ago by Level 4 User (5.7k points)

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