Bolts and nuts

Algebra Level 2

A container holds an equal number of nuts and bolts. The mass of the bolts is twice the mass of the nuts. When half the nuts are removed, the mass of the contents is 90 grams. What will be the mass of the contents (in grams) when half the bolts are then removed from the 90 grams content?

36 45 54 72

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1 solution

Let b b and n n be the mass of bolts and nuts, respectively, that is initially inside the container.

Initially, we have

b = 2 n b=2n

When half of the nuts are removed, we have

1 2 n + b = 90 \dfrac{1}{2}n+b=90

Substituting, we have

1 2 n + 2 n = 90 \dfrac{1}{2}n+2n=90

n = 36 n=36

It follows that b = 72 b=72 .

When half the bolts are then removed. We have

1 2 n + 1 2 b = 1 2 ( 36 ) + 1 2 ( 72 ) = 18 + 36 = \dfrac{1}{2}n+\dfrac{1}{2}b=\dfrac{1}{2}(36)+\dfrac{1}{2}(72)=18+36= 54 \boxed{54}

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This should be mentioned that 1/2 bolts are removed after removing 1/2 nuts plzz clarify

Saksham Jain - 3 years, 7 months ago

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