Bunuel wrote:

A black and white drawing shows four circles of equal radius. Three of the circles are each divided into k equal parts, where k > 5. The fourth circle is divided into k - 4 equal parts. A child colors one of the k parts in each of the three circles, and one of the k - 4 parts in the fourth circle. What part of a whole circle did the child color?

(A) \(\frac{4k - 12}{k(k - 4)}\)

(B) \(\frac{4k - 8}{k(k - 4)}\)

(C) \(\frac{4k - 4}{k(k - 4)}\)

(D) \(\frac{2k - 4}{k(k - 4)}\)

(E) \(\frac{k + 4}{k(k - 4)}\)

+1 kudos to

pushpitkc , whose answer I did not see

while I was formalizing what I drew for this problem.

Attachment:

circparts.png [ 19.23 KiB | Viewed 429 times ]
Use k = 6, and sketch (it's fast - a little over 1 min to solve)

(6-4) leaves more than one part for the last circle.

Draw four circles.

Three circles are divided into k = 6 parts

Divide three circles into sixths

Shade one part of each. Shaded part =

\(\frac{1}{6}\) of a whole circle

The fourth circle is divided into (k-4) = (6-4) = 2 parts

Divide the fourth circle in half. Shade one half.

The one shaded part =

\(\frac{1}{2}=\frac{3}{6}\) of a whole circle

The

total of the shaded areas is

• 3 circles:

\((\frac{1}{6} + \frac{1}{6} + \frac{1}{6}) = \frac{3}{6}\)• Fourth circle:

\(\frac{1}{2} = \frac{3}{6}\)• Combined:

\((\frac{3}{6} + \frac{3}{6}) = \frac{6}{6}= 1\) whole circle

Using k = 6, find the answer that yields 1(A) \(\frac{4k - 12}{k(k - 4)}\)

\(=\frac{(24-12)}{6(2)}=\frac{12}{12}=1\) CORRECT(B) \(\frac{4k - 8}{k(k - 4)}\)

=\(\frac{(24 - 8)}{6(2)} = \frac{16}{12}\) REJECT(C) \(\frac{4k - 4}{k(k - 4)}\)

=\(\frac{(24 - 4)}{6(2)} = \frac{20}{12}\) REJECT(D) \(\frac{2k - 4}{k(k - 4)}\)

=\(\frac{(12 - 4)}{6(2)} = \frac{8}{12}\) REJECT(E) \(\frac{k + 4}{k(k - 4)}\)

=\(\frac{(6 + 4)}{6(2)} = \frac{10}{12}\) REJECTAnswer A

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