The Enigma of the Erasers (Homeschool Math Challenging Puzzles)

Dan has 745 red, green, blue and yellow erasers.

If the number of red erasers is tripled, the number of green erasers is halved, the number of blue erasers is decreased by 39, and the number of yellow erasers is increased by 48, there will be an equal number of erasers for each of the colors.

How many erasers are there for each of the colors?


This is part of a series on Homeschool Math Challenging Puzzles, suitable for Grades 2-4. (Of course, students of other grades are also welcome to try them out.) The questions are suitable for:

  • Homeschooling for gifted kids
  • Preparation for GEP (Gifted Education Programme) screening and selection tests
  • Preparation for Math Olympiad
  • Puzzles for kids interested in math but find school work too easy.

Solution

Let the number of red erasers be 1 unit:

R -> 1u

3R -> 3u

This means that the number of green erasers must be 3u x 2 = 6u.

G -> 6u

(This is so that after the green erasers are halved, it is also 3u.)

B -> 3u+39

Y -> 3u-48

Hence,

R+G+B+Y = 13u – 9 = 745

13u = 754

1u = 754/13 = 58.

Hence, the number of erasers for each color are:

R = 1u = 58

G = 6u = 348

B = 3u+39 = 213

Y = 3u-48 = 126

Author: mathtuition88

http://mathtuition88.com

One thought on “The Enigma of the Erasers (Homeschool Math Challenging Puzzles)”

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