A Fractional Pink Shade

What fraction of this figure is shaded with the pink color?

Get the answer here!


Author : Ed Southall of Solve My Maths

Area Of A Fractional Pink Shade

A look at the question first!

Let's first recall the formula for the calculation of area of a triangle.

Area of triangle =  1/2 x Base x Height

Let's assume the side of the square is 1.

Now the triangle with the pink shade & triangle opposite to it are similar triangles. Similar triangles are triangle whose sides are in proportion with each other.

Since here base of pink triangle is double of un shaded opposite triangle, the height of pink triangle must be double of that smaller triangle.

But together, heights of both triangles must be equal to side of the square i.e. 1.

And hence, height of smaller triangle must be 1/3 & that of pink 2/3. (h + 2h = 3 ; h = 1/3).

So,

Area of Pink Triangle = 1/2 x Base x Height = 1/2 x 1 x 2/3 = 1/3.

So the area of pink shaded part is 1/3rd of total area occupied by square.



 

The CryptArithmetic Problem

Can you solve the below alphametic riddle by replacing letters of words by a numbers so that the below equation holds true?

BASE +
BALL
---------
GAMES
----------

 
Find numbers replaced letters here! 

Source 

The CryptArithmetic Problem's Solution

What was the problem?

Let's first recall the given equation.

  BASE +
  BALL
---------
GAMES
----------

We are assuming repeating the numbers are not allowed. 

Let's first take last 2 digits operation into consideration i.e. SE + LL = ES or 1ES (carry in 2 digit operation can't exceed 1). For a moment, let's assume no carry generated.

10S + E + 10L + L = 10E + S .....(1)

9 (E - S) = 11L

To satisfy this equation L must be 9 and (E - S) must be equal to 11. But difference between 2 digits can't exceed 9. Hence, SE + LL must have generated carry.So rewriting (1),

10S + E + 10L + L = 100 + 10E + S

9 (E - S) + 100 = 11L

Now if [9 (E - S)] exceeds 99 then L must be greater than 9. But L must be digit from 0 to 9. Hence, [9 (E - S)] must be negative bringing down LHS below 100. Only value of E - S to satisfy the given condition is -5 with L = 5. Or we can say, S - E = 5.

Now, possible pairs for SE are (9,4), (8,3), (7,2), (6,1), (5,0). Out of these only (8,3) is pair that makes equation SE + LL = SE + 55 = 1ES i.e. 83 + 55 = 138. Hence, S = 8 and E = 3.

Replacing letters with numbers that we have got so far.

    1---------
  BA83 +
  BA55
---------
GAM38
----------

Now, M = 2A + 1. Hence, M must be odd number that could be any one among 1,7,9 (since 3 and 5 already used for E and L respectively).

If M = 1, then A = 0 and B must be 5. But L = 5 hence M can't be 1

If M = 7, then A = 3 or A = 8. If A = 3 then B = 1.5 and that's not valid digit. And if A = 8 then it generates carry 1 and 2B + 1 = 8 again leaves B = 3.5 - not a perfect digit.

If M = 9, then A = 4 (A = 9 not possible as M = 9) and B must be 7 with carry G = 1.Hence for first 2 digits we have 74 + 74 + 1 = 149.

Finally, rewriting the entire equation with numbers replacing digits as -

    1
---------
  7483 +
  7455
---------
14938
----------

So numbers for letters are S = 8, E = 3, L = 5, A = 4, B = 7, M = 9 and G = 1.