Albert had 152 more marbles than Bernard at first. After Albert gave away 3/4 of his marbles and Bernard gave away 2/3 of his marbles, Bernard had 9 more marbles than Albert. How many marbles did Albert have left?
+313 Then the number of marbles Albert has is given by;
nA = x - 152
Now when Bernard give away 2/3rd of his marble, he will be left with ;
(1 - 2/3)x = x/3
-marbles.
And when Alfred give away 3/4th of his marbles, he will be left with ;
(1 - 3/4)(x + 152) = \({ 1\over 4}{}{}\) (x + 152)
-marbles.
Now it is given that at this stage Bernard will have 9 more marbles than Alfred. i e ;
1/3x = 1/4 (x + 152)+9
Now solving the above equation for x as follows;
1/3x=1/4(x + 152)+9
\({x\over 3}\frac{}{}\)\(= {(x +152) + 9⋅4 \over 4}\frac{}{}\)
=\( {x +188 \over 4}{}{}\)
x ⋅ 4 = 3 (x + 188)
= 3x + 3 ⋅ 188
4x=3x + 564
4x - 3x = 564
x = 564
Therefore initial number of marbles Bernard had was 564 marbles. Therefore number of marbles Alfred had is given by;
x + 152=564 + 152
=716
Therefore Alfred had 716 marbles. And he gave away 3/4th of his marbles. ie;
\({3 \over 4}{}{}\)⋅ 716 = 537
Therefore Alfred gave away 537 marbles
And the number of marbles left with him is given by;
716−537=179
Therefore Alfred is left with 179 marbles.
