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?

Guest Aug 22, 2021

#1**+2 **

Then the number of marbles Albert has is given by;

n_{A} = x - 152

Now when Bernard give away 2/3^{rd} of his marble, he will be left with ;

(1 - 2/3)x = x/3

-marbles.

And when Alfred give away 3/4^{th }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 3/4^{th} 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.

apsiganocj Aug 22, 2021