Question

three bags of sweets weigh 27/4 kg. two of them have the same weight and the third bag is heavier than each of the bags of equal weight by 6/5 kg. find the weight of each bag. i do not get it please explain

Answer 1

I'm here buddy,

so, let's take the value of the two bags with equal weight as x.

=     x + x + (x + $\frac{6}{5}$) = $\frac{27}{4}$

=     3x + $\frac{6}{5}$ = $\frac{27}{4}$

=     3x = $\frac{27}{4}$ - $\frac{6}{5}$

( let's take the LCM of 4 and 5 = 20

=     3x = $\frac{135}{20}$ - $\frac{24}{20}$

=     3x = $\frac{111}{20}$

=       x = $\frac{111}{20}$ ÷ $\frac{3}{1}$ = $\frac{111}{20}$ × $\frac{1}{3}$ = $\frac{37}{20}$

So, the weight of the equal bags are $\frac{37}{20}$ and the weight of the third bag ( heavy one ) is $\frac{37}{20}$ + $\frac{6}{5}$ = $\frac{37}{20}$ + $\frac{24}{20}$ = $\frac{61}{20}$

1st bag =     $\frac{37}{20}$ kg

2nd bag =  $\frac{37}{20}$ kg

3rd bag =   $\frac{61}{20}$ kg

Hope it helps...