Question #dfbed

Question

2846 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-24T02:28:33

So, the dog will eat $(7 + \frac{7 d}{3})$ $(7+(7d)/3)$ cans in $(3 + d)$ $(3+d)$ days.

Let $x$ $x$ be the number of cans that can be eaten by it in $(3 + d)$ $(3+d)$ days.

Here the ratio number of cans $= 7 : x$ $= 7:x$

The ratio of number of days $= 3 : (3 + d)$ $= 3:(3+d)$

Here the number of cans and days are in direct proportion.

$:. 7:x=3:(3+d)$

or, $7/x=3/(3+d)$

or, $3x=7(3+d) rarr$ Cross multiply.

or, $3x=21+7d$

or, $(3x)/3=(21+7d)/3 rarr$ Dividng both sides by 3.

or, $x=21/3+(7d)/3 rarr$ Split RHS

or, $x=7+(7d)/3$

So, the dog will eat $(7+(7d)/3)$ cans in $(3+d)$ days.