Robins eat 12 worms in the morning on an average sunny day. On rainy days, robins average 4 fewer worms per morning. Last week, there were 3 sunny days and 4 rainy days. How many worms did a robin eat that week?
We shall split this into two parts, one is solving the number of worms a robin eats on a sunny day and one is is solving the number of worms a robin eats on a rainy day.
Part 1 - Sunny Day
12 x 3 = 36
Part 2 - Rainy Day
(12 - 4) x 4
= 8 x 4
= 32
Then, we add 36 + 32 = 68 worms.
Hope this helps!
(3 x 12) + (4 x 8) = 68 worms
That's correct! Another way to approach the problem is by using the distributive property of multiplication:
12 worms/day x (3 sunny days + 4 rainy days) = 12 x 7 = 84 worms
Then, we subtract the 4 fewer worms on rainy days:
84 - 4 x 4 = 68 worms
So, the robin ate 68 worms that week.
To find out how many worms a robin ate last week, we need to calculate the total number of worms eaten on sunny days and the total number of worms eaten on rainy days.
On sunny days, a robin eats 12 worms in the morning. There were 3 sunny days last week, so the total number of worms eaten on sunny days is 12 worms/day * 3 days = 36 worms.
On rainy days, a robin eats 4 fewer worms than on sunny days. So, on rainy days, they eat 12 worms/day - 4 worms/day = 8 worms/day. There were 4 rainy days last week, so the total number of worms eaten on rainy days is 8 worms/day * 4 days = 32 worms.
To find the total number of worms eaten last week, we add the number of worms eaten on sunny days and the number of worms eaten on rainy days: 36 worms + 32 worms = 68 worms.
Therefore, a robin ate a total of 68 worms last week.