Question

Ivan caught a total of 7 2/5 pounds of fish one day. Of the fish caught, 4 5/8 pounds were sea bass and the rest were mackerel. He gave away 1 7/8 pounds of mackerel. How many pounds of mackerel did he have left.

Answer 1

Given:

Total fish (Sea bass and mackerel) = $7\dfrac{2}{5}$ pounds

Sea bass = $4\dfrac{5}{8}$ pounds

He gave away $1\dfrac{7}{8}$ pounds of mackerel.

To find:

The remaining mackerel.

Solution:

We know that,

Mackerel = Total fish - Sea bass

$\text{Mackerel}=7\dfrac{2}{5}-4\dfrac{5}{8}$

$\text{Mackerel}=\dfrac{37}{5}-\dfrac{37}{8}$

$\text{Mackerel}=\dfrac{296-185}{40}$

$\text{Mackerel}=\dfrac{111}{40}$

He gave away $1\dfrac{7}{8}$ pounds of mackerel. So, the remaining mackerel is:

$\text{Remaining Mackerel}=\dfrac{111}{40}-1\dfrac{7}{8}$

$\text{Remaining Mackerel}=\dfrac{111}{40}-\dfrac{15}{8}$

$\text{Remaining Mackerel}=\dfrac{111-75}{40}$

$\text{Remaining Mackerel}=\dfrac{36}{40}$

$\text{Remaining Mackerel}=\dfrac{9}{10}$

Therefore, the remaining Mackerel is $\dfrac{9}{10}$ pounds or 0.9 pounds.

Answer 2

Answer:

The amount of Mackerel left is 9/10.

Step-by-step explanation:

total fish = 7 2/5 pounds

sea bass = 4 5/8

The amount of mackerel =

$7\frac{2}{5}-4\frac{5}{8}\\\\=\frac{37}{5}-\frac{37}{8}\\\\=\frac{296-185}{40}\\\\=2 \frac{31}{40}$

Mackerel left =

$2 \frac{31}{40}-1\frac{7}{8}\\\\= \frac{111}{40}-\frac{15}{8}\\\\=\frac{111-75}{40}\\\\=\frac{36}{40}\\\\=\frac{9}{10}$