Know Relative Sizes of Measurement Units: CCSS.Math.Content.4.MD.A.1

Help Questions

4th Grade Math › Know Relative Sizes of Measurement Units: CCSS.Math.Content.4.MD.A.1

Questions 1 - 10

1

Fill in the missing piece of the table.

Screen shot 2015 09 01 at 10.27.21 am

Explanation

To solve this problem we can set up a proportion and cross multiply to solve for our unknown.

$\frac{1min}{60 sec}=\frac{3min}{x}$

First we cross multiply.

Then we divide each side by to isolate the .

$\frac{1min(x)}{1min}=\frac{3min(60sec)}{1min}$

2

Fill in the missing piece of the table.

Screen shot 2015 09 01 at 9.24.25 am

$300\textup,000$

$3000\textup,000$

$30\textup,000$

$3\textup,000$

Explanation

To solve this problem we can set up a proportion and cross multiply to solve for our unknown.

$\frac{1km}{100\textup,000 cm}=\frac{3km}{x}$

First we cross multiply.

$1km(x)=3km(100\textup,000cm)$

Then we divide each side by to isolate the .

$\frac{1km(x)}{1km}=\frac{3km(100\textup,000cm)}{1km}$

$x=300\textup,000 cm$

3

Fill in the missing piece of the table.

Screen shot 2015 09 01 at 3.29.15 pm

Explanation

To solve this problem we can set up a proportion and cross multiply to solve for our unknown.

$\frac{1lb}{16oz}=\frac{7lb}{x}$

First we cross multiply.

Then we divide each side by to isolate the .

$\frac{1lb(x)}{1lb}=\frac{7lb(16oz)}{1lb}$

4

Fill in the missing piece of the table.

Screen shot 2015 09 01 at 10.22.55 am

Explanation

To solve this problem we can set up a proportion and cross multiply to solve for our unknown.

$\frac{1hr}{60 min}=\frac{3hr}{x}$

First we cross multiply.

Then we divide each side by to isolate the .

$\frac{1hr(x)}{1hr}=\frac{3hr(60m)}{1hr}$

5

Fill in the missing piece of the table.

Screen shot 2015 09 01 at 9.27.44 am

$3\textup,000$

$30\textup,000$

$300\textup,000$

Explanation

To solve this problem we can set up a proportion and cross multiply to solve for our unknown.

$\frac{1km}{1\textup,000 m}=\frac{3km}{x}$

First we cross multiply.

$1km(x)=3km(1\textup,000m)$

Then we divide each side by to isolate the .

$\frac{1km(x)}{1km}=\frac{3km(1\textup,000m)}{1km}$

$x=3\textup,000 m$

6

Fill in the missing piece of the table.

Screen shot 2015 09 01 at 10.22.43 am

Explanation

To solve this problem we can set up a proportion and cross multiply to solve for our unknown.

$\frac{1hr}{60 min}=\frac{4hr}{x}$

First we cross multiply.

Then we divide each side by to isolate the .

$\frac{1hr(x)}{1hr}=\frac{4hr(60m)}{1hr}$

7

Fill in the missing pieces of the table.

Screen shot 2020 04 08 at 10.28.46 am

and $14\textup,400$

and $14\textup,000$

and $14\textup,040$

and $140\textup,000$

and $14\textup,004$

Explanation

Looking at the hour side if the table, the hour increases by ; thus, the missing hour is .

Nexst, we can set up a proportion and cross multiply to solve for our unknown.

$\frac{1hr}{360 sec}=\frac{4hr}{x}$

First we cross multiply.

Then we divide each side by to isolate the .

$\frac{1hr(x)}{1hr}=\frac{4hr(360sec)}{1hr}$

$x=14\textup,400 sec$

8

Fill in the missing piece of the table.

Screen shot 2015 09 01 at 3.30.09 pm

Explanation

To solve this problem we can set up a proportion and cross multiply to solve for our unknown.

$\frac{1lb}{16oz}=\frac{3lb}{x}$

First we cross multiply.

Then we divide each side by to isolate the .

$\frac{1lb(x)}{1lb}=\frac{3lb(16oz)}{1lb}$

9

Fill in the missing piece of the table.

Screen shot 2015 09 01 at 11.53.55 am

Explanation

To solve this problem we can set up a proportion and cross multiply to solve for our unknown.

$\frac{1g}{0.001 kg}=\frac{2g}{x}$

First we cross multiply.

Then we divide each side by to isolate the .

$\frac{1g(x)}{1g}=\frac{2g(0.001kg)}{1g}$

10

Fill in the missing piece of the table.

Screen shot 2015 09 01 at 3.33.38 pm

Explanation

To solve this problem we can set up a proportion and cross multiply to solve for our unknown.

$\frac{1oz}{0.0625lb}=\frac{2oz}{x}$

First we cross multiply.

Then we divide each side by to isolate the .

$\frac{1oz(x)}{1oz}=\frac{2oz(0.0625lb)}{1oz}$

Page 1 of 8

Return to subject