Common Core: 4th Grade Math : Common Core Math: Grade 4

Study concepts, example questions & explanations for Common Core: 4th Grade Math

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Example Questions

Example Question #21 : Know Relative Sizes Of Measurement Units: Ccss.Math.Content.4.Md.A.1

Fill in the missing piece of the table. 

Screen shot 2015 09 01 at 11.53.22 am

Possible Answers:

\(\displaystyle 0.5\)

\(\displaystyle 005\)

\(\displaystyle 0.005\)

\(\displaystyle 0.05\)

\(\displaystyle 05\)

Correct answer:

\(\displaystyle 0.005\)

Explanation:

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

\(\displaystyle \frac{1g}{0.001 kg}=\frac{5g}{x}\)

First we cross multiply. 

\(\displaystyle 1g(x)=5g(0.001kg)\) 

Then we divide each side by \(\displaystyle 1g\) to isolate the \(\displaystyle x\).

\(\displaystyle \frac{1g(x)}{1g}=\frac{5g(0.001kg)}{1g}\)

\(\displaystyle x=0.005kg\)

Example Question #2771 : Operations

Fill in the missing piece of the table. 


Screen shot 2015 09 01 at 11.53.35 am

Possible Answers:

\(\displaystyle 0.4\)

\(\displaystyle 04\)

\(\displaystyle 0.004\)

\(\displaystyle 004\)

\(\displaystyle 0.04\)

Correct answer:

\(\displaystyle 0.004\)

Explanation:

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

\(\displaystyle \frac{1g}{0.001 kg}=\frac{4g}{x}\)

First we cross multiply. 

\(\displaystyle 1g(x)=4g(0.001kg)\) 

Then we divide each side by \(\displaystyle 1g\) to isolate the \(\displaystyle x\).

\(\displaystyle \frac{1g(x)}{1g}=\frac{4g(0.001kg)}{1g}\)

\(\displaystyle x=0.004kg\)

Example Question #971 : Common Core Math: Grade 4

Fill in the missing piece of the table. 

Screen shot 2015 09 01 at 11.53.45 am

Possible Answers:

\(\displaystyle 03\)

\(\displaystyle 0.003\)

\(\displaystyle 003\)

\(\displaystyle 0.03\)

\(\displaystyle 0.3\)

Correct answer:

\(\displaystyle 0.003\)

Explanation:

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

\(\displaystyle \frac{1g}{0.001 kg}=\frac{3g}{x}\)

First we cross multiply. 

\(\displaystyle 1g(x)=3g(0.001kg)\) 

Then we divide each side by \(\displaystyle 1g\) to isolate the \(\displaystyle x\).

\(\displaystyle \frac{1g(x)}{1g}=\frac{3g(0.001kg)}{1g}\)

\(\displaystyle x=0.003kg\)

Example Question #972 : Common Core Math: Grade 4

Fill in the missing piece of the table. 

Screen shot 2015 09 01 at 11.53.55 am

Possible Answers:

\(\displaystyle 0.002\)

\(\displaystyle 02\)

\(\displaystyle 0.02\)

\(\displaystyle 0.2\)

\(\displaystyle 002\)

Correct answer:

\(\displaystyle 0.002\)

Explanation:

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

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

First we cross multiply. 

\(\displaystyle 1g(x)=2g(0.001kg)\) 

Then we divide each side by \(\displaystyle 1g\) to isolate the \(\displaystyle x\).

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

\(\displaystyle x=0.002kg\)

Example Question #25 : Know Relative Sizes Of Measurement Units: Ccss.Math.Content.4.Md.A.1

Fill in the missing piece of the table. 

Screen shot 2015 09 01 at 3.29.15 pm

 

Possible Answers:

\(\displaystyle 110\)

\(\displaystyle 111\)

\(\displaystyle 112\)

\(\displaystyle 106\)

\(\displaystyle 107\)

Correct answer:

\(\displaystyle 112\)

Explanation:

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

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

First we cross multiply. 

\(\displaystyle 1lb(x)=7lb(16oz)\) 

Then we divide each side by \(\displaystyle 1lb\) to isolate the \(\displaystyle x\).

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

\(\displaystyle x=112oz\)

Example Question #26 : Know Relative Sizes Of Measurement Units: Ccss.Math.Content.4.Md.A.1

Fill in the missing piece of the table. 

Screen shot 2015 09 01 at 3.29.49 pm

 

Possible Answers:

\(\displaystyle 76\)

\(\displaystyle 80\)

\(\displaystyle 82\)

\(\displaystyle 78\)

\(\displaystyle 84\)

Correct answer:

\(\displaystyle 80\)

Explanation:

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

\(\displaystyle \frac{1lb}{16oz}=\frac{5lb}{x}\)

First we cross multiply. 

\(\displaystyle 1lb(x)=5lb(16oz)\) 

Then we divide each side by \(\displaystyle 1lb\) to isolate the \(\displaystyle x\).

\(\displaystyle \frac{1lb(x)}{1lb}=\frac{5lb(16oz)}{1lb}\)

\(\displaystyle x=80oz\)

Example Question #321 : How To Multiply

Fill in the missing piece of the table. 

Screen shot 2015 09 01 at 3.29.39 pm

Possible Answers:

\(\displaystyle 97\)

\(\displaystyle 99\)

\(\displaystyle 98\)

\(\displaystyle 100\)

\(\displaystyle 96\)

Correct answer:

\(\displaystyle 96\)

Explanation:

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

\(\displaystyle \frac{1lb}{16oz}=\frac{6lb}{x}\)

First we cross multiply. 

\(\displaystyle 1lb(x)=6lb(16oz)\) 

Then we divide each side by \(\displaystyle 1lb\) to isolate the \(\displaystyle x\).

\(\displaystyle \frac{1lb(x)}{1lb}=\frac{6lb(16oz)}{1lb}\)

\(\displaystyle x=96oz\)

Example Question #321 : How To Multiply

Fill in the missing piece of the table. 


Screen shot 2015 09 01 at 3.29.59 pm

Possible Answers:

\(\displaystyle 68\)

\(\displaystyle 64\)

\(\displaystyle 62\)

\(\displaystyle 70\)

\(\displaystyle 66\)

Correct answer:

\(\displaystyle 64\)

Explanation:

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

\(\displaystyle \frac{1lb}{16oz}=\frac{4lb}{x}\)

First we cross multiply. 

\(\displaystyle 1lb(x)=4lb(16oz)\) 

Then we divide each side by \(\displaystyle 1lb\) to isolate the \(\displaystyle x\).

\(\displaystyle \frac{1lb(x)}{1lb}=\frac{4lb(16oz)}{1lb}\)

\(\displaystyle x=64oz\)

Example Question #322 : How To Multiply

Fill in the missing piece of the table. 

Screen shot 2015 09 01 at 3.30.09 pm

Possible Answers:

\(\displaystyle 50\)

\(\displaystyle 42\)

\(\displaystyle 46\)

\(\displaystyle 48\)

\(\displaystyle 44\)

Correct answer:

\(\displaystyle 48\)

Explanation:

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

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

First we cross multiply. 

\(\displaystyle 1lb(x)=3lb(16oz)\) 

Then we divide each side by \(\displaystyle 1lb\) to isolate the \(\displaystyle x\).

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

\(\displaystyle x=48oz\)

Example Question #32 : Measurement & Data

Fill in the missing piece of the table. 

Screen shot 2015 09 01 at 3.30.19 pm

Possible Answers:

\(\displaystyle 32\)

\(\displaystyle 36\)

\(\displaystyle 40\)

\(\displaystyle 38\)

\(\displaystyle 34\)

Correct answer:

\(\displaystyle 32\)

Explanation:

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

\(\displaystyle \frac{1lb}{16oz}=\frac{2lb}{x}\)

First we cross multiply. 

\(\displaystyle 1lb(x)=2lb(16oz)\) 

Then we divide each side by \(\displaystyle 1lb\) to isolate the \(\displaystyle x\).

\(\displaystyle \frac{1lb(x)}{1lb}=\frac{2lb(16oz)}{1lb}\)

\(\displaystyle x=32oz\)

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