randFromArray([ 5, 8 ]) randRange( 3, D - 1 ) randRange( 1, 5 ) * D N / D * INVITEES

person( 1 ) invited INVITEES friends to his birthday party. Some people had other plans and could not attend, but \frac{N}{D} of the people person( 1 ) invited were able to attend.

person( 1 ) invited INVITEES friends to her birthday party. Some people had other plans and could not attend, but \frac{N}{D} of the people person( 1 ) invited were able to attend.

How many people went to person( 1 )'s birthday party?

SOLUTION people

We need to figure out what \dfrac{N}{D} of INVITEES is to find out how many people attended the party.

We can find \dfrac{N}{D} of INVITEES by multiplying \dfrac{N}{D} and INVITEES.

\color{BLUE}{\dfrac{N}{D}} \times \color{ORANGE}{INVITEES} = \color{GREEN}{\text{?}}

\color{BLUE}{\dfrac{N}{D}} \times \color{ORANGE}{\dfrac{INVITEES}{1}} = \dfrac{\color{BLUE}{N} \cdot \color{ORANGE}{INVITEES}} {\color{BLUE}{D} \cdot \color{ORANGE}{1}} = \color{GREEN}{\text{?}}

\dfrac{\color{BLUE}{N} \cdot \color{ORANGE}{INVITEES}} {\color{BLUE}{D}} = \color{GREEN}{\dfrac{N * INVITEES}{D}} = \color{GREEN}{SOLUTION}

We can also visually see that \dfrac{N}{D} of INVITEES is SOLUTION:

init({ range: [ [ -0.05, 1 ], [ 0, 3 ] ],scale: [ 450, 30 ] }); rectchart( [ N, D - N ], [ BLUE, "#ccc" ], 2 ); rectchart( [ INVITEES, 0 ], [ ORANGE, "#999" ], 1 ); rectchart( [ SOLUTION, INVITEES - SOLUTION ], [ GREEN, BACKGROUND ], 0 ); style({ color: BLUE }, function() { label([ 0, 2.5 ], "\\frac{" + N + "}{" + D + "}", "left" ); }); style({ color: ORANGE }, function() { label([ 0, 1.5 ], INVITEES, "left" ); }); style({ color: GREEN }, function() { label([ 0, 0.5 ], SOLUTION, "left" ); });

SOLUTION people attended person( 1 )'s party.

(randRange( 2, 10 )*D) randFromArray([ 2, 8 ]) randRange( 1, D - 1 ) N / D * AMOUNT

After saving up for a while, person( 1 ) had $AMOUNT.00 in his piggy bank, and he spent \frac{N}{D} of that money on books at the bookstore.

After saving up for a while, person( 1 ) had $AMOUNT.00 in her piggy bank, and she spent \frac{N}{D} of that money on books at the bookstore.

How much money did person( 1 ) spend?

\$SOLUTION

We need to figure out what \dfrac{N}{D} of $AMOUNT.00 is to find out how much person(1) spent.

We can find \dfrac{N}{D} of $AMOUNT.00 by multiplying \dfrac{N}{D} and AMOUNT.

\color{BLUE}{\dfrac{N}{D}} \times \color{ORANGE}{AMOUNT} = \color{GREEN}{\text{?}}

\color{BLUE}{\dfrac{N}{D}} \times \color{ORANGE}{\dfrac{AMOUNT}{1}} = \dfrac{\color{BLUE}{N} \cdot \color{ORANGE}{AMOUNT}} {\color{BLUE}{D} \cdot \color{ORANGE}{1}} = \color{GREEN}{\text{?}}

\dfrac{\color{BLUE}{N} \cdot \color{ORANGE}{AMOUNT}} {\color{BLUE}{D}} = \color{GREEN}{\dfrac{N * AMOUNT}{D}} = \color{GREEN}{SOLUTION}

We can also visually see that \dfrac{N}{D} of AMOUNT is SOLUTION:

init({ range: [ [ -0.05, 1 ], [ 0, 3 ] ],scale: [ 450, 30 ] }); rectchart( [ N, D - N ], [ BLUE, "#ccc" ], 2 ); rectchart( [ AMOUNT, 0 ], [ ORANGE, "#999" ], 1 ); rectchart( [ SOLUTION, AMOUNT - SOLUTION ], [ GREEN, BACKGROUND ], 0 ); style({ color: BLUE }, function() { label([ 0, 2.5 ], "\\frac{" + N + "}{" + D + "}", "left" ); }); style({ color: ORANGE }, function() { label([ 0, 1.5 ], AMOUNT, "left" ); }); style({ color: GREEN }, function() { label([ 0, 0.5 ], SOLUTION, "left" ); });

person( 1 ) spent $SOLUTION.00 on books.

(randRange( 2, 10 )*D) randFromArray([ 8, 6 ]) randRange( 1, D - 1 ) roundTo( 2, N / D * AMOUNT )

Every day person( 1 ) put the extra change from his pockets into a glass jar. After randRange( 10, 30 ) weeks, he had saved up $AMOUNT.00. person( 1 ) decided to use \frac{N}{D} of the money from the jar to buy canned food for a homeless shelter.

Every day person( 1 ) put the extra change from her pockets into a glass jar. After randRange( 10, 30 ) weeks, she had saved up $AMOUNT.00. person( 1 ) decided to use \frac{N}{D} of the money from the jar to buy canned food for a homeless shelter.

How much money did person( 1 ) spend on canned food?

\$SOLUTION

We need to figure out what \dfrac{N}{D} of $AMOUNT.00 is to find out how much person(1) spent.

We can find \dfrac{N}{D} of $AMOUNT.00 by multiplying \dfrac{N}{D} and AMOUNT.

\color{BLUE}{\dfrac{N}{D}} \times \color{ORANGE}{AMOUNT} = \color{GREEN}{\text{?}}

\color{BLUE}{\dfrac{N}{D}} \times \color{ORANGE}{\dfrac{AMOUNT}{1}} = \dfrac{\color{BLUE}{N} \cdot \color{ORANGE}{AMOUNT}} {\color{BLUE}{D} \cdot \color{ORANGE}{1}} = \color{GREEN}{\text{?}}

\dfrac{\color{BLUE}{N} \cdot \color{ORANGE}{AMOUNT}} {\color{BLUE}{D}} = \color{GREEN}{\dfrac{N * AMOUNT}{D}} = \color{GREEN}{SOLUTION}

We can also visually see that \dfrac{N}{D} of AMOUNT is SOLUTION:

init({ range: [ [ -0.05, 1 ], [ 0, 3 ] ],scale: [ 450, 30 ] }); rectchart( [ N, D - N ], [ BLUE, "#ccc" ], 2 ); rectchart( [ AMOUNT, 0 ], [ ORANGE, "#999" ], 1 ); rectchart( [ SOLUTION, AMOUNT - SOLUTION ], [ GREEN, BACKGROUND ], 0 ); style({ color: BLUE }, function() { label([ 0, 2.5 ], "\\frac{" + N + "}{" + D + "}", "left" ); }); style({ color: ORANGE }, function() { label([ 0, 1.5 ], AMOUNT, "left" ); }); style({ color: GREEN }, function() { label([ 0, 0.5 ], SOLUTION, "left" ); });

person( 1 ) spent $SOLUTION.00 on canned food for the homeless shelter.

randRange( 10, 20 ) randFromArray([ 8, 6, 10 ]) randRange( 1, D - 1 ) N / D * GALLONS

Before leaving on a road trip, person( 1 ) filled up his gas tank, which holds GALLONS gallons of gas. After 0.5 * randRange( 3 / 0.5, 10 / 0.5 ) hours, person( 1 ) noticed that the gas tank was \frac{N}{D} full.

Before leaving on a road trip, person( 1 ) filled up her gas tank, which holds GALLONS gallons of gas. After 0.5 * randRange( 3 / 0.5, 10 / 0.5 ) hours, person( 1 ) noticed that the gas tank was \frac{N}{D} full.

How many gallons of gas were left in the tank?

SOLUTION gallons

Since a fraction of the gas in his tank was left, we just need to figure out what \dfrac{N}{D} of GALLONS gallons is to find out how much gas was left in the tank.

Since a fraction of the gas in her tank was left, we just need to figure out what \dfrac{N}{D} of GALLONS gallons is to find out how much gas was left in the tank.

We can find \dfrac{N}{D} of GALLONS gallons by multiplying \dfrac{N}{D} and GALLONS.

\color{BLUE}{\dfrac{N}{D}} \times \color{ORANGE}{GALLONS} = \color{GREEN}{\text{?}}

\color{BLUE}{\dfrac{N}{D}} \times \color{ORANGE}{\dfrac{GALLONS}{1}} = \dfrac{\color{BLUE}{N} \cdot \color{ORANGE}{GALLONS}} {\color{BLUE}{D} \cdot \color{ORANGE}{1}} = \color{GREEN}{\text{?}}

\dfrac{\color{BLUE}{N} \cdot \color{ORANGE}{GALLONS}} {\color{BLUE}{D}} = \color{GREEN}{\dfrac{N * GALLONS}{D}} = \color{GREEN}{SOLUTION}

We can also visually see that \dfrac{N}{D} of GALLONS is SOLUTION:

init({ range: [ [ -0.05, 1 ], [ 0, 3 ] ],scale: [ 450, 30 ] }); rectchart( [ N, D - N ], [ BLUE, "#ccc" ], 2 ); rectchart( [ GALLONS, 0 ], [ ORANGE, "#999" ], 1 ); rectchart( [ SOLUTION, GALLONS - SOLUTION ], [ GREEN, BACKGROUND ], 0 ); style({ color: BLUE }, function() { label([ 0, 2.5 ], "\\frac{" + N + "}{" + D + "}", "left" ); }); style({ color: ORANGE }, function() { label([ 0, 1.5 ], GALLONS, "left" ); }); style({ color: GREEN }, function() { label([ 0, 0.5 ], SOLUTION, "left" ); });

person( 1 ) had SOLUTION gallons of gas left in his tank when he checked.person( 1 ) had SOLUTION gallons of gas left in her tank when she checked.

randRange( 15, 40 ) randFromArray([ 8, 6, 10 ]) randRange( 1, D - 1 ) N / D * ATTENDEES

ATTENDEES people had a picnic in the park. \frac{N}{D} of the people at the picnic were adults.

How many adults were at the picnic?

SOLUTION adults

We need to figure out what \dfrac{N}{D} of ATTENDEES is to find out how many people at the picnic were adults.

We can find \dfrac{N}{D} of ATTENDEES by multiplying \dfrac{N}{D} and ATTENDEES.

\color{BLUE}{\dfrac{N}{D}} \times \color{ORANGE}{ATTENDEES} = \color{GREEN}{\text{?}}

\color{BLUE}{\dfrac{N}{D}} \times \color{ORANGE}{\dfrac{ATTENDEES}{1}} = \dfrac{\color{BLUE}{N} \cdot \color{ORANGE}{ATTENDEES}} {\color{BLUE}{D} \cdot \color{ORANGE}{1}} = \color{GREEN}{\text{?}}

\dfrac{\color{BLUE}{N} \cdot \color{ORANGE}{ATTENDEES}} {\color{BLUE}{D}} = \color{GREEN}{\dfrac{N * ATTENDEES}{D}} = \color{GREEN}{SOLUTION}

We can also visually see that \dfrac{N}{D} of ATTENDEES is SOLUTION:

init({ range: [ [ -0.05, 1 ], [ 0, 3 ] ],scale: [ 450, 30 ] }); rectchart( [ N, D - N ], [ BLUE, "#ccc" ], 2 ); rectchart( [ ATTENDEES, 0 ], [ ORANGE, "#999" ], 1 ); rectchart( [ SOLUTION, ATTENDEES - SOLUTION ], [ GREEN, BACKGROUND ], 0 ); style({ color: BLUE }, function() { label([ 0, 2.5 ], "\\frac{" + N + "}{" + D + "}", "left" ); }); style({ color: ORANGE }, function() { label([ 0, 1.5 ], ATTENDEES, "left" ); }); style({ color: GREEN }, function() { label([ 0, 0.5 ], SOLUTION, "left" ); });

SOLUTION people at the picnic were adults.

(randRange( 1,5 )*D) randFromArray([ 2,8 ]) randRange( 1, D - 1 ) N / D * BATCHES

person(1) decided to bake cookies for the school bake sale. He found a recipe that called for \frac{N}{D} of a cup of chocolate chips.

person(1) decided to bake cookies for the school bake sale. She found a recipe that called for \frac{N}{D} of a cup of chocolate chips.

To have enough cookies for the bake sale, person(1) needed to make BATCHES batches of cookies.

How many cups of chocolate chips did person(1) need in total?

SOLUTION cups

We can multiply \dfrac{N}{D} cup by BATCHES batches to find out how many cups of chocolate chips person(1) needed.

\color{BLUE}{\dfrac{N}{D}} \times \color{ORANGE}{BATCHES} = \color{GREEN}{\text{?}}

\color{BLUE}{\dfrac{N}{D}} \times \color{ORANGE}{\dfrac{BATCHES}{1}} = \dfrac{\color{BLUE}{N} \cdot \color{ORANGE}{BATCHES}} {\color{BLUE}{D} \cdot \color{ORANGE}{1}} = \color{GREEN}{\text{?}}

\dfrac{\color{BLUE}{N} \cdot \color{ORANGE}{BATCHES}} {\color{BLUE}{D}} = \color{GREEN}{\dfrac{N * BATCHES}{D}} = \color{GREEN}{SOLUTION}

person(1) needed SOLUTION cup of chocolate chips to make enough cookies for the bake sale.person(1) needed SOLUTION cups of chocolate chips to make enough cookies for the bake sale.