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Word Problem Involving Add or Subtract Fractions With Different Denominators
  • 时间:2024-09-17

Word Problem Involving Add or Subtract Fractions With Different Denominators


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Jamie bought a box of fruit weighing 3$frac{2}{5}$ kilograms. If she bought a second box that weighed 7$frac{1}{3}$ kilograms, what is the combined weight of both boxes?

Solution

Step 1:

Weight of the first box of fruit = 3$frac{2}{5}$ kilograms

Weight of the second box of fruit = 7$frac{1}{3}$ kilograms

The combined of the two boxes of fruit = 3$frac{2}{5}$ + 7$frac{1}{3}$ = $frac{17}{5}$ + $frac{22}{3}$

Step 2:

The denominators are different. So the LCD of the fractions or LCM of denominators 3 and 5 is 15.

Rewriting to get equivalent fractions with LCD as denominator

$frac{17×3}{5×3}$ + $frac{22×5}{3×5}$ = $frac{51}{15}$ + $frac{110}{15}$ = $frac{(51+110)}{15}$ = $frac{161}{15}$ = 10$frac{11}{15}$

During the weekend, Nancy spent a total 5$frac{1}{3}$ hours studying. If she spent 3$frac{1}{4}$ hours studying on Saturday, how long did she study on Sunday?

Solution

Step 1:

Time spent studying on the weekend = 5$frac{1}{3}$ hours

Time spent studying on Saturday = 3$frac{1}{4}$ hours

Time spent studying on Sunday =

Time spent studying on the weekend − Time spent studying on Saturday

= 5$frac{1}{3}$ − 3$frac{1}{4}$ = $frac{16}{3}$$frac{13}{4}$

Step 2:

LCD of the fractions or the LCM of the denominators 3 and 4 is 12

Rewriting to get equivalent fractions with LCD as denominator

$frac{16×4}{3×4}$$frac{13×3}{4×3}$ = $frac{64}{12}$$frac{39}{12}$ = $frac{64−39}{12}$ = $frac{25}{12}$ = 2$frac{1}{12}$ hours

So, the time spent studying on Sunday = 2$frac{1}{12}$ hours

Marcos bought apples that weighed 6$frac{2}{3}$ kilograms. If he gave away 3$frac{1}{5}$ kilograms of apples to his friends, how many kilograms of apples does he have left?

Solution

Step 1:

Weight of the apples bought = 6$frac{2}{3}$ kilograms

Weight of the apples given to friends = 3$frac{1}{5}$ kilograms

Weight of the apples left =

Weight of the apples bought − Weight of the apples given to friends

= 6$frac{2}{3}$ − 3$frac{1}{5}$ = $frac{20}{3}$$frac{16}{5}$

Step 2:

LCD of the fractions or LCM of the denominators 3 and 5 is 15

Rewriting to get equivalent fractions with LCD as denominator

$frac{20×5}{3×5}$$frac{16×3}{5×3}$ = $frac{100}{15}$$frac{48}{15}$ = $frac{100−48}{15}$ = $frac{52}{15}$ = 3$frac{7}{15}$ kilograms

So, the weight of the apples left = 3$frac{7}{15}$ kilograms

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