Question

1. Liezl bought 2 14kilograms of pork, 5 12 kilograms of chicken and 1 13 kilograms of beef. How many kilograms of meat did she buy in all?
2. Roel travelled 1 12 km while Arnel 1 12 km more than the distance covered by Roel. How many kilometers did the two travel together?

3. Kenneth and Ian have 20 kilograms of fruits. Kenneth has 9 58 kilograms of fruits. How many kilograms of fruits does Ian have?
4. Father has a 17 37 m of nylon string. He gave 10 m of it for his brother’s fishing net. How long is the nylon string that is left?
5. Althea is working in a bakeshop. She put two cheesecakes on display. The first cheesecake was cut into 8 slices and there are 5 slices left. The other was cut into 12 slices and there are 4 slices left. How much cheesecake is left on display?

Answer 1

Answer:

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Answer 2

Given:

1. Amount of pork Liezl bought = $2\frac{1}{4}kg$

Amount of chicken Liezl bought = $5\frac{1}{2} kg$

Amount of beef Liezl bought = $1\frac{1}{3}kg$

2. Distance travelled by Roel = $1\frac{1}{2}km$

Distance travelled by Anel = $1\frac{1}{2} +1\frac{1}{2} km$

3. Weight of fruits that Kenneth and Ian have = $20kg$

Weight of fruits Kenneth has = $9\frac{5}{8}kg$

4. Length of string father has = $17\frac{3}{7}m$

Length of string he gave to brother = $10m$

5. No. of slices Althea cut for the first cheesecake = $8$

No. of slices left from the first cheesecake = $5$

No. of slices Althea cut for the second cheesecake = $12$

No. of slices left from the first cheesecake = $4$

To find:

1. Total amount of meat bought by Liezl.

2. Total distance travelled by Roel and Arnel together.

3. Weight of fruits that Ian has.

4. Length of nylon string remaining.

5. Total slices of cheesecake remaining on display.

Solution:

1. Total amount of meat purchased = Weight of pork + Weight of chicken + Weight of beef

$Total=2\frac{1}{4}+5\frac{1}{2}+1\frac{1}{3}$

Convert these mixed fractions into improper fractions.

$2\frac{1}{4}=\frac{(2*4)+1}{4}=\frac{9}{4}$

$5\frac{1}{2}=\frac{(5*2)+1}{2}=\frac{11}{2}$

$1\frac{1}{3}=\frac{(1*3)+1}{3}=\frac{4}{3}$

$Total=\frac{9}{4}+\frac{11}{2}+\frac{4}{3}$

$LCM(4,2,3)=12$

$\frac{9}{4}=\frac{9}{4}*\frac{3}{3}=\frac{27}{12}$

$\frac{11}{2}=\frac{11}{2}*\frac{6}{6} =\frac{66}{12}$

$\frac{4}{3}=\frac{4}{3}*\frac{4}{4}=\frac{16}{12}$

Total meat purchased = $\frac{27}{12}+\frac{66}{12}+\frac{16}{12}=\frac{27+66+16}{12}=\frac{109}{12}=9\frac{1}{12}kg$

2. Distance travelled by Arnel = Distance travelled by Roel + Extra distance travelled by Arnel

Distance travelled by Arnel = $1\frac{1}{2} + 1\frac{1}{2}$

$1\frac{1}{2} =\frac{3}{2}$

Distance travelled by Arnel = $\frac{3}{2} + \frac{3}{2} =\frac{3+3}{2}=\frac{6}{2}=3km$

Distance travelled together = Distance travelled by Roel + Distance travelled by Arnel

Distance travelled together = $1\frac{1}{2} +3=\frac{3}{2}+3=\frac{3}{2}+\frac{3}{1}$

$LCM(2,1)=2$

$\frac{3}{2}=\frac{3}{2}*\frac{1}{1}=\frac{3}{2}$

$\frac{3}{1}= \frac{3}{1}*\frac{2}{2}= \frac{6}{2}$

Distance travelled together = $\frac{3}{2}+ \frac{6}{2}=\frac{3+6}{2}= \frac{9}{2}=4\frac{1}{2}km$

3. Both Kenneth and Ian have $20kg$ fruits out of which Kenneth's weight of fruits is $9\frac{5}{8}kg$.

$9\frac{5}{8}=\frac{(9*8)+5}{8}= \frac{77}{8}$

Weight of fruits Ian has = Total weight of fruits - Weight of fruits Kenneth has

Weight of fruits Ian has = $20-9\frac{5}{8}=20-\frac{77}{8}=\frac{20}{1} -\frac{77}{8}$

$LCM(1,8)=8$

$\frac{20}{1}=\frac{20}{1}* \frac{8}{8}=\frac{160}{8}$

$\frac{77}{8}= \frac{77}{8}* \frac{1}{1}= \frac{77}{8}$

Weight of fruits Ian has = $\frac{160}{8}- \frac{77}{8}= \frac{160-77}{8}=\frac{83}{8} =10\frac{3}{8}kg$

4. To find the remaining length of string, subtract the length of string he gave to his brother from the total length.

Remaining length of string = Total length of the string - Length of string father gave to brother

$17\frac{3}{7}=\frac{(17*7)+3}{7} =\frac{122}{7}$

Remaining length of string = $17\frac{3}{7}-10=\frac{122}{7}-10=\frac{122}{7}-\frac{10}{1}$

$LCM(1,7)=7$

$\frac{122}{7}= \frac{122}{7}* \frac{1}{1} =\frac{122}{7}$

$\frac{10}{1}= \frac{10}{1}* \frac{7}{7}= \frac{70}{7}$

Remaining length of string = $\frac{122}{7}-\frac{70}{7}=\frac{122-70}{7}=\frac{52}{7}=7\frac{3}{7}m$

5. The amount of first cheesecake remaining on the display is $\frac{5}{8}$

The amount of second cheesecake remaining on the display is $\frac{4}{12}$.

Cheesecake remaining on the display = Amount remaining from first cheesecake + Amount remaining from second cheesecake

Total cheesecake remaining on the display = $\frac{5}{8} +\frac{4}{12}$

$LCM(8,12)=24$

$\frac{5}{8}= \frac{5}{8}* \frac{3}{3}= \frac{15}{24}$

$\frac{4}{12}= \frac{4}{12}* \frac{2}{2}= \frac{8}{24}$

Total cheesecake remaining on the display = $\frac{15}{24}+ \frac{8}{24}= \frac{15+8}{24}= \frac{23}{24}$

1. Liezl purchased $9\frac{1}{12}kg$ weight of meat in total.

2. Roel and Arnel travelled a distance of $4\frac{1}{2}km$ together.

3. Ian has $10\frac{3}{8}kg$ fruits.

4. The remaining length of the string is $7\frac{3}{7}m$.

5. The amount of cheesecakes left on the display is $\frac{23}{24}$ of both the cheesecakes.