# STUCK ON HOMEWORK?

## ASK FOR HELP FROM OUR KIASUPARENTS.COM COMMUNITY!

# Question

# Answer

getmaven

Basically, to answer this question, you need to understand a few concepts:

- The volume of the water in the cylinder is the same at 25cm
^{3} - Therefore you want to first simplify the problem by comparing something that has one less or one more item. In this case, that would be 2
^{nd}and 3^{rd}or the 2^{nd}and 4^{th}. We shall go with the 2^{nd}and 3^{rd}option here. The only difference is that there is no R in the third. Therefore, R must be the differenceR: 82-71 = 11cm

^{3}. - Now that you know R is 11cm
^{3}, you can get Q by subtracting R and the volume of the water from the 4^{th}cyclinderQ: 58 – 25 – 11 = 22cm

^{3} - Finally, since you have these 2, you can find the final one by subtracting all the three values (including the water from point 1) from the 2
^{nd}jarP: 82 – 25 – 11 – 22 = 24cm

^{3}Alternatively, since you have both Q + R with water already in the 4

^{th}jar (which is 58cm^{3)},

you can simply take the 2^{nd}jar minus that valueP: 82 – 58 = 24cm

^{3}(I prefer this method because less working means less careless mistakes) - I hope this helps explain the problem very clearly.

Adwin

Water = 25 cm^{3}

P + Q + R + Water = 82 cm^{3}

P + Q + Water = 71 cm^{3}

Q + R + Water = 58 cm^{3}

P + Q + R = 57 cm^{3}

P + Q = 46 cm^{3
}**R = 57 cm ^{3} – 46 cm^{3} = 11 cm^{3}**

Q + R = 33 cm^{3
}**Q = 33 – 11 = 22 cm ^{3}**

P + Q + R = 57 cm^{3
}**P = 57 – 22 – 11 = 24 cm ^{3}**

**Find Tuition/Enrichment Centres**