# Sarin learns rate of change in school Part 5 (Math Question)

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The blog postings are about the Singapore Math. The readers can learn from the postings about Solving Singapore Primary School Mathematics. The blog presenting the Math Concept, Math Questions with solutions that teaches in Singapore Primary Schools. You or the kids can learn the skills of dealing with Math Modeling, Problem Sum from Lower Primary School to Upper Primary School level after reading the postings. This posting is an upper primary school math question on Volume, Flow Rate(Speed) and Problem Sum.

Read the posting on Sarin learns rate of change in school (Math Concept) to learn about concept of rate of change.

You can read the posting on Sarin learns Perimeter, Area and Volume in school(Math Concept) to learn the concept of volume.

Challenge yourself with the question before look for the given solution‼!

Upper primary school mathematics question UPQ257

The figure below shows 2 completely-filled tanks being emptied of the water from 2 different taps.

The taps at Tank A and Tank B were turned on at 7 am and 8.30 am respectively, until both tanks were completely empty. At 11 am, the water level in both the tanks was the same. At 12.30 pm, Tank B was completely empty and Tank A was only completely empty at 1 pm. If the rate of the flow of water from each tap was constant throughout, what was the height of Tank A?

Solution:

Using Portions of level per hour

The flow rates of both taps are constant and the tanks are in regular shapes.

The change of water levels of both tanks are proportional to their per hour flow rates.

So the water levels change according to the number of hours the taps turned on.

Draw the model water levels with respect to number of hours the taps turned on.

The water levels at 11am are same and can determine the relation of the flow rates of the two tanks

From the model

1 unit = 5cm

The height of Tank A = 5 × 9 = 45cm

Alternative solution:

Using equations

Equations:

The height (HA) of tank A = HA1 + HA2

The height (HB) of tank B = HB1 + HB2 = HA – 5

At 11 am, the water levels the same

HA2 = HB2

Volume of tank (V) = Height (H) × Base Area (A)

Flow rate (S) × Time = Volume (V)

For tank A

Set equation from 11am to 1pm (2h)

2SA = HA2 × AA è12SA = 6HA2 × AA

Set equation from 7am to 1pm (6h)

(HA1 + HA2) × AA = 6SA   ==>  (HA1 + HA2) × AA = 12SA

Slove the two equations

6HA2 = 2(HA1 + HA2) ==> 6HA2 = 2HA ==>8HA2 = 16HA

For tank B

Set equation from 11am to 12.30pm (1.5h)

1.5SB = HB2 × AB ==> 6SB = 4HB2 × AB

Set equation from 8.30am to 12.30pm (4h)

(HB1 + HB2) × AB = 4SB  ==>1.5(HB1 + HB2) × AB = 6SB

Slove the two equations

4HB2 = 1.5(HB1 + HB2) ==>  8HB2 = 3(HB1 + HB2) = 3 (HA – 5) == > 8HB2 = 18(HA – 5)

Hence

16HA = 18HA – 90

2HA = 90

HA = 45cm

The height of tank A is 45cm