# Sarin learns Perimeter, Area and Volume in school part 116 (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 presents the Math Concept, the Math Questions with solutions that teaches in Singapore Primary Schools. You or the kids can acquire the skill of dealing with the Math Modeling, the Math Problem Solving and the Problem Sum from Lower Primary School to Upper Primary School level after reading the blog postings. This posting is an upper primary school math question on RatioSolid and Volume.

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

Read also the posting on Sarin learns Shapes and Solids in school (Math Concept) to know about solid.

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

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

Upper primary school mathematics question UPQ562

Tank A contains some water with a water level of 51 cm. Tank B is empty. The length of Tank A is half that of Tank B and the breadth of Tank B is three times that of Tank A. What is the water level in Tank B if all the water in Tank A is pour into Tank B?

Solution:

The ratio of the length,

AL : BL = 1 : 2

The ratio of the breadth,

AB : BB = 1 : 3

Since the base area is determine by length and breadth,

The ratio of the Base area,

Base A : Base B = 1×1 : 2×3

Base A : Base B = 1 : 6

The height of the water level is inverse proportional to the base area when volume is constant.

High A : h = 6 : 1

High A = 51 cm

The water level in Tank B = 51 ÷ 6 = 8.5 cm

Alternative solution:

The volume, V = Height × Base area

For Tank A,

V = H × AL × AB

H = 51 cm,

V = 51 × AL × AB

For Tank B,

V = H × BL × AL

H = h

BL is 2 times of AL,

BL = 2 × AL

BB is 3 times of AB,

BB = 3 × AB

V = h × 2 × AL × 3 × AB

V = 6 × h × AL × AB

The volume is the same as water pour from Tank A to Tank B,

51 × AL × AB = 6 × h × AL × AB

51 = 6 × h

The height of Tank B, h = 51 ÷ 6 = 8.5 cm

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