Jump to
- Rational Numbers
- Exponents
- Squares and Square Roots
- Cubes and Cube Roots
- Playing with Numbers
- Operations on Algebraic Expressions
- Factorisation
- Linear Equations
- Percentage
- Profit and Loss
- Compound Interest
- Direct and Inverse Proportions
- Time and Work
- Polygons
- Parallelograms
- Construction of Quadrilaterals
- Area of a Trapezium and a Polygon
- Three-Dimensional Figures
- Volume and Surface Area of Solids
- Data Handling
- Introduction to Coordinate Geometry
- Line Graphs and Linear Graphs
- Pie Charts
- Probability
RS Aggarwal Solutions Class 8 Mathematics Solutions for Volume and Surface Area of Solids Exercise 20A in Chapter 20 - Volume and Surface Area of Solids
Prev
A cardboard box is 1.2m long, 72cm wide, and 54cm high. How many bars of soap
can be put it into it if each bar measures 6cm × 4.5 cm × 4 cm?
Answer:
We know that volume of cuboid = length × breadth × height
V = 120 × 72 × 54 = 466560 cm3
Volume of each bar soap = 6 × 4.5 × 4 = 108 cm3
Therefore, total number of soaps can be accommodated in a box is
Volume of the box/ volume of each soap
= 466560/108
= 4320 bars
Related Questions
Facebook
Whatsapp
Copy Link
Exercises
Chapters
Lido
Courses
Quick Links
Terms & Policies
Terms & Policies
Connect with us on social media!
2022 © Quality Tutorials Pvt Ltd All rights reserved