Exercise 1.3 deals with estimation. In other words, it includes rounding off large numbers and dealing with relatively quick calculations. In this exercise you will see how to estimate sums, differences, and products of large numbers through rounding. Do keep in mind that estimation is just to help you calculate on a time basis when you don't necessarily require exact answers.
The idea of estimation is not only great for maths exam but also applicable in the real world, for example when shopping, budgeting or travelling. The NCERT Solutions of this exercise follows the latest NCERT syllabus and includes the type questions you may come across in your CBSE exams. The solutions are explained in a step by step manner through simple words for better understanding.
Completing this exercise will help with your number mental ability as it will quicken your problem solving capability, it will be particularly useful for maths olympiad preparation or for just improving your logical thinking skill.
The NCERT Solutions for Class 7 Maths Chapter 1: Exercise 1.3 focuses on estimating large numbers by rounding them off and they provide clear and simple steps for every question. Click below to download the free PDF and practice with ease!
The key concerts covered in this exercise are:
1. For the numbers in the previous exercise, find out how to get each number by making the smallest number of button clicks and write the expression.
Sol. (a) 8300
(8×1000)+(3×100)=8300
8×(+1000)=8000 ( 8 clicks)
3×(+100)=300 ( 3 clicks)
Total clicks: 8+3=11
(b) 40629
(4×10000)+(6×100)+(2×10)+
(9×1)=40629
4×(+10000)=40000 (4 clicks)
6×(+100)=600 ( 6 clicks)
2×(+10)=20 ( 2 clicks)
9×(+1)=9 ( 9 clicks)
Total clicks: 4+6+2+9=21
(c) 56354
(5×10000)+(6×1000)+(3×100)
+(5×10)+(4×1)=56354
5×(+10000)=50000 ( 5 clicks)
6×(+1000)=6000 ( 6 clicks)
3×(+100)=300 ( 3 clicks)
5×(+10)=50 ( 5 clicks)
4×(+1)=4 ( 4 clicks)
Total clicks: 5+6+3+5+4=23
(d) 66666
(6×10000)+(6×1000)+(6×100)
+(6×10)+(6×1)=66666
6×(+10000)=60000 ( 6 clicks)
6×(+1000)=6000 ( 6 clicks)
6×(+100)=600 ( 6 clicks)
6×(+10)=60 ( 6 clicks)
6×(+1)=6 ( 6 clicks)
Total clicks: 6+6+6+6+6=30
(e) 367813
(3×100000)+(6×10000)+(7×
1000)+(8×100)+(1×10)+(3×1)
=367813
3×(+100000)=300000 ( 3 clicks)
6×(+10000)=60000 ( 6 clicks)
7×(+1000)=7000 ( 7 clicks)
8×(+100)=800 ( 8 clicks)
1×(+10)=10 ( 1 click)
3×(+1)=3 ( 3 dicks)
Total clicks: 3+6+7+8+1+3=28
2. Do you see any connection between each number and the corresponding smallest number of button clicks?
Sol. The smallest number of button clicks depends on the place value of the digit in the number. If the digit in a particular place is small, then less number of clicks are required and if it is big, then more number of clicks are required.
3. If you notice, the expressions for the least button clicks also give the Indian place value notation of the numbers. Think about why this is so.
Sol. When we write a number using the least button clicks, we divide it into parts like lakh, thousand, hundred, ten, and one. These are the same place values we use in the Indian place value system, so both look the same.
(Session 2026 - 27)