Class 7 Maths Chapter 2 Exercise 2.6 Pdf Notes NCERT Solutions
Class 7 Maths Chapter 2 Fractions and Decimals Exercise 2.6 pdf notes:-
Exercise 2.6 Class 7 maths Chapter 2 Pdf Notes:-
To see video Solution Of This Exercise Click Here
Ncert Solution for Class 6 Maths Chapter 2 Fractions and decimals Exercise 2.6 Tips:-
DIVISION OF DECIMAL NUMBERS
Savita was preparing a design to decorate her classroom. She needed a few coloured
strips of paper of length 1.9 cm each. She had a strip of coloured paper of length 9.5 cm.
How many pieces of the required length will she get out of this strip
Take 31.5 ÷ 10 = 3.15. In 31.5 and 3.15, the digits are
same i.e., 3, 1, and 5 but the decimal point has shifted in the
quotient. To which side and by how many digits? The decimal
point has shifted to the left by one place. Note that 10 has one
zero over 1.
Consider now 31.5 ÷ 100 = 0.315. In 31.5 and 0.315 the
digits are same, but what about the decimal point in the quotient?
It has shifted to the left by two places. Note that 100 has two zeros over1.
So we can say that, while dividing a number by 10, 100 or 1000, the digits of the
number and the quotient are same but the decimal point in the quotient shifts to the
left by as many places as there are zeros over 1. Using this observation let us now
quickly find: 2.38 ÷ 10 = 0.238, 2.38 ÷ 100 = 0.0238, 2.38 ÷ 1000 = 0.00238
You get 324. There are two digits to the right of the decimal in
12.96. Making similar placement of the decimal in 324, you will get 3.24.
Note that here and in the next section, we have considered only those
divisions in which, ignoring the decimal, the number would be completely
divisible by another number to give remainder zero. Like, in 19.5 ÷ 5, the
number 195 when divided by 5, leaves remainder zero.
However, there are situations in which the number may not be completely
divisible by another number, i.e., we may not get remainder zero. For example, 195 ÷ 7.
We deal with such situations in later classes.
.gif)