Express \(0.6 + 0.\overline{7}+ 0.4\overline{7}\) in the form p/q , where p and q are integers and q ≠ 0
Solution:
Given, the expression is \(0.6 + 0.\overline{7}+ 0.4\overline{7}\)
We have to write the given term in p/q form.
Let x = 0.6
Multiplying both sides by 10,
10x = 10(0.6)
10x = 6
x = 6/10
x = 3/5
A recurring decimal is a decimal representation of a number whose digits are repeating its values at regular intervals.
Let \(y = 0.\overline{7}\)
So, y = 0.7777 -------------------- (1)
Multiplying both sides by 10,
10y = 10(0.7777)
10y = 7.7777 ---------------------- (2)
Subtracting (1) and (2),
10y - y = 7.7777 - 0.7777
9y = 7.0000
9y = 7
y = 7/9
Let \(a = 0.4\overline{7}\)
A recurring decimal is a decimal representation of a number whose digits are repeating its values at regular intervals.
So, a = 0.47777 --------------------- (3)
Multiplying by 10 on both sides,
10a = 10(0.47777)
10a = 4.7777 ------------------------- (4)
Subtracting (3) and (4),
10a - a = 4.7777 - 0.4777
9a = 4.3
a = 4.3/9
a = 43/90
Now, \(0.6 + 0.\overline{7}+ 0.4\overline{7}\) = 3/5 + 7/9 + 43/90
= [3(18) + 7(10) + 43]/90
= (54 + 70 + 43)/90
= 167/90
Therefore, \(0.6 + 0.\overline{7}+ 0.4\overline{7}\) = 167/90
✦ Try This: Express \(0.6 + 0.\overline{8}+ 0.7\overline{8}\) in the form p/q , where p and q are integers and q ≠ 0
☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 1
NCERT Exemplar Class 9 Maths Exercise 1.4 Problem 1
Express \(0.6 + 0.\overline{7}+ 0.4\overline{7}\) in the form p/q , where p and q are integers and q ≠ 0
Summary:
p/q form of the expression \(0.6 + 0.\overline{7}+ 0.4\overline{7}\) is 167/90, where p and q are integers and q ≠ 0
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