Ncert solutions

Ncert solutions

Grade 8

Linear Equations in One Variable | Exercise 2.5

Question 10

Q10) Simplify and solve the following linear equation.

0.25\left(4f-3\right)=0.05\left(10f-9\right)

Solution:

0.25\left(4f-3\right)=0.05\left(10f-9\right)

\Rightarrow\ 1.00f-0.75=0.50f-0.45

\Rightarrow\ 1.00f-0.50f=-0.45+0.75

\Rightarrow\ 0.50f=0.3

\Rightarrow\ f=\frac{0.3}{0.50}

\Rightarrow\ f=0.6

To check:

0.25\left(4f-3\right)=0.05\left(10f-9\right)

\Rightarrow\ 0.25\left(4\times0.6-3\right)=0.05\left(10\times0.6-9\right)

\Rightarrow\ 0.25\left(2.4-3\right)=0.05\left(6.0-9\right)

\Rightarrow\ 0.25\times\left(-0.6\right)=0.05\times\left(-3\right)

\Rightarrow\ -0.150=-0.150

L.H.S. = R. H. S.

Therefore, it is correct.

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subject-cta

Question 10

Q10) Simplify and solve the following linear equation.

0.25\left(4f-3\right)=0.05\left(10f-9\right)

Solution:

0.25\left(4f-3\right)=0.05\left(10f-9\right)

\Rightarrow\ 1.00f-0.75=0.50f-0.45

\Rightarrow\ 1.00f-0.50f=-0.45+0.75

\Rightarrow\ 0.50f=0.3

\Rightarrow\ f=\frac{0.3}{0.50}

\Rightarrow\ f=0.6

To check:

0.25\left(4f-3\right)=0.05\left(10f-9\right)

\Rightarrow\ 0.25\left(4\times0.6-3\right)=0.05\left(10\times0.6-9\right)

\Rightarrow\ 0.25\left(2.4-3\right)=0.05\left(6.0-9\right)

\Rightarrow\ 0.25\times\left(-0.6\right)=0.05\times\left(-3\right)

\Rightarrow\ -0.150=-0.150

L.H.S. = R. H. S.

Therefore, it is correct.

Still have questions? Our expert teachers can help you out

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