Question

10. 0.25(4-3) = 0.05(10f – 9)

Answer 1

Answer:

Given,0.25(4f−3)=0.05(10f−9)

⟹

100

25

(4f−3)=

100

5

(10f−9)

⟹25(4f−3)=5(10f−9)[Onmultiplyingbothsidesby100]

⟹5(4f−3)=10f−9[]Ondividingbothsidesby5]

⟹20f−15=10f−9

⟹20f−10f=−9+15[Transposing10ftoLHSand−15toRHS]

⟹10f=6

⟹f=

10

6

or

5

3