Question
Simplify and solve the linear equation $0.25(4f – 3) = 0.05 (10f – 9).$
✓
Answer
$0.25(4f – 3) = 0.05 (10f – 9)$
$\therefore f – 0.75 = 0.5f – 0.45$
$\therefore f – 0.5f = 0.45 + 0.75 ... [$Transposing $0.5f$ to $L.H.S.$ and $– 0.75$ to $R.H.S.]$
$\therefore 0.5f = 0.30$
$\therefore f = \frac{{0.30}}{{0.5}} ... [$Dividing both sides by $ 0.5]$
$\therefore f = 0.6$ this is the required solution.
$\therefore f – 0.75 = 0.5f – 0.45$
$\therefore f – 0.5f = 0.45 + 0.75 ... [$Transposing $0.5f$ to $L.H.S.$ and $– 0.75$ to $R.H.S.]$
$\therefore 0.5f = 0.30$
$\therefore f = \frac{{0.30}}{{0.5}} ... [$Dividing both sides by $ 0.5]$
$\therefore f = 0.6$ this is the required solution.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Find all possible values of $y$ for which the number $53y1$ is divisible by $3.$ Also, find each such number.
→Marble Game
Pramod is babysitting his little sister Monika and her two friends, Puja and Jyoti. Monika is wearing red, Puja is wearing blue and Jyoti is wearing green coloured clothes.
Pramod fills a bucket with 12 red (R) marbles. 8 blue (B) marbles and 4 green (G) marbles. He tells the giris that they will play a game. He will reach into the bucket and pull out a marble at random. The girl whose clothes match the colour of the marble scores 1 point.
(i) What is the probability of each girl scoring 1 point on the first draw?
Monika:
Puja:
Jyoti :
(ii) What is the probability of not drawing a green marble on the first draw?
(iii) If two marbles of each colour are added to the bucket, do the probabilities in part (a) change? Explain your answer.
(iv) If the number of each colour is doubled, do the probabilities in part (a) change? Explain why or why not.
Pramod is babysitting his little sister Monika and her two friends, Puja and Jyoti. Monika is wearing red, Puja is wearing blue and Jyoti is wearing green coloured clothes.
Pramod fills a bucket with 12 red (R) marbles. 8 blue (B) marbles and 4 green (G) marbles. He tells the giris that they will play a game. He will reach into the bucket and pull out a marble at random. The girl whose clothes match the colour of the marble scores 1 point.
(i) What is the probability of each girl scoring 1 point on the first draw?
Monika:
Puja:
Jyoti :
(ii) What is the probability of not drawing a green marble on the first draw?
(iii) If two marbles of each colour are added to the bucket, do the probabilities in part (a) change? Explain your answer.
(iv) If the number of each colour is doubled, do the probabilities in part (a) change? Explain why or why not.
Using the given pattern, find the missing numbers:
$1^2+2^2+2^2=3^2 $
$ 2^2+3^2+6^2=7^2$
$ 3^2+4^2+12^2=13^2 $
$4^2+ 5^2+\_^2= 21^2$
$5^2+\_^2+ 30^2 = 31^2$
$6^2+ 7^2+\_\_\_^2=\_\_\_^2$
→$1^2+2^2+2^2=3^2 $
$ 2^2+3^2+6^2=7^2$
$ 3^2+4^2+12^2=13^2 $
$4^2+ 5^2+\_^2= 21^2$
$5^2+\_^2+ 30^2 = 31^2$
$6^2+ 7^2+\_\_\_^2=\_\_\_^2$
Simplify:
$ (3 x+2 y)^2-(3 x-2 y)^2 $
→$ (3 x+2 y)^2-(3 x-2 y)^2 $
Simplify each of the following and write the name of property used for simplifying
(i) $\left[\frac{5}{9} \times\left(\frac{-6}{7}\right)\right] \times\left[\frac{9}{5} \times\left(\frac{21}{12}\right)\right]$
(ii) $\frac{3}{2} \times \frac{15}{11}+\frac{7}{3}-\frac{15}{11} \times\left(\frac{-19}{21}\right)$
(iii) $\frac{-4}{17} \times \frac{21}{33} \times\left(\frac{-51}{8}\right) \times\left(\frac{-11}{7}\right)$
→(i) $\left[\frac{5}{9} \times\left(\frac{-6}{7}\right)\right] \times\left[\frac{9}{5} \times\left(\frac{21}{12}\right)\right]$
(ii) $\frac{3}{2} \times \frac{15}{11}+\frac{7}{3}-\frac{15}{11} \times\left(\frac{-19}{21}\right)$
(iii) $\frac{-4}{17} \times \frac{21}{33} \times\left(\frac{-51}{8}\right) \times\left(\frac{-11}{7}\right)$
Express the following rational numbers in standard form. $\frac{-36}{-63}$
→The value of a machine depreciates every year by $5\%.$ If the present value of the machine be $Rs. 100000,$ what will be its value after $2$ years$?$
→Find two number betwen $\frac{1}{5}$ and $\frac{1}{2}$.
→Show that: $\frac{\sqrt[3]{729}}{\sqrt[3]{1000}}=\frac{\sqrt[3]{729}}{\sqrt[3]{1000}}$
→Express the product of $3.2 \times 10^6$ and $4.1 \times 10^{-1}$ in the standard form.
→