The price of a car is Rs 3 25 000 the value of the car depreciates by 20% in the first year and after that, it depreciates by 25% every year. What will be the car's value after three years
Answers
Question:-
The price of a car is 3,25,000. The value of the car depreciates by 20% in the first year and after that it depreciates 25% every year. What will be the car's value after three years.
Required Answer:-
Given:-
- The price of a car is 3,25,000
- The value depreciates by 20% in the first year.
- The value depreciates 25% in second year.
- The value depreciates 25% in second year.
To Find:-
- The value of the car after 3 years.
Solution:-
First, let's see what is "Depreciation"
Depreciation refers to when the value of something goes down over time.
The value of a car is usually decreases in value with time. Therefore its value is said to depreciate.
Now, we have to find the value of the car each year respectively.
- Value depreciation on 1st year is by 20%
Value of the car in 1st year:-
=> 3,25,000 - (20% of 3,25,000)
=> 3,25,000 - (20/100 × 3,25,000)
=> 3,25,000 - 65,000
=> 2,60,000
Hence, the value of the car after 1st year is Rs. 2,60,000
Again,
- Value depreciation on 2nd year is 25%
Value of the car in 2nd year:-
=> 2,60,000 - (25% of 2,60,000)
=> 2,60,000 - (25/100 × 2,60,000)
=> 2,60,000 - 65,000
=> 1,95,000
Hence, the value of the car after 2nd year is Rs. 1,95,000
Now,
- Value depreciation on 3rd year is 25%
Value of the car in 3rd year:-
=> 1,95,000 - (25% of 1,95,000)
=> 1,95,000 - (25/100 × 1,95,000)
=> 1,95,000 - 48,750
=> 1,46,250
Hence, the value of the car in 3rd year is Rs. 1,46,250
Therefore, the value of the car after 3 years will be Rs. 1,46,250
Answer:
\begin{gathered} \begin{gathered}\\\large{\underline{\underline{\sf{\pmb{Question:}}}}}\end{gathered} \end{gathered}Question:Question:
An article is sold at a profit of 10%. Had it been sold for 30 more, the profit would have been 25%, find the C.P.
\begin{gathered} \begin{gathered}\\\large{\underline{\underline{\sf{\pmb{Required \: Answer:}}}}}\end{gathered} \end{gathered}RequiredAnswer:RequiredAnswer:
\begin{gathered} \begin{gathered} \\ \underline{\underline{\sf{\pmb{Given:}}}}\end{gathered}\end{gathered}Given:Given:
\begin{gathered} \begin{gathered}\\\sf{\rightarrow{The \: article \: is \: sold \: at \: a \: profit \: of \: 10\%}}\end{gathered}\end{gathered}→Thearticleissoldataprofitof10%
\sf{\rightarrow{It \: would \: have \: been \: a \: profit \: of \: 25\% \: if \: it \: had \: been \: sold \: for \: 30 \: more.}}→Itwouldhavebeenaprofitof25%ifithadbeensoldfor30more.
\begin{gathered} \begin{gathered} \\ \underline{\underline{\sf{\pmb{To \: Find:}}}}\end{gathered} \end{gathered}ToFind:ToFind:
\begin{gathered} \begin{gathered} \\ \sf{\rightarrow{The \: cost \: price \: of\: the \: article.}}\end{gathered}\end{gathered}→Thecostpriceofthearticle.
\begin{gathered} \begin{gathered} \\ \underline{\underline{\sf{\pmb{Solution:}}}}\\\end{gathered} \end{gathered}Solution:Solution:
Let us assume that , the cost price of the article is x
Now, let's take a careful look at the question. We are told that the article was sold at a gain of 10% . Which means the gain was 10% on the cost price. Again, it would be gain of 25% if it had been sold for 30 more. This means, the difference of the sum of money between the gain of 10% and the gain of 25% is 30.
_________________________
Let's solve this!
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\sf{In \: 10\% \: gain, \: the \: cost \: price \: is = \: \bold{x \times 10\% }}In10%gain,thecostpriceis=x×10%
\sf{\rightarrow{In \: 10\% \: gain, \: the \: cost \: price \: is = \: \bold{x \times \dfrac{10}{100} }}}→In10%gain,thecostpriceis=x×10010
\begin{gathered} \begin{gathered} \sf{\rightarrow{In \: 10\% \: gain, \: the \: cost \: price \: is = \: \bold{ \dfrac{10x}{100}} \blue{ - - - - - - - - - - - (1)} }}\\\end{gathered} \end{gathered}→In10%gain,thecostpriceis=10010x−−−−−−−−−−−(1)
Again,
If the gain was 25%,
\sf{In \: 25\% \: gain, \: the \: cost \: price \: is = \: \bold{x \times 25\% }}In25%gain,thecostpriceis=x×25%
\sf{\rightarrow{In \: 25\% \: gain, \: the \: cost \: price \: is = \: \bold{x \times \dfrac{25}{100} }}}→In25%gain,thecostpriceis=x×10025
\begin{gathered} \begin{gathered} \sf{\rightarrow{In \: 25\% \: gain, \: the \: cost \: price \: is = \: \bold{ \dfrac{25x}{100} \blue{ - - - - - - - - - - (2)}}}}\\\end{gathered}\end{gathered}→In25%gain,thecostpriceis=10025x−−−−−−−−−−(2)
From equation 1 and 2 , we can write:-
\bold{\tt{ \dfrac{25x}{100} - \dfrac{10x}{100} = 30}}10025x−10010x=30
\begin{gathered}\begin{gathered}\\\implies{\tt{ \dfrac{25x - 10x}{100} = 30}}\end{gathered}\end{gathered}⟹10025x−10x=30
\begin{gathered} \begin{gathered} \\ \implies{\tt{25x - 10x = 3000}} \: \: \: \: \: \: \boxed{\rm{Multiplying \: both \: sides \: by \: 100}}\end{gathered}\end{gathered}⟹25x−10x=3000Multiplyingbothsidesby100
\begin{gathered} \begin{gathered}\\\implies{\tt{15x = 3000}}\end{gathered} \end{gathered}⟹15x=3000
\begin{gathered} \begin{gathered} \\ \implies{\tt{x = \dfrac{3000}{15} }} \: \: \: \: \: \: \: \boxed{\rm{Dividing \: both \: sides \: by \: 15}}\end{gathered} \end{gathered}