Question

the present value of a machine is 32400.it decreases at the rate of 10% p.a. What was the value of a machine 2 years ago​

Answer 1

❏ Question:-

@The present value of a machine is 32400.it decreases at the rate of 10% p.a. What was the value of a machine 2 years ago

❏ Solution:-

➤ Given:-

• Present price of the machine($\bf P_c$)=32,400 Rs.

• Decreasing Rate (r) = 10% / year.

• Time (t)= 2 years

➤ To Find :-

• The value of a machine 2 years ago=?

➤ Solution :-

Let, Value of the machine(2 yrs ago)was= P Rs.

Now , applying the Formula of Uniform Rate of decrease.

$\sf\longrightarrow\boxed{ P_c=P(1-\frac{r}{100}){}^{n\times t}}$

$\sf\longrightarrow 32,400=P(1-\frac{10}{100}){}^{1\times 2}$

$\sf\longrightarrow 32,400=P(\frac{90}{100}){}^{ 2}$

$\sf\longrightarrow 32,400=P\times\frac{90}{100}\times\frac{90}{100}$

$\sf\longrightarrow 32,400\times\frac{100}{110}\times\frac{100}{110}=P$

$\sf\longrightarrow P=32,4\cancel{00}\times\frac{100}{9\cancel0}\times\frac{100}{9\cancel0}$

$\sf\longrightarrow P=\cancel{324}\times\frac{100}{\cancel{9}}\times\frac{100}{\cancel{9}}$

$\sf\longrightarrow\boxed{\large{\red{ P=40,000\:\: Rs} }}$

$\bf\therefore$ The value of a machine 2 years ago was= 40,000 Rs.

━━━━━━━━━━━━━━━━━━━━━━━

❏ Used ForMuLaS:-

✰INTEREST✰

➔ For Compound interest:-

$\sf\longrightarrow\boxed{ P_c=P(1+\frac{r}{100}){}^{n\times t}}$

$\sf\longrightarrow\boxed{I_c=P[(1+\frac{r}{100}){}^{n\times t}-1]}$

➔ For Simple interest:-

$\sf\longrightarrow\boxed{I_s=\frac{Prt}{100}}$

$\sf\longrightarrow\boxed{P_s=(P+\frac{Prt}{100})}$

Where ➔ • $P_c$= Total amount after

compound interest .

• $P_s$= Total amount after

Simple interest .

• P= principal amount.

• t= time.

• r= rate of interest.

• n= No. of interest cycle per year .

Ex:-

➝n=1 (when compounded annually)

➝n=2 (when compounded half yearly)

➝n=4 (when compounded quarterly)

━━━━━━━━━━━━━━━━━━━━━━━

$\underline{ \huge\mathfrak{hope \: this \: helps \: you}}$

✦ Please rate it if it helps✦

Answer 2

❏ Question:-

@The present value of a machine is 32400.it decreases at the rate of 10% p.a. What was the value of a machine 2 years ago

❏ Solution:-

➤ Given:-

• Present price of the machine($\bf P_c$)=32,400 Rs.

• Decreasing Rate (r) = 10% / year.

• Time (t)= 2 years

➤ To Find :-

• The value of a machine 2 years ago=?

➤ Solution :-

Let, Value of the machine(2 yrs ago)was= P Rs.

Now , applying the Formula of Uniform Rate of decrease.

$\sf\longrightarrow\boxed{ P_c=P(1-\frac{r}{100}){}^{n\times t}}$

$\sf\longrightarrow 32,400=P(1-\frac{10}{100}){}^{1\times 2}$

$\sf\longrightarrow 32,400=P(\frac{90}{100}){}^{ 2}$

$\sf\longrightarrow 32,400=P\times\frac{90}{100}\times\frac{90}{100}$

$\sf\longrightarrow 32,400\times\frac{100}{110}\times\frac{100}{110}=P$

$\sf\longrightarrow P=32,4\cancel{00}\times\frac{100}{9\cancel0}\times\frac{100}{9\cancel0}$

$\sf\longrightarrow P=\cancel{324}\times\frac{100}{\cancel{9}}\times\frac{100}{\cancel{9}}$

$\sf\longrightarrow\boxed{\large{\red{ P=40,000\:\: Rs} }}$

$\bf\therefore$ The value of a machine 2 years ago was= 40,000 Rs.

━━━━━━━━━━━━━━━━━━━━━━━

❏ Used ForMuLaS:-

✰INTEREST✰

➔ For Compound interest:-

$\sf\longrightarrow\boxed{ P_c=P(1+\frac{r}{100}){}^{n\times t}}$

$\sf\longrightarrow\boxed{I_c=P[(1+\frac{r}{100}){}^{n\times t}-1]}$

➔ For Simple interest:-

$\sf\longrightarrow\boxed{I_s=\frac{Prt}{100}}$

$\sf\longrightarrow\boxed{P_s=(P+\frac{Prt}{100})}$

Where ➔ • $P_c$= Total amount after

compound interest .

• $P_s$= Total amount after

Simple interest .

• P= principal amount.

• t= time.

• r= rate of interest.

• n= No. of interest cycle per year .

Ex:-

➝n=1 (when compounded annually)

➝n=2 (when compounded half yearly)

➝n=4 (when compounded quarterly)

━━━━━━━━━━━━━━━━━━━━━━━

$\underline{ \huge\mathfrak{hope \: this \: helps \: you}}$

❤️❤️Follow me❤️❤️