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3 years ago

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Bhairav collected Rs.600 in his piggy bank by putting in Rs.2 and Rs.5 coins. The number of Rs.5 coins are twice as many as Rs.2

coins. Find the number of Rs.2 and Rs.5 coins.

1 answer:

Anna [14]3 years ago

7 0

Answer:

100 Rs 5 coins and 50 Rs 2 coins

Step-by-step explanation:

Bhairav collected Rs.600 in his piggy bank by putting in Rs.2 and Rs.5 coins. The number of Rs.5 coins are twice as many as Rs.2 coins

Let x be the number of 5 coins

and y be the number of 2 coins

The number of Rs.5 coins are twice as many as Rs.2 coins

$x=2y$

$5x+2y=600$

Replace x with 2y

$5x+2y=600\\5(2y)+2y=600\\10y+2y=600\\12y=600\\y=50$

$x=2y\\x=2(50)\\x=100$

So 100 Rs 5 coins and 50 Rs 2 coins

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(a) $\log_3(\dfrac{81}{3})=3$

(b) $\log_5(\dfrac{625}{25})=2$

(c) $\log_2(\dfrac{64}{8})=3$

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Step-by-step explanation:

Let as consider the given equations are $\log_3(\dfrac{81}{3})=?,\log_5(\dfrac{625}{25})=?,\log_2(\dfrac{64}{8})=?,\log_4(\dfrac{64}{16})=?,\log_6(36^4)=?,\log(100^3)=?$ .

(a)

$\log_3(\dfrac{81}{3})=\log_3(27)$

$\log_3(\dfrac{81}{3})=\log_3(3^3)$

$\log_3(\dfrac{81}{3})=3$ $[\because \log_aa^x=x]$

(b)

$\log_5(\dfrac{625}{25})=\log_5(25)$

$\log_5(\dfrac{625}{25})=\log_5(5^2)$

$\log_5(\dfrac{625}{25})=2$ $[\because \log_aa^x=x]$

(c)

$\log_2(\dfrac{64}{8})=\log_2(8)$

$\log_2(\dfrac{64}{8})=\log_2(2^3)$

$\log_2(\dfrac{64}{8})=3$ $[\because \log_aa^x=x]$

(d)

$\log_4(\dfrac{64}{16})=\log_4(4)$

$\log_4(\dfrac{64}{16})=1$ $[\because \log_aa^x=x]$

(e)

$\log_6(36^4)=\log_6((6^2)^4)$

$\log_6(36^4)=\log_6(6^8)$

$\log_6(36^4)=8$ $[\because \log_aa^x=x]$

(f)

$\log(100^3)=\log((10^2)^3)$

$\log(100^3)=\log(10^6)$

$\log(100^3)=6$ $[\because \log10^x=x]$

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