
A money box contains one rupee and two-rupee coins in the ratio \[5:6\]. If the total value of the coins in the money box is \[Rs.85\]. Find the number of two-rupee coins.
Answer
593.7k+ views
Hint: A ratio says how much of one thing there is compared to another thing.
Example
There are \[3\] shaded parts to \[1\] unshaded parts.
1. Ratio of problems are solved by taking the ratio be ‘\[x\]’.
2. Then multiply this ‘\[x\]’ on both the numbers which are present in ratio\[(a:b)\].
Like: One number become \[ = ax\]
Another number become \[ = bx\]
Complete step-by-step solution:
Given ratio \[6:6\] of one-rupee coins to two-rupee coins.
Let us assume the ratio as ‘\[x\]’.
Then, the number of one-rupee coins be \[ = 5x\]
That of two-rupee coins \[ = 6x\]
Now, Value of \[5x\] one-rupee coins \[ = Rs.1 \times 5x\] \[ = Rs.5x\].
In the same way the value of \[6x\] two-rupee coins \[ = Rs.2 \times 6x\] \[ = Rs.12x\].
As given in question total value of the coins in the money box \[Rs.85\]
So, we added both the value of one- and two-rupee coins, and put it equal to the total value.
Therefore,
\[ \Rightarrow 5x + 12x = 85\]
\[ \Rightarrow 17x = 85\]
\[ \Rightarrow x = 5\]
So, The number of two-rupee coins \[ = 6 \times 5\] \[ = 30\]
Hence, we can say that there are total \[30\] two-rupee coins in the money box.
Note: Alternate method-
Let one-rupee coins be \[ = x\] and two-rupee coins be \[ = y\]
According to the question
$\Rightarrow$ \[\dfrac{x}{y} = \dfrac{5}{6}\]
$\Rightarrow$ \[6x = 5y\]
$\Rightarrow$ \[x = \dfrac{{5y}}{6}\] ……… (i)
Total value of the coins \[ = Rs.85\]
Worth of \[x\] one-rupee coins \[ = 1 \times x\]
Worth of \[y\] two-rupee coins \[ = 2 \times y\]
Adding the worth of one- and two-rupee coins put to total value.
$\Rightarrow$ \[x + 2y = 85\]
$\Rightarrow$ \[\dfrac{{5y}}{6} + 2y = 85\] [Putting value of \[x\]from eq. (i)]
$\Rightarrow$ \[\dfrac{{5y + 12y}}{6} = 85\]
$\Rightarrow$ \[17y = 85 \times 6\]
$\Rightarrow$ \[y = \dfrac{{85 \times 6}}{{17}}\]
$\Rightarrow$ \[y = 30\]
So, no. of \[2\] rupee coin in many box \[ = 30\]
Example
There are \[3\] shaded parts to \[1\] unshaded parts.
1. Ratio of problems are solved by taking the ratio be ‘\[x\]’.
2. Then multiply this ‘\[x\]’ on both the numbers which are present in ratio\[(a:b)\].
Like: One number become \[ = ax\]
Another number become \[ = bx\]
Complete step-by-step solution:
Given ratio \[6:6\] of one-rupee coins to two-rupee coins.
Let us assume the ratio as ‘\[x\]’.
Then, the number of one-rupee coins be \[ = 5x\]
That of two-rupee coins \[ = 6x\]
Now, Value of \[5x\] one-rupee coins \[ = Rs.1 \times 5x\] \[ = Rs.5x\].
In the same way the value of \[6x\] two-rupee coins \[ = Rs.2 \times 6x\] \[ = Rs.12x\].
As given in question total value of the coins in the money box \[Rs.85\]
So, we added both the value of one- and two-rupee coins, and put it equal to the total value.
Therefore,
\[ \Rightarrow 5x + 12x = 85\]
\[ \Rightarrow 17x = 85\]
\[ \Rightarrow x = 5\]
So, The number of two-rupee coins \[ = 6 \times 5\] \[ = 30\]
Hence, we can say that there are total \[30\] two-rupee coins in the money box.
Note: Alternate method-
Let one-rupee coins be \[ = x\] and two-rupee coins be \[ = y\]
According to the question
$\Rightarrow$ \[\dfrac{x}{y} = \dfrac{5}{6}\]
$\Rightarrow$ \[6x = 5y\]
$\Rightarrow$ \[x = \dfrac{{5y}}{6}\] ……… (i)
Total value of the coins \[ = Rs.85\]
Worth of \[x\] one-rupee coins \[ = 1 \times x\]
Worth of \[y\] two-rupee coins \[ = 2 \times y\]
Adding the worth of one- and two-rupee coins put to total value.
$\Rightarrow$ \[x + 2y = 85\]
$\Rightarrow$ \[\dfrac{{5y}}{6} + 2y = 85\] [Putting value of \[x\]from eq. (i)]
$\Rightarrow$ \[\dfrac{{5y + 12y}}{6} = 85\]
$\Rightarrow$ \[17y = 85 \times 6\]
$\Rightarrow$ \[y = \dfrac{{85 \times 6}}{{17}}\]
$\Rightarrow$ \[y = 30\]
So, no. of \[2\] rupee coin in many box \[ = 30\]
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