If $ 4.5m $ of a uniform rod weighs $ 17.1kg $ . What is the weight of $ 12m $ of such a rod?
A. $ 51.2kg $
B. $ 53kg $
C. $ 45.6kg $
D. $ 56kg $
Answer
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Hint: In order to find the weight of $ 12m $ rod. First find the weight of $ 1m $ uniform rod by dividing $ 17.1kg $ by $ 4.5m $ as because $ 4.5m $ of a uniform rod weighs $ 17.1kg $ . Then multiply the value of $ 12m $ to $ 1m $ , which would result in the weight of $ 12m $ rod.
Complete step by step solution:
It’s given that $ 4.5m $ of a uniform rod weighs $ 17.1kg $ .
Which can be written as: $ 4.5m = 17.1kg $
Dividing both the sides by $ 4.5 $ in order to find the value of $ 1m $ , and we get:
$
\dfrac{{4.5m}}{{4.5}} = \dfrac{{17.1}}{{4.5}}kg \\
\Rightarrow 1m = \dfrac{{17.1}}{{4.5}}kg \;
$
Since, we got the weight of the rod for $ 1m $ , and we need the weight of the rod for $ 12m $ .
So, multiplying both the sides of the above equation $ 1m = \dfrac{{17.1}}{{4.5}}kg $ by $ 12m $ , and we get:
$
1m \times 12 = \dfrac{{17.1}}{{4.5}} \times 12kg \\
\Rightarrow 12m = \dfrac{{17.1}}{{4.5}} \times 12kg \;
$
Solving the right side of the equation:
$
\Rightarrow 12m = \dfrac{{17.1}}{{4.5}} \times 12kg \\
\Rightarrow 12m = \dfrac{{205.2}}{{4.5}}kg \;
$
Since, there is a decimal point on the right most of the numerator and denominator. So, multiplying the numerator and denominator by $ 10 $ in order to remove the decimal, and we get:
$
\Rightarrow 12m = \dfrac{{205.2 \times 10}}{{4.5 \times 10}}kg \\
\Rightarrow 12m = \dfrac{{2052}}{{45}}kg \;
$
Now, simply dividing the term on the right-hand side and we get:
$
\Rightarrow 12m = \dfrac{{2052}}{{45}}kg \\
\Rightarrow 12m = 45.6kg \;
$
Which matches with option C.
Hence, if $ 4.5m $ of a uniform rod weighs $ 17.1kg $ , then the weight of $ 12m $ of such a rod is $ 45.6kg $ .
Therefore, the Option C is correct, that is $ 45.6kg $ .
So, the correct answer is “Option C”.
Note: We can leave the value of the result in fraction but it’s preferred to convert it in simplest form or in the form of decimal.
Similarly, if we want to find the rod of different length, we just need to multiply $ 1m $ with that value. For example: if we want to find the value $ 8m $ , then multiply both sides of
$ 1m = \dfrac{{17.1}}{{4.5}}kg $ , and then simplify.
Complete step by step solution:
It’s given that $ 4.5m $ of a uniform rod weighs $ 17.1kg $ .
Which can be written as: $ 4.5m = 17.1kg $
Dividing both the sides by $ 4.5 $ in order to find the value of $ 1m $ , and we get:
$
\dfrac{{4.5m}}{{4.5}} = \dfrac{{17.1}}{{4.5}}kg \\
\Rightarrow 1m = \dfrac{{17.1}}{{4.5}}kg \;
$
Since, we got the weight of the rod for $ 1m $ , and we need the weight of the rod for $ 12m $ .
So, multiplying both the sides of the above equation $ 1m = \dfrac{{17.1}}{{4.5}}kg $ by $ 12m $ , and we get:
$
1m \times 12 = \dfrac{{17.1}}{{4.5}} \times 12kg \\
\Rightarrow 12m = \dfrac{{17.1}}{{4.5}} \times 12kg \;
$
Solving the right side of the equation:
$
\Rightarrow 12m = \dfrac{{17.1}}{{4.5}} \times 12kg \\
\Rightarrow 12m = \dfrac{{205.2}}{{4.5}}kg \;
$
Since, there is a decimal point on the right most of the numerator and denominator. So, multiplying the numerator and denominator by $ 10 $ in order to remove the decimal, and we get:
$
\Rightarrow 12m = \dfrac{{205.2 \times 10}}{{4.5 \times 10}}kg \\
\Rightarrow 12m = \dfrac{{2052}}{{45}}kg \;
$
Now, simply dividing the term on the right-hand side and we get:
$
\Rightarrow 12m = \dfrac{{2052}}{{45}}kg \\
\Rightarrow 12m = 45.6kg \;
$
Which matches with option C.
Hence, if $ 4.5m $ of a uniform rod weighs $ 17.1kg $ , then the weight of $ 12m $ of such a rod is $ 45.6kg $ .
Therefore, the Option C is correct, that is $ 45.6kg $ .
So, the correct answer is “Option C”.
Note: We can leave the value of the result in fraction but it’s preferred to convert it in simplest form or in the form of decimal.
Similarly, if we want to find the rod of different length, we just need to multiply $ 1m $ with that value. For example: if we want to find the value $ 8m $ , then multiply both sides of
$ 1m = \dfrac{{17.1}}{{4.5}}kg $ , and then simplify.
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