Question

The price of gold is directly proportional to the square of its weight. A person broke down the gold in the ratio of $3:2:1$ and faces a loss of $Rs.4620$. Find the initial price of gold.

Answer 1

Hint: First, the gold broke down into three ratios $3:2:1$. That means the total gold given is $6$ (since the ration is given as $3:2:1$ then add the three values we get, $3 + 2 + 1 = 6$)
But since the price of the gold is directly proportional to the square of gold's weight, which means the price of the gold is given as weight squared into times of price.

Complete step-by-step solution:
Since from the given, we generalized the price of the one gold is weight squared into the price is written as $P = k \times {w^2}$.
The total gold from the ration is given as six ($3 + 2 + 1 = 6$)
Now substitute the value from the ratio into the generalized formula we get, $P = k \times {w^2} \Rightarrow k \times {6^2} \Rightarrow 36k$ which is the original price of the gold with the weights.
After the broke into three parts, now we will calculate separately for the given ratio $3:2:1$
For the price of three (weight) gold, $P(3) = k \times {3^2}$ (where P is the price and three square is the weighted square with directly proportional)
Further solving we get, $P(3) = k \times {3^2} \Rightarrow 9k$ (by the multiplication operation)
For the price of two (weight) gold, $P(2) = k \times {2^2}$ (where P is the price and two square is the weighted square with directly proportional)
Further solving we get, $P(2) = k \times {2^2} \Rightarrow 4k$ (by the multiplication operation)
For the price of one (weight) gold, $P(1) = k \times {1^2}$ (where P is the price and one square is the weighted square with directly proportional)
Further solving we get, $P(1) = k \times {1^2} \Rightarrow k$ (by the multiplication operation)
Hence adding all the ratio, we get, $P(1) + P(2) + P(3) = 9 + 4 + 1 \Rightarrow 14$ which is the new weight
Hence from the original gold, we subtract the new gold weight to get the value of the k (faces a loss of $Rs.4620$),
Thus, we get, $P(6) - [P(1) + P(2) + P(3)]$ (where $P(6)$ the original price of the gold with the weights)
Hence, we get, $P(6) - [P(1) + P(2) + P(3)] = 36k - 14k \Rightarrow 22k$
Now evaluate the value k with the faces a loss of $Rs.4620$, then we get $22k = 4620$
Further solving this we get, $22k = 4620 \Rightarrow k = 210$ (by division operation)
Thus, the original price can be calculated by $P = k \times {w^2}$($k \times {6^2} \Rightarrow 36k$ which is the original price of the gold with the weights).
Applying the value of k, thus we get, $P = 210 \times 36 \Rightarrow 7560$
Thus, the initial amount is $Rs.7560$.

Note: If the gold price is not given as directly proportional to the weighted square, then the answer is different.
That is, we need to calculate the ratio first, $3 + 2 + 1 = 6$and $P = k \times w \Rightarrow k \times 6 \Rightarrow 6k$ (original price)
The ratio is the quantitative in the given relation, that between two or more than two amounts that shows the number of times one value contains or with others.
The weight of the gold is not changing but the price with the weight will be changed daily for the gold, so calculate according to a given question.