A lump of two metals weighing 18 gms. is worth Rs.87 but if their weight be interchanged, it would be worth Rs.78.60. If the price of one metal be Rs.6.70 per gm., find the weight of the other metal in the mixture.
Answers
the 2 metal= 18 gms of ₹87 but in 78.60₹
1 metal per gm= 6.70₹
=6.70₹ ×18=1 metal
120.40 for 1 metal
120.40×2= two metal
answer=1240.80₹
Solution: Let the weight of first metal be x. Then the weight for other metal would be (18-x).
Also, let the price of other metal be Rs. K per gram. We know that price for the first metal is Rs. 6.7. Thus
Now, Price for first metal = Rs. 6.7 per gram
Price for second metal = Rs. K per gram
Originally, Weight of first metal = x gm and weight of second metal = (18-x) gm
Also, originally worth of the lump = Rs. 87
Therefore, 6.7x + K (18-x) = 87
When the weights are interchanged, Weight of the first metal = (18-x) gm and Weight of second metal = x gm
Therefore, 6.7(18-x) + Kx = 78.6
Adding the two equations above:
6.7x + 18K – Kx + (6.7*18) -6.7x +Kx = 87+ 78.6
=>18K + 120.6 = 165.6
=>18k = 45
=>K= 45/18 = 5/2
Substituting value of K= 5/2 in equation 1
6.7x + (5/2 * (18 – X)) = 87
=> x = 10
Thus, weight of other metal in the mixture = (18-10) gm = 8 gm