MISCELLANEOUS PROBLEMS –
Example.1) A jeweler has bars of 18 carat gold and 6 carat gold. How much of each must be melted together to obtained a bar of 12 carat gold weighing 160 gm ? it is given that pure gold is 24 carat.
18
Ans.) Percentage of gold in 18 carat gold = ( ------- X 100 ) % = 75%
24
6
Percentage of gold in 6 carat gold = ( ------- X 100 ) % = 25 %
24
12
Percentage of gold in 12 carat gold = ( ------- X 100 ) % = 50 %
24
Let x gm of 18 carat gold and y gm of 16 carat gold be mixed
Then, x + y = 160 ………………………. (i)
And, 75% of x + 25 % of y = 50% of 160
75 x 25 y 50
=> -------- + -------- = -------- X 160
100 100 100
=> 75 x + 25 y = 50 X 160
=> 3 x + y = 320 …………..……….(ii)
Now, we will subtract (i) from (ii), and we get –
3x + y = 320
x + y = 160
-----------------
2x = 160
=> x = 80
Now we will substitute x = 80 in (i), and we get –
x + y = 160
=> 80 + y = 160
=> y = 80
Hence 80 gm of 18-carat gold should be mixed with 80 gm of 6-carat gold to obtain 120 gm of 12-carat gold. (Ans.)
Example.2) A chemist has one solution containing 75% syrup and a second one containing 50% syrup. How much of each should be used to make 25 liters of a 60% syrup solution?
Ans.) Let x liters of 75% solution be mixed with y liters of 50% syrup
So, as per the given condition –
x + y = 25 ………………………..(i)
and, 75% of x + 50% of y = 40% of 15
75 x 50 y 60
=> -------- + --------- = -------- X 25
100 100 100
=> 75 x + 50 y = 60 X 25
=> 3 x + 2 y = 60 ……………………………(ii)
Now, we will multiply (i) by 3, and we will get –
3 x + 3 y = 75 ……..……………(iii)
Now, we will substitute (ii) from (iii), and we will get –
3 x + 3 y = 75
3 x + 2 y = 60
------------------
y = 15
now we will substitute y = 15 in (i), and we will get –
x + y = 25
=> x + 15 = 25
=> x = 10
Hence, 10 liters of 75% solution is to be mixed with 15 liters of 50% solution to make 25 liters of 60% solution. (Ans.)