[নোটঃ এই আর্টিকেলটি (4) Decimals, Fraction, Percentage বিভাগের অধীনে 〈5〉Mixture and Coin Problems চ্যাপ্টারের অন্তর্গত, যা 〈4.5.a〉চিহ্ন দিয়ে প্রকাশ করা হয়েছে]
〈4.5.a〉বিভিন্ন ধরণের মিশ্রণ এবং মুদ্রা সম্পর্কিত সমস্যা
Chemical Mixtures
A \(\)500\(\) mL solution is \(\)20\(\)% alcohol by volume. If \(\)100\(\) mL of water is added, what is the new concentration of alcohol, as a percent of volume?
এ ধরণের অঙ্ক সমাধানের জন্যে mixture dilution সূত্র প্রয়োগ করতে হবে
Mixture dilution:
The concentration of a mixture may be expressed with percentage. The concentration always falls if water is added to this. We need to learn how this type of calculations is made:
\(\)C1V1=C2V2\(\)
Here,
\(\)C1\(\)= initial concentration
\(\)V1\(\)= initial volume
\(\)C2\(\)= final concentration
\(\)V2\(\)= final volume
Example:
Jack has \(\)3\(\) gallons of \(\)40\(\)% syrup preparation. How much water does he need to add there to make the concentration \(\)30\(\)%?
Solution: Here
\(\)C1=40\(\)%
\(\)V1=3\(\) gal
\(\)C2=30\(\)%
\(\)V2\(\) = ?
From the formula,\(\){ V }_{ 2 }=\frac { ({ C }_{ 1 }{ V }_{ 1 }) }{ { C }_{ 2 } } \(\)
or, \(\)V2=(40\(\)%\(\)*3\(\)gal)/\(\)30\(\)%=\(\)4\(\)gal
So, he needs \(\)(4-3)\(\)gal = \(\)1\(\)gal water. (Ans.)
Coin problems:
The key to these problems is to keep the quantity of coins distinct from the value of the coins. An example will illustrate.
Laura has \(\)20\(\) coins consisting of quarters and dimes. If she has a total of $\(\)3.05\(\), how many dimes does she have?
(A) 3
(B) 7
(C) 10
(D) 13
(E) 16
[Dime = 10 cent; Quarter = 25 cent]
Let \(\)D\(\) stand for the number of dimes, and let \(\)Q\(\) stand for the number of quarters. Since the total number of coins in 20, we get \(\)D+Q=20\(\), or \(\)Q=20-D\(\). Now, each dime is worth \(\)10\(\)¢, so the value of the dimes is \(\)10\(\)D. Similarly, the value of the quarters is \(\)25\(\)Q = \(\)25(20-D\(\)). Summarizing this information in a table yields:
Notice that the total value entry in the table was converted from $\(\)3.05\(\) to \(\)30\(\)5¢. Adding up the value of the dimes and the quarters yields the following equation:
\(\)10D+25(20-D)=305\(\)
or, \(\)10D+500-25D=305\(\)
or, \(\)-15D=-95\(\)
so, \(\)D=13\(\)
Hence, there are \(\)13\(\) dimes, and the answer is (\(\)D\(\)).
