Lesson What type of problems are these?

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What type of problems are these?


Problem 1

There are  100  candy bars.  Some cost  75 cents;  some cost  $1.  The total cost was  $80.  How many of  $1  bars?

Solution

Let us assume for a minute that all the bars are priced  $1  each.
Then the total cost would be  $100,  in  $20 more than  $80.
It is clear that the difference is due to presence of  $0.75  bars that we intently counted as  $1.
It is also clear that the number of these  $0.75  bars is  20%2F%281.00-0.75%29 = 20%2F0.25 = 80.

It gives the answer:  80 bars in  $0.75  and  20  bars in  $1.


Problem 2

One roll cost  4$,  another cost 6$;  total sold was 480;  total amount collected  $2340.00.  How many of each were sold?

Answer

%28480%2A6+-+2340%29%2F%286-4%29 = 270 rolls by $4.

Solution

Let us assume for a minute that all the rolls are priced $6 each.
Then the total cost would be  480*6 = 2880,  in  2880-2340 = 540 dollars more than  $2340.
It is clear that the difference is due to presence of  $4  rolls that we intently counted as  $6.
It is also clear that the number of these  $4  rolls is  540%2F%286-4%29 = 540%2F2 = 270 to compensate the difference.

Thus it gives the answer:  270 in  $4  and  480-270 = 210  in  $6.


Problem 3

A daily wager is paid Rs 15 for every day he works and fines Rs 8 for each day he is absent. In October he earned Rs 258.
How many days was he absent?

Solution

One can solve the problem by reducing it to a system of two equations or to a single equation.
More elegant way is to use the common sense and mental math.

The solution is %2815%2A31+-+258%29%2F%2815%2B8%29 = %28465-238%29%2F23 = 207%2F23 = 9 days.

What is 15*31 in this formula? - It is the amount the worker had earned if he worked every day of 31 days of October.

But he earned only Rs 258. It is because he was absent on some days in October. It is the same as if he returned Rs 15
for each day he was absent, and returned also Rs 8 as the fines. 15 + 8 = 23, and it is the denominator.

Now, the numerator is exactly the amount he "returned" or loose for his absent days.

Thus the ratio in the formula is exactly the number of days the worker was absent.

The problem is solved.

The answer is: 9 days.


Problem 4

Betty buys combination of stamps of  45 cents and  65 cents.  She spent $24.50 on  50  stamps.
How many of each did she buy?

Solution

Let us assume for a minute that all the stamps cost  45 cents each.
Then the total cost would be  50*45 = 2250 cents = $22.50,  which is exactly in  $2 less than  $24.50.
It is clear that the difference is due to presence of  $0.65  stamps that we intently counted as  $0.45.
It is also clear that the number of these  $0.65  stamps is  2%2F%280.65-0.45%29 = 2%2F0.20 = 10 to compensate the difference.

It gives the answer:  10 stamps in  65 cents  and  50-10 = 40  stamps in  45 cents.


For more problems that can be solved using this approach see the lessons
    Problem on animals at a farm
    Problem on pills in containers     and
    Solving coin problems without using equations
under the current topic  Miscellaneous Word Problems  and under the topic  Coin problems  of the section  Word problems  in this site.

Use this file/link  ALGEBRA-I - YOUR ONLINE TEXTBOOK  to navigate over all topics and lessons of the online textbook  ALGEBRA-I.

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