SOLUTION: The monthly telephone bill has a fixed tariff of Rs 300 for upto 50 outgoing calls. For over 50 calls there is a charge of Rs 1.5 per call. the ratio of bills paid by A and B for
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Question 1110806: The monthly telephone bill has a fixed tariff of Rs 300 for upto 50 outgoing calls. For over 50 calls there is a charge of Rs 1.5 per call. the ratio of bills paid by A and B for a particular month is 3:4 and the number of outgoing calls made by A is 90, then what is the number of outgoing calls made by B?
Found 2 solutions by mananth, greenestamps:
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
Bill for A = 300+(90-50)*1.5= 360
Bill for B
300+(x-50)1.5
= 300 +1.5x -75
225+1.5x
360/(225+1.5x) =3/4
360*4 = 3(225+1.5x)
1440= 675 -4.5x
1440-675= 4.5x
765= 4.5x
x=170 calls total
Answer by greenestamps(13198) (Show Source): You can put this solution on YOUR website!
The response from the other tutor shows a good algebraic solution, starting with expressions for the bills A and B pay, then using those expressions to set up a proportion using the information that the ratio of the bills is 3:4.
When the given information in a problem includes the ratio of two numbers, often an easier path to the solution is to start with that ratio.
Let's see how the amount of work required with that method compares with the method used by the other tutor.
Given the ratio 3:4...
let 3x = amount paid by A
let 4x = amount paid by B
A paid the flat rate of Rs 300 for the first 50 calls plus Rs 1.5 for each of the remaining (90-50)=40 calls:
3x = 300+1.5(40) = 360
x = 120
The amount B paid was 4x = 480.
If the number of calls above 50 B made is y, then
300+1.5y = 480
1.5y = 180
y = 180/1.5 = 120
The number of calls B made was 50+120 = 170.
ANSWER: 170
In this problem, the calculations required to reach the answer seem a bit easier than those required with the other method....
Is this method better? Not necessarily. The purpose of my response showing a very different path to the solution is not to show you a better method, but rather to show you that, when solving ANY problem (math problem, or any problem you encounter in your life!) you should always consider different possible ways of solving the problem.
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