**Category : **9th Class

**Word Problems Based on Mathematics**

In this chapter we will learn life problems related to mathematics.

**Problems Based on Time and Work**

Time and work are always indirectly proportional.

\[\therefore \] A person can take a time to do any work \[\alpha \frac{1}{capability\text{ }of\text{ }person\text{ }to\text{ }do\text{ }that\text{ }work}\]

Thus, we know that if a person have lot of capability to do a work, then he takes less time to do that work and if a person have less capability to do a work, then he takes more time to do that work.

Note: It is assumed that the person works at uniform rate until and unless specified.

**Problems Based on Time and Distance**

Length of the path covered by an object is called distance. Speed is the distance travelled by an object in unit time.

\[\therefore \text{ Speed = }\frac{distance}{time}\]

Average speed is the ratio of the total distance and the total time taken by an object to cover that distance.

\[\therefore \text{ Average speed = }\frac{\text{total distance covered}}{\text{total time taken}}\]

Note:

(a) When two trains are moving with velocities a km/hr and b km/hr respectively, then relative speed will be (where a > b)

i. (a - b) km/hr, if they are moving in same direction.

ii. (a + b) km/hr, if they are moving in opposite direction.

(b) When the speed of boat in still water is a km/hr and speed of stream is b km/hr then

i. speed in downstream = (a + b) km/hr.

ii. speed in upstream = (a - b) km/hr.

**Problems Based on Pipes and Cisterns**

Suppose three pipes A, B and C takes a, b and c hour respectively to fill/empty the cistern, then

(a) The part of the cistern filled/emptied by pipe A in \[1h=\frac{1}{a}\] similar for pipe B and C.

(b) The part of the cistern filled/emptied by A, B and C in \[1hr=\left( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right)part\]

**Problems Based on Profit, Loss and Discount**

For these types of problems, formulas related to profit, loss and discount are given below:

- If profit is x%, then \[S.P.=\left( \frac{100+x}{100} \right)\times C.P.\]
- \[\operatorname{Profit}/Loss%=\left( \frac{\text{Amount of profit/loss}}{\text{C}\text{.P}\text{.}}\times \text{ }100 \right)%\]
- Loss = C.P - S.P.
- Profit = S.P - C.P.
- If loss is y% then \[S.P.=\left( \frac{100-y}{100} \right)\times C.P.\]
- If a person buys two articles at total cost of Rs. R and sells one at a loss of y% and other at a profit of x%, then

\[\text{cost of one article}=\text{ }\frac{\text{C}\text{.P}\text{. of both}\times y}{x+y}\], \[\text{cost of second article}=\frac{\text{C}\text{.P}\text{. of both}\times x}{x+y}\]

Note: Resultant profit on getting two successive profits of \[{{x}_{1}}\]% and \[{{x}_{2}}\]% is \[\left( {{x}_{1}}+{{x}_{2}}+\frac{{{x}_{1}}{{x}_{2}}}{100} \right)%\]

** **

**Problems Based on Alligation or Mixture**

When two or types of quantities of things are mixed, a mixture is produced. Alligation is a rule that helps us to find the proportion in which two or more ingredient of the given price must be mixed to produce a mixture at a given price.

**Rule of Alligation**

\[\frac{\text{Quantity of cheaper ingredient}}{\text{Quantity of dearer ingredient}}=\frac{\text{CP of dearer}-\text{Mean price}}{\text{Mean price}-\text{CP of dearer}}\]

Note: A vessel contains x litre of liquid A. From this vessel, y litre are withdrawn and replaced by y litre another liquid B. Next y litre of this mixture is withdrawn and replaced by y litre of liquid B. This operation is repeated times.

\[Then,\text{ }\frac{Quantity\text{ }of\text{ }liquid\text{ }A\text{ }left\text{ }after\text{ }{{n}^{th}}\text{ }operation}{Quantity\text{ }of\text{ }liquid\text{ }A\text{ }initially\text{ }present}={{\left( \frac{x-y}{x} \right)}^{n}}\]

- Example:

Six women or four men can complete a piece of work in 24 days. 5n days will 3 women and 10 men together complete the same piece of work?

Solution: \[\because \] Six women = four men

\[\Rightarrow \] 4 men = 6 women \[\Rightarrow \text{ }10\text{ }men=\frac{6}{4}\times 10=15\text{ }women\]

\[\therefore \] 10 men + 3 women = (15 + 3) women

\[\therefore \] Required number of days\[=\frac{6\times 24}{18}=8\text{ }days\].

- Example:

A train covered a distance of 1235 KM in 19 hours. Also, the average speed of a car is six-fifth the average speed of the train. How much distance will the car cover in 24 hours?

Solution: Average speed of the train \[=\frac{1235}{19}=65\text{ }km/hr\]

Speed of the car \[=\frac{6}{5}\times 65=78\text{ }km/hr\]

Required distance\[=78\times 24=1872\text{ }km\].

- Example:

Mr. Smith purchased 100 of an article at the rate of Rs. 480 per piece. He then listed the price so as to gain a profit of 25%. While the articles, he a discount of 5%. Find the percentage of profit in the deal.

(a) 16.85% (b) 18.75%

(c) 20.25% (c) 25%

(e) None of these

Ans. (b)

Explanation: Profit percent \[=\left( x+y+\frac{xy}{100} \right)%\]

\[=\left( 25-5-\frac{25\text{ }x\text{ }5}{100} \right)%=18.75%\]

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