# 9th Class Mathematics Word Problems Based on Mathematics Word Problems Based on Mathematics

## Word Problems Based on Mathematics

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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