Arithmetic
Category : 9th Class
ARITHMETIC
Learning Objectives
Percentage
Percentage
Percentage is a fraction whose denominator is 100. The numerator of the such fraction is called the rate percent.
For example: 15 percent means \[\frac{15}{100}\]and denoted by 15 %.
sign of % or mathematically we can write \[\frac{P}{q}=\left( \frac{P}{q}\times 100 \right)%.\]
Application Based Problem on Percentage
The following are the points to remember to solve the problem related to variation in the price of an article.
expenditure remains unaffected, is \[\left( \frac{X}{100+X}\times 100 \right)%\]
expenditure remains unaffected, is \[\left( \frac{X}{100X}\times 100 \right)%\]
Problem Based on the population of a Locality
Suppose the present population of a locality be ‘A’ and let it increases by x % per annum then
Ratio and Proportion
Ratio
A ratio is a relation between two quantities of same kind. Comparison is made between the two quantities by considering what part of one quantity is that of the other quantity. The two quantities are called the terms of ratio.
If x and y are two quantities of same kind then the ratio of x to y is x/y or x : y. It is represented by x : y.
Important Points Related to Ratio
\[\frac{a}{b}=\frac{c}{d}=\frac{e}{f}=\]………then each ratio is equal to \[\frac{a+c+e........}{b+d+f........}\]
Proportion
= product of means. Four quantities p, q, r, s are in proportion if ps = qr.
Important Points Related to Proportion
If \[\frac{a}{b}=\frac{c}{d}\] then
\[\frac{a}{b}=\frac{b}{c}\Rightarrow {{b}^{2}}=ac\sqrt{ac}\]hence, b is called mean.
Profit and Loss
Cost Price
It is the price of an article at which the shopkeeper purchases the goods from manufacturer or wholesaler, in short it can be written as C.R
Selling Price
It is price of the article at which it is sold by the shopkeeper to the customer. In short it can be written as S.P.
Profit and Profit Percent
If the S.P. of an article is greater than the CR then profit will occur and it is the difference between S.P. and C.P.
i.e. Profit = S.P. – C.P. and profit percent is written as: \[\operatorname{Profit}%=\frac{Profit}{C.P.}\times 100\]
Loss and Loss Percent
If the selling price of an article is less then the cost price then the difference between the cost price and the selling price is called loss.
i.e. Loss = C.P. – S.P.
The loss percent is the loss that would be made on a cost price of? 100.
i.e. \[\operatorname{Loss}%\frac{Loss}{C.P.}\times 100\]
Relation between Profit and Loss
(A) \[\operatorname{Profit}=\frac{profit%\times C.P.}{100}\]
(B) \[\operatorname{Profit}=\frac{Loss%\times C.P.}{100}\]
(A) \[\operatorname{S}.P.=C.P.\times \left( \frac{100+Profit%}{100} \right)\]
(B) \[\operatorname{S}.P.=C.P.\times \left( \frac{100Loss%}{100} \right)\]
(A) \[\operatorname{C}.P.=\frac{S.P.\times 100}{100+Profit%}\]
Discount
SP = Marked price – Discount.
Time and Work
In our daily life we come across many problems which are based on time and work. The term time and work are interrelated with each other. Time and work are directly proportional to each other. The amount of work done increases with the time and the amount of work left decreases with the number of labourers or workers and time. If the number of workers increases then the time taken to complete the work will decrease. Thus the number of workers and time are inversely proportional to each other We normally solve the problems related to the time and work using unitary method.
Important Formulae for Work Related Problems
If A can do a piece of work in ‘n’ number of days, then
Work done by A in 1 day \[=\frac{1}{n}.\]
Or conversely, if A can do the work in one day is \[=\frac{1}{n}.\]
Then A can complete the work in \[\frac{\frac{1}{1}}{n}=n\] days.
Pipes and Cisterns
As you know that a cistern or a water tank is always connected with two types of pipes. One which fills it up and the other which empties it out. The pipe which fills up the cistern is called an inlet and the one which empties it is called an outlet.
Inlet: A pipe connected with a tank or reservoir for filling is called inlet.
Outlet: A pipe connected with a tank and used for emptying is called outlet.
Shortcut Formula
Time taken to fill the tank, when both the pipes are opened: \[\left( \frac{xy}{yx} \right)\]
So time to fill the tank will be: \[\left( \frac{xy}{x+y} \right)\]
So time taken to fill the tank: \[\left( \frac{xyz}{yz+zxxy} \right)\]
Time and Distance
Time and distance and the conversions of these units also includes questions on boats in streams and jets with tailwind. It is an important topic and an aspirants definitely have to master it.
Speed
In this section we introduce the idea of speed, considering both instantaneous speed and average speed.
Instantaneous speed = speed at any instant in time
Average speed \[=\,\,\,\,\,\,\,\,\,\,\frac{distance\text{ }travelled}{time\text{ }taken}\]
If a car travels 100 miles in 2 hours,
average speed \[\begin{array}{*{35}{l}}
~\frac{100}{2}=50 \\
~ \\
\end{array}\] mph
The following table lists units in common use for speed and their abbreviations
Distance 
Time 
Speed 
Abbreviation 
mile 
hours 
miles per hour 
mph 
kilometres 
hours 
kilometres per hour 
km/h 
metres 
hours 
metres per hour 
m/h 
metres feet 
seconds 
metres per second feet per 
m/s 
centimetres 
seconds 
second centimetres per 
f.p.s. or ft. per sec. 

seconds 
second 
cm/sec or cm/s 
Calculating Speed, Distance and Time
\[Speed\,=\frac{\operatorname{Distance}}{Time\,}\]
\[\operatorname{Distance}=Speed\,\times Time\]
\[\operatorname{Time}=\,\frac{Distance}{Speed}\]
Commonly Asked Questions
(a) 300 (b) 350
(c) 400 (d) 100
(e) None of the
Answer: (a)
Explanation: According to the question,
\[60 % M = 20 % N\]
\[\frac{60}{100}M=\frac{20}{100}N\]
\[3M=N\]
\[\therefore \,\,N=\frac{3\times 100}{100}\,M\,=300%.\]
(a) \[0:\,\,1\] (b) \[2\,\,:\,\,1\]
(c) \[~1:0\] (d) \[5:0\]
(e) None of these
Answer: (b)
Explanation: \[\frac{4x+3y}{6x+5y}=\frac{11}{17}\Rightarrow \,17(4x+3y)=11(6x+5y)\]
\[\Rightarrow \,\,68x+\,\,51y=66x\,+55y\Rightarrow 68x66x=55y51y\]
Find the loss percent.
(a) 13% (b) 16%
(c) 15% (d) 18%
(e) None of these
Answer: (b)
Explanation: Let the S.P. of 1 orange = Rs.1
C P. of 42 oranges \[= S.P.+ loss \,=\,\,Rs.42+Rs.8=Rs.50\]
\[\operatorname{Loss}%=\frac{loss}{C.P.}\times 100=\,\,\frac{8}{50}\times 100=16%\]
(a) Rs. 895 (b) Rs. 656
(c) Rs. 785 (d) Rs. 799
(e) None of these
Answer: (d)
Explanation: M.P. of a shirt \[=Rs.940, rate of discount = 15%\]
Discount \[=15% of\,Rs.940= \frac{15}{100}\times \,\,940=Rs.141\]
S.P of the shirt = M.P. – Discount = Rs.940 – Rs.141 = Rs.799
(a) \[10\frac{1}{2}\] days
(b) \[6\frac{2}{3}\] days
(c) \[\frac{4}{9}\] days
(d) \[4\frac{4}{9}\]
Answer: (d)
Explanation: Work done by Robert in one day \[=\frac{1}{8}\]
Work done by James in one day \[=\frac{1}{10}\]
Work done by both in one day \[=\frac{9}{40}\]
Therefore, time taken by them to finish the work \[=\frac{40}{9}\]
(a) 10 minutes (b) 12 minutes
(c) 15 minutes (d) 13 minutes
(e) None of these
Answer: (a)
Explanation: Work done by the waste pipe in 1 minute
\[=\frac{1}{20}\left( \frac{1}{12}+\frac{1}{15} \right)=\frac{1}{10}\] [– ve sing means emptying]
\[\therefore \] Waste pipe will empty the full cistern in 10 minutes.
(a) 50 m (b) 80 m
(c) 72 m (d) 82 m
(e) None of these
Answer: (a)
Explanation: Let the length of each train be x metres. Then, the total distance covered \[=\left( x+x \right) m= 2x m\]
Relative speed \[=\left( 4636 \right)=10km/h\,=\frac{10\times 5}{18}m/s\]
Now, \[36=\frac{2x\times 18}{50}\,or\,x=50m\].
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