11th Class Mental Ability Arithmetic Notes - Arithmetic

Notes - Arithmetic

Category : 11th Class


Learning Objectives

  • Percentage
  • Ratio & Proportion
  • Terms Related to Profit and Loss
  • Profit and Loss
  • Time and Work
  • Pipes and Cisterns
  • Time and Distance





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 %.

  • % of A means \[\frac{A}{100}\] and simplifying it. Example: \[45\,%=\frac{45}{100}=\frac{9}{20}\]
  • For conversion of fraction \[\frac{p}{q}\] as percentage, we simply multiply it by 100 and out the 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.

  • If the price of an article increases by x % then the reduction in consumption, so that expenditure remains unaffected, is\[\left( \frac{x}{100+x}\times 100 \right)%\]
  • If the price of an article decreases by x % then the increase in consumption, so that expenditure remains unaffected, is \[\left( \frac{x}{100-x}\times 100 \right)%\]


Problem Based on the Population of a Locality

Suppose the present population of a locality be W and let it increases by x % per annum then

  • Population after y years \[=A{{\left( 1+\frac{x}{100} \right)}^{y}}\]
  • Population before y years =\[\frac{A}{{{\left( 1+\frac{x}{100} \right)}^{y}}}\]


Ratio and Proportion

In this chapter we will study about the comparison of two or more quantities. When we compare only two quantities of same kind, it is called ratio and more than two quantities is called proportion.



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

  • The first term of ratio is called antecedent and the second term is called the consequent.
  • If \[\frac{a}{b}=\frac{c}{d}=\frac{e}{f}=.......\]then each ratio is equal to \[\frac{a+c+e.......}{b+d+f......}\]
  • Multiplication and division by the same number (except zero) with antecedent and consequent of the ratios are equal in values, the resultant ratio remains unchanged.



It is the equality of two ratios i.e. if a : b = c : d, then ad = cd that implies product of extremes = product of means. Four quantities p, q, r, s are in proportion if ps = qr.


Important Points Related to Proportion


  • \[\frac{a+b}{b}=\frac{c+d}{d}\](componendo)
  • \[\frac{a-b}{b}=\frac{c-d}{d}\](dividendo)
  • \[\frac{a+b}{a-b}=\frac{c+d}{c-d}\](componendo and dividendo)
  • If three numbers a, b, c are in continued proportion and written as a : b :: b : c then\[\frac{a}{b}=\frac{b}{c}\Rightarrow {{b}^{2}}=ac\Rightarrow b=\sqrt{ac}\] 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 CP.


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 -s the difference between S.P. and C.P. i.e. Profit = S.P. - C.P. -and profit percent is written as: \[Profit%=\frac{Profit}{C.P.}\times 100\]

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 Rs 100.

i.e. \[Loss%=\frac{Loss}{C.P.}\times 100\]


Relation between Profit and Loss

  • To find the profit and loss when C.P. , Profit % or Loss % are given:

      (a) \[Profit=\frac{Profit%\text{ }x\text{ }C.P.}{100}\]

      (b) \[Loss=\frac{Loss%\times C.P.}{100}\]

  • To find S.P. when C.P. & profit % or Loss % are given:

      (a) \[S.P.=C.P.\times \left( \frac{100+\Pr ofit%}{100} \right)\]

      (b) \[S.P.=C.P.\times \left( \frac{100-Loss%}{100} \right)\]


  • To find C.P when S.P & profit % are given:

      (a) \[C.P.=\frac{S.P.\times 100}{100+\Pr ofit%}\]

  • To find C.P when S.P and loss % are given:

      (b) \[C.P.=\frac{S.P.\times 100}{100-Loss%}\]



  • Discount means reduction in the price. This reduction is always given on the marked price or list price.
  • When discount is offered on an article, then we calculate the selling price (S.P) as:

      SP = Marked price – Discount.

  • Discount = M.P - S.P. = Marked price - Selling price.
  • Discount % =\[\frac{Discount}{M.P}\times 100\]
  • \[~S.P=M.P.-Discount=M.P.-\frac{Discount%\times M.P.}{100}\]
  • \[~S.P=M.P.\times \left\{ \frac{100-Discount%}{100} \right\}\]
  • \[M.P.=\frac{100\times S.P.}{100-Discount\,%}\]
  • Two successive discounts of x% allowed on an item are equivalent to a single discount of \[\left( x+y-\frac{xy}{100} \right)%\]which is less than the sum of individual discounts.


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

  • If a pipe can fill a tank in x hours/then the part filled in 1 hour = \[\frac{1}{x}\]
  • If a pipe can fill a tank in x hours and another pipe can empty the full tank in y hours, then the net part filled in 1 hour, when both the pipes are opened:

            \[\left( \frac{1}{x}-\frac{1}{y} \right)\]

Time taken to fill the tank, when both the pipes are opened: \[\left( \frac{xy}{y-x} \right)\]

  • If a pipe can fill a tank in x hours and another fill the same tank in y hours, then the net part filled in 1 hr, when both pipes are opened: \[\left( \frac{1}{x}+\frac{1}{y} \right).\]

      So time to fill the tank will be: \[\left( \frac{xy}{x+y} \right)\]

  • If a pipe fills a tank in x hrs and another fills the same tank in y hrs, but a third empties the full tank in z hrs and all of them are opened together, the net part filled in 1 hr:

      \[\left( \frac{1}{x}+\frac{1}{y}-\frac{1}{z} \right)\]

            So time taken to fill the tank: \[\left( \frac{xyz}{yz+zx-xy} \right)\]


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.



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                =         \[\frac{distance\text{ }travelled}{time\,\,taken}\]

If a car travels 100 miles in 2 hours,

Average speed \[=\frac{100}{2}=50\text{ }mph\]

The following table lists units in common use for speed and their abbreviations:








miles per hour




kilometres per hour




metres per hour




metres per second




feet per second

f. p. s. or ft. per sec.



centimetres per second

cm/sec or cm/s


Calculating Speed, Distance and Time

In this section we extend the ideas of speed for calculating distances and times, using the following formula:          


\[Distance\,\,\,\,\,\,\,\,\,\,=\,\,\,\,\,\,\,\,\,\,Speed\times Time\]



Commonly Asked Questions

  1. The two numbers M and N are such that 60% of M is equal to the 20% of N, then N =? % of

(a) 300                                                  (b) 350

(c) 400                                                  (d) 100

(e) None of these

Ans.     (a)

Explanation: According to the question,

\[60%\text{ }M=20%\text{ }N\]



\[\therefore \,\,N=\frac{3\times 100}{100}M=300%\]


  1. If\[\mathbf{4x+3y:6x+5y=}\frac{\mathbf{11}}{\mathbf{17}}\mathbf{,}\]then find\[\mathbf{x:y}\].

(a) \[0:1\]                                               (b) \[2:1\]

(c) \[1:0\]                                               (d) \[5:0\]

(e) None of these

Ans.     (b)

Explanation: \[\frac{4x+3y}{6x+5y}=\frac{11}{17}\Rightarrow 17(4x+3y)=11(6x+5y)\]

\[\Rightarrow 68x+51y=66x+55y\Rightarrow 68x-66x=55y-51y\]

\[\Rightarrow 2x=4y\Rightarrow \frac{x}{y}=\frac{4}{2}\Rightarrow x:y=2:1\]


  1. By selling 42 oranges, a person losses a sum equal to the selling price of 8 oranges. Find the loss percent.

(a) 13 %                                                (b) 16 %

(c) 15%                                                 (d) 18%

(e) None of these

Ans.     (b)

Explanation: Let the S.P. of 1 orange = Rs 1

  1. P of 42 oranges \[=Rs\text{ }1\times 42=42\]

Losses = S. P. to 8 oranges = \[Rs\,\,1\times 8=\,\,Rs\,\,8\]

C P. of 42 oranges = S.P. + loss = Rs 42 + Rs 8 = Rs 50

\[Loss%=\frac{loss}{C.P}\times 100=\frac{8}{50}\times 100=16%\]


  1. The marked price of a shirt is Rs 940 and the shopkeeper allows a discount of 15% on it. Find the selling price of the shirt.

(a) Rs 895                                             (b) Rs 656

(c) Rs 785                                              (d) Rs 799

(e) None of these

Ans.     (d)

Explanation: M.P. of a shirt = Rs 940, rate of discount = 15%

Discount = 15% of \[Rs\text{ }940=\frac{15}{100}\times 940=Rs\,\,141\]

S.P of the shirt = M.P. - Discount = Rs 940 - Rs 141 = Rs 799


  1. Robert can finish the writing of the book in 8 days while James can finish the same work in 10 days. If they work together then how long they will take to finish the same work?

(a) \[10\frac{1}{2}days\]                                                 (b) \[6\frac{2}{3}days\]

(c) \[\frac{4}{9}days\]                                                    (d) \[4\frac{4}{9}\]

(e) None of these

Ans.     (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 = \[\frac{40}{9}\]


  1. A cistern has two taps which fill it in 12 minutes and 15 minutes respectively. There is also a waste pipe in the cistern. When all the three are opened, the empty cistern is full in 20 minutes.             How long will the waste pipe take to empty the full cistern?

(a) 10 minutes                                        (b) 12 minutes

(c) 15 minutes                                        (d) 13 minutes

(e) None of these

Ans.     (a)

Explanation: Work done by the waste pipe in 1 minute

\[=\frac{1}{20}-\left( \frac{1}{12}+\frac{1}{15} \right)=\frac{-1}{10}\] [- we sign means emptying]

\[\therefore \] Waste pipe will empty the full cistern in 10 minutes.


  1. Two trains of equal lengths are running on parallel tracks in the same direction at 46 km/h and 36 km/h, respectively. The faster train passes the slower train in 36 sec. The length of           each train is:

(a) 50 m                                                            (b) 80 m

(c) 72 m                                                            (d) 82 m

(e) None of these

Ans.     (a)

Explanation: Let the length of each train be x metres. Then, the total distance covered \[=\left( x+x \right)\text{ }m=2x\text{ }m\]

Relative speed \[=\left( 46-36 \right)=10\text{ }km/h=\frac{10\times 5}{18}\text{ }m/s\]

Now, \[36=\frac{2x\times 18}{50}\,\,or\,\,x=50\,m.\]

Other Topics

Notes - Arithmetic

You need to login to perform this action.
You will be redirected in 3 sec spinner