some useful quicker methods

QUICKER METHODS

. Number System

(i) Remainder Rule is applied to find the remainder for thesmaller division, when the same number is divided by thetwo different divisors such that one divisor is a multiple ofthe other divisor and also the remainder for the greaterdivisor is known.

If the remainder for the greater divisor = r and thesmaller divisor = d, then the remainder rule states that,when r > d the required remainder for the smaller divisorwill be the remainder found out by dividing the r by d,and when r < d, then the required remainder is r itself.

(ii) If two different numbers a and b, on being divided by thesame divisor leave remainders r1 and r2 respectively, thentheir sum (a + b), if divided by the same divisor will leaveremainder R as given below:R = (r1 + r2) Divisor = (Sum of remainders) DivisorNote: If R becomes negative in the above equation, then

the required remainder will be the sum of theremainders. That is, the required remainder = sumof remainders.

(iii) When two numbers after being divided by the same divisorleave the same remainder, then the difference of those twonumbers must be exactly divisible by the same divisor.

(iv) If a given number is divided successively by the differentfactors of the divisor leaving remainders r1, r2 and r3respectively, then the true remainder (ie remainder whenthe number is divided by the divisor) can be obtained byusing the following formula:

True remainder = (First remainder) + (Second remainder First divisor) + (Third remainder First divisor Seconddivisor).

2 Magical Book on Arithmetical Formulae

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(v) When (x + 1)n is divided by x, then the remainder is always1; where x and n are natural numbers.

(vi) When (x 1)n is divided by x, then the remainder will be 1,if n is an even natural number. But the remainder will be(x 1), if n is an odd natural number.

(vii) The sum of the digits of two-digit number is S. If the digitsare reversed, the number is decreased by N, then thenumber is as given below:

Number = 5 NS9

+

12

NS2

= 5DecreaseSum of digits

9

+12

DecreaseSum of digits9

Note:If after reversing the digits, the number is increased by N,then the number is as given below:

Number = 5 9

NS + 12 9

NS

= 5DecreaseSum of digits

9

+ 12

DecreaseSum of digits9

+ 2

1

(viii) When the difference between two-digit number and thenumber obtained by interchanging the digits is given, thenthe difference of the two digits of the two-digit number isas given below:

Difference of two digits =

Difference in original andint erchanged numbers

9Note: We cannot get the sum of two digits of the given

two-digit numbers.

(ix) A number on being divided by d1 and d2 successively leavesthe remainders r1 and r2 respectively. If the number is divided

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by d1 d2, then the remainder is given by (d1 r2 + r1).

(x) When the sum of two-digit number and the number obtainedby interchanging the digits number is as given below:Sum of two digits

= Sum of original and int erchanged numbers

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Highest Common Factor

(i) To find the greatest number that will exactly divide x, yand z.Required number = HCF of x, y and z

(ii) To find the greatest number that will divide x, y and zleaving remainders a, b and c respectively.Required number = HCF of (x a), (y b) and (z c)

(iii) To find the greatest number that will divide x, y and zleaving the same remainder r in each case.Required number = HCF of (x r), (y r) and (z r)

(iv) To find the greatest number that will divide x, y and z livingthe same remainder in each case.Required number = HCF of |(x y)|, |(y z)| and |(z x)|

(v) To find the all possible numbers, when the product of twonumbers and their HCF are given, we follow the followingsteps:

Step I: Find the value of 2Product(HCF) .

Step II: Find the possible pairs of value got in step I.Step III: Multiply the HCF with the pair of prime factors

obtained in step II.


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Least Common Multiple

(i) To find the least number which is exactly divisible by x, yand z.Required number = LCM of x, y and z

(ii) To find the least number which when divided by x, y and zleaves the remainders a, b and c respectively. It is alwaysobserved that, (x a) = (y b) = (z c) = K (say) Required number = (LCM of x, y and z) K

(iii) To find the least number which, when divided by x, y and zleaves the same remainder r in each case.Required number = (LCM of x, y and z) + r

(iv) To find the n-digit greatest number which, when divided byx, y and z,(1) leaves no remainder (ie exactly divisible) Following step-wise methods are adopted:

Step I: LCM of x, y and z = LStep II: L) n-digit greatest number (

Remainder (R)

Step III: Required number = n-digit greatest number R(2) leaves remainder K in each case

Following step-wise method is adopted:Step I: LCM of x, y and z = LStep II: L) n-digit greatest number (

Remainder (R)Step III: Required number = (n-digit greatest number

R) + K

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(v) To find the n-digit smallest number which, when divided byx, y and z.(1) leaves no remainder (ie exactly divisible) Following steps are followed:

Step I: LCM of x, y and z = LStep II: L) n-digit smallest number (

Remainder (R)

Step III: The required number = n-digit smallestnumber + (L R)

(2) leaves remainder K in each case.First two steps are the same as in the case of (1).Step III: Required number = n-digit smallest number

+ (L R) + K

(vi) To find the least number which on being divided by x, y andz leaves in each case a remiander R, but when divided by Nleaves no remainder, following step-wise methods areadopted:

Step I: Find the LCM of x, y and z say (L).Step II: Required number will be in the form of (LK + R); where K is a positive integer.Step III: N) L (Quotient (Q) (

Remainder (R0)

L = N Q + R0Now, put the vaue of L into the expression obtained instep II. required number will be in the form of (N Q + R0) K + Ror, (N Q K) + (R0K + R)Clearly, N Q K is always divisible by N.

Step IV: Now make (R0 K + R) divisible by N by putting the least value of K. Say, 1, 2, 3, 4 ......

Now, put the value of K into the expression (LK + R) whichwill be the required number.


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Exponents and SurdsLaws of Integral ExponentsFor all real numbers a and b, if m and n are positive integers, then

(i) am an = am+nFor example, 23 24 = 23+4 = 27 = 128

(ii) (am)n = amn

For example, [(2)2]3 = (2)23 = (2)6 = 61

( 2) = 164

(iii) (ab)m = ambmFor example, (2 3)4 = 24 34 = 16 81 = 1296

(iv)

m ma bb a

For example,

52 2 5 10 103 3 3 44 4 4 3

(v) m

m n m nn

aa a aa

For example, 37 34= 374 = 33 =27

(vi) 0 1 m

m m m mm

aa a a aa

For example, 75 75 = 755 = 70 =1

(vii)

n n

n

a ab b

For example, 4 4

4

2 2 163 3 81

Laws of Surds(i) For any positive integer n and a positive rational number

a, nn a a .(ii) If n is a positive integer and a, b are rational numbers,

then n n na b ab .(iii) If n is a positive integer and a, b are rational numbers,

then n

nn

a abb .

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(iv) If m and n are positive integers and a is a positive rationalnumber, then

m nn mn ma a a .(v) If m and n are positive integers and a is a positive rational

number, then

mp n p mn pmn m a a aFor example,

1

5 4 43 3 35 555 4 2 2 2 8

Average

(i) If the average age of m boys is x and the average age of nboys out of them (m boys) is y then the average age of the

rest of the boys is mx nym n

; where m > n.

(ii) If the average of n quantities is equal to x. When a quantityis removed or added the average becomes y. Then the valueof removed or added quantity is [n (x y) + y].In other words, it may be written asvalue of new entrant (or removed quantity) = Number of oldmembers Increase in average + New average.

(iii) The average weight of n persons is increased by x kg whensome of them [n1, n2, ... n, where n1 + n2 + ... < n] who weigh[y1 + y2 + ... where, y1 + y2 + ... = y kg] are replaced by thesame number of persons. Then the weight of the new personsis (y + nx).Weight of new persons = Weight of removed person + Num-bers of persons Increase in average.

(iv) The average age of n persons is decreased by x years whensome of them [n1, n2 ... n; where n1 + n2 + ...


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Age of new persons = Age of removed persons Number ofpersons Decrease in average

(v) The average of marks obtained by n candidates in a certainexamination is T. If the average marks of passed candi-dates is P and that of the failed candidates is F. Then thenumber of candidates who passed the examination is

n T FP F

.

Number of passed candidates

= Total candidates (Total Average Failed Average)

Passed Average Failed Average

(vi) If the average of n results (where n is an odd number) is a

and the average of first 1

2n

results is b and that of last

12

n is c, then

12

n th result is

12

n b c na .

(vii) If the average of n results (where n is an odd number) is a

and the average of first 1

2n

th result is b and that of

last 1

2n

th results is c , then 1

2n

th results is

12

nna b c .

(viii)If a batsman in his nth innings makes a score of x, andthereby increases his average by y, then the average aftern innings is [x y (n 1)].

(ix) If a cricketer has completed n innings and his average is xruns,then the number of runs, he must make in his nextinnings so as to raise his average to y are [n (y x) + y].

(x) If average of n consecutive odd numbers is x, then thedifference between the smallest and the largest numbers is

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given by 2(n 1).Note: We see that the above formula is independent of x.That means, this formula always holds good irrespective ofthe value of x.

(xi) Average of first n multiple of a number x is ( )2

x x n .

Percentage

(i) If two values are respectively x% and y% more than a third

value, then the first is the 100 100%100

xy

of the second.

(ii) If two values are respectively x% and y% more than a third

value, then the second is the 100 100%100

yx

of the first.

(iii) If two values are respectively x% and y% less than a third

value, then the second is the 100 100%100

yx

of the first.

(iv) If two values are respectively x% and y% less than a third

value, then the first is the 100 100%100

xy

of the second.

(v) If A is x% of C and B is y% of C, then A is 100%xy

of B.

(vi) x% of a quantity is taken by the first, y% of the remaining istaken by the second and z% of the remaining is taken bythird person. Now, if Rs A is left in the fund, then there was

100 100 100

100 100 100A

x y z in the beginning.


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(vii) If initial quantity is A and x% of the quantity is taken bythe first, y% of the remaining was taken by the second andz% of the remaining is taken by third person; then

100 100 100100 100 100

A x y z

is left in the fund.

(viii) x% of a quantity is added. Again, y% of the increased quantityis added. Again z% of the increased quantity is added.Now, it becomes A, then the initial amount is given by

100 100 100

100 100 100A

x y z .

(ix) If initial quantity is A and x% of the initial quantity is added.Again y% of the increased quantity is added. Again z% ofthe increased quantity is added, then initial quantity

becomes 100 100 100

100 100 100A x y z

.

(x) If the price of a commodity increases by r%, then thereduction in consumption so as not to increase the

expenditure is 100 %100r

r .

(xi) If the price of a commodity decreases by r%, then increasein consumption so as not to decrease expenditure on this

item is 100 %

100r

r

.

(xii) If first value is r% more than the second value, then the

second is 100 %100r

r

less than the first value.

(xiii) If the first value is r% less than the second value, then the

second value is 100 %100r

r

more than the first value.

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(xiv) If the value of a number is first increased by x% and laterdecreased by x%, then net change is always a decrease

which is equal to x% of x or 2

100x

.

(xv) If the value is first increased by x% and then decreased by

y%, then there is %100xyx y increase or decrease,

according to the +ve or ve sign respectively.

(xvi) If the value is increased successively by x% and x%, then

the final increase is given by 2

2 %100xx

.

(xvii) If the value is increased successively by x% and y%, then

the final increase is given by %100xyx y .

(xviii) If the value is decreased successively by x% and y%, then

the final decrease is given by %100xyx y .

(xix) If the value is decreased successively by x% and x%, then

the final decrease is given by 2

2 %100xx

.

(xx) (a) If the one factor is decreased by x% and the otherfactor is increased by y%.

(b) or, if the one factor is increased by x% and the otherfactor is decreased by y%, then the effect on theproduct = Increase % value Decrease % value

Increase % value Decrease % value

100

and the value

is increased or decreased according to the +ve or vesign obtained.Note: The above written formula is the general form


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of both the cases.

For Case (a) it becomes: 100yxy x

Whereas for Case (b) it becomes: 100xyx y

Thus, we see that it is more easy to remember the generalformula which works in both the cases equally.

(xxi) The pass marks in an examination is x%. If a candidatewho secures y marks fails by z marks, then the maximum

marks is given by 100 y zx

.

(xxii) A candidate scoring x% in an examination fails by a marks,while another candidate who scores y% marks gets bmarks more than the minimum required pass marks.Then the maximum marks for that examination are

M 100 a b

y x

.

(xxiii) In measuring the sides of a rectangle, one side is takenx% in excess and the other y% in deficit. The error percent in area calculated from the measurement is

100xyx y in excess or deficit, according to the +ve

or ve sign. In another form this may be written as %

error = % excess % deficit % excess % deficit

100

(xxiv) If one of the sides of a rectangle is increased by x% andthe other is increased by y%, then the per cent value by

which area changes is given by %100xyx y increase

.(xxv) If one of the sides of a rectangle is decreased by x% and

the other is decreased by y% then the per cent value by

which area changes is given by %100xyx y decrease.

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(xxvi) In an examination x% failed in English and y% failed inmaths. If z% of students failed in both the subjects, thepercentage of students who passed in both the subjectsis zyx 100 .

(xxvii) A man spends x% of his income. His income is increasedby y% and his expenditure also increases by z%, thenthe percentage increase in his savings is given by

100 %100

y xzx

.

(xxviii) A solution of salt and water contains x% salt by weight.Of it A kg water evaporates and the solution nowcontains y% of salt. The original quantity of solution is

given by yA

y x kg. In other words, it may be rewritten

as the original quantity of solution = Quantity of

evaporated water Final % of salt% Diff. of salt

.

(xxix) When a certain quantity of goods B is added to changethe percentage of goods A in a mixture of A and B, thenthe quantity of B to be added is

Previous % value of A Mixture Quantity Mixture Quantity

Changed % value of A

(xxx) If the original price of a commodity is Rs X and new priceof the commodity is Rs Y, then the decrease or increasein consumption so as not to increase or decrease the

expenditure respectively is 100 %Y X

Y

,

ie Difference in price

100 %New price

.

(xxxi) To split a number N into two parts such that one part is

p% of the other. The two split parts are 100

100N

p

and


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100p N

p

.

(xxxii) If X litres of oil was poured into a tank and it was still x%empty, then the quantity of oil that must be poured into

the tank in order to fill it to the brim is 100X x

x

litres.

(xxxiii)If X litres of oil was poured into a tank and it was still x%

empty, then the capacity of the tank is 100

100X

x

litres

.(xxxiv) If a number is successively increased by x%, y% and z%,

then single equivalent increase in that number will be

2 %100 100xy yz zx xyzx y z

.

(xxxv) A person spends x% of his monthly income on item Aand y% of the remaining on the item B. He saves theremaining amount. If the savings amount is Rs S, then

(a) the monthly income of person = Rs 2(100)

(100 )(100 )S

x y

(b) the monthly amount spent on the item

A = Rs 100

(100 )(100 )S x

x y

(c) the monthly amount spent on the item

B = Rs (100 )y S

y

Note: Here S = Savings per month.

(xxxvi) When the price of an item was increased by x%, a familyreduced its consumption in such a way that theexpenditure on the item was only y% more than before.If W kg were consumed per month before, then the new

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monthly consumption is given by 100100

yx

W kg.

(xxxvii)If the price of an item is increased by x% and a housewifereduced the consumption of that item by x%, then her

expenditure on that item decreases by 2

10x

%. Or, in

words it can be written as the following:Per cent Expenditue Change

= 2Common increase or decrease

10

%.

Note:Here -ve sign shows the decrease in expenditure,ie in the above case there is always decrease inthe expenditure.

Ratio and Proportion

(i) If two numbers are in the ratio of a : b and the sum of

these numbers is x, then these numbers will be ax

a b and

bxa b respectively.

(ii) To find the number of coins.

Number of each type of coins = Amount in rupees

Value of coins in rupees

(iii) To find the strength to milk strength of milk in the mixture

= Quantity of Milk

Total Quantity of Mixture

(iv) The contents of two vessels containing water and milk are

in the ratio 1 1x : y and x : y2 2 are mixed in the ratio x : y.The resulting mixture will have water and milk in the ratio

of 1 2 2 2 1 1 1 2 2 2 1 1:xx x y yx x y xy x y yy x y


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(v) If two numbers are in the ratio of a : b and the differencebetween these numbers is x, then these numbers will be

(a) ax

a b and bx

a b respectively. (where, a > b)

(b) ax

b a and bx

b a respectively (where a < b)

(vi) If three numbers are in the ratio of a : b : c and the sum ofthese numbers is x, then these numbers will be

,ax bxa b c a b c and

cxa b c respectively.

(vii) If the ratio between the first and the second quantities isa : b and the ratio between the second and the thirdquantities is c : d, then the ratio among first, second andthird quantities is given by ac : bc : bd. The above ratio canbe represented diagrammatically as

(viii) If the ratio between the first and the second quantities isa : b; the ratio between the second and the third quantitiesis c : d and the ratio between the third and the fourthquantities is e : f then the ratio among the first, second,third and fourth quantities is given by

(ix) If in x litres mixture of milk and water, the ratio of milkand water is a : b, the quantity of water to be added in

order to make this ratio c : d is

x ad bcc a b

.

(x) A mixture contains milk and water in the ratio a : b. If

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x litres of water is added to the mixture, milk and waterbecome in the ratio a : c. Then the quantity of milk in the

mixture is given by ax

c b and that of water is given by

bxc b .

(xi) If two quantities X and Y are in the ratio x : y. ThenX + Y : X Y : : x + y : x y

(xii) In any two two-dimensional figure, if the correspondingsides are in the ratio a : b, then their areas are in the ratioa2 : b2.

(xiii) In any two 3-dimensional figures, if the corresponding sidesor other measuring lengths are in the ratio a : b, then theirvolumes are in the ratio a3 : b3.

(xiv) The ratio between two numbers is a : b. If each number beincreased by x, the ratio becomes c : d. Then, the two

numbers are given as xa c d

ad bc

and

xb c dad bc

; where

c a d b

(xv) The incomes of two persons are in the ratio a : b and theirexpenditures are in the ratio c : d. If each of them saves

Rs X, then their incomes are given by Xa d c

ad bc

and

Xb d cad bc

.

(xvi) The incomes of two persons are in the ratio a : b and theirexpenditures are in the ratio c : d. If each of them saves

Rs X, then their expenditures are given by Xc b a

ad bc

and

Xd b aad bc

.


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(xvii) Two candles of the same height are lighted at the sametime. The first is consumed in T1 hours and the second inT2 hours. Assuming that each candle burns at a constantrate, the time after which the ratio of first candle to second

candle becomes x : y is given by

1 2

1 2

1xT Ty

x T Ty

hours.

Partnership

(i) If investments are in the ratio of a : b : c and the timing oftheir investments in the ratio of x : y : z then the ratio oftheir profits are in the ratio of ax : by : cz.

(ii) If investments are in the ratio a : b : c and profits in the ratio

p : q : r, then the ratio of time = p q r: :a b c .

(iii) Three partners invest their capitals in a business. If theratio of their periods of investments are t1 : t2 : t3 and theirprofits are in the ratio of a : b : c, then the capitals will be in

the ratio of 1 2 3

: :a b ct t t .

Profit and Loss

(i) If certain article is bought at the rate of A for a rupee, thento gain x%, the article must be sold at the rate of

100100 x A for a rupee (Remember the rule of fraction).

(ii) If a man purchases x items for Rs y and sells y items forRs x, then the profit or loss [depending upon the respective(+ve) or (ve) sign in the final result] made by him is

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2 2

2 100 %x y

y

.

(iii) If a man purchases a items for Rs b and sells c items forRs d, then the gain or loss [depending upon the respective(+ve) or (ve) sign in the final result] made by him is

100 %ad bcbc

.

(iv) Problems Based on Dishonest Dealer

% gain = Error 100

True value Error

or, % gain = True weight False weight 100

False weight

(v) (a) When there are two successive profits of x% and y%,then the resultant profit per cent is given by

100xyx y .

(b) When there is a profit of x% and loss of y% in atransaction, then the resultant profit or loss per cent is

given by 100xyx y according to the + ve and the -ve

signs respectively.

(c) When there are two successive loss of x% and y%, then

the resultant loss per cent is given by 100xyx y .

(vi) If an article is sold at a profit of x% and if both the costprice and selling price are Rs A less, the profit would be y%

more, then the cost price is Ax y

y

. In other words,

cost price = Initial Profit % Increase in profit % A

Increase in profit %


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(vii) If cost price of x articles is equal to the selling price of y

articles, then the profit percentage = 100%x y

y

.

(viii) (a) A person buys certain quantity of an article for Rs A. Ifhe sells mth part of the stock at a profit of x% and theremaining nth part at y% profit, then the per cent profit

in this transaction is mx nym n

or

First part % profit on first part Second part % profit on second part

Total of two parts

.

(b) If x part is sold at m% profit and the rest y part is soldat n% loss and Rs P is earned as overall profit, then

the value of the total consignment is Rs P 100xm ny

.

(ix) If a man buys two items A and B for Rs P and sells one itemA so as to lose x% and the other item B so as to gain y%,and on the whole he neither gains nor loses, then

(a) the cost of the item A is Py

x + y and

(b) the cost of the item B is Px

x + y .

(x) (a) By selling a certain item at the rate of X items a rupee,a man loses x%. If he wants to gain y%, then the number

of items should be sold for a rupee is 100100

x Xy

.

(b) By selling an article for Rs A, a dealer makes a profit ofx%. If he wants to make profit of y%, then he should

increase his selling price by Rs 100y x A

x

and the

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selling price is given by Rs 100100

y Ax

.

(c) By selling an article for Rs A, a dealer makes a loss ofx%. If he wants to make a profit of y%, then he should

increase his selling price by Rs 100x y A

x

and the

selling price is given by Rs 100100

y Ax

.

(xi) When each of the two commodities is sold at the sameprice Rs A, and a profit of P% is made on the first and a lossof L% is made on the second, then the percentage gain or

loss is

100 2

100 100P L PLP L

according to the +ve or ve sign.

Note: (a) In the special case when P = L we have2 2100 0 2

200 100P p

Since the sign is ve, there is always loss and the

value is given as 2% value

100.

(b) When each of the two commodities is sold at thesame price Rs A, and a profit of P% is made on thefirst and a profit of L% is made on the second,

then the percentage gain is

100 2

100 100P L PLP L

.

(xii) If a merchant, by selling his goods, has a gain of x% of theselling price, then his real gain per cent on the cost price is

100 %100

xx

.

Note: Real profit per cent is always calculated on cost priceand real profit per cent is always more than the %profit on selling price.

(xiii) If a merchant, by selling his goods, has a loss of x%, of theselling price, then his real loss per cent on the cost price is


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100100

xx

%.

(xiv) If an item is bought at the rate of X items for a rupee, thenthe number of items sold for a rupee in order to gain x% is

100100

Xx

.

Discount

(i) If a tradesman marks his goods at x% above his cost priceand allows purchasers a discount of y% for cash, then there

is 100xyx y % profit or loss according to +ve or ve sign

respectively.

Note: When x = y, then formula becomes 2

100x

. ve sign

indicates that there will be always loss.

(ii) A person marks his goods x% above the cost price but allowsy% discount for cash payment. If he sells the article for

Rs X, then the cost price is 100 100100 100

Xx y

.

(iii) If a trader buys an article at x% discount on its originalprice and sells it at y% increase on the price he buys it,then the percentage of profit he makes on the original price

is 100xyy x .

(iv) A dealer buys an item at x% discount on its original price. Ifhe sells it at a y% increase on the original price, then the

per cent profit he gets is 100100y x

x .

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(v) A businessman marks an article at Rs A and allows x%discount (on the marked price). He gains y%. If the costprice of the article is Rs B, then the selling price of thearticle can be calculatd from the equation given below

100 100100 100

A x B y = selling price.

(vi) If a person buys an article with x per cent discount on themarked price and sells the article with y per cent profit onthe marked price, then his per cent profit on the price he

buys the article is given by 100100

x yx

per cent.

(vii) A person sells articles at Rs A each after giving x% discounton marked price. Had he not given the discount, he wouldhave earned a profit of y% on the cost price. Then the cost

price of each article is given by Rs 2100

100 100A

x y

.

(viii) A certain company declares x per cent discount for wholesalebuyers. If a person buys articles from the company for Rs Aafter getting discount. He fixed up the selling price of thearticles in such a way that he earned a profit y% on originalcompany price. Then the total selling price is given by

Rs 100100

yAx

.

(ix) A shopkeeper sold an article for Rs A after giving x% discounton the labelled price and made y% profit on the cost price.Had he not given the discount, the percentage profit would

have been 100100x y

x

per cent.

(x) (a) Equivalent discount of two successive discounts x% and

y% = %100xyx y .

(b) Equivalent discount of three successive discounts x%,

y% and z% = 2 %100 (100)xy yz zx xyzx y z

.


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Simple Interest

(i) If a person deposits Rs x1 in a bank at r1% per annum andRs x2 in another bank at r2% per annum, then the rate of

interest for the whole sum is 1 1 2 2

1 2

x r x rx x

.

(ii) If the simple interest on a sum of money is n1

of the

principal, and the number of years is equal to the rate per

cent per annum, then the rate per cent is 1100n

%.

(iii) If the simple interest on certain sum P is I and thenumber of years is equal to the rate per cent per annum,

then the rate per cent or time is given by 100 I

P

.

(iv) The annual payment that will discharge a debt of Rs A duein t years at the rate of interest r% per annum is

100

1100

2

Art t

t

.

(v) If a sum of money becomes x times in t years at SI, the

rate of interest is given by 100 1 %xt

.

(vi) A certain sum is invested for certain time. It amounts toRs A1 at r1% per annum. But when invested at r2% perannum, it amounts to Rs A2, then the time is given by

1 2

2 1 1 2

100A A

A r A r

years.

(vii) A certain sum is invested for certain time. It amounts to

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Rs A1 at r1% per annum. But when invested at r2% perannum, it amounts to Rs A2, then the sum is given by Rs

2 1 1 2

1 2

A r A rr r

.

(viii) A sum was put at SI at a certain rate for t years.Had it been put at x% higher rate, it would have

fetched Rs A more, then the sum is Rs 100A

t x

or

More Interest 100Time More Rate

.

(ix) If a certain sum of money amounts to Rs A1 in t1 years and

to Rs A2 in t2 years, then the sum is given by 2 1 1 2

1 2

A t A tt t

.

(x) The simple interest on a sum of money will be Rs x after tyears. If in the next t years principal becomes n times,then the total interest at the end of the 2tth year is givenby Rs [(n + 1) x].

(xi) The simple interest on a sum of money will be Rs x aftert1 years. If in the next t2 years principal becomes n times,then the total interest at the end of (t1+

t2)th year is given

by Rs 21

1 tx nt

.

(xii) A sum of Rs X is lent out in n parts in such a way thatthe interest on first part at r1% for t1 years, the intereston second part at r2% for t2 years the interest on third partat r3% for t3 years, and so on, are equal, the ratioin which the sum was divided in n parts is given by

1 1 2 2 3 3

1 1 1 1: : : ....n nr t r t r t r t

.

(xiii) If a sum of money becomes n times at the simple interestrate of r% per annum, then it will become m times at the


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simple interest rate of 11

m rn

per cent.

(xiv) When different amounts mature to the same amountat simple rate of interest, the ratio of the amountsinvested are in inverse ratio of (100 + time rate).That is, the ratio in which the amounts are invested is

1 1 2 2

1 1:100 100r t r t 3 3

1 1: : .... :100 100 n nr t r t

.

(xv) There is a direct relationship between the principal and

the amount and is given by sum = 100

100Amount

rt

.

(xvi) A person lent a certain sum of money at r% simple interestand in t years the interest amounted to Rs A less than

the sum lent, then the sum lent is given by Rs 100

100Art

.(xvii) If a sum of money becomes n times in t years at a simple

interest, then the time in which it will amount to m times

itself is given by 11

mn

t years.

(xviii) If the simple interest on Rs P1 is less than the interest onRs P2 at r% simple interest by Rs A, then the time is

given by 2 1100A

r P P

years.

(xix) Two equal amounts of money are deposited at r1% and r2%for t1 and t2 years respectively. If the difference between

their interests is Id then the sum = 1 1 2 2

100dIr t r t

.

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(xx) If the difference between the interest receivedfrom two different banks on RsX for t years is Rs Id,then the difference between their rates is given by

100dIX t

per cent.

(xxi) If a sum amounts to Rs A1 in t1 years and Rs A2 in t2 yearsat simple rate of interest, then rate per annum

2 1

1 2 2 1

100 A AA t A t

.

(xxii) A person invested 1

1n of his capital at x1%, 2

1n at x2% and

the remainder 3

1n at x3%. If his annual income is Rs A,

the capital is given by Rs21 3

1 2 3

100Axx x

n n n

.

(xxiii) The time in which a sum of money becomes n timesitself at r% per annum simple interest is given by

1 100nr years.

Compound Interest

(i) A sum of money, placed at compound interest, becomes ntimes in t years and m times in x years. We calculate the

value of x from the equation given below: 1 1t xn m .

(ii) If the compound interest on a certain sum for2 years is Rs C and simple interest is Rs S, then


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the rate of interest per annum is 2 100 %C SS

or 2 Difference 100

SI

per cent.

(iii) When difference between the compound interest and simpleinterest on a certain sum of money for 2 years at r% rate isRs x, then the sum is given by

Sum = Difference 100 100

Rate Rate =

2 22

100 100x xr r

(iv) If the difference between CI and SI on a certain sum for 3

years at r% is Rs x, then the sum will be

3

2

Difference 100300r r

.

(v) If an amount of money grows upto Rs A1 in n years and uptoRs A2 in (n + 1) years on compound interest, then the rate

per cent is given by 2 1

1

100A AA

or

Difference of amount after years and ( 1) years 100Amount after years

n nn

.

(vi) If the compound rate of interest for the first t1years is r1%, for the next t2 years is r2%, for the next t3years is r3%, ... and the last tn years is rn%, thencompound interest on Rs x for (t1 + t2 + t3.....tn) years is

21

211 1 .... 1100 100 100

ntt tnrr rx x

.

(vii)If a sum of money, say Rs x, is divided among n parts insuch a manner that when placed at compound interest,amount obtained in each case remains equal while therate of interest on each part is r1, r2, r3, ..., rn respectivelyand time period for each part is t1 , t2, t3, ..., tn respectively,then the divided parts of the sum will in the ratio of

1 2 3

1 2 3

1 1 1 1: : : ... :1 1 11

100 100 100100

nt t t tnr r r r

.

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Mixture and Alligation

(i) The proportion in which rice at Rs x per kg must be mixedwith rice at Rs y per kg, so that the mixture be worth Rs z a

kg, is given by y zz x

.

(ii) A mixture of a certain quantity of milk with l litres of wateris worth Rs x per litre. If pure milk be worth Rs y per litre,

then the quantity of milk is given by l x

y x litres.

(iii)n gm of sugar solution has x% sugar in it. The quantity ofsugar should be added to make it y% in the solution is given

by 100y xn

y gm.

or Quantity of sugar added =

required % valueSolution

present % value100 required % value

(iv) In a group, there are some 4-legged creaturesand some 2-legged creatures. If heads are counted,there are x and if leggs are counted there are y,then the number of 4-legged creatures are given by

22

y x or

Total legs 2 Total heads2

and the number

of 2-legged creatures are given by 4

2x y

or

4 Total heads Total legs2

.

(v) If x glasses of equal size are filled with a mixture ofspirit and water. The ratio of spirit and water in each

glass are as follows: xx bababa :...,:,: 2211 . If the contents of


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all the x glasses are emptied into a single vessel, thenproportion of spirit and water in it is given by

21

1 1 2 2

... :xx x

aa aa b a b a b

21

1 1 2 2

... xx x

bb ba b a b a b

(vi) If x glasses of different sizes, say S1, S2, S3, ... Sx, are filledwith a mixture of spirit and water. The ratio of spirit andwater in each glass are as follows, a1 : b1, a2 : b2, a3 : b3, ....,ax : bx. If the contents of all the glasses are emptied into asingle vessel, then proportion of spirit and water in it is givenby

2 21 1 3 3

1 1 2 2 3 3

... :x xx x

a Sa S a S a Sa b a b a b a b

2 21 1 3 3

1 1 2 2 3 3

... x xx x

b Sb S b S b Sa b a b a b a b

Time and Work

(i) If M1 persons can do W1 works in D1 days and M2 personscan do W2 works in D2 days, then we have a very generalformula in the relationship of M1D1W2 = M2D2W1.

(ii) If M1 persons can do W1 works in D1 days working T1 hoursa day and M2 persons can do W2 works in D2 days workingT2 hours a day, then we have a very general formula in therelationship of M1 D1 T1 W2 = M2 D2 T2 W1.

(iii) If A and B can do a piece of work in x days, B and C in ydays, C and A in z days, then (A + B + C) working together

will do the same work in 2xyz days

xy yz xz

.

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Let 2xyz

xy yz xz be r, then

A alone will do the same work in yr

y r days or

2xyzxy yz zx

days,

B alone will do the same work in zr

z r days or

2xyzyz zx xy

days and

C alone will do the same work in xr

x r days or

2xyzxz xy yz

days.

(iv) If A can do a piece of work in x days and B can do it in ydays, then A and B working together will do the same work

in xy

x y days.

(v) If A, B and C can do a work in x, y and z days respectively,then all of them working together can finish the work in

xyzxy yz xz

days.

(vi) If A and B together can do a piece of work in x days and Aalone can do it in y days, then B alone can do the work in

xyy x days.

(vii) If x1 men or y1 women can reap a field in D days, then x2


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men and y2 women take to reap it 1 1

2 1 1 2

D x yx y x y

days.

(viii) If a1 men and b1 boys can do a piece of work in x days anda2 men and b2 boys can do it in y days, then the following

relationship is obtained: 1 man = 2 1

1 2

yb xbxa ya

boys..

(ix) A certain number of men can do a work in D days. If therewere x men less it could be finished in d days more,

then the number of men originally are x D d

d

.

(x) If x1 men or x2 women or x3 boys can do a work in D days,then the number of days in which 1 man, 1 woman and 1boy do the same work is given by the following formula:

Number of required days =

1 2 3

1 2 1 3 2 3

D x x xx x x x x x days.

(xi) A certain number of men can do a work in D days. If therewere x men more it could be finished in d days less,

then the number of men originally are

x D dd .

OrNumber of more workers Number of days taken by the second group

Number of less days

(xii) If A working alone takes x days more than A and B, and Bworking alone takes y days more than A and B together,then the number of days taken by A and B working to-

gether is given by xy days.

(xiii) If A and B can do a work in x and y days respectively and Aleaves the work after doing for a days, then B does the

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remaining work in

x a yx days.

(xiv) If A and B can do a work in x and y days respectively, andB leaves the work after doing for a days, then A does the

remaining work in

y a x

y days.

(xv) A and B can do a piece of work in x and y days respectivelyand both of them starts the work together. If B leaves thework a days before the completion of work, then the total

time in which the whole work is completed

y a xx y

days.

(xvi) A and B can do a piece of work in x and y days respectivelyand both of them starts the work together. If A leaves thework a days before the completion of the work, then the

total time in which the whole work is completed

x a yx y

days.

(xvii) A can do a piece of work in x days. If A does the work onlyfor a days and the remaining work is done by B in b days,

the B alone can do the work in

xbx a days.

(xviii) A and B can do a piece of work in x and y days respectively.Both starts the work together. But due to some problemsA leaves the work after some time, and B does the re-maining work in a days, then the time after which A leaves

the work is given by

y a xx y days.

(xix) A completes a work in x days. B completes the same workin y days. A started working alone and after a days B


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joined him. Then the time in which they will take together

to complete the remaining work is given by

x a yx y .

Work and Wages

(i) A can do a work in x days and B can do the same work in ydays. If the contract for the work is Rs X and both of themwork togehter, then the share of A and B is given by

Rs X y

x y

and X x

x y

respectively.

(ii) A, B and C can do a work in x, y and z days respectively. Ifdoing that work together they get an amount of Rs X, then the

Share of A = Rs Xyz

xy xz yz , Share of B = Rs

Xxzxy xz yz

and Share of C = Rs Xxy

xy xz yz and ratio of their shares is

given by A : B : C = yz : xz : xy

(iii) A, B and C contract a work for Rs X. If together A and B are

supposed to do xy of the work, then the share of C is given by

1 xRs Xy

.

Pipes and Cisterns

(i) If a pipe can fill a tank in x hours and another pipe canempty the full tank in y hours, then the net part filled in 1

hour, when both the pipes are opened

1 1x y .

time (T) taken to fill the tank, when both the pipes are

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opened = xy

y xNote: If T is (+ve), then cistern gets filled up and if T is (-ve),then cistern gets emptied.

(ii) If a pipe fills a tank in x hours and another fills the sametank in y hours, but a third one empties the full tank in zhours, and all of them are opened together, the net part

filled in 1 hour

1 1 1x y z

time taken to fill the tank

xyzyz xz xy hours.

(iii) Two pipes A and B can fill a tank in x minutes and y minutesrespectively. If both the pipes are opened simultaneously,then the time after which pipe B should be closed, so that

the tank is full in t minutes, is

ty 1x minutes.

(iv) Two pipes P and Q will fill a cistern in x hours and y hoursrespectively. If both pipes are opened together, then thetime after which the first pipe must be turned off, so that

the cistern may be just filled in t hours, is

tx 1y hours.

(v) A cistern is normally filled in x hours but takes t hourslonger to fill because of a leak in its bottom. If the cistern is

full, the leak will empty it in

x x tt hours.

Time and Diatance

(i) If a certain distance is covered at x km/hr and the samedistance is covered at y km/hr, then the average speed

during the whole journey is 2xyx y km/hr.

(ii) A person is walking at a speed of x km/hr. After every


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kilometre, if he takes rest for t hours, then the time he will

take to cover a distance of y km is 1y y tx

hour. Or,

in other words, required time Distance to be covered= +

Speed

Number of rest Time for each rest

(iii) A person covers a certain distance between two points.Having an average speed of x km/hr, he is late by x1 hours.However, with a speed of y km/hr he reaches his destinationy1 hours earlier. The distance between the two points is

given by

1 1xy x yy x

km. Or, Required distance =

Product of two speeds Difference between arrival timesDifference of two speeds

(iv) A person goes to a destination at a speed of x km/hr andreturns to his place at a speed of y km/hr. If he takesT hours in all, the distance between his place and

destination is xy T

x y

km. In other words,

Required distance =

Product of the two speedsTotal time takenAddition of the two speeds

(v) If a person does a journey in T hours and the first half atS1 km/hr and the second half at S2 km/hr, then the distance

= 1 2

1 2

2 Time S SS S

Where, S1 = Speed during first half and S2 = Speed during second half of journey

(vi) If the new speed of a person is ab of the usual speed,

then the change in time taken to cover the same

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distance is 1ba

usual time or, usual time is given by

Change in timeb 1a

hours.

(vii) If two persons A and B start from a place walking at x km/hrand y km/hr respectively, at the end of t hours, when theyare moving in same direction and x < y, they will be(y x)t km apart.

(viii) If two persons A and B start from a place walking at x km/hrand y km/hr respectively, at the end of t hours, when theyare moving in opposite directions, they will be (x + y) t kmapart.

(ix) If two runners A and B cover the same distance at the rateof x km/hr and y km/hr respectively, then the distance

travelled, when A takes t hours longer than B, is xy t

y x

km. Or

Distance = Multiplication of speeds

Difference of Speeds Difference in time to

cover the distance

(x) A person takes x hours to walk to a certain place and rideback. However, he could have gained t hours, if he hadcovered both ways by riding, then the time taken by him towalk both ways is (x + t) hours.Or,Both ways walking = One way walking and one way ridingtime + gain in time

(xi) A man takes x hours to walk to a certain place and rideback. However, if he walks both ways he needs t hours more,then the time taken by him to ride both ways is (x t) hours.


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(xii) A person A leaves a point P and reaches Q in x hours. Ifanother person B leaves the point Q, t hours later than Aand reaches the point P in y hours, then the time in which

A meets to B is (y + t)

yxx

hours.

(xiii) A person A leaves a point P and reaches Q in x hours. Ifanother person B leaves the point Q, t hours earlier than Aand reaches the point P in y hours, then the time in which

A meets to B is (y t)

yxx

hours.

(xiv) Speed and time taken are inversely proportional.Therefore, S1T1 = S2T2 = S3T3 .....Where, S1, S2, S3 ... are the speeds and T1, T2, T3 .... are thetime taken to travel the same distance.

(xv) A thief is spotted by a policeman from a distance of d km.When the policeman starts the chase, the thief also startsrunning. Assuming the speed of the thief x kilometres anhour, and that of the policeman y kilometres an hour, then

the thief will run before he is overtaken = xd

y x

km.

Or,The distance covered by the thief before he gets caught

Lead of distance= Speed of thiefRelative speed

Trains

(i) Two trains are moving in the same direction at x km/hr andy km/hr (where x > y). If the faster train crosses a man inthe slower train in t seconds, then the length of the faster

train is given by 5

18x y t metres.

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(ii) Two trains are moving in opposite directions at x km/hrand y km/hr (where x > y), if the faster train crosses a manin the slower train in t seconds, then the length of the

faster train is given by 5

18x y t metres.

(iii) A train running at x km/hr takes t1 seconds to pass aplatform. Next it takes t2 seconds to pass a man walking aty km/hr in the opposite direction, then the length of the

train is 25

18x y t metres and that of the platform is

1 2 25

18x t t yt metres.

(iv) A train running at x km/hr takes t1 seconds to passa platform. Next it takes t2 seconds to pass a manwalking at y km/hr in the same direction, then the length

of the train is 25

18x y t metres and that of the

platform is 1 2 25

18x t t yt metres.

(v) L metres long train crosses a bridge of length L1 metres inT seconds. Time taken by the train to cross a platform of L2

metres is given by 21

L LT

L L

seconds.

(vi) If L metres long train crosses a bridge or a platform of lengthL1 metres in T seconds, then the time taken by train to

cross a pole is given by 1

L TL L

seconds.

(vii) Two trains start at the same time from A and B and proceedtowards each other at the rate of x km/hr and y km/hrrespectively. When they meet it is found that one train hastravelled d km more than the other. Then the distance

between A and B is x yx y

d km. Or


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Distance = Difference in distance Sum of speeds

Difference in speeds .

(viii) A train passes by a stationary man standing onthe platform or a pole in t1 seconds and passes bythe platform completely in t2 seconds. If the lengthof the platform is p metres, then the length of

the train is 1

2 1

t pt t

metres and the speed of the

train is 2 1

pt t

m/sec. Or Length of the train

Length of the platformDifference in time

Time taken to cross astationary pole or man and

speed of the train = Length of the platform

Difference in time .

(ix) Two trains of the same length but with different speedspass a static pole in t1 seconds and t2 seconds respectively.They are moving in the same direction. The time they will

take to cross each other is given by 1 2

2 1

2t tt t

seconds.

(x) Two trains of the same length but with different speedspass a static pole in t1 seconds and t2 seconds respectively.They are moving in the opposite directions. The time they

will take to cross each other is given by 1 2

1 2

2t tt t

seconds.

(xi) Two trains of the length l1 and l2 m respectively with differentspeeds pass a static pole in t1 seconds and t2 secondsrespectively. When they are moving in the same direction,

they will cross each other in 1 2 1 2

2 1 1 2

l l t tt l t l

seconds.

(xii) Two trains of the length l1 m and l2 m respectively withdifferent speeds pass a static pole in t1 seconds and t2

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seconds respectively. When they are moving in the opposite

direction they will cross each other in 1 2 1 2

2 1 1 2

l l t tt l t l

seconds.

(xiii) Two trains of length l1 m and l2 m respectively run onparallel lines of rails. When running in the samedirection the faster train passes the slower onein t1 seconds, but when they are running in oppositedirections with the same speeds as earlier, theypass each other in t2 seconds. Then the speed of each trainis given as the following. Speed of the faster

train = 1 2 1 2

1 22l l t t

t t

m/sec and the speed of the slower

train = 1 2 1 2

1 22l l t t

t t

m/sec. Thus a general formula

for the speed is given as Average length of two trains

1 1

Opposite direction's time Same direction's time

(xiv) If a train crosses L1 m and L2 m long bridge or platform ortunnel in T1 seconds and T2 seconds respectively, then the

length of the train is 1 2 2 1

1 2

L T L TT T

m and the speed of the

train is 1 2

1 2

L LT T

m/sec.

(xv) A goods train and a passenger train are running on paralleltracks in the same or in the opposite direction. The driverof the goods train observes that the passenger train comingfrom behind overtakes and crosses his train completely inT1 seconds. Whereas a passenger on the passenger trainmarks that he crosses the goods train in T2 seconds. If thespeeds of the trains be in the ratio of a : b, then the ratio of

their lengths is given by 2

1 2

TT T

.

(xvi) Two trains A and B start from P and Q towards Q and Prespectively. After passing each other they take T1 hoursand T2 hours to reach Q and P respectively. If the train


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from P is moving x km/hr, then the speed of the other train

is 1

2

TxT

km/hr. Or Speed of the first train

Time taken by first train after meetingTime taken by second train after meeting

Boats and Streams

(i) If x km be the rate of stream and a man takes n times aslong to row up as to row down the river, then the rate of the

man in still water is given by x 11

nn

km/hr.

(ii) If the speed of the boat in still water is x km/hr and the rateof current is y km/hr, then the distance travelled downstreamin T hours is (x + y)T km, ie Distance travelled downstream= Downstream Rate Time. And the distance travelledupstream in T hours is (x y)T km, ie Distance travelledupstream = Upstream Rate Time.

(iii) A man can row x km/hr in still waters. If in a stream whichis flowing at y km/hr, it takes him z hours to row to a place

and back, the distance between the two places is 2 2

2z x y

x

.(iv) A man rows a certain distance downstream in x hours and

returns the same distance in y hours. If the stream flows atthe rate of z km/hr, then the speed of the man in still water

is given by z x yy x

km/hr. Or, Speed in still water

= Rate of stream Sum of upstream and downstream time

Differenceof upstreamanddownstreamtime

km/hr

.(v) If a man can row at a speed of x km/hr in still water to a

certain upstream point and back to the starting point in ariver which flows at y km/hr, then the averge speed for total

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journey (up + down) is given by x y x y

x

km/hr.

Plane Mensuration

(1) To find the area of an equilateral triangle if its height isgiven.

Area of the equilateral triangle = 2Height

3

(2) To find the area of a rectangle when its perimeter and diago-nal are given.

Area of a rectangle = 2 2Perimeter Diagonal

8 2

sq units.

(3) To find the perimeter of a square if its diagonal is given.Perimeter of the square = 2 2 Diagonal

(4) If the diagonal of a square becomes x times, then the areaof the square becomes x2 times.

(5) If the ratio of the areas of square A and square B is a : b,then(i) the ratio of their sides = :a b ,(ii) the ratio of their perimeters = :a b and(iii)the ratio of their diagonals = :a b .

(6) If the perimeter of a square is equal to the perimeter of a

circle, then the side of the square is 2r and radius of

the circle is 2x

. Where, x is the side of the square and r

is the radius of the circle.

(7) To find the area of a parallelogram, if the lengths of the twoadjacent sides and the length of the diagonal connectingthe ends of the two sides are given. (see the figure).


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Where, a and b are the two adjacent sides and D is thediagonal connecting the ends of the two sides.

Area of a parallelogram = 2 s s a s b s D and

S = 2a b D

(8) In a parallelogram, the sum of the squares of the diagonals= 2 (the sum of the squares of the two adjacent sides) Or,

2 2 2 21 2 2D D a b Where, D1 and D2 are the diagonals and a and b are theadjacent sides.

(9) To find the sides of a parallelogram if the distance betweenits opposite sides and the area of the parallelogram is given.

Here, ABCD is a parallelogram, h1 and h2 are the distancebetween opposite sides, l and b are the sides of theparallelogram. A is area of the parallelogram.

A = 1 2lh bh

l = 1

Ah and b = 2

Ah

(10) To find the area of a trapezium, when the lengths of parallelsides and non-parallel sides are given.

Area of a trapezium = a b s s k s c s d

k

where,

k = (a b), ie the difference between the parallel sides and cand d are the two non-parallel sides of the trapezium. And

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2k c ds .

(11) To find the perpendicular distance between the two parallelsides of the trapezium.

Perpendicular distance = 2 s s k s c s dk

where,

k = (a b), ie the difference between the parallel sides andc and d are the two non-parallel sides of the trapezium.

And s = 2k c d

(12) There are two concentric circles of radii R and r respectively.Now consider the following cases.Case I: If larger circle makes n revolutions to cover a

certain distance, then the smaller circle makes nrevolutions to cover the same distance.

Case II: If smaller circle makes Rr

n revolutions to cover

a certain distance, then the larger circle makes

r nR

revolutions to cover the same distance.

(13) Length of a carpet d m wide, required to cover the floor of a

room which is x m long and y m broad is given by xyd

m.

Or Length required = Length of room Breadth of room

Width of carpet

(14) A d m wide carpet is used to cover the floor of a roomwhich is x m long and y m broad. If the carpet is availableat Rs A per metre, then the total amount required to

cover the floor of the room is given by Rs xyAd

. Or

Amount required

= Rs length of room breadth of room

Rate per metrewidth of carpet


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Note: Length of the carpet = xyd or,

length of room breadth of roomwidth of carpet

(15) Number of tiles, each measuring d1 m d2 m, required topave a rectangular courtyard x m long and y m wide are

given by 1 2

x yd d

.

Or

Number of tiles required = length breadth of courtyardlength breadth of each tile

(16) Certain number of tiles, each measuring d1 m d2 m, arerequired to pave a rectangular courtyard x m long and y mwide. If the tiles are available at Rs A per piece, then the

amount needs to be spent is given by Rs 1 2

x yAd d

.

Or, Amount required

= price per tile length breadth of courtyardlength breadth of each tile

(17) A room x m long and y m broad is to be paved withsquare tiles of equal sizes. The largest possible tile sothat the tiles exactly fit is given by HCF of length andbreadth of the room and the number of tiles required are

2x y

HCF of x and y

.

(18) If a square hall x metres long is surrounded by a verandah(on the outside of the square hall) d metres wide, then thearea of the verandah is given by 4d(x + d) sq metres.

(19) If a square plot is x m long. It has a gravel path d m wide allround it on the inside, then the area of the path is given by4d(x d) sq m.

(20) If a rectangular hall x m long and y m broad, is surrounded

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by a verandah (on the outside of the rectangular hall) d mwide, then the area of the verandah is given by2d[(x + y) + 2d] m2. OrArea of verandah = 2(width of verandah) [length + breadthof room + 2 (width of verandah)]

(21) If a rectangular plot is x m by y m. It has a gravel path dm wide all round it on the inside, then the area of the pathis given by 2d(x + y 2d) sq m.

(22) A rectangular garden is x metres long and y metres broad.It is to be provided with pavements d metres wide all roundit both on its outside as well as inside. Then the total areaof the pavement is given by 4d(x + y) sq m.

(23) A square garden is x metres long. It is to be provided withpavements d metres wide all round it both on its outsideas well as inside. Then the total area of the pavement isgiven by (8dx) sq metres.

(24) An oblong piece of ground measures x m by y m. From thecentre of each side a path d m wide goes across to thecentre of the opposite side.I. Area of the path = d(x + y d)

= (width of path) (length + breadth of park width ofpath)II. Area of the park minus the path = (x d) (y d)

= (length of park width of path) (breadth of park width of the path)

(25) There is a square garden of side x metres. From the centreof each side a path d metres wide goes across to the centreof opposite side.I. Area of the path = d(2x d) sq metres.II. Area of the garden minus the path = (x d)2 sqmetres.

(26) To find the area of a rhombus if one side and one diagonalare given.

Area of a rhombus = diagonl 2

2 diagonalside2


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(27) To find the other diagonal of a rhombus, if perimeter ofrhombus and one of its diagonals are given. Other diagonal

= 2 2

2 diagonalside2

; where side = Perimeter

4

(28) The radius of a circular wheel is r m. The number of

revolutions it will make in travelling d km is given by d

2 r .

Or

Number of revolutions = Distance

2 r

(29) The circumference of a circular garden is c metres. Insidethe garden a road of d metres width runs round it. Thearea of the ring-shaped road is given by d (c d) or d(2r d) [ c = 2r], where r = radius of the circle.Area of ring-shaped road = width of ring (circumference ofthe circle width of ring). Or width of ring (2 radius of the circle width of ring)

OAC is a circle of radius = r, there is pathway, inside thecircle of width = d.

(30) The circumference of a circular garden is c metres. Outsidethe garden, a road of d m width runs round it. The area ofthe ring-shaped road is given by d(c + d) sq metres. Ord(2r + d) [ c = 2r] where r = radius of the circleArea of ring-shaped road = width of ring (circumference + width of ring)

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OAC is a circle of radius = r, there is pathway, outside thecircle of width d.

(31) A circular garden has ring-shaped road around it both on itsinside and outside, each of width d units. If r is the radiusof the garden, then the total area of the path is (4dr) squnits or 2Cd [where C = perimeter = 2r]

(32) To find the area of the shaded portion of the following figure:

Area of the shaded portion ABCD = 2 21 2360 r r

(33) There is an equilateral triangle of which each side is x m.With all the three corners as centres, circles are described

each of radius 2x

m. The area common to all the circles and

the triangle is 218

x or 12 (radius)2 and the area of the

remaining portion (shaded portion) of the triangle is

23 radius2 or (0.162) (radius)

2 or, (0.0405)x2.

(34) The diameter of a coin is x cm. If four of these coins beplaced on a table so that the rim of each touches that of theother two, then the area of the unoccupied space between


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them is 24

4x or

2314

x or (0.215)x2 sq cm and area of

each sector is given by 21

16x sq cm.

(35) If the length of a rectangle is increased by x%, then thepercentage decrease in width, to maintain the same area,

is given by 100100x

x .

(36) If the length of a rectangle is increased by x%, then thepercentage decrease in width, to reduce the area by y%, is

given by 100100x y

x .

(37) If the length of a rectangle is increased by x%, then thepercentage decrease in width, to increase the area by y%,

is given by Difference in and 100

100x yx

.

(38) If the length of a rectangle is decreased by x%, then thepercentage increase in width, to increase the area by y%, is

given by 100100x y

x .

(39) If the length of a rectangle is decreased by x%, then thepercentage increase in width, to maintain the same area, is

given by 100100

xx

.

(40) If length and breadth of a rectangle is increased x and y per

cent respectively, then area is increased by %100xyx y .

Note:

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If any of the two measuring sides of rectangle is decreasedthen put negative value for that in the given formula.

(41) If all the measuring sides of any two dimensional figure is

changed by x%, then its area changes by 2

2 %100xx

.

(42) If all the measuring sides of any two-dimensional figure arechanged (increased or decreased) by x% then its perimeteralso changes by the same, ie x%.

(43) If all sides of a quadrilateral are increased by x%, then itscorresponding diagonals also increased by x%.

(44) The area of the largest triangle inscribed in a semi-circle ofradius r is r2.

(45) The area of the largest circle that can be drawn in a square

of side x is 2

2x

.

(46) To find the area of the quadrilateral when its any diagonaland the perpendiculars drawn on this diagonal from othertwo vertices are given.

Area of the quadrilateral = 12 any diagonal (sum of

perpendiculars drawn on diagonal from two vertices)

(47) The area of a circle circumscribing an equilateral triangle of

side x is 23x . (See figure)


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(48) The area of a circle inscribed in an equilateral triangle of

side x is 212x . (See figure)

(49) An equilateral triangle is circumscribed by a circle andanother circle is inscribed in that triangle, then the ratio ofthe areas of the two circles is 4 : 1. (See figure)

(50) There is a relation between the number of sides and thenumber of diagonals in a polygon. The relationship is given

below. Number of diagonals = 3

2n n

; where, n = number

of sides in the polygon.

(51) The area of a rectangular plot is x times its breadth. If the

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difference between the length and breadth is y metres,then the breadth is given by (x y) metres.

Solid Mensuration

(1) To find volume of a cuboid if its area of base or top, area ofside face and area of other side face are given.

Volume of the cuboid = 1 2 3A A A

= area of base or top area of one face

area of the other face

Where, A1 = area of base or top, A2 = area of one side face and A3 = area of other side face.

(2) To find total surface area of a cuboid if the sum of all threesides and diagonal are given. Total surface area= (Sum of all three sides)2 (Diagonal)2

(3) To find number of bricks when the dimensions of brick andwall are given.

Required number of bricks = Volume of wall

Volume of one brick

(4) To find capacity, volume of material and weight of materialof a closed box, when external dimensions (ie length, breadthand height) and thickness of material of which box is made,are given.(i) Capacity of box = (External length 2 thickness)

(External breadth 2 thickness) (External height 2 thickness)

(ii) Volume of material = External Volume Capacity(iii) Weight of wood = Volume of wood Density of wood.

(5) To find the volume of a cube if the surface area of the cubeis given.

Volume of cube =

3Surface area

6


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(6) To find the volume of rain water at a place if the annualrainfall of that place is given.Volume of rain water = Height (or level) of water (ie Annualrainfall) Base area (ie area of the place)

(7) Total volume of a solid does not change even when its shapechanges. Old volume = New volume

(8) To find the number of possible cubes when disintegrationof a cube into identical cubes.

Number of cubes = 3

Original length of sideNew length of side

(9) A hollow cylindrical tube open at both ends is made of athick metal. If the internal diameter or radius and length ofthe tube are given, then the volume of metal is given by [ height (2 Internal radius + thickness) thickness] cuunits.Note: In the given formula, we can write 2 intenal radius = internal diameter

(10) A hollow cylindrical tube open at both ends is made of athick metal. If the internal and external diameter or radiusof the tube are given, then the volume of metal is given by

height 2 2External radius Internal radius cu cm.

(11) A hollow cylindrical tube open at both ends is made of athick metal. If the external diameter or radius and length ofthe tube are given, then the volume of metal is given by

height 2 outer radius thickness thickness cuunits.Note: In the given formula, we can write 2 outer radius

= outer diameter.

(12) If a rectangular sheet is rolled into a cylinder so thatthe one side becomes the height of the cylinder thenthe volume of the cylinder so formed is given by

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2height other side of the sheet4

.

(13) If a sphere of certain diameter or radius is drawn into acylinder of certain diameter or radius, then the length or

height of the cylinder is given by

3

24 (radius of sphere)

3 radius of cylinder

.

(14) If length, breadth and height of a cuboid is increased by x%,y% and z% respectively, then its volume is increased by

2%

100 100xy xz yz xyzx y z

.

(15) If side of a cube is increased by x%, then its volume

increases by 2 3

2

33 %

100 100x xx

or

3

1 1 100%100

x

.

(16) If side of a cube is increased by x%, then its surface area

increases by 2

2100xx

per cent.

(17) If the radius (or diameter) of a sphere or a hemisphereis changed by x% then its volume changes by

2 3

2

33

100 100x xx

% or

3

1 1 100%100

x

.

(18) If the radius (or diameter) of a sphere or a hemisphere ischanged by x% then its curved surface area changes by

2

2100xx

per cent.

(19) If height of a right circular cylinder is changed by x% andradius remains the same then its volume changes by x%.


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(20) If radius of a right circular cylinder is changed by x% andheight remains the same the volume changes by

2

2 %100xx

Or,

2

1 1 100%100

x

.

(21) If radius of a right circular cylinder is changed by x% andheight is changed by y% then volume changes by

2 2

2

22 %

100 100x xy x yx y

.

(22) If height and radius of a right circular cylinder both changes

by x% then volume changes by 2 3

2

33 %

100 100x xx

.

(23) If the radius of a right circular cylinder is changed by x%and height is changed by y% then curved surface area

changes by 100xyx y per cent.

(24) If height and radius of a right-circular cone both change by

x% then volume changes by 2 3

2

33 %

100 100x xx

.

(25) If the ratio of surface areas of the two spheres are given,then the ratio of their volumes will be obtained from thefollowing result:(Ratio of the surface areas)3 = (Ratio of volumes)2

(26) If the ratio of the radii of two spheres are given, then theratio of their surface areas will be obtained from the followingresult:(Ratio of radii)2 = Ratio of surface areas.

(27) If the ratio of the radii of two spheres are given, then theratio of their volumes will be obtained from the following

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result:(Ratio of radii)3 = Ratio of volumes

(28) If the ratio of the heights of two circular cylinders of equalvolume are given, then the ratio of their radii is given by thefollowing result:Ratio of radii = inverse ratio of heights

(29) If the ratio of curved surface areas of two circular cylindersof equal volume are given, then the ratio of their heights isgiven by the following result:Ratio of curved surface areas = ratio of heights .

(30) If the ratio of radii of two circular cylinders of equal volumeare given, then the ratio of their curved surface areas aregiven by the following result:Ratio of curved surface areas = inverse ratio of radii.

(31) If the ratio of heights of two circular cylinders of equal radii are given then the ratio of their volumes are given by the following

result:Ratio of volumes = Ratio of heights.

(32) If the ratio of heights of two circular cylinders of equal radiiare given then the ratio of their curved surface areas aregiven by the following result:Ratio of curved surface areas = Ratio of heights.

(33) If the ratio of volumes of two circular cylinders of equalradii are given then the ratio of their curved surface areasare given by the following result:Ratio of volumes = Ratio of curved surface areas.

(34) If the ratio of radii of two circular cylinders of equal heightsare given, then the ratio of their volumes is given by thefollowing result:Ratio of volumes = (Ratio of radii)2

(35) If the ratio of radii of two circular cylinders of equal heights


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are given, then the ratio of their curved surface areas isgiven by the following results:Ratio of curved surface areas = ratio of radii

(36) If the ratio of curved surface areas of two circular cylindersof equal heights are given, then the ratio of their volumesis given by the following results:Ratio of volumes = (Ratio of curved surface areas)2

(37) If the ratio of radii of two circular cylinders of equal curvedsurface areas are given, then the ratio of volumes iscalculated from the following result:Ratio of volumes = Ratio of radii

(38) If the ratio of heights of two circular cylinders of equal curvedsurface areas are given, then the ratio of their volumes iscalculated from the following result:Ratio of volumes = inverse ratio of heights.

(39) If the ratio of heights of two circular cylinders of equal curvedsurface areas is given, then the ratio of their radii is givenby the following result:Ratio of radii = Inverse ratio of heights.

(40) If the ratio of slant heights of two right circular cones ofequal curves surface areas is given, then the ratio of theirradii is given by the following result:Ratio of radii = inverse ratio of slant heights.

(41) If the ratio of sides of two cubes is given, then the ratio oftheir volumes is calculated from the following result:Ratio of volumes = (Ratio of sides)3

(42) If the ratio of sides of two cubes is given, then the ratio oftheir surface areas is given by the following result:Ratio of surface areas = (Ratio of sides)2

(43) If the ratio of volumes of two cubes is given, then the ratioof their surface areas is given from the following result:

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(ratio of surface areas)3 = (ratio of volumes)2

(44) If the ratio of heights (not slant height) and the ratio ofdiameters or radii of two right circular cones are given, thenthe ratio of their volumes can be calculated by the givenformula:Ratio of volumes = (Ratio of radii)2 (Ratio of heights)

(45) If the ratio of radii and the ratio of volumes of two rightcircular cones are given, then the ratio of their heights canbe calculated by the following result:

Ratio of heights = (inverse ratio of radii)2 (ratio of volumes)(46) If the ratio of volumes and the ratio of heights of two right

circular cones are given, then the ratio of their radii is givenby the following result:

ratio of radii = ratio of volumes inverse ratioof heights

(47) If the ratio of heights and the ratio of radii of two circularcylinders are given, then the ratio of their curved surfaceareas is given by (ratio of radii) (ratio of heights).

(48) If the ratio of radii and the ratio of curved surface areas oftwo circular cylinders are given then the ratio of theirheights are given by (Ratio of curved surface areas) (Inverseratio of radii).

(49) If the ratio of heights and the ratio of curved surface areasof two circular cylinders are given, then the ratio of theirradii is given by (Ratio of curved surface areas) (Inverseratio of heights)

(50) If a cylinder, a hemisphere and a cone stand on the samebase and have the same heights, then(a) The ratio of their volumes = 3 : 2 : 1 and

(b) The ratio of their curve surface areas = 2 : 2 : 1.

(51) The ratio of the volumes of a cube to that of the sphere


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which will fit inside the cube is (6 : ).

(52) If a cube of maximum volume (each corner touching thesurface from inside is cut from a sphere, then the ratio of

the volumes of the cube and the sphere is (2 : 3 ).

(53) The curved surface area of a sphere and that of a cylinder

which circumscribes the sphere is the same.

Problems Based on Ages

(i) If t years earlier the fathers age was x times that of his son.At present the fathers age is y times that of his son. Then

the present ages of the son and the father are 1t xx y

and

1t xy

x y

respectively.

(ii) If the present age of the father is x times the age of his son. tyears hence, the fathers age becomes y times the age of hisson. Then the present ages of the father and his son are

1y txx y

and 1y tx y

years respectively.

(iii) If t1 years earlier the age of the father was x times theage of his son. t2 years hence, the age of the fatherbecomes y times the age of his son. Then the present

ages of the son and the father are

2 11 1t y t x

x y

and

21Son's age 1 12 2 2

ttx + y x + y years respectively.

Note: When t1 = t2 = t, then the formulae become the following:

(a) present age of the son =

2t x yx y

years and

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(b) present age of the father

= Son's age

2 2tx y x y years

(iv) If t years earlier, the fathers age was x times that of his son.At present the fathers age is y times that of his son. Thenthe sum total of the age of the father and the son is

1 1t x yx y

years.

(v) If the present age of the father is x times that the age of hisson. t years hence, the fathers age becomes y times the ageof his son. Then the sum of the present ages of father and

his son is t y 1 x 1x y

years.

(vi) If the sum of the present ages of A and B is x years. t yearsago, the age of A was y times the age of the B. Then thepresent ages of A and B are as follows:

(a) Age of B = 1

1x t y

y

(b) Age of A = 11

xy t yy

(vii) If the sum of the present ages of A and B is x years. Aftert years, the age of A will be y times that of B. Then the agesof A and B are as given below:

(a) The age of A = 11

xy t yy

and

(b) The age of B = 1

1x t y

y

years.

(viii) If the ratio of the ages of A and B at present is a : b. After Tyears the ratio will become c : d. Then the present ages of Aand B are as follows:

Age of A =

difference of cross productsT c d T c d

a = aad bc


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Age of B =

difference of cross productsT c d T c d

b = bad bc

(ix) If the ratio of the ages of A and B at present is a : b. T yearsearlier, the ratio was c : d. Then the present

(a) Age of A =

difference of cross productsT c dT(c d)a = a

bc ad

(b) Age of B =

difference of cross productsT c dT(c d)b = b

bc ad

(x) If the ratio of the ages of A and B at present is a : b. After Tyears the ratio will become c : d. Then the sum of presentages of A and B is

T c d a bad bc

or

difference of cross productsT c d

a + b

.

(xi) If the product of the present ages of A and B is x years andthe ratio of the present ages of A and B is a : b. Then thepresent

(a) Age of A = xa

ab years and

(b) Age of B = xb

ab years

(xii) If the ratio of the ages of A and B at present is a : b. T yearsearlier, the ratio was c : d, then the sum of the present agesof A and B is

ordifference of cross products

T c d T c da + b a + b

bc ad

.

(xiii) If a mans age is x% of what it was t1 years ago, but y% ofwhat it will be after t2 years. Then his present age is

1 2xt ytx y

years.

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Note: If t1 = t2 = t, then formula will become x y tx y

years.

(xiv) The ratio of As and Bs ages is a : b. If the difference betweenthe present ages of A and B t years hence is x then

(a) the present age of A (younger) = 1

t xba

years

(b) the present age of B (older) =

1

bt xa

ba

years and

(c) the sum of the present ages of A and B

= 1

1

t x bb aa

years.

Note: Here A is always younger than B and a : b is

younger : older. Hence older

youngerba

.

(xv) The difference of present ages of A and B is x years. If tyears back their ages were in the ratio a : b, then

(a) the age of 1

xA t xab

years

(b) the age of 1

xB t ab

years and

(c) the total Age of A and B = 21

xt xab

years.

Note: Here A > B, ie A is older than B. Hence a > b

some useful quicker methods

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number systemi remainder