QUICKER METHODS
QUICKER METHODS
..
Number System
Number System
(i
(i)) Remainder Rule is applied to find the remainder for theRemainder Rule is applied to find the remainder for the smaller division, when the same number is divided by the smaller division, when the same number is divided by the two different divisors such that one divisor is a multiple of two different divisors such that one divisor is a multiple of the other divisor and also the remainder for the greater the other divisor and also the remainder for the greater divisor is known.
divisor is known.
If the remainder for the greater divisor =
If the remainder for the greater divisor = r r and the and the smaller divisor =
smaller divisor = d d , then the remainder rule states that,, then the remainder rule states that, when
when r r >> d d the required remainder for the smaller divisorthe required remainder for the smaller divisor will be the remainder found out by dividing the ‘
will be the remainder found out by dividing the ‘ r r ’ by ‘ ’ by ‘ d d ’,’, and
and whenwhen r r << d d ,, then the required remainder is ‘ then the required remainder is ‘ r r ’ itself.’ itself. (ii)
(ii) If two different numbersIf two different numbers a a and and b b , on being divided by the, on being divided by the same divisor leave remainders
same divisor leave remainders r r 11 and and r r 22 respectively, then respectively, then their sum (a + b), if divided by the same divisor will leave their sum (a + b), if divided by the same divisor will leave remainder R as given below:
remainder R as given below: R =
R = ((r r 11 + + r r 22) – Divisor) – Divisor =
= (Sum (Sum of of remainders) remainders) – – DivisorDivisor Note
Note:: If R becomes negative in thIf R becomes negative in the above equation, the above equation, thenen the required remainder will be the sum of the the required remainder will be the sum of the remainders. That is, the required remainder = sum remainders. That is, the required remainder = sum of remainders.
of remainders.
(iii)
(iii) When two numbers after being divided by the same divisorWhen two numbers after being divided by the same divisor leave the same remainder, then the difference of those two leave the same remainder, then the difference of those two numbers must be exactly divisible by the same divisor. numbers must be exactly divisible by the same divisor.
(iv)
(iv) If a given number is divided successively by the differentIf a given number is divided successively by the different factors of the divisor leaving remainders
factors of the divisor leaving remainders r r 11,, r r 22 and and r r 33 respectively, then the true remainder (ie remainder when respectively, then the true remainder (ie remainder when the number is divided by the divisor) can be obtained by the number is divided by the divisor) can be obtained by using the following formula:
using the following formula: True
True remainder remainder = = (First re(First re mainder) mainder) + + (Second (Second remainder remainder ×× First divisor) + (Third remainder × First divisor × Second First divisor) + (Third remainder × First divisor × Second divisor).
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K KUNDAN
(v
(v)) When (When (x x + + 1)1)nn is divided by is divided by x x , then the remainder is always, then the remainder is always
1; where
1; where x x and and n n are natural numbers. are natural numbers.
(vi)
(vi) When (When (x x – – 1)1)nn is divided by is divided by x x , then the remainder will be 1,, then the remainder will be 1,
if
if n n is an even natural number. But the remainder will be is an even natural number. But the remainder will be ((x x – 1), if – 1), if n is an odd natural number.n is an odd natural number.
(vii)
(vii) The sum The sum of the of the digits of digits of two-digit numbetwo-digit numbe r is r is S. If S. If the digitsthe digits are reversed, the number is decreased by N, then the are reversed, the number is decreased by N, then the number is as given below:
number is as given below: Number = 5 Number = 5 SS NN 9 9 + + 1 1 2 2 N N S S 2 2 = 5
= 5 SuSumm ofof didigigitstsDecreaseDecrease99 ++11 2 2
Decrease Decrease Su
Summ ofof didigigitsts
9 9 Note Note::
If after reversing the digits, the number is increased by N, If after reversing the digits, the number is increased by N, then the number is as given below:
then the number is as given below: Number = 5 Number = 5 S S 99 + + 11 2 2 99 S S = 5
= 5 SuSumm ofof didigigitsts DecreaseDecrease 9 9 ++ 1 1 2 2 Decrease Decrease Su
Summ ofof didigigitsts
9 9 ++ 22 1 1 (viii)
(viii) When the difference between two-digit number and theWhen the difference between two-digit number and the number obtained by interchanging the digits is given, then number obtained by interchanging the digits is given, then the difference of the two digits of the two-digit number is the difference of the two digits of the two-digit number is as given below:
as given below:
Difference of two digits = Difference of two digits =
D
Diifffferereennccee iinn ooririggiinnalal anandd in
intt ererchchanangedged numnumberberss 9
9 Note:
Note: We cannot get the sum of two digits of the givenWe cannot get the sum of two digits of the given two-digit numbers.
two-digit numbers.
(ix)
(ix) A number on being divided byA number on being divided by d d 11 and andd d 22 successively leaves successively leaves the remainders
by
by d d 11 × × d d 22, then the remainder is given by (, then the remainder is given by (d d 11 × × r r 22 + + r r 11).).
(x
(x)) When the sum of two-digit number and the number obtainedWhen the sum of two-digit number and the number obtained by interchanging the digits number is as given below: by interchanging the digits number is as given below: Sum of two digits
Sum of two digits =
= SSumum ofof ororigigininaall aandnd iinntercterchahangngeded nunumbmbererss 11
11
Highest Common Factor
Highest Common Factor
(i
(i )) To To find find the the greategreate st st numbenumbe r r that that will will exactly exactly dividedivide x x ,, y y and
and z z ..
Required number = HCF of
Required number = HCF of x x ,, y y and and z z
(ii)
(ii) To To find find the the gregreatest atest numnumber ber that that will will dividedivide x x ,, y y and and z z leaving remainders
leaving remainders a a ,, b b and and c c respectively. respectively. Required number
Required number = = HCF of (HCF of (x x – – a a ), (), (y y – – b b ) and () and (z z – – c c ))
(iii)
(iii) To To find find the the gregreatest atest numnumber ber that that will will dividedivide x x ,, y y andand z z leaving the same remainder ‘
leaving the same remainder ‘ r r ’ in each case.’ in each case. Required number = HCF of (
Required number = HCF of (x x – – r r ), (), (y y – – r r ) and () and (z z – – r r ))
(iv)
(iv) To find the greatest number that will divide To find the greatest number that will divide x x ,, y y and and z livingz living the same remainder in each case.
the same remainder in each case. Required number = HCF of |(
Required number = HCF of |(x x – – y y )|, |()|, |(y y – – z z )| and |()| and |(z z – – x x )|)|
(v
(v )) To find To find the the all possible all possible numbers, numbers, when when the the product product of of twotwo numbers and their HCF are given, we follow the following numbers and their HCF are given, we follow the following steps:
steps: Step I:
Step I: Find the value ofFind the value of 22
Pr Pr oductoduct
(HCF)
(HCF) ..
Step II
Step II:: Find Find the posthe possiblsible paie pairs ors of valuf value got e got in stin step I.ep I. Step III
Step III:: Multiply Multiply the HCF with tthe HCF with the pair of he pair of prime factorprime factorss obtained in step II.
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M a g i c a l BM a g i c a l Bo oo ok k o n Ao n Ar i tr i t h m e t i c a l F o r m u lh m e t i c a l F o r m u l a e a eLeast Common Multiple
Least Common Multiple
(i
(i)) To To find find the the least least number number which which is is exactly exactly divisible divisible byby x x ,, y y and
and z z ..
Required number = LCM of
Required number = LCM of x x ,, y y and and z z
(ii)
(ii) To To find find the the least least number number which wwhich w hen hen divided divided byby x x ,, y y and and z z leaves the remainders
leaves the remainders a a ,, b b andand c c respectively. It is always respectively. It is always observed that, (
observed that, (xx – – a a ) = ) = ((y y – – b b ) = ) = ((z z – – c c ) = K (say)) = K (say)
Required number = (LCM of Required number = (LCM of x x ,, yy andand z z ) – ) – KK (iii)(iii) To find To find the the least numleast num ber ber which, when which, when divided bydivided by x x ,, y y and and z z leaves the same remainder
leaves the same remainder r r in each case. in each case. Required number = (LCM of
Required number = (LCM of x, yx, y andand z) + r z) + r
(iv)
(iv) To To find find thethe n n -digit greatest number which, when divided by -digit greatest number which, when divided by x
x ,, y y and andz z ,, (1)
(1) leaves no remainder (ie exactly divisible) leaves no remainder (ie exactly divisible) Following step-wise methods are adopted: Following step-wise methods are adopted: Step I
Step I: LCM of: LCM of x x ,, y y and andz z = L = L Step II
Step II: : L)L) n n -digit greatest number (-digit greatest number ( Remainder (R) Remainder (R)
Step III
Step III: Required number =: Required number = n n -digit greatest number – R-digit greatest number – R (2)
(2) leaves remainder K in each caseleaves remainder K in each case
Following step-wise method is adopted: Following step-wise method is adopted: Step I
Step I: LCM of: LCM of x x ,, y y and andz z = L = L Step II:
Step II: L)L) n n -digit greatest number (-digit greatest number ( Remainder (R) Remainder (R) Step III
Step III: Required number = (: Required number = (n n -digit greatest number-digit greatest number – R) – R) + K+ K
(v
(v)) To find To find thethe n n -digit smallest number which, when divided by -digit smallest number which, when divided by x
x ,, y y and andz z .. (1)
(1) leaves no remainder (ie exactly divisible) leaves no remainder (ie exactly divisible) Following steps are followed:
Following steps are followed: Step I
Step I: LCM of: LCM of x x ,, y y and andz z = = L L Step II
Step II: L): L) n n -digit smallest number (-digit smallest number ( Remainder (R) Remainder (R) Step III
Step III: The required number =: The required number = n n -digit smallest-digit smallest
number + (L – R) number + (L – R) (2)
(2) leaves remainder K in each case.leaves remainder K in each case.
First two steps are the same as in the case of (1). First two steps are the same as in the case of (1). Step III
Step III: Required number =: Required number = n n -digit smallest number-digit smallest number + (L – R) + K + (L – R) + K (vi)
(vi) To find the least number To find the least number which on being divided bywhich on being divided by x, y x,y and and z
z leaves in each case a remiander R, but when divided by N leaves in each case a remiander R, but when divided by N leaves no remainder, following step-wise methods are leaves no remainder, following step-wise methods are adopted:
adopted: Step I
Step I: Find the LCM of: Find the LCM of x x ,, yy andand z z say (L). say (L). Step II
Step II: Required number will be in the form of : Required number will be in the form of (LK + R); where K is a positive integer. (LK + R); where K is a positive integer. Step III
Step III: N) L (Quotient (Q) (: N) L (Quotient (Q) ( Remainder (R Remainder (R00))
L = N × Q + R L = N × Q + R00
Now, put the vaue of L into the expression obtained in Now, put the vaue of L into the expression obtained in step II.
step II.
required number will be in the form of (N × Q + R required number will be in the form of (N × Q + R00) K + R) K + R or, (N × Q × K) + (R
or, (N × Q × K) + (R00K + R)K + R)
Clearly, N × Q × K is always divisible by N. Clearly, N × Q × K is always divisible by N.
Step IV
Step IV: Now make (R: Now make (R00 K + R) divisible by N by putting K + R) divisible by N by putting the least value of K. Say, 1, 2, 3, 4 ... the least value of K. Say, 1, 2, 3, 4 ... Now, put the value of K into the expression (LK + R) which Now, put the value of K into the expression (LK + R) which will be the required number.
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K KUNDAN
Exponents and Surds
Exponents and Surds
Laws of Integral Exponents
Laws of Integral Exponents
For all real numbersFor all real numbers a a and and b b , if, if m m and and n n are positive integers, then are positive integers, then (i
(i )) a a m m × ×a a n n = =a a m m ++n n
For example, 2
For example, 233 × 2 × 244 = 2 = 23+43+4 = 2 = 277 = 128 = 128
(ii)
(ii) ((a a m m ))n n = =a a mn mn
For example, [(–2) For example, [(–2)22]] –3 –3 = (–2) = (–2)2×–32×–3 = (–2) = (–2) –6 –6 = = 6 6 1 1 (( 2)2) = = 1 1 64 64 (iii)
(iii) ((ab ab ))m m = = a a m m b b m m
For example, (2 × 3) For example, (2 × 3)44 = 2 = 244 × 3 × 344 = 16 × 81 = = 16 × 81 = 12961296 (iv) (iv) m m m m a a b b b b a a For example, For example, 5 5 2 2 22 55 1100 1100 3 3 33 33 44 4 4 44 44 33 (v (v )) m m m m nn mm nn n n a a a a aa a a a a For example, 3 For example, 377 ÷ 3 ÷ 344= 3= 37–47–4 = 3 = 333 =27 =27 (vi) (vi) 00 11 m m m m mm mm mm m m a a a a aa aa aa a a For example, 7 For example, 755 ÷ 7 ÷ 755 = 7 = 75–55–5 = 7 = 700 =1 =1 (vii) (vii) n n n n n n a a a a b b b b For example, For example, 4 4 44 4 4 2 2 22 1166 3 3 33 8811
Laws of Surds
Laws of Surds
(i(i )) For any positive integer ‘ For any positive integer ‘ n n ’ and a positive rational number’ and a positive rational number ‘
‘ a a ’,’,
n n n n a
a a a .. (ii)
(ii) If If ‘ ‘ n n ’ is a positive integer and ‘ ’ is a positive integer and ‘ a ’, a ’, ‘ ‘ b b ’ ’ are are rational rational numbers,numbers, then
then nnaa
nnbb
n n aabb ..(iii)
(iii) If If ‘ ‘ n n ’ is a positive integer and ‘ ’ is a positive integer and ‘ a a ’, ’, ‘ ‘ b b ’ ’ are are rational rational numbers,numbers, then then n n n n n n a a a a b b b b ..
(iv)
(iv) If ‘ If ‘ m m ’ and ‘ ’ and ‘ n n ’ are positive integers and ‘ ’ are positive integers and ‘ a a ’ is a positive rational’ is a positive rational number, then number, then m mnnaa mnmnaa n n mma a .. (v
(v )) If ‘ If ‘ m m ’ and ‘ ’ and ‘ n n ’ are positive integers and ‘ ’ are positive integers and ‘ a a ’ is a positive rational’ is a positive rational number, then number, then
p p m m nn pp mnmn pm pm n n m m aa aa a a For example, For example,
1 1 5 5 44 44 3 3 55 33 55 33 55 5 5 44 22 22 22 88Average
Average
(i(i)) If the average age of ‘ If the average age of ‘ m m ’ boys is ‘ ’ boys is ‘ x x ’ and the average age of ‘ ’ and the average age of ‘ n n ’ ’ boys out of them (
boys out of them (mm boys) isboys) is ‘y ‘y ’ then the average age of the’ then the average age of the rest of the boys is
rest of the boys is mmxx nny y m m n n ; where; where m m > > n n .. (ii)
(ii) If the average ofIf the average of n n quantities is equal to quantities is equal to x x . When a quantity . When a quantity is removed or added the average becomes ‘
is removed or added the average becomes ‘ y y ’. Then the value’. Then the value of removed or added quantity is [
of removed or added quantity is [n n ( (x x – – y y ) +) + y y ].]. In other words, it may be written as
In other words, it may be written as
value of new entrant (or removed quantity) = Number of old value of new entrant (or removed quantity) = Number of old members × Increase in average + New average.
members × Increase in average + New average.
(iii)
(iii) The average weight The average weight of ‘ of ‘ n n ’ persons is increased by ‘ ’ persons is increased by ‘ x x ’ kg when’ kg when some of them [
some of them [n n 11,, n n 22, ..., ...n n , where, where n n 11 + + n n 22 + ... < + ... < n n ] who weigh] who weigh [[y y 11 ++ y y 22+ ... where,+ ... where, y y 11 ++ y y 22 + ... = + ... = y y kg] are replaced by the kg] are replaced by the same number of persons. Then the weight of the new
same number of persons. Then the weight of the new personspersons is (
is (y y + + nx nx ).).
Weight of new persons = Weight of removed person + Weight of new persons = Weight of removed person + Num-bers of persons × Increase in average.
bers of persons × Increase in average. (iv)
(iv) The average age of ‘ The average age of ‘ n n ’ persons is decreased by ‘ ’ persons is decreased by ‘ x x ’ years when’ years when some of them [
some of them [n n 11,, n n 22 ... ... n n ; where; where n n 11 + +n n 22 + ... < + ... <n n ] aged [] aged [y y 11 + +y y 22 + ... where,
+ ... where, y y 11 ++ y y 22 + ... = + ... = y y years] are replaced by the same years] are replaced by the same number of persons. Then the age of the new persons is number of persons. Then the age of the new persons is [[y y – – nx nx ].].
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M a g i c a l BM a g i c a l Bo oo ok k o n Ao n Ar i tr i t h m e t i c a l F o r m u lh m e t i c a l F o r m u l a e a eK KUNDAN
K KUNDAN
Age of new persons = Age of removed persons – Number of Age of new persons = Age of removed persons – Number of persons × Decrease in average
persons × Decrease in average
(v
(v)) The average of The average of marks obtained by marks obtained by ‘ ‘ n n ’ candidates in a certain’ candidates in a certain examination is ‘
examination is ‘ T T ’. If the average marks of passed candi-’. If the average marks of passed candi-dates is ‘
dates is ‘ P P ’ and that of the failed candidates is ‘ ’ and that of the failed candidates is ‘ F F ’. Then the’. Then the number of candidates who passed the examination is number of candidates who passed the examination is
n n TT F F P P F F ..Number of passed candidates Number of passed candidates
=
= Total Total candidatcandidates (TotPassed Average – Failed AveragePassed Average – Failed Averagees (Total Averaal Average – Failed ge – Failed Average)Average)
(vi)
(vi) If the average ofIf the average of n n results (where results (where n n is an odd number) is is an odd number) is ‘a ‘a ’ ’ and the average of first
and the average of first 11
2 2 n n
results is ‘ results is ‘ b b ’ and that of last’ and that of last
1 1 2 2 n n is is ‘ ‘ c c ’, then’, then 1 1 2 2 n n th result isth result is
1 1 2 2 n n b b cc nna a .. (vii)(vii) If the average ofIf the average of n n results (where results (where n n is an odd number) is ‘ is an odd number) is ‘ a a ’ ’ and the average of first
and the average of first 11 2 2 n n
th result is ‘ th result is ‘ b b ’ and that of ’ and that of
last last 11 2 2 n n
th results isth results is ‘c ‘c ’, then’, then
1 1 2 2 n n th results isth results is
1 1 2 2 n n n naa bb cc .. (viii)(viii) If a batsman in hisIf a batsman in his n n th innings makes a score of ‘ th innings makes a score of ‘ x x ’, and’, and thereby increases his average by
thereby increases his average by ‘y ‘y ’, then the average after’, then the average after ‘n
‘n ’ innings is [x –’ innings is [x – y y ((n n – 1)]. – 1)].
(ix)
(ix) If a cricketer has completed ‘ If a cricketer has completed ‘ n n ’ innings and his average is ‘ ’ innings and his average is ‘ x x ’ ’ runs,then the number of runs, he must make in his next runs,then the number of runs, he must make in his next innings so as to raise his average to ‘
innings so as to raise his average to ‘ y y ’ are [’ are [n n ((y y –– x x ) +) + y y ].].
(x
(x)) If average of ‘ If average of ‘ n n ’ consecutive odd numbers is ‘ ’ consecutive odd numbers is ‘ x x ’, then the’, then the difference between the smallest and the largest numbers is difference between the smallest and the largest numbers is
given by 2( given by 2(n n – 1). – 1).
Note
Note:: We see that tWe see that the above fohe above formula irmula is indeps independent ofendent of x x .. That m
That m eans, eans, this this formula formula always always holds holds good good irrespective irrespective of of the value of
the value of x x ..
(xi)
(xi) Average of firstAverage of first n n multiple of a number multiple of a number x x is is (( )) 2 2 x x xx n n ..
Percentage
Percentage
(i(i)) If two values are respectivelyIf two values are respectively x x % and% and y y % more than a third% more than a third value, then the first is the
value, then the first is the 100100 100%100% 100 100 x x y y
of the second. of the second.
(ii)
(ii) If two values are respectivelyIf two values are respectively x x % and% and y y % more than a third% more than a third value, then the second is the
value, then the second is the 100100 100%100% 100 100 y y x x
of the first.of the first.
(iii)
(iii) If two values are respectively If two values are respectively x x % and% and y y % less than a third% less than a third value, then the second is the
value, then the second is the 100100 100%100% 100 100 y y x x
of the first. of the first.
(iv)
(iv) If two values are respectivelyIf two values are respectively x x % and% and y y % less than a third% less than a third value, then the first is the
value, then the first is the 100100100100 y y x x 100%100%
of the second. of the second.
(v
(v)) If A isIf A is x x % of C and B is% of C and B is y y % of C, then A is% of C, then A is x x 100%100% y
y of B. of B. (vi)
(vi) x x % of a quantity is taken by the first,% of a quantity is taken by the first, y y % of the remaining is% of the remaining is taken by the second and
taken by the second and z z % of the remaining is taken by % of the remaining is taken by third person. Now, if Rs A is left in the fund, then there was third person. Now, if Rs A is left in the fund, then there was
1 10000 110000 110000 1 10000 110000 110000 A A x x yy z z 10
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K KUNDAN
(vii)
(vii) If initial quantity is A andIf initial quantity is A and x x % of the quantity is taken by % of the quantity is taken by the first,
the first, y y % of the remaining was taken by the second and% of the remaining was taken by the second and z
z % of the remaining is taken by third person; then% of the remaining is taken by third person; then
110000
101000
110000
1 10000 110000 110000 A A xx yy zz is left in the fund. is left in the fund.
(viii)
(viii) x x % % of a quantity is added. Again,of a quantity is added. Again,y y % % of the increased quantity of the increased quantity is added. Again
is added. Again z z % of the increased quantity is added.% of the increased quantity is added. Now, it becomes A, then the initial amount is given by Now, it becomes A, then the initial amount is given by
1 10000 110000 110000 1 10000 110000 110000 A A x x yy z z .. (ix)(ix) If initial quantity is A andIf initial quantity is A and x x % of % of the initial quantity is added.the initial quantity is added. Again
Again y y % of the increased quantity is added. Again% of the increased quantity is added. Again z z % % of of the increased quantity is added, then initial quantity the increased quantity is added, then initial quantity becomes becomes
101000
101000
110000
1 10000 110000 110000 A A xx yy zz .. (x(x)) If the price of a commodity increases byIf the price of a commodity increases by r r %, then the%, then the reduction in consumption so as not to increase the reduction in consumption so as not to increase the expenditure is expenditure is 110000 %% 100 100 r r r r .. (xi)
(xi) If the price of a commodity decreases byIf the price of a commodity decreases by r r %, then increase%, then increase in consumption so as not to decrease expenditure on this in consumption so as not to decrease expenditure on this item is item is
100100r r r r
110000 %% .. (xii)(xii) If first value isIf first value is r r % more than the second value, then the% more than the second value, then the second is
second is 100100r r r r 110000 %% less than the first value. less than the first value. (xiii)
(xiii) If the first value isIf the first value is r r % less than the second value, then the% less than the second value, then the second value is second value is 110000 %% 100 100 r r r r
(xiv)
(xiv) If the value of a number is first increased byIf the value of a number is first increased by x x % and later% and later decreased by
decreased by x x %, then net change is always a decrease%, then net change is always a decrease which is equal to
which is equal to x x % % of of x x or or
2 2 100 100 x x .. (xv)
(xv) If the value is first increased byIf the value is first increased by x x % and then decreased by % and then decreased by y
y %, then there is%, then there is %% 100 100 xy xy x x y y
increase or decrease, increase or decrease,
according to the +ve or –ve sign respectively. according to the +ve or –ve sign respectively.
(xvi)
(xvi) If the value is increased successively byIf the value is increased successively by x x % and% and x x %, then%, then the final increase is given by
the final increase is given by
2 2 2 2 %% 100 100 x x x x .. (xvii)
(xvii) If the value is increased successively byIf the value is increased successively by x x % and% and y y %, then%, then the final increase is given by
the final increase is given by %%
100 100 xy xy x x y y .. (xviii)
(xviii) If the value is decreased successively by If the value is decreased successively by x x % and% and y y %, then%, then the final decrease is given by
the final decrease is given by
100 100 xy xy x x y y (xix)
(xix) If the value is decreased successively byIf the value is decreased successively by x x % and% and x x %, then%, then the final decrease is given by
the final decrease is given by
2 2 2 2 %% 100 100 x x x x ..
((xxxx) ) ((aa)) If the one factor is decreased byIf the one factor is decreased by x x % and the other% and the other factor is increased by
factor is increased by y y %.%. (b)
(b) or, if the one factor is increased byor, if the one factor is increased by x x % and the other% and the other factor is decreased by
factor is decreased by y y %, then the effect on the%, then the effect on the product = Increase % value – Decrease % value product = Increase % value – Decrease % value –
– IncIncrearease se % val% valueue DecDecreasrease e % va% valuelue 100
100
and the value and the value is increased or decreased according to the +ve or –ve is increased or decreased according to the +ve or –ve sign obtained.
sign obtained. Note
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M a g i c a l BM a g i c a l Bo oo ok k o n Ao n Ar i tr i t h m e t i c a l F o r m u lh m e t i c a l F o r m u l a e a eK KUNDAN
K KUNDAN
of both the cases. of both the cases. For Case (a) it becomes: For Case (a) it becomes:
100 100 yx yx y y x x Whereas for Case (b) it becomes: Whereas for Case (b) it becomes:
100 100 xy xy x x y y Thus, we see that it is more
Thus, we see that it is more easy to rememeasy to remem ber the geneber the gene ralral formula which works in both the cases equally.
formula which works in both the cases equally.
(xxi)
(xxi) The The pass pass marks marks in in an an examination examination isis x x %. If a candidate%. If a candidate who secures
who secures y y marks fails by marks fails by z z marks, then the maximum marks, then the maximum marks is given by marks is given by 100100 y
y z z
x x .. (xxii)(xxii) A candidate scoringA candidate scoring x x % % in an examination fails byin an examination fails by‘a ‘a ’ marks,’ marks, while another candidate who scores
while another candidate who scores y y % marks gets ‘ % marks gets ‘ b’ b’ marks more than the minimum required pass marks. marks more than the minimum required pass marks. The
The n n ththe e mama ximxim um um mama rks rks fofo r r thth at at exex amamininatatioion n areare M M 100100 a
a b b
y y x x .. (xxiii)(xxiii) In measuring the sides of a rectangle, one side is takenIn measuring the sides of a rectangle, one side is taken x
x % in excess and the other% in excess and the other y y % in deficit. The error per% in deficit. The error per cent in area calculated from the measurement is cent in area calculated from the measurement is
100 100 xy xy x
x y y in excess or deficit, according to the +ve in excess or deficit, according to the +ve or –ve sign. In another form this may be written as % or –ve sign. In another form this may be written as % error = % excess – % deficit –
error = % excess – % deficit – %% eexxcceessss %% ddeeffiicciitt 100
100
(xxiv)
(xxiv) If one of the sides of a rectangle is increased byIf one of the sides of a rectangle is increased by x x % and% and the other is increased by
the other is increased by y y %, then the per cent value by %, then the per cent value by which area changes is given by
which area changes is given by %%
100 100 xy xy x x y y increase increase .. (xxv)
(xxv) If one of the sides of a rectangle is decreased byIf one of the sides of a rectangle is decreased by x x % and% and the other is decreased by
the other is decreased by y y % then the per cent value by % then the per cent value by which area changes is given by
which area changes is given by %%
100 100 xy xy x x y y decrease. decrease.
(xxvi)
(xxvi) In an examinationIn an examination x x % failed in English and% failed in English and y y % failed in% failed in maths. If
maths. If z z % of students failed in both the subjects, the% of students failed in both the subjects, the percentage of students who passed in both the subjects percentage of students who passed in both the subjects is
is 100100
x x y y z z
..(xxvii)
(xxvii) A man spendsA man spends x x % of his income. His income is increased% of his income. His income is increased by
by y y % and his expenditure also increases by% and his expenditure also increases by z z %, then%, then the percentage increase in his savings is given by the percentage increase in his savings is given by
100 100 % % 100 100 y y xz xz x x .. (xxviii)
(xxviii) A solution of salt and water containsA solution of salt and water contains x x % salt by weight.% salt by weight. Of it ‘A’ kg water evaporates and the solution now Of it ‘A’ kg water evaporates and the solution now contains
contains y y % of salt. The original quantity of solution is% of salt. The original quantity of solution is given by
given by A A yyy y x x
kg. In other words, i kg. In other words, it may be t may be rewrittenrewritten
as the original quantity of solution = Quantity of as the original quantity of solution = Quantity of evaporated water ×
evaporated water × FFiin%% DnaDiifff.all %% of. ooff soff ssasaallttalltt
.. (xxix)
(xxix) When a certain quantity of goods B is added to changeWhen a certain quantity of goods B is added to change the percentage of goods A in a mixture of A and B, then the percentage of goods A in a mixture of A and B, then the quantity of B to be added is
the quantity of B to be added is
Previous % value of A Previous % value of A
×
× MixMixturture Quae Quantintityty –– MixMixtuture Qure Quantantitity y Changed % value o Changed % value of f AA (xxx)
(xxx) If the original price of a commodity isIf the original price of a commodity is Rs XRs X and new priceand new price of the commodity is
of the commodity is Rs Rs Y Y , then the decrease or increase, then the decrease or increase in consumption so as not to increase or decrease the in consumption so as not to increase or decrease the expenditure respectively is
expenditure respectively is YYY Y X X 110000 %%
,, ie
ie Difference in priceDifference in priceNew priceNew price 110000 %%
.. (xxxi)
(xxxi) To split To split a numbea numbe r r N into N into two parts two parts such that such that one one part ispart is p% of the other. The two split parts are
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M a g i c a l BM a g i c a l Bo oo ok k o n Ao n Ar i tr i t h m e t i c a l F o r m u lh m e t i c a l F o r m u l a e a e 100 100 p p p p .. (xxxii)(xxxii) IfIfX X litres of oil was poured into a tank and it was still litres of oil was poured into a tank and it was still x x %% empty, then the quantity of oil that must be poured into empty, then the quantity of oil that must be poured into the tank in order to fill it to the brim is
the tank in order to fill it to the brim is
100 100 X X x x x x litres. litres. (xxxiii)
(xxxiii)IfIf X X litres of oil was poured into a tank and it was still litres of oil was poured into a tank and it was still x x %% empty, then the capacity of the tank is
empty, then the capacity of the tank is 100100X X 100100x x
litres litres .. (xxxiv)
(xxxiv) If a number is successively increased byIf a number is successively increased by x x %,%, y y % and% and z z %,%, then single equivalent increase in that number will be then single equivalent increase in that number will be
22 %% 1 10000 101000 x xyy yyzz zzxx xxyyz z x x yy z z .. (xxxv)(xxxv) A person spendsA person spends x x % of his monthly income on item ‘A’ % of his monthly income on item ‘A’ and
and y y % of the remaining on the item ‘B’. He saves the% of the remaining on the item ‘B’. He saves the remaining amount. If the savings amount is Rs ‘S’, then remaining amount. If the savings amount is Rs ‘S’, then (a)
(a) the monthly the monthly income of person income of person ==Rs Rs
2 2 (100) (100) ((110000 ))((110000 )) S S x x y y (b)
(b) the monthly amount spent on the item the monthly amount spent on the item A = A = Rs Rs 100 100 ((110000 ))((110000 )) S S x x x x y y (c)
(c) the monthly amount spent on the item the monthly amount spent on the item B = Rs B = Rs ((110000 )) y y S S y y Note
Note: Here ‘S’ = Savings per month.: Here ‘S’ = Savings per month.
(xxxvi)
(xxxvi) When the price of an item was increased byWhen the price of an item was increased by x x %, a %, a family family reduced its consumption in such a way that the reduced its consumption in such a way that the expenditure on the item was only
expenditure on the item was only y y % more than before.% more than before. If ‘W’ kg were consumed per month before, then the new If ‘W’ kg were consumed per month before, then the new
K KUNDAN
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monthly consumption is given by monthly consumption is given by 100100
100 100 y y x x W W kg. kg. (xxxvii)
(xxxvii)If the price of an item is increased byIf the price of an item is increased by x x % and a housewife% and a housewife reduced the consumption of that item by
reduced the consumption of that item by x x %, then her%, then her expenditure on that item decreases by
expenditure on that item decreases by
2 2 10 10 x x %. Or, in%. Or, in
words it can be written as the following: words it can be written as the following: Per cent Expenditue Change
Per cent Expenditue Change =
=
2 2
Com
Commonmon inincrcreaseasee oror decdecreareasese 10 10 %.%. Note
Note:: Here -ve sign shows the decrease iHere -ve sign shows the decrease in expenditure,n expenditure, ie in the above case there is always decrease in ie in the above case there is always decrease in the expenditure.
the expenditure.
Ratio and Proportion
Ratio and Proportion
(i
(i)) If two numbers are in the ratio of If two numbers are in the ratio of aa :: b b and the sum of and the sum of these numbers is
these numbers is x x , then these numbers will be, then these numbers will be ax ax a ab b and and bx bx a a b b respectively. respectively. (ii)
(ii) To find To find the the number number of of coins.coins. Number of each type of coins =
Number of each type of coins = Amount in rupeesAmount in rupees
Value of coins in rupees Value of coins in rupees
(iii)
(iii) To find the strength to To find the strength to milk strength of milk in the milk strength of milk in the mixturemixture =
= QuantitQuantity of y of MilkMilk
Total
Total QuantitQuantity of Mixty of Mixtureure
(iv)
(iv) The contents The contents of of two two vessels vessels containing water containing water and mand milk areilk are in the ratio
in the ratio xx :: y 11 y 11 and and xx22 :: y y 2 2 are mixed in the ratio x : y. are mixed in the ratio x : y.
The resulting m
The resulting m ixture will ixture will have water have water and milk in and milk in the ratiothe ratio of
16
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M a g i c a l BM a g i c a l Bo oo ok k o n Ao n Ar i tr i t h m e t i c a l F o r m u lh m e t i c a l F o r m u l a e a e (v(v)) If two numbers are in the ratio of a : b and the differenceIf two numbers are in the ratio of a : b and the difference between these numbers is x, then these numbers will be between these numbers is x, then these numbers will be (a)
(a) ax ax a
a b b and and bx bx a
ab b respectively. (where, respectively. (where, a a >> b b )) (b)
(b) ax ax b
ba a and and bx bx b
ba a respectively (where respectively (where a a << b b )) (vi)
(vi) If three numbers are in the ratio ofIf three numbers are in the ratio of aa :: bb :: c c and the sum of and the sum of these numbers is
these numbers is x x , then these numbers will be, then these numbers will be ,, a axx bbxx a b a b cc aa bbcc andand cx cx a a bbc c respectively. respectively. (vii)
(vii) If the ratio between the first and the second quantities isIf the ratio between the first and the second quantities is a
a :: b b and the ratio between the second and the third and the ratio between the second and the third quantities is
quantities is c c :: d d , then the ratio among first, second and, then the ratio among first, second and third quantities is given by
third quantities is given by ac ac : : bc bc : : bd bd . The above ratio can. The above ratio can be represented diagrammatically as
be represented diagrammatically as
(viii)
(viii) If the ratio between the first and the second quantities isIf the ratio between the first and the second quantities is a
a :: b b ; the ratio between the second and the third quantities; the ratio between the second and the third quantities is
is cc :: d d and the ratio between the third and the fourth and the ratio between the third and the fourth quantities is
quantities is ee :: f f then the ratio among the first, second, then the ratio among the first, second, third and fourth quantities is given by
third and fourth quantities is given by
(ix)
(ix) If inIf in xx litres mixture of milk and water, the ratio of milklitres mixture of milk and water, the ratio of milk and water is
and water is aa :: b,b, the quantity of water to be added in the quantity of water to be added in order to make this ratio
order to make this ratio cc :: d d is is
x x aadd bbc c cc aa b b .. (xx
x litres of water is added to the mixture, milk and water litres of water is added to the mixture, milk and water become in the ratio
become in the ratio aa :: c c . Then the quantity of milk in the. Then the quantity of milk in the mixture is given by
mixture is given by ax ax
ccb b and that of water is given by and that of water is given by bx
bx cc b b .. (xi)
(xi) If two quantities X and Y are in the ratioIf two quantities X and Y are in the ratio xx :: y y . Then. Then X + Y
X + Y : : X – Y X – Y : : : : x + yx + y :: x – y x – y
(xii)
(xii) In any two two-dimensional figure, if the correspondingIn any two two-dimensional figure, if the corresponding sides are in the ratio
sides are in the ratio a a : : b , then their areas are in the ratio b , then their areas are in the ratio a
a 22 : :b b 22..
(xiii)
(xiii) In any two 3-dimensional figures, if the corresponding sidesIn any two 3-dimensional figures, if the corresponding sides or other measuring lengths are in the ratio
or other measuring lengths are in the ratio a a : : b b , then their, then their volumes are in the ratio
volumes are in the ratio a a 33 : : b b 33..
(xiv)
(xiv) The ratio betwee The ratio betwee n two numn two num bers isbers is aa :: b b . If each number be. If each number be increased by
increased by x x , the ratio becomes, the ratio becomes c c :: d d . Then, the two. Then, the two numbers are given as
numbers are given as xxa ca c
d d
a add bbc c and and
x xb cb c d d a add bbc c ; where; where c – a c – a d – d – b b (xv)(xv) The incomes The incomes of of two persons two persons are in are in the ratiothe ratio aa :: bb and theirand their expenditures are in the ratio
expenditures are in the ratio cc :: d d . If each of them saves. If each of them saves Rs X, then their incomes are given by
Rs X, then their incomes are given by XXa da d
c c
a add bbc c andand
X Xb db d c c a add bbc c .. (xvi)(xvi) The incomes The incomes of of two persons two persons are in are in the ratiothe ratio a a :: b b and their and their expenditures are in the ratio
expenditures are in the ratio cc :: d d . If each of them saves. If each of them saves Rs
Rs X X , then their expenditures are given by, then their expenditures are given by XXc bc b
a a
a add bbc c andand
X Xd bd b a a a add bbc c ..18
18
M a g i c a l BM a g i c a l Bo oo ok k o n Ao n Ar i tr i t h m e t i c a l F o r m u lh m e t i c a l F o r m u l a e a e (xvii)(xvii) Two Two candles candles of of the the same same height height are are lighted lighted at at the the samesame time. The first is consumed in T
time. The first is consumed in T 11 hours and the second in hours and the second in T
T 22 hours. Assuming that each candle burns at a constant hours. Assuming that each candle burns at a constant rate, the time after which the ratio of first candle to second rate, the time after which the ratio of first candle to second
candle becomes
candle becomes xx :: y y is given byis given by
1 1 22 1 1 22 1 1 x x T T T T y y x x T T T T y y hours. hours.
Partnership
Partnership
(i(i)) If investments are in the ratio ofIf investments are in the ratio of aa :: bb :: c c and the timing of and the timing of their investments in the ratio of
their investments in the ratio of x x :: yy :: z z then the ratio of then the ratio of their profits are in the ratio of
their profits are in the ratio of axax :: byby :: cz cz .. (ii)
(ii) If investments are in the ratioIf investments are in the ratio aa :: b b :: c c and profits in the ratio and profits in the ratio p
p : : q q : : r r , then the ratio of time =, then the ratio of time = p p:: qq : : r r a
a bb c c ..
(iii)
(iii) Three Three partnpartners ers inveinve st st thetheir ir capitcapitals als in in a a busibusinesness. s. If If thethe ratio of their periods of investments are
ratio of their periods of investments are t t 11 :: t t 2 2 :: t t 3 3 and their and their profits are in the ratio of
profits are in the ratio of a a :: b b :: c c , then the capitals will be in, then the capitals will be in the ratio of the ratio of 1 1 22 33 :: :: a a bb c c tt tt t t ..
Profit and Loss
Profit and Loss
(i
(i)) If certain article is bought at the rate of ‘A’ for a rupee, thenIf certain article is bought at the rate of ‘A’ for a rupee, then to gain
to gain x x %, the article must be sold at the rate of %, the article must be sold at the rate of 100 100 100 100 x x
× A for a rupee (Remember the rule of fraction). × A for a rupee (Remember the rule of fraction).
(ii)
(ii) If a man purchases ‘ If a man purchases ‘ x x ’ items for Rs ‘ ’ items for Rs ‘ y y ’ and sells ‘ ’ and sells ‘ y y ’ items for’ items for Rs ‘
Rs ‘ x x ’, then the profit or loss [depending upon the respective’, then the profit or loss [depending upon the respective (+ve) or (–ve) sign in the final result] made by him is (+ve) or (–ve) sign in the final result] made by him is
2 2 22 2 2 110000 %% x x y y y y .. (iii)
(iii) If a man purchases ‘ If a man purchases ‘ a a ’ items for Rs ‘ ’ items for Rs ‘ b b ’ and sells ‘ ’ and sells ‘ c c ’ items for’ items for Rs ‘
Rs ‘ d d ’, then the gain or loss [depending upon the respective’, then the gain or loss [depending upon the respective (+ve) or (–ve) sign in the final result] made by him is (+ve) or (–ve) sign in the final result] made by him is
1 10000 %% a add bbc c bc bc .. (iv)
(iv) Problems Based on Dishonest DealerProblems Based on Dishonest Dealer % gain =
% gain = True va True value –ErrorErrorlue – ErrorError100100 or, % gain =
or, % gain = True weig True weightFalse weightFalse weightht –– False weFalse weightight 100100 ( v
( v) ) ((aa)) When there are two successive profits ofWhen there are two successive profits of x x % and% and y y %% ,, then the resultant profit per cent is given by then the resultant profit per cent is given by
100 100 xy xy x x y y .. (b)
(b) When there is a profit ofWhen there is a profit of x x % and loss of% and loss of y y % in a% in a transaction, then the resultant profit or loss per cent is transaction, then the resultant profit or loss per cent is given by
given by xx y y 100100xy xy according to the + ve and the -ve according to the + ve and the -ve signs respectively.
signs respectively. (c)
(c) When there are two successive loss ofWhen there are two successive loss of x x % and% and y y %, then%, then the resultant loss per cent is given by
the resultant loss per cent is given by
100 100 xy xy x x y y .. (vi)
(vi) If an article is sold at a profit ofIf an article is sold at a profit of x x % and if both the cost% and if both the cost price and selling price are Rs A less, the profit would be price and selling price are Rs A less, the profit would be y y %% more, then the cost price is
more, then the cost price is
xx y y
AA y y . In other words,. In other words,
cost price =
cost price =
InInititiaial l PrProfofit it % % IncrIncreasease e in in prprofofit it %%
AA Increase in profit %Increase in profit %
20
20
M a g i c a l BM a g i c a l Bo oo ok k o n Ao n Ar i tr i t h m e t i c a l F o r m u lh m e t i c a l F o r m u l a e a e (vii)(vii) If cost price of If cost price of x x articles is equal to the selling price of articles is equal to the selling price of y y articles, then the profit percentage =
articles, then the profit percentage = xxy y y y 100%100%
..
(viii)(a)
(viii)(a) A person buys certain quantity of an article for Rs A. If A person buys certain quantity of an article for Rs A. If he sells
he sells m m th part of the stock at a profit ofth part of the stock at a profit of x x % and the% and the remaining
remaining n n th part atth part at y y % profit, then the per cent profit% profit, then the per cent profit in this transaction is in this transaction is mmxx nny y m m n n or or First
First part part % % profit profit on on first first part part Second Second partpart % profit on s
% profit on secec ond partond part Total
Total of two partsof two parts
.. (b)
(b) IfIf x x part is sold at m% profit and the rest part is sold at m% profit and the rest y y part is sold part is sold at
at n n % loss and Rs% loss and Rs P P is earned as overall profit, then is earned as overall profit, then the value of the total consignment is Rs
the value of the total consignment is Rs PP ×10×100 0 x xmm –– nny y .. (ix)
(ix) If a man buys two items A and B for Rs P and sells one itemIf a man buys two items A and B for Rs P and sells one item A so as to lose
A so as to lose x x % and the other item B so as to gain% and the other item B so as to gain y y %,%, and on the whole he neither gains nor loses, then
and on the whole he neither gains nor loses, then (a)
(a) the cost of the item A is the cost of the item A is xx +Py Py + y y
andand (b)
(b) the cost of the item B is the cost of the item B is Px Px x x ++ y y ..
((x ) x ) ((aa)) By selling a certain item at the rate of ‘ By selling a certain item at the rate of ‘ X’ X’ items a rupee, items a rupee, a man loses
a man loses x x %. If he wants to gain%. If he wants to gain y y %, %, then the numbthen the numberer of items should be sold for a rupee is
of items should be sold for a rupee is 100100100100 y y x x X X
.. (b)
(b) By selling an article for RsBy selling an article for Rs A A , a dealer makes a profit of , a dealer makes a profit of x
x %. If he wants to make profit of%. If he wants to make profit of y y %, then he should%, then he should increase his selling price by Rs
increase his selling price by Rs 100100yy x x x x A A
selling price is given by Rs
selling price is given by Rs 100100100100 y y x x A A
.. (c)
(c) By selling an article for Rs By selling an article for Rs A A , a dealer makes a loss of , a dealer makes a loss of x
x %. If he wants to make a profit of%. If he wants to make a profit of y y %, then he should%, then he should increase his selling price by Rs
increase his selling price by Rs
100 100 x x y y A A x x
and the and the
selling price is given by Rs
selling price is given by Rs 100100100100 y y x x A A
.. (xi)
(xi) When each of the two commodities is sold at the sameWhen each of the two commodities is sold at the same price
price Rs A Rs A , and a profit of P% is made on the first and a loss, and a profit of P% is made on the first and a loss of
of L L % is made on the second, then the percentage gain or% is made on the second, then the percentage gain or loss is
loss is
101110000000
PPPP
LL
110
00022PPLLL L
according to the +ve or –ve sign. according to the +ve or –ve sign.
Note
Note:: (a)(a) In the special case whenIn the special case when P P = L = L we have we have
2 2 22 1 10000 00 22 2 20000 110000 P P p p
Since the sign is –ve, there is always loss and the Since the sign is –ve, there is always loss and the value is given as value is given as
2 2 % value % value 100 100 .. (b)(b) When each of the two commodities is sold at theWhen each of the two commodities is sold at the same price Rs A,
same price Rs A, and a profit ofand a profit of P P % is made on the% is made on the first and a profit of
first and a profit of L L % is made on the second,% is made on the second, then the percentage gain is
then the percentage gain is
1 10000 22 1 10000 110000 P P LL PPLL P P L L .. (xii)(xii) If a merchant, by selling his goods, has a gain ofIf a merchant, by selling his goods, has a gain of x x % of the% of the selling price, then his real gain per cent on the cost price is selling price, then his real gain per cent on the cost price is
1 10000 %% 100 100 x x x x .. Note
Note: Real profit per cent is always calculated on cost price: Real profit per cent is always calculated on cost price and real profit per cent is always more than the % and real profit per cent is always more than the % profit on selling price.
profit on selling price. (xiii)
(xiii) If a merchant, by selling his goods, has a loss ofIf a merchant, by selling his goods, has a loss of x x %, of the%, of the selling price, then his real loss per cent on the cost price is selling price, then his real loss per cent on the cost price is
