Gyan on web : Math Shortcuts for Competitive Examinations

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To find the number of factors of a given number, express the number as a product of powers of prime numbers.
In this case, 48 can be written as 16 * 3 = (2⁴ * 3)
Now, increment the power of each of the prime numbers by 1 and multiply the result.
In this case it will be (4 + 1)*(1 + 1) = 5 * 2 = 10 (the power of 2 is 4 and the power of 3 is 1)
Therefore, there will 10 factors including 1 and 48. Excluding, these two numbers, you will have 10 – 2 = 8 factors.
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The sum of first n natural numbers = n (n+1)/2
The sum of squares of first n natural numbers is n (n+1)(2n+1)/6
The sum of first n even numbers= n (n+1)
The sum of first n odd numbers= n^2
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To find the squares of numbers near numbers of which squares are known
To find 41^2 , Add 40+41 to 1600 =1681
To find 59^2 , Subtract 60^2-(60+59) =3481
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If an equation (i:e f(x)=0 ) contains all positive co-efficient of any powers of x , it has no positive roots then.
eg: x^4+3x^2+2x+6=0 has no positive roots .
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For an equation f(x)=0 , the maximum number of positive roots it can have is the number of sign changes in f(x) ; and the maximum number of negative roots it can have is the number of sign changes in f(-x) .
Hence the remaining are the minimum number of imaginary roots of the equation(Since we also know that the index of the maximum power of x is the number of roots of an equation.)
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For a cubic equation ax^3+bx^2+cx+d=o
sum of the roots = – b/a
sum of the product of the roots taken two at a time = c/a
product of the roots = -d/a
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For a biquadratic equation ax^4+bx^3+cx^2+dx+e = 0
sum of the roots = – b/a
sum of the product of the roots taken three at a time = c/a
sum of the product of the roots taken two at a time = -d/a
product of the roots = e/a
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If for two numbers x+y=k(=constant), then their PRODUCT is MAXIMUM if
x=y(=k/2). The maximum product is then (k^2)/4
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If for two numbers x*y=k(=constant), then their SUM is MINIMUM if
x=y(=root(k)). The minimum sum is then 2*root(k) .
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|x| + |y| >= |x+y| (|| stands for absolute value or modulus )
(Useful in solving some inequations)
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Product of any two numbers = Product of their HCF and LCM .
Hence product of two numbers = LCM of the numbers if they are prime to each other
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For any regular polygon , the sum of the exterior angles is equal to 360 degrees
hence measure of any external angle is equal to 360/n. ( where n is the number of sides)
For any regular polygon , the sum of interior angles =(n-2)180 degrees

So measure of one angle in
Square                    =90
Pentagon                =108
Hexagon                 =120
Heptagon                =128.5
Octagon                  =135
Nonagon                 =140
Decagon                  = 144
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If any parallelogram can be inscribed in a circle , it must be a rectangle.
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If a trapezium can be inscribed in a circle it must be an isosceles trapezium (i:e oblique sides equal).
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For an isosceles trapezium , sum of a pair of opposite sides is equal in length to the sum of the other pair of opposite sides .(i:e AB+CD = AD+BC , taken in order) .
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Area of a regular hexagon: root (3)*3/2*(side)*(side)
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For any 2 numbers a>b

a>AM>GM>HM>b (where AM, GM ,HM stand for arithmetic, geometric , harmonic menasa respectively)

(GM)^2 = AM * HM

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For three positive numbers a, b ,c

(a+b+c) * (1/a+1/b+1/c)>=9

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For any positive integer n

2<= (1+1/n)^n <=3
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a^2+b^2+c^2 >= ab+bc+ca
If a=b=c , then the equality holds in the above.

a^4+b^4+c^4+d^4 >=4abcd

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(n!)^2 > n^n (! for factorial)

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If a+b+c+d=constant , then the product a^p * b^q * c^r * d^s will be maximum
if a/p = b/q = c/r = d/s .
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Consider the two equations

a1x+b1y=c1
a2x+b2y=c2

Then ,
If a1/a2 = b1/b2 = c1/c2 , then we have infinite solutions for these equations.
If a1/a2 = b1/b2 <> c1/c2 , then we have no solution for these equations.(<> means not equal to )
If a1/a2 <> b1/b2 , then we have a unique solutions for these equations..
++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ For any quadrilateral whose diagonals intersect at right angles , the area of the quadrilateral is
0.5*d1*d2, where d1,d2 are the lenghts of the diagonals.

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Problems on clocks can be tackled as assuming two runners going round a circle , one 12 times as fast as the other . That is ,
the minute hand describes 6 degrees /minute
the hour hand describes 1/2 degrees /minute .

Thus the minute hand describes 5(1/2) degrees more than the hour hand per minute .
The hour and the minute hand meet each other after every 65(5/11) minutes after being together at midnight.
(This can be derived from the above) .

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If n is even , n(n+1)(n+2) is divisible by 24

If n is any integer , n^2 + 4 is not divisible by 4
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Given the coordinates (a,b) (c,d) (e,f) (g,h) of a parallelogram , the coordinates of the meeting point of the diagonals can be found out by solving for
[(a+e)/2,(b+f)/2] =[ (c+g)/2 , (d+h)/2]
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Area of a triangle
1/2*base*altitude = 1/2*a*b*sinC = 1/2*b*c*sinA = 1/2*c*a*sinB = root(s*(s-a)*(s-b)*(s-c)) where s=a+b+c/2
=a*b*c/(4*R) where R is the CIRCUMRADIUS of the triangle = r*s ,where r is the inradius of the triangle .
In any triangle
a=b*CosC + c*CosB
b=c*CosA + a*CosC
c=a*CosB + b*CosA

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If a1/b1 = a2/b2 = a3/b3 = ………….. , then each ratio is equal to
(k1*a1+ k2*a2+k3*a3+…………..) / (k1*b1+ k2*b2+k3*b3+…………..) , which is also equal to
(a1+a2+a3+…………./b1+b2+b3+……….)

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(7)In any triangle
a/SinA = b/SinB =c/SinC=2R , where R is the circumradius

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++x^n -a^n = (x-a)(x^(n-1) + x^(n-2) + …….+ a^(n-1) ) ……Very useful for finding multiples .For example (17-14=3 will be a multiple of 17^3 – 14^3)

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e^x = 1 + (x)/1! + (x^2)/2! + (x^3)/3! + ……..to infinity
2 < e < 3
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log(1+x) = x – (x^2)/2 + (x^3)/3 – (x^4)/4 ………to infinity [ Note the alternating sign . .Also note that the ogarithm is with respect to base e ]

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In a GP the product of any two terms equidistant from a term is always constant .

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For a cyclic quadrilateral , area = root( (s-a) * (s-b) * (s-c) * (s-d) ) , where s=(a+b+c+d)/2

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For a cyclic quadrilateral , the measure of an external angle is equal to the measure of the internal opposite angle.

(m+n)! is divisible by m! * n! .

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If a quadrilateral circumscribes a circle , the sum of a pair of opposite sides is equal to the sum of the other pair .
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The sum of an infinite GP = a/(1-r) , where a and r are resp. the first term and common ratio of the GP .
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The equation whose roots are the reciprocal of the roots of the equation ax^2+bx+c is cx^2+bx+a
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The coordinates of the centroid of a triangle with vertices (a,b) (c,d) (e,f)
is((a+c+e)/3 , (b+d+f)/3) .

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The ratio of the radii of the circumcircle and incircle of an equilateral triangle is 2:1 .

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Area of a parallelogram = base * height
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APPOLLONIUS THEOREM:
In a triangle , if AD be the median to the side BC , then
AB^2 + AC^2 = 2(AD^2 + BD^2) or 2(AD^2 + DC^2) .
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for similar cones , ratio of radii = ratio of their bases.

The HCF and LCM of two nos. are equal when they are equal .
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Volume of a pyramid = 1/3 * base area * height

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In an isosceles triangle , the perpendicular from the vertex to the base or the angular bisector from vertex to base bisects the base.
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In any triangle the angular bisector of an angle bisects the base in the ratio of the
other two sides.

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++The quadrilateral formed by joining the angular bisectors of another quadrilateral is
always a rectangle.

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Roots of x^2+x+1=0 are 1,w,w^2 where 1+w+w^2=0 and w^3=1

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|a|+|b| = |a+b| if a*b>=0
else |a|+|b| >= |a+b|

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2<= (1+1/n)^n <=3

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WINE and WATER formula:

If Q be the volume of a vessel
q qty of a mixture of water and wine be removed each time from a mixture
n be the number of times this operation be done
and A be the final qty of wine in the mixture

then ,
A/Q = (1-q/Q)^n

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Area of a hexagon = root(3) * 3 * (side)^2

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(1+x)^n ~ (1+nx) if x<<<1

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Some pythagorean triplets:

3,4,5 (3^2=4+5)
5,12,13 (5^2=12+13)
7,24,25 (7^2=24+25)
8,15,17 (8^2 / 2 = 15+17 )
9,40,41 (9^2=40+41)
11,60,61 (11^2=60+61)
12,35,37 (12^2 / 2 = 35+37)
16,63,65 (16^2 /2 = 63+65)
20,21,29(EXCEPTION)
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Appolonius theorem could be applied to the 4 triangles formed in a parallelogram.

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Area of a trapezium = 1/2 * (sum of parallel sids) * height = median * height
where median is the line joining the midpoints of the oblique sides.

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when a three digit number is reversed and the difference of these two numbers is taken , the middle number is always 9 and the sum of the other two numbers is always 9 .

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ANy function of the type y=f(x)=(ax-b)/(bx-a) is always of the form x=f(y) .

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Let W be any point inside a rectangle ABCD .
Then
WD^2 + WB^2 = WC^2 + WA^2

Let a be the side of an equilateral triangle . then if three circles be drawn inside
this triangle touching each other then each’s radius = a/(2*(root(3)+1))
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Let ‘x’ be certain base in which the representation of a number is ‘abcd’ , then the decimal value of this number is a*x^3 + b*x^2 + c*x + d
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when you multiply each side of the inequality by -1, you have to reverse the direction of the inequality.
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To find the squares of numbers from 50 to 59
For 5X^2 , use the formulae
(5X)^2 = 5^2 +X / X^2
Eg ; (55^2) = 25+5 /25
=3025
(56)^2 = 25+6/36
=3136
(59)^2 = 25+9/81
=3481
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many of u must b aware of this formula, but the ppl who don’t know it must b useful for them.
a+b+(ab/100)
this is used for succesive discounts types of sums.
like 1999 population increses by 10% and then in 2000 by 5%
so the population in 2000 now is 10+5+(50/100)=+15.5% more that was in 1999
and if there is a decrease then it will be preceeded by a -ve sign and likeiwse

Math Shortcuts for Competitive exams – Part 2

Numbers

1) 2^2n-1 is always divisible by 3

2^2n-1 = (3-1)^2n -1
= 3M +1 -1
= 3M, thus divisible by 3

2) What is the sum of the divisors of 2^5.3^7.5^3.7^2?

ANS : (2^6-1)(3^8-1)(5^4-1)(7^3-1)/2.4.6
Funda : if a number ‘n’ is represented as
a^x * b^y * c^z ….
where, {a,b,c,.. } are prime numbers then

Quote:

(a) the total number of factors is (x+1)(y+1)(z+1) ….
(b) the total number of relatively prime numbers less than the number is n * (1-1/a) * (1-1/b) * (1-1/c)….
(c) the sum of relatively prime numbers less than the number is n/2 * n * (1-1/a) * (1-1/b) * (1-1/c)….
(d) the sum of factors of the number is {a^(x+1)} * {b^(y+1)} * …../(x*y*…)

3) what is the highest power of 10 in 203!ANS : express 10 as product of primes; 10 = 2*5

divide 203 with 2 and 5 individually
203/2 = 101
101/2 = 50
50/2 = 25
25/2 = 12
12/2 = 6
6/2 = 3
3/2 = 1
thus power of 2 in 203! is, 101 + 50 + 25 + 12 + 6 + 3 + 1 = 198

divide 203 with 5
203/5 = 40
40/5 = 8
8/5 = 1

thus power of 5 in 203! is, 49

so the power of 10 in 203! factorial is 49

4) x + y + z = 7 and xy + yz + zx = 10, then what is the maximum value of x? ( CAT 2002 has similar question )

ANS: 49-20 = 29, now if one of the y,z is zero, then the sum of other 2 squares shud be equal to 29, which means, x can take a max value at 5

5) In how many ways can 2310 be expressed as a product of 3 factors?

ANS: 2310 = 2*3*5*7*11
When a number can be expressed as a product of n distinct primes,
then it can be expressed as a product of 3 numbers in (3^(n+1) + 1)/2 ways

6) In how many ways, 729 can be expressed as a difference of 2 squares?

ANS: 729 = a^2 – b^2
= (a-b)(a+b),
since 729 = 3^5,
total ways of getting 729 are, 1*729, 3*243, 9*81, 27*27.
So 4 ways
Funda is that, all four ways of expressing can be used to findout distinct a,b values,
for example take 9*81
now since 9*81 = (a-b)(a+b) by solving the system a-b = 9 and a+b = 81 we can have 45,36 as soln.

7) How many times the digit 0 will appear from 1 to 10000

ANS: In 2 digit numbers : 9,
In 3 digit numbers : 18 + 162 = 180,
In 4 digit numbers : 2187 + 486 + 27 = 2700,
total = 9 + 180 + 2700 + 4 = 2893

8 ) What is the sum of all irreducible factors between 10 and 20 with denominator as 3?

ANS :
sum = 10.33 + 10.66 + 11.33 + 11.66 + 12.33 + 12.66 + 13.33 + 13.66…….
= 21 + 23 + ……
= 300

9) if n = 1+x where x is the product of 4 consecutive number then n is,
1) an odd number,
2) is a perfect square

SOLN : (1) is clearly evident
(2) let the 4 numbers be n-2,n-1,n and n+1 then by multing the whole thing and adding 1 we will have a perfect square

10) When 987 and 643 are divided by same number ‘n’ the reminder is also same, what is that number if the number is a odd prime number?

ANS : since both leave the same reminder, let the reminder be ‘r’,
then, 987 = an + r
and 643 = bn + r and thus
987 – 643 is divisible by ‘r’ and
987 – 643 = 344 = 86 * 4 = 43 * 8 and thus the prime is 43
hence ‘r’ is 43

11) when a number is divided by 11,7,4 the reminders are 5,6,3 respectively. what would be the reminders when the same number is divided by 4,7,11 respectively?

ANS : whenever such problem is given,
we need to write the numbers in top row and rems in the bottom row like this

11 7 4
| \ \
5 6 3

( coudnt express here properly

)
now the number is of the form, LCM ( 11,7,4 ) + 11*(3*7 + 6) + 5
that is 302 + LCM(11,7,4) and thus the rems when the same number is divided by 4,7,11 respectively are,

302 mod 4 = 2
75 mod 7 = 5
10 mod 11 = 10

12) a^n – b^n is always divisible by a-b

13) if a+b+c = 0 then a^3 + b^3 + c^3 = 3abc

EXAMPLE: 40^3-17^3-23^3 is divisble by
since 40-23-17 = 0, 40^3-17^3-23^3 = 3*40*23*17 and thus, the number is divisible by 3,5,8,17,23 etc.

14) There is a seller of cigerette and match boxes who sits in the narrow lanes of cochin. He prices the cigerattes at 85 p, but found that there are no takers. So he reduced the price of cigarette and managed to sell all the cigerattes, realising Rs. 77.28 in all. What is the number of cigerattes?

a) 49
b) 81
c) 84
d) 92

ANS : (d)
since 77.28 = 92 * 84, and since price of cigarette is less than 85, we have (d) as answer

Quote:

i have given this question to make the funda clear

15) What does 100 stand for if 5 X 6 = 33
ANS : 81
SOLN : this is a number system question,
30 in decimal system is 33 in some base ‘n’, by solving we will have n as 9
and thus, 100 will be 9^2 = 81

16) In any number system 121 is a perfect square,
SOLN: let the base be ‘n’
then 121 can be written as n^2 + 2*n + 1 = (n+1)^2
hence proved

17) Most of you ppl know these, anyways, just in case

Quote:

(a) sum of first ‘n’ natural numbers – n*(n+1)/2
(b) sum of the squares of first ‘n’ natural numbers – n*(n+1)*(2n+1)/6
(c) sum of the cubes of first ‘n’ natural numbers – n^2*(n+1)^2/4
(d) total number of primes between 1 and 100 – 25

18 ) See Attachment to know how to find LCM, GCF of Fractions

Quote:

CAT 2002 has 2 questions on the above simple concept

19) Converting Recurring Decimals to Fractions

let the number x be 0.23434343434……..

thus 1000 x = 234.3434343434……
and 10 x = 2.3434343434………
thus, 990 x = 232
and hence, x = 232/990

20) Reminder Funda

(a) (a + b + c) % n = (a%n + b%n + c%n) %n

EXAMPLE: The reminders when 3 numbers 1221, 1331, 1441 are divided by certain number 9 are 6, 8, 1 respectively. What would be the reminder when you divide 3993 with

9? ( never seen such question though

)
the reminder would be (6 + 8 + 1) % 9 = 6

(b) (a*b*c) % n = (a%n * b%n * c%n) %n

EXAMPLE: What is the remainder left when 1073 * 1079 * 1087 is divided by 119 ? ( seen this kinda questions alot

)
1073 % 119 = ?
since 1190 is divisible by 119, 1073 mod 119 is 2
and thus, “the remainder left when 1073 * 1079 * 1087 is divided by 119 ” is 2*8*16 mod 119 and that is 256 mod 119 and that is (238 + 18 ) mod 119 and that is 18

Glossary : % stands for reminder operation

find the number of zeroes in 1^1* 2^2* 3^3* 4^4………….. 98^98* 99^99* 100^100

the expression can be rewritten as (100!)^100 / 0!* 1!* 2!* 3!….99!

Now the numerator has 2400 zeros

the formula for finding number of zeros in n! is

[n/5]+[n/5^2]…[n/5^r]
where r is such that 5^r<=n<5^(r+1)

and [..] is the grestest integer function

for the numerator find the number of zeros using the above formulae..

for 0!…4! number of zeros ..0
5!…9!.number os zeros ..1
9!…14!… 2
15!..19!………………3
20!..24!………………4!
now at 25! the series makes a jump to 6
25!…29!……………..6
30!…34!……………..7
this goes on and again makes a jump at 50!
and then at 75!

so the number of zeros is…

5(1+2….19) + 25+ 50+ 75

the last 3 terms 25 50 and 75 are because of the jumps..

this gives numerator has 1100 zeros

now total number of zeros in expression is no of zeros in denominator – no of zeros in numerator
2400 – 1100

the Answer 1300

Math Shortcuts for Competitive Examinations – Part 3

इस पृष्ठ का हिन्दी अनुवाद

By morgan ⋅ September 21, 2009 ⋅ Post a comment

To find out if a number is divisible by seven:
Take the last digit, double it, and subtract it from the rest of the
number.
If the answer is more than a 2 digit number perform the above
again.
If the result is 0 or is divisible by 7 the original number is also
divisible by 7.

Example 1 ) 259
9*2= 18.
25-18 = 7 which is divisible by 7 so 259 is also divisible by 7.

Example 2 ) 2793
3*2= 6
279-6= 273

now 3*2=6
27-6= 21 which is divisible by 7 so 2793 is also divisible by 7 .

Now find out if following are divisible by 7

1) 2841
2) 3873
3) 1393
4) 2877

TO FIND SQUARE OF A NUMBER BETWEEN 40 to 50
Sq (44) .

1) Subtract the number from 50 getting result A.
2) Square A getting result X.
3) Subtract A from 25 getting result Y
4) Answer is xy

EXAMPLE 1 : 44
50-44=6
Sq of 6 =36
25-6 = 19
So answer 1936

EXAMPLE 2 : 47
50-47=3
Sq 0f 3 = 09
25-3= 22
So answer = 2209

NOW TRY To Find Sq of 48 ,26 and 49

TO FIND SQUARE OF A 3 DIGIT NUMBER :

LET THE NUMBER BE XYZ

SQ (XYZ) is calculated like this

STEP 1. Last digit = last digit of SQ(Z)
STEP 2. Second Last Digit = 2*Y*Z + any carryover from STEP 1.
STEP 3. Third Last Digit 2*X*Z+ Sq(Y) + any carryover from STEP
2.
STEP 4. Fourth last digit is 2*X*Y + any carryover from STEP 3.
STEP 5 . In the beginning of result will be Sq(X) + any carryover
from Step 4.

EXAMPLE :

SQ (431)

STEP 1. Last digit = last digit of SQ(1) =1
STEP 2. Second Last Digit = 2*3*1 + any carryover from STEP
1.= 6
STEP 3. Third Last Digit 2*4*1+ Sq(3) + any carryover from STEP
2.= 2*4*1 +9= 17. so 7 and 1 carryover
STEP 4. Fourth last digit is 2*4*3 + any carryover (which is 1) . =
24+1=25. So 5 and carry over 2.
STEP 5 . In the beginning of result will be Sq(4) + any carryover
from Step 4. So 16+2 =18.

So the result will be 185761.

If the option provided to you are such that the last two digits are
different, then you need to carry out first two steps only , thus
saving time. You may save up to 30 seconds on each
calculations and if there are 4 such questions you save 2
minutes which may really affect UR Percentile score.

PYTHAGORAS THEROEM :

In any given exam there are about 2 to 3 questions based on pythagoras theorem. Wouldn’t it be nice that you remember some of the pythagoras triplets thus saving up to 30 seconds in each question. This saved time may be used to attempt other questions. Remember one more right question may make a lot of difference in UR PERCENTILE score.

The unique set of pythagoras triplets with the Hypotenuse less than 100 or one of the side less than 20 are as follows :

(3,4,5), (5, 12, 13), (8, 15, 17), (7, 24, 25), (20, 21, 29), (12, 35, 37), (9, 40, 41), (28, 45, 53), (11, 60, 61), (33, 56, 65), (16, 63, 65), (48, 55, 73), (36, 77, 85), (13, 84, 85), (39, 80, 89), and (65, 72, 97).
(15,112,113), (17,144,145), (19,180,181), (20,99,101)

If you multiply the digits of the above mentioned sets by any constant you will again get a pythagoras triplet .

Example : Take the set (3,4,5).
Multiply it by 2 you get (6,8,10) which is also a pythagoras triplet.
Multiply it by 3 you get ( 9,12,15) which is also a pythagoras triplet.
Multiply it by 4 you get (12,16,20) which is also a pythagoras triplet.
You may multiply by any constant you will get a pythagoras triplet

Take another example (5,12,13)
Multiply it by 5,6 and 7 and check if you get a pythagoras triplet.

TIPS FOR SMART GUESSING :

You will notice that in any case, whether it is a unique triplet or it is a derived triplet (derived by multiplying a constant to a unique triplet), all the three numbers cannot be odd.

In case of unique triplet , the hypotenuse is always odd and one of the remaining side is odd the other one is even.

Below are the first few unique triplets with first number as Odd.

3    4    5
5    12   13
7    24   25
9    40   41
11   60   61

You will notice following trend for unique triplets with first side as odd.

Hypotenuse = (Sq(first side) +1) / 2
Other side = Hypotenuse -1

Example : First side = 3 ,
so hypotenuse = (3*3+1)/2= 5 and other side = 5-1=4

Example 2: First side = 11
so hypotenuse = (9*9+1)/2= 41 and other side = 41-1=40

Please note that the above is not true for a derived triplet for example 9,12 and 15, which has been obtained from multiplying 3 to the triplet of 3,4,5. You may check for other derived triplets.

Below are the first few unique triplets with first number as Even .

4    3    5
8    15   17
12   35   37
16   63   65
20   99   101

You will notice following trend for unique triplets with first side as Even.

Hypotenuse = Sq( first side/ 2)+1
Other side = Hypotenuse-2

Example 1. First side =8
So hypotenuse = sq(8/2) +1= 17
Other side = 17-2=15

Example 2. First side = 16
So hypotenuse = Sq(16/2) +1 =65
Other side = 65-2= 63

PROFIT AND LOSS : In every exam there are from one to three
questions on profit and loss, stating that the cost was first
increased by certain % and then decreased by certain %. How
nice it would be if there was an easy way to calculate the final
change in % of the cost with just one formula. It would really help
you in saving time and improving UR Percentile. Here is the
formula for the same :

Suppose the price is first increase by X% and then decreased
by Y% , the final change % in the price is given by the following
formula

Final Difference % = X- Y – XY/100.

EXAMPLE 1. : The price of T.V set is increased by 40 % of the
cost price and then decreased by 25% of the new price . On
selling, the profit for the dealer was Rs.1,000 . At what price was
the T.V sold.

From the above mentioned formula you get :
Final difference % = 40-25-(40*25/100)= 5 %.

So if 5 % = 1,000
then 100 % = 20,000.
C.P = 20,000
S.P = 20,000+ 1000= 21,000.

EXAMPLE 2 : The price of T.V set is increased by 25 % of cost
price and then decreased by 40% of the new price . On selling,
the loss for the dealer was Rs.5,000 . At what price was the T.V
sold.

From the above mentioned formula you get :
Final difference % = 25-40-(25*45/100)= -25 %.

So if 25 % = 5,000
then 100 % = 20,000.
C.P = 20,000
S.P = 20,000 – 5,000= 15,000.

Now find out the difference in % of a product which was :
First increased by 20 % and then decreased by 10 %.
First Increased by 25 % and then decrease by 20 %.
First Increased by 20 % and then decrease by 25 %.
First Increased by 10 % and then decrease by 10 %.
First Increased by 20 % and then decrease by 15 %.

TIPS TO IMPROVE UR PERCENTILE :

HOW ABOUT SOLVING THE FOLLOWING QUESTION IN JUST
10 SECONDS

Ajay can finish work in 21 days and Blake in 42 days. If Ajay,
Blake and Chandana work together they finish the work in 12
days. In how many days Blake and Chandana can finish the
work together ?

(21*12 )/(24-12) = (21*12)/9= 7*4= 28 days.

NOW CAREFULLY READ THE FOLLOWING TO SOLVE THE
TIME AND WORK PROBLEMS IN FEW SECONDS.

TIME AND WORK :

1. If A can finish work in X time and B can finish work in Y time
then both together can finish work in (X*Y)/ (X+Y) time.

2. If A can finish work in X time and A and B together can finish
work in S time then B can finish work in (XS)/(X-S) time.

3. If A can finish work in X time and B in Y time and C in Z time
then they all working together will finish the work in
(XYZ)/ (XY +YZ +XZ) time

4. If A can finish work in X time and B in Y time and A,B and C
together in S time then :
C can finish work alone in (XYS)/ (XY-SX-SY)
B+C can finish in (SX)/(X-S)
and A+ C can finish in (SY)/(Y-S)

Here is another shortcut

TYPE 1 : Price of a commodity is increased by 60 %. By how
much % should the consumption be reduced so that the
expense remain the same.

TYPE 2 : Price of a commodity is decreased by 60 %. By how
much % can the consumption be increased so that the expense
remain the same.

Solution :
TYPE1 : (100* 60 ) / (100+60) = 37.5 %
TYPE 2 : (100* 60 ) / (100-60) = 150 %