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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)
For a cubic equation ax^3+bx^2+cx+d=o
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If any parallelogram can be inscribed in a circle , it must be a rectangle.
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 .
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.)
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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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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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
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) .
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)
(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
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)
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
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.
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 .
if a/p = b/q = c/r = d/s .
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Consider the two equations
a1x+b1y=c1
a2x+b2y=c2
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.
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 .
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) .
(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]
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 .
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
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+……….)
(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
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
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 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) .
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 .
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) .
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.
other two sides.
++++++++++++++++++++++++++++++++++++++++++++++++++++++++++The quadrilateral formed by joining the angular bisectors of another quadrilateral is
always a rectangle.
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
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
|a|+|b| = |a+b| if a*b>=0
else |a|+|b| >= |a+b|
else |a|+|b| >= |a+b|
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2<= (1+1/n)^n <=3
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WINE and WATER formula:
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
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
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.
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.
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) .
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
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))
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)
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
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
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