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Hi, I have compiled some math formulas from various sources. This may help some.
THEORY OF EQUATIONS:
Shortcuts to getting information about the roots
-----------------------------
(1) If an equation contains all positive co-efficients of any powers of x, then it has no positive roots.[/i]
e.g. \(x^4+3x^2+2x+6=0\) has no positive roots .
(2) If all the even powers of x have same sign coefficients and all the odd powers of x have the opposite sign coefficients, then the equation has no negative roots.
e.g. \(x^2 - x +2 = 0\)
(3) Summarizing DESCARTES rules of signs:
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).
(4)Consider the two equations
ax + by = c
dx + ey = f
Then, If \(\frac{a}{d} = \frac{b}{e} = \frac{c}{f}\), then we have infinite solutions for these equations.
If \(\frac{a}{d} = \frac{b}{e} \neq \frac{c}{f}\) , then we have no solution for these equations.
If \(\frac{a}{d} \neq \frac{b}{e}\) , then we have a unique solutions for these equations.
(5) Complex roots occur in pairs, hence if one of the roots of an equation is 2+3i , another has to be 2-3i and if there are three possible roots of the equation, we can conclude that the last root is real. This real roots could be found out by finding the sum of the roots of the equation and subtracting (2+3i)+(2-3i)=4 from that sum.
(6) If an equation f(x)= 0 has only odd powers of x and all these have the same sign coefficients or if f(x) = 0 has only odd powers of x and all these have the same sign coefficients then the equation has no real roots in each case, except for x=0 in the second case.
(7) Besides Complex roots, even irrational roots occur in pairs. Hence if 2+root(3) is a root, then even 2-root(3) is a root . (All these are very useful in finding number of positive, negative, real, complex etc roots of an equation )
(8) |x| + |y| >= |x+y| (|| stands for absolute value or modulus ) (Useful in solving some inequations)
(9) 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
(10) 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
Geometry
General Notions and useful shortcuts:
Polygons:
(1) For any regular polygon, the sum of the interior angles is equal to 360 degrees
(2) If any parallelogram can be inscribed in a circle , it must be a rectangle.
(2.1)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]
(3) If a trapezium can be inscribed in a circle it must be an isosceles trapezium (i:e oblique sies equal).
(4) 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) .
Triangles
(1) In an isosceles triangle , the perpendicular from the vertex to the base or the angular bisector from vertex to base bisects the base.
(2) In any triangle the angular bisector of an angle bisects the base in the ratio of the other two sides.
(3) The ratio of the radii of the circumcircle and incircle of an equilateral triangle is 2:1 .
(4.1)In any triangle
a=b*CosC + c*CosB
b=c*CosA + a*CosC
c=a*CosB + b*CosA
(4.2)In any triangle
a/SinA = b/SinB =c/SinC=2R , where R is the circumradius
cosC = (a^2 + b^2 - c^2)/2ab
sin2A = 2 sinA * cosA
cos2A = cos^2(A) - sin^2 (A)
(5.1)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) .
(5.2) Appolonius theorem could be applied to the 4 triangles formed in a parallelogram.
(6) 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) .
(7) 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))
(8) Let W be any point inside a rectangle ABCD .
Then WD^2 + WB^2 = WC^2 + WA^2
(9) 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)
Quadrilateral
(1) For a cyclic quadrilateral , the measure of an external angle is equal to the measure of the
internal opposite angle.
(2) If a quadrilateral circumscribes a circle , the sum of a pair of opposite sides is equal
to the sum of the other pair .
(3) the quadrilateral formed by joining the angular bisectors of another quadrilateral is
always a rectangle.
Areas:
(1)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
(2.1) For a cyclic quadrilateral , area = root( (s-a) * (s-b) * (s-c) * (s-d) ) , where s=(a+b+c+d)/2
(2.2) 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.
(3.1) Area of a regular hexagon : root(3)*3/2*(side)*(side)
(3.2) Area of a hexagon = root(3) * 3 * (side)^2
(4) Area of a parallelogram = base * height
(5) 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.
Stereometry
(1) for similar cones , ratio of radii = ratio of their bases.
(2) Volume of a pyramid = 1/3 * base area * height
Number properties
(1) 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 .
(2) The HCF and LCM of two nos. are equal when they are equal .
(3) For any 2 numbers a>b
a>AM>GM>HM>b (where AM, GM ,HM stand for arithmetic, geometric , harmonic menasa
respectively)
(4) (GM)^2 = AM * HM
(5) For three positive numbers a, b ,c
(a+b+c) * (1/a+1/b+1/c)>=9
(6) For any positive integer n
2<= (1+1/n)^n <=3
(7) a^2+b^2+c^2 >= ab+bc+ca
If a=b=c , then the equality holds in the above.
(8) a^4+b^4+c^4+d^4 >=4abcd
(9) 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
(10) (m+n)! is divisible by m! * n! .
(11.1)If n is even , n(n+1)(n+2) is divisible by 24
(11.2)If n is any integer , n^2 + 4 is not divisible by 4
(12) 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)
(13) 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 .
(14) 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
(15) 2<= (1+1/n)^n <=3
(16) (1+x)^n ~ (1+nx) if x<<<1
(17) |a|+|b| = |a+b| if a*b>=0 else |a|+|b| >= |a+b|
(18) In a GP (Geometric Progression?) the product of any two terms equidistant from a term is always constant .
(19)The sum of an infinite GP = a/(1-r) , where a and r are resp. the first term and common ratio of the GP .
(20)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+..........)
Life application problems
Mixture Problems
(10) 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
Some neat shortcuts on Simple/Compound Interest.
Shortcut #1:
-------------
We all know the traditional formula to compute interest...
CI = P*(1+R/100)^N - P
The calculation get very tedious when N>2 (more than 2 years). The method suggested below is elegant way to get CI/Amount after 'N' years. You need to recall the good ol' Pascal's Triange in following way:
Code:
Number of Years (N)
-------------------
1 1
2 1 2 1
3 1 3 3 1
4 1 4 6 4 1
. 1 .... .... ... ... 1
Example: P = 1000, R=10 %, and N=3 years. What is CI & Amount?
Step 1: 10% of 1000 = 100, Again 10% of 100 = 10 and 10% of 10 = 1
We did this three times b'cos N=3.
Step 2:
Now Amount after 3 years = 1 * 1000 + 3 * 100 + 3 * 10 + 1 * 1 = Rs.1331/-
The coefficents - 1,3,3,1 are lifted from the pascal's triangle above.
Step 3:
CI after 3 years = 3*100 + 3*10 + 3*1 = Rs.331/- (leaving out first term in step
2)
If N =2, we would have had, Amt = 1 * 1000 + 2 * 100 + 1 * 10 = Rs. 1210/-
CI = 2 * 100 + 1* 10 = Rs. 210/-
This method is extendable for any 'N' and it avoids calculations involving higher
powers on 'N' altogether!
A variant to this short cut can be applied to find depreciating value of some
property. (Example, A property worth 100,000 depreciates by 10% every year, find
its value after 'N' years).
Shortcut #2:
-------------
(i) When interest is calculated as CI, the number of years for the Amount to double (two times the principal) can be found with this following formula:
P * N ~ 72 (approximately equal to).
Exampe, if R=6% p.a. then it takes roughly 12 years for the Principal to double itself.
Note: This is just a approximate formula (when R takes large values, the error % in
formula increases).
(ii) When interest is calculated as SI, number of years for amt to double can be
found as:
N * R = 100 . BTW this formula is exact!
Adding to what 'Peebs' said, this shortcut does work for any P/N/R.
Basically if you look closely at this method, what I had posted is actually derived
from the Binomial expansion of the polynomial -- (1+r/100)^n but in a more
"edible" format digestable by us! BTW herez one shortcut on recurring decimals to fractions ...Its more easier to explain with an example..
Eg: 2.384384384 ....
Step 1: since the 3 digits '384' is recurring part, multiply 2.384 by 1000 = so we
get 2384.
Next '2' is the non recurring part in the recurring decimal so subtract 2 from 2384
= 2382.
If it had been 2.3848484.., we would have had 2384 - 23 = 2361. Had it been
2.384444.. NR would be 2384 - 238 = 2146 and so on.
We now find denominator part .......
Step 3: In step 1 we multiplied 2.384384... by 1000 to get 2384, so put that first.
Step 4: next since all digits of the decimal part of recurring decimal is recurring,
subtract 1 from step 3. Had the recurring decimal been 2.3848484, we need to
subtract 10. If it had been 2.3844444, we needed to have subtracted 100 ..and so
on...
Hence here, DR = 1000 - 1 = 999
Hence fraction of the Recurring decimal is 2382/999!!
Some more examples ....
1.56787878 ... = (15678 - 156) / (10000 - 100) = 15522/9900
23.67898989... = (236789 - 2367) / (10000 - 100) = 234422/9900
124.454545... = (12445 - 124) / (100 - 1) = 12321/99
Clock problems
(4) 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 .
(5) 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) .
- - Theory of equations
- Geometry
- - General Notions and Shortcuts
- - - Polygons
- - - Triangles
- - - Quadrilaterals
- - Areas
- Number properties
- Life-application problems
THEORY OF EQUATIONS:
Shortcuts to getting information about the roots
-----------------------------
(1) If an equation contains all positive co-efficients of any powers of x, then it has no positive roots.[/i]
e.g. \(x^4+3x^2+2x+6=0\) has no positive roots .
(2) If all the even powers of x have same sign coefficients and all the odd powers of x have the opposite sign coefficients, then the equation has no negative roots.
e.g. \(x^2 - x +2 = 0\)
(3) Summarizing DESCARTES rules of signs:
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).
(4)Consider the two equations
ax + by = c
dx + ey = f
Then, If \(\frac{a}{d} = \frac{b}{e} = \frac{c}{f}\), then we have infinite solutions for these equations.
If \(\frac{a}{d} = \frac{b}{e} \neq \frac{c}{f}\) , then we have no solution for these equations.
If \(\frac{a}{d} \neq \frac{b}{e}\) , then we have a unique solutions for these equations.
(5) Complex roots occur in pairs, hence if one of the roots of an equation is 2+3i , another has to be 2-3i and if there are three possible roots of the equation, we can conclude that the last root is real. This real roots could be found out by finding the sum of the roots of the equation and subtracting (2+3i)+(2-3i)=4 from that sum.
(6) If an equation f(x)= 0 has only odd powers of x and all these have the same sign coefficients or if f(x) = 0 has only odd powers of x and all these have the same sign coefficients then the equation has no real roots in each case, except for x=0 in the second case.
(7) Besides Complex roots, even irrational roots occur in pairs. Hence if 2+root(3) is a root, then even 2-root(3) is a root . (All these are very useful in finding number of positive, negative, real, complex etc roots of an equation )
(8) |x| + |y| >= |x+y| (|| stands for absolute value or modulus ) (Useful in solving some inequations)
(9) 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
(10) 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
Geometry
General Notions and useful shortcuts:
Polygons:
(1) For any regular polygon, the sum of the interior angles is equal to 360 degrees
(2) If any parallelogram can be inscribed in a circle , it must be a rectangle.
(2.1)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]
(3) If a trapezium can be inscribed in a circle it must be an isosceles trapezium (i:e oblique sies equal).
(4) 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) .
Triangles
(1) In an isosceles triangle , the perpendicular from the vertex to the base or the angular bisector from vertex to base bisects the base.
(2) In any triangle the angular bisector of an angle bisects the base in the ratio of the other two sides.
(3) The ratio of the radii of the circumcircle and incircle of an equilateral triangle is 2:1 .
(4.1)In any triangle
a=b*CosC + c*CosB
b=c*CosA + a*CosC
c=a*CosB + b*CosA
(4.2)In any triangle
a/SinA = b/SinB =c/SinC=2R , where R is the circumradius
cosC = (a^2 + b^2 - c^2)/2ab
sin2A = 2 sinA * cosA
cos2A = cos^2(A) - sin^2 (A)
(5.1)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) .
(5.2) Appolonius theorem could be applied to the 4 triangles formed in a parallelogram.
(6) 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) .
(7) 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))
(8) Let W be any point inside a rectangle ABCD .
Then WD^2 + WB^2 = WC^2 + WA^2
(9) 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)
Quadrilateral
(1) For a cyclic quadrilateral , the measure of an external angle is equal to the measure of the
internal opposite angle.
(2) If a quadrilateral circumscribes a circle , the sum of a pair of opposite sides is equal
to the sum of the other pair .
(3) the quadrilateral formed by joining the angular bisectors of another quadrilateral is
always a rectangle.
Areas:
(1)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
(2.1) For a cyclic quadrilateral , area = root( (s-a) * (s-b) * (s-c) * (s-d) ) , where s=(a+b+c+d)/2
(2.2) 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.
(3.1) Area of a regular hexagon : root(3)*3/2*(side)*(side)
(3.2) Area of a hexagon = root(3) * 3 * (side)^2
(4) Area of a parallelogram = base * height
(5) 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.
Stereometry
(1) for similar cones , ratio of radii = ratio of their bases.
(2) Volume of a pyramid = 1/3 * base area * height
Number properties
(1) 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 .
(2) The HCF and LCM of two nos. are equal when they are equal .
(3) For any 2 numbers a>b
a>AM>GM>HM>b (where AM, GM ,HM stand for arithmetic, geometric , harmonic menasa
respectively)
(4) (GM)^2 = AM * HM
(5) For three positive numbers a, b ,c
(a+b+c) * (1/a+1/b+1/c)>=9
(6) For any positive integer n
2<= (1+1/n)^n <=3
(7) a^2+b^2+c^2 >= ab+bc+ca
If a=b=c , then the equality holds in the above.
(8) a^4+b^4+c^4+d^4 >=4abcd
(9) 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
(10) (m+n)! is divisible by m! * n! .
(11.1)If n is even , n(n+1)(n+2) is divisible by 24
(11.2)If n is any integer , n^2 + 4 is not divisible by 4
(12) 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)
(13) 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 .
(14) 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
(15) 2<= (1+1/n)^n <=3
(16) (1+x)^n ~ (1+nx) if x<<<1
(17) |a|+|b| = |a+b| if a*b>=0 else |a|+|b| >= |a+b|
(18) In a GP (Geometric Progression?) the product of any two terms equidistant from a term is always constant .
(19)The sum of an infinite GP = a/(1-r) , where a and r are resp. the first term and common ratio of the GP .
(20)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+..........)
Life application problems
Mixture Problems
(10) 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
Some neat shortcuts on Simple/Compound Interest.
Shortcut #1:
-------------
We all know the traditional formula to compute interest...
CI = P*(1+R/100)^N - P
The calculation get very tedious when N>2 (more than 2 years). The method suggested below is elegant way to get CI/Amount after 'N' years. You need to recall the good ol' Pascal's Triange in following way:
Code:
Number of Years (N)
-------------------
1 1
2 1 2 1
3 1 3 3 1
4 1 4 6 4 1
. 1 .... .... ... ... 1
Example: P = 1000, R=10 %, and N=3 years. What is CI & Amount?
Step 1: 10% of 1000 = 100, Again 10% of 100 = 10 and 10% of 10 = 1
We did this three times b'cos N=3.
Step 2:
Now Amount after 3 years = 1 * 1000 + 3 * 100 + 3 * 10 + 1 * 1 = Rs.1331/-
The coefficents - 1,3,3,1 are lifted from the pascal's triangle above.
Step 3:
CI after 3 years = 3*100 + 3*10 + 3*1 = Rs.331/- (leaving out first term in step
2)
If N =2, we would have had, Amt = 1 * 1000 + 2 * 100 + 1 * 10 = Rs. 1210/-
CI = 2 * 100 + 1* 10 = Rs. 210/-
This method is extendable for any 'N' and it avoids calculations involving higher
powers on 'N' altogether!
A variant to this short cut can be applied to find depreciating value of some
property. (Example, A property worth 100,000 depreciates by 10% every year, find
its value after 'N' years).
Shortcut #2:
-------------
(i) When interest is calculated as CI, the number of years for the Amount to double (two times the principal) can be found with this following formula:
P * N ~ 72 (approximately equal to).
Exampe, if R=6% p.a. then it takes roughly 12 years for the Principal to double itself.
Note: This is just a approximate formula (when R takes large values, the error % in
formula increases).
(ii) When interest is calculated as SI, number of years for amt to double can be
found as:
N * R = 100 . BTW this formula is exact!
Adding to what 'Peebs' said, this shortcut does work for any P/N/R.
Basically if you look closely at this method, what I had posted is actually derived
from the Binomial expansion of the polynomial -- (1+r/100)^n but in a more
"edible" format digestable by us! BTW herez one shortcut on recurring decimals to fractions ...Its more easier to explain with an example..
Eg: 2.384384384 ....
Step 1: since the 3 digits '384' is recurring part, multiply 2.384 by 1000 = so we
get 2384.
Next '2' is the non recurring part in the recurring decimal so subtract 2 from 2384
= 2382.
If it had been 2.3848484.., we would have had 2384 - 23 = 2361. Had it been
2.384444.. NR would be 2384 - 238 = 2146 and so on.
We now find denominator part .......
Step 3: In step 1 we multiplied 2.384384... by 1000 to get 2384, so put that first.
Step 4: next since all digits of the decimal part of recurring decimal is recurring,
subtract 1 from step 3. Had the recurring decimal been 2.3848484, we need to
subtract 10. If it had been 2.3844444, we needed to have subtracted 100 ..and so
on...
Hence here, DR = 1000 - 1 = 999
Hence fraction of the Recurring decimal is 2382/999!!
Some more examples ....
1.56787878 ... = (15678 - 156) / (10000 - 100) = 15522/9900
23.67898989... = (236789 - 2367) / (10000 - 100) = 234422/9900
124.454545... = (12445 - 124) / (100 - 1) = 12321/99
Clock problems
(4) 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 .
(5) 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) .
Few tricks
1) 0.6666...
to convert 0.6666... to fraction, we know 6 is repeated every time.i.e there is only 1 digit repetition so I will divide 6 by single digit 9
so, i can write it as 6/9
2) 0.46464646...
here 46 is repeated so there are 2 digits
fraction is 46/99
3)0.23434343434...
here we can see that after 0.2 , 34 is repeated
fraction = ( decimal - non repeated decimal)/(9 digits equal to no of repeated decimals * 10^(no of non repeated decimal))
looks complicated, but it is actually not Wink
fraction = (234 - 2)/(99 * 10^1) = 232/990
4)0.1234343434.....
fraction = (1234-12)/(99 * 10^2) = 1222/9900
i guess this method is easy for finding the fraction of a recurring decimal.
1) 0.6666...
to convert 0.6666... to fraction, we know 6 is repeated every time.i.e there is only 1 digit repetition so I will divide 6 by single digit 9
so, i can write it as 6/9
2) 0.46464646...
here 46 is repeated so there are 2 digits
fraction is 46/99
3)0.23434343434...
here we can see that after 0.2 , 34 is repeated
fraction = ( decimal - non repeated decimal)/(9 digits equal to no of repeated decimals * 10^(no of non repeated decimal))
looks complicated, but it is actually not Wink
fraction = (234 - 2)/(99 * 10^1) = 232/990
4)0.1234343434.....
fraction = (1234-12)/(99 * 10^2) = 1222/9900
i guess this method is easy for finding the fraction of a recurring decimal.
Great compilation but it would be even better if split into categories/formatted a bit to make it friendly to the eye (or maybe I am just old 
)
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General Discussion
Posts: 24
Good one mate,
A very useful post.
A very useful post.
Very usefull thanks a lot!
Yes BB,
I thought the same. I tried to make it a little bit more formatted but cannt do much
I thought the same. I tried to make it a little bit more formatted but cannt do much
Timetrader, is this something you can lend your hand to?
P.S. Looks above and beyond what the GMAT asks
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P.S. Looks above and beyond what the GMAT asks
_________________
Posts: 97
I'll try to do some editing.. I'll save the reserve copy.. If you'd like to reverse everything just let me now 
Great thing, Thanks!!!
I'm noticing that some of these equations may require a calculator...which is beyond what the GMAT requires imo. Am I correct?
Posts: 71
Good collection of the basic principles of Math....thanks!!!
Great collection thank you!
bandit, this is a great compilation! Thanks.
One section looks incorrect (or am I misreading?)
(11.2)If n is any integer , n^2 + 4 is not divisible by 4
if I take n=6, (6^2 + 4)=40; divisible by 4.
One section looks incorrect (or am I misreading?)
(11.2)If n is any integer , n^2 + 4 is not divisible by 4
if I take n=6, (6^2 + 4)=40; divisible by 4.
gmattokyo
My thoughts exactly! I think the rule should read
(11.2)If n is any odd integer , n^2 + 4 is not divisible by 4
bgs7b6
And maybe even a step further,
(11.2)If n is any odd integer, n^2 = Odd Number. An Odd number + An Even Number = Odd Number
Great job bandit. Thanks a lot.
but i didn't understand this one, could you clarify?
(8) |x| + |y| >= |x+y| (|| stands for absolute value or modulus ) (Useful in solving some inequations)
Ex:
Let's say, x=-6 and y=1 then, |-6|+|1|=7 but |-6+1|=5
Have i skipped smth?
but i didn't understand this one, could you clarify?
(8) |x| + |y| >= |x+y| (|| stands for absolute value or modulus ) (Useful in solving some inequations)
Ex:
Let's say, x=-6 and y=1 then, |-6|+|1|=7 but |-6+1|=5
Have i skipped smth?
Posts: 65
Quote:
How would you derive clock problem's solution from the above ? Can you explain please.
Posts: 65
bandit
For any regular polygon, the sum of the interior angles is equal to 360 degrees . It is not correct i beleive.
It is (n-2)180 where n >3 . For e.x. if n =5 (lets say a regular pentagon) , sum is 540 degree.
Posts: 1
" (5) For three positive numbers a, b ,c
(a+b+c) * (1/a+1/b+1/c)>=9"
+1 for the wonderful collection, really helpful.
Can you please make me understand with the help of an example may be, how does the above relation hold true? I am a bit confused with the (1/a+1/b+1/c).
Thanks
(a+b+c) * (1/a+1/b+1/c)>=9"
+1 for the wonderful collection, really helpful.
Can you please make me understand with the help of an example may be, how does the above relation hold true? I am a bit confused with the (1/a+1/b+1/c).
Thanks
Posts: 39
very nice compilation.
Thanks for sharing
Thanks for sharing
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