General Formula for Work-Time relationship
The inverse of time it takes working together equals the sum of the inverse of the time it takes each individually.
If A takes a hours and B takes b hours to finish a task, then A and B together takes t hours which can be calculated from the following formula:
\(\dfrac{1}{t} = \dfrac{1}{a} + \dfrac{1}{b}\)
Showing posts with label KeyFormulas. Show all posts
Showing posts with label KeyFormulas. Show all posts
Sunday
Saturday
MetroEye01:Rate of Work Problem
General Formula for work-person-rate problem
\(\dfrac{W_1}{M_1\cdot H_1}\) = \(\dfrac{W_2}{M_2\cdot H_2}\)
where W is the work to be done
M is number of persons required
H is the time required.
\(\dfrac{W_1}{M_1\cdot H_1}\) = \(\dfrac{W_2}{M_2\cdot H_2}\)
where W is the work to be done
M is number of persons required
H is the time required.
Progression03: Sum of Odd and Even Numbers
Sum of first n odd numbers
S = \( n^2\)Number of terms(n), if nth term of odd-number series is \( t_n\)
n = \(\dfrac{t_n+1}{2}\)Sum of first n even numbers
S = n(n+1)Number of terms(n), if nth term of even-number series is \( t_n\)
n = \(\dfrac{t_n}{2}\)
Friday
Progression02:Sum of first n natural numbers
Sum of first n natural numbers 1,2,3,4,....... n
nth term/number of terms if sum of n terms is given
S = \(\dfrac{n(n+1)}{2}\)
nth term/number of terms if sum of n terms is given
n = \(\dfrac{\sqrt{1+8S} -1}{2}\)
Progression01: Sum and nth term of a series
Any series a, a+d, a+2d, a+3d, ..........
where a --> First term
d --> difference between the terms
n --> number of terms
nth term of the series
First term -> a
Second term -> a+1d
Third term -> a+2d
.
.
.
nth term -> a+(n-1)d
n'th term of the series = a + (n-1)d
Sum of the first n terms
S = a+a+d+a+2d+a+3d+.........+a+(n-1)d
= na+d+2d+3d+.....(n-1)d
= na+d(1+2+3+......+(n-1))
= na+d((n-1)(n-1+1)/2)
= n(a+(n-1)d/2)
Sum of the first n terms = \(\dfrac{n}{2}\left(2a+\left(n-1\right)d\right)\)
Number of terms in the series
Number of terms = \(\dfrac{\left(LastTerm-FirstTerm\right)}{d}+1\)
where a --> First term
d --> difference between the terms
n --> number of terms
nth term of the series
First term -> a
Second term -> a+1d
Third term -> a+2d
.
.
.
nth term -> a+(n-1)d
n'th term of the series = a + (n-1)d
Sum of the first n terms
S = a+a+d+a+2d+a+3d+.........+a+(n-1)d
= na+d+2d+3d+.....(n-1)d
= na+d(1+2+3+......+(n-1))
= na+d((n-1)(n-1+1)/2)
= n(a+(n-1)d/2)
Sum of the first n terms = \(\dfrac{n}{2}\left(2a+\left(n-1\right)d\right)\)
Number of terms in the series
Number of terms = \(\dfrac{\left(LastTerm-FirstTerm\right)}{d}+1\)
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