TIME AND WORK

Time and work

Men days:

It is defined as the men multiplied by the day.

MxD= constant

Note:  If M1 men can do W1 work in D1 days H1 hours per day and M2 men can do W2 work in D2 days working H2 hours per day then

M1xD1xH1/W1=M2xD2xH2/W2

LCM AND HCF

LCM AND HCF

LCM: LCM is the least common multiple of two or more numbers is the smallest number that is divisible by each of these two given numbers.

HCF: HCM is the highest common multiple of two or more numbers is the largest number of the given numbers.

For example:

LCM of (3,6,9) is 18 and HCF of ( 3,6,9) is 3.

AREAS

AREAS OF RECTANGLE

A plan of a small garden is drawn to scale of 1 : 25. A pond covers 24cm2 on the plan. What is the area in m2 of the real pond.”

scale is given in length

so the area of pond will be A=24*(25)^2 cm^2

A=15000cm2

but 1m=100cm

so A=15000/(100*100)=1.5m^2

SIMPLE INTEREST AND COMPOUND INTEREST

SIMPLE INTEREST AND COMPOUND INTEREST

Principal:  The amount of money deposited or borrowed is called Principal.

Interest: The amount earned or paid for the use of money is called interest

SIMPLE INTEREST:  The interest that is earned only on the principal is called simple interst.

SI=(P*R*T)/100

where P=principal
R=rate of interest
T=time in year

Compound interest: The principal which is paid not ony pricipal but also on the interest is called compound interest.

The amount at the end of n years is given by

A=P(1+r/100)ⁿ

P=Principal

r=rate of interest

I=A-P

ADDITION AND SUBTRACTION

Addition and Subtraction

Addition:- Addition can explain by following example

Lets ram has 2 apple and john has 3 Apple

So if ram gives all 2 apple to john then john will have total 2+3=5

i.e. 5 Apple.

 So addition means 2+3=5

Addition method on number line
 lets say we have to add 2 and 3
then draw number line

—1—2—3—4—5—6—7—8-
So after 2 on the no. line we have to add 3, so after 2-3, 3-4 and 4-5 so after 3 interval or addition of 3 on 2 we get 5.

Subtraction: Subtraction is loosing of quantity.

Lets say amy has 5 mangoes and she gives 2 mangoes to john then amy will now remaining 5-2=3 mangoes.

On the number line:

—1—2—3—4—5—6—7—8—9—10-

On the number line if the 5-4 and 4-3 so the total reduction after 2 interval is 3.

COMPLEX NUMBER

Complex no has both real and imaginary part.

It is represent by A+Bi

where A is real part and B is imaginary part.

i is the iota and its value i=√-1

Complex no. also represent by e^iθ or cosθ+isinθ

1. Addition of complex no.

 z1=x+yi and z2=x1+y1i

then

z1+z2=x+yi+x1+y1i

z1+z2=x+x1+yi+y1i

z1+z2=x+x1+yi+y1i

z1+z2=x+x1+(yi+y1)i

2. Subtraction of two complex no.

 z1=x+yi and z2=x1+y1i

then

z1-z2=x+yi-(x1+y1i)

z1-z2=x-x1+yi-y1i

z1-z2=x-x1+yi-y1i

z1-z2=x-x1+(yi-y1)i

3. Product of two complex no.

 z1=x+yi and z2=x1+y1i

then

Z1.Z2=(x+yi).(x1+y1i)

Z1.Z2=(xx1+x1yi+xy1i+yy1i)

Z1.Z2=(xx1+(x1y+xy1)i+yy1i²)

Z1.Z2=(xx1+(x1y+xy1)i-yy1)

Z1.Z2=(xx1-yy1+(x1y+xy1)i

4. Product of two complex no.

 z1=x+yi and z2=x1+y1i

then

z1/z2=(x+yi)/(x1+y1i)

RATIO AND PROPORTION

Ratio and ProportionA ratio is a comparison of two or more quantities of similar kind.
It is represent by A:B or A/B
Properties:

i) If both the terms of a ratio are multiplied or divided by same non zero quantity then the value of ratio remains same.
If A/B=p/q then A/B=kp/kq

ii) If the ratio is square or multiply by a constant then ratio will be remains same.

iii) 

PARTIAL FRACTIONS

FIND THE PARTIAL FRACTION OF (x+14)/(x-4)(x+2)

(x+14)/(x-4)(x+2)=a/(x-4)+b/(x+2).......1.

(x+14)/(x-4)(x+2)=⟦a(x+2)+b(x-4)⟧(x-4)(x+2)

or

(x+14)=⟦a(x+2)+b(x-4)⟧

put 

x=-2

-2+14=⟦a(-2+2)+b(-2-4)⟧

12=b(-6)

b=-2

and

put

x=4

4+14=⟦a(4+2)+b(4-4)⟧

18=6a

a=3

hence

from 1.

(x+14)/(x-4)(x+2)=3/(x-4)-2/(x+2) 

Simplify (1/16x)+(1/8)) / ((1/4)+(1/8x))?

((1/16x)+(1/8)) / ((1/4)+(1/8x)) =y assume y=N/D

N=numerator=((1/16x)+(1/8))

and

D=(1/4)+(1/8x)

N=((1/16x)+(1/8))=

Now solve this

N=(1+2x)/16x

and

D=(1/4)+(1/8x)

D=(2x+1)/8x

Now y=N/D=(1+2x)/16x/(2x+1)/8x=8x/16x=1/2

SPEED, TIME AND DISTANCE

The time taken by a train running at 30kmph to pass a man walking in the same direction at a speed of 6 kmph is 3/7 th of the time taken by the train to cross a platform. If the length of the train is 96 mts. What is the length of the platform?

let the time taken for train to cross the man is t1

then 

relative speed=30-6=24kmph=6.66mps

t1=96/6.66

and

t2=time taken for train to cross the platform is t2

speed of train=30kmph=8.33mps

t2=(96+l)

l=length of train

Now according to given condition

t1=(3/7)t2

96/6.66=(3/7)(96+l)/8.33

solve for l

l=184m

A boat takes twice as much time going upstream as it takes to go downstream. While covering the same distance. If the speed of the boat in still water is 6 km/hr. What is the speed of the current?

———→ downstream 

Assume speed of current is V.

Relative velocity= V+6 and time= T1

and 

←———upstream

Relative velocity=6-V  and time T2

Note: Relative velocity=6-V not V-6 becuse if V>6 i.e. velocity of current is greater then velocity of boat then boat will not go to upstream.

Now,

i.e.T2=2T1

d/(6-V)=2d/(V+6)

V+6=2(6-V)

V+6=12-2V

3V=6

V=2kmph

The speed of the boat in still water is 22 kmph. It takes the same time for the boat to travel 1.2 km upstream as it does to travel 32km downstream. What is the speed of the current?

———→ downstream

Assume speed of current is V.

Relative velocity= V+22 and time= T1 

Distance=32km

and 

←———upstream

Relative velocity=22-V  and time T2

Distance=1.2km

Note: Relative velocity=22-V not V-22 becuse if V>22 i.e. velocity of current is greater then velocity of boat then boat will not go to upstream.

Now,

i.e.T2=T1

1.2/(22-V)=32/(V+22)

(V+22)1.2=32(22-V)

1.2V+22×1.2=32×22-32V

V=20.4kmph

SURDS AND INDICES

How do I solve 4x−2x−2=0?

4x−2x−2=0

2^2X-2^X-2=0

Assume 2^x=t

then

above equation become

t²-t-2=0

t²-2t+t-2=0

from this

t=2 and t=-1

but 2^x=t

2^x=2

x=1 

and

2^x=-1

for above condition x value not possible

hence only solution x=1

How can I simplify ab√√ab√3 with exponent properties?

y=√ab/∛a∛b

y=√a√b/∛a∛b     because √ab=√a√b

y=a^(1/2-1/3)×b^(1/2-1/3)  because a^x/a^y=a^x-y

y=a^(5/6)×b^(5/6)      because a^x×b^x=(ab)^x

y=(ab)^(5/6)

If the value of (a + 1/a) is equal to 3, then what will be the value of a2 + 1/a2?

a+1/a=3

squering both sides

a²+1/a²+2×a×(1/a)=9

a²+1/a²+2=9  use formula (x+y)²=x²+y²+2xy

a²+1/a²=7

If (8x²)^8-r(2x)^-r=2^ax^{b   then find a and b in terms of r ?}
(8x²)^8-r(2x)^-r=2^ax^b

(2^³x²)^8-r(2x)^-r=2^ax^b

2^3(8-r)x^2(8-r)=2^ax^b

2^(24-3r)x^(16-2r)=2^ax^b

^{now compare the power of 2 and x}

a=(24-3r)

and b=16-2r

IF (8-√18)/√2=a+b√2 then find a and b ?
we have to eliminate √2 denominator.

for this multiply and divide by √2

i.e.

(8-√18)/√2=(8-√18)×√2/(√2×√2) 

                   =(8-√18)×√2/2

                   =(8√2-√18√2)/2

                   =(8√2-√36)/2

                   =(8√2-2√3)/2

                    =4√2-√9

                      =-3+4√2

a+b√2=-3+4√2

a=-3 and b=4
What is the value of 22014−22013

2^2014 - 2^2013

= 2^(2013+1) - 2^2013

= 2^2013 (2^1- 1)

= 2^ 2013 * (2–1)

= 2^ 2013 * 1

= 2^2013