[Decent_flirt]

Originally Posted by manjushaa

hey how do i do these?
1.the amt of work in a leather factory is increased by 50%.By what % is it neccessary to increase the no of workers to complete the new amt of work in previously planned time,if the productivity of the new labour is 25% more(40%)
2.a shopkeeper first raises the price of jewellery by x% then he decreases the new price by x %.after 1 such up down cycle,the price of a jewellery is decreased by rs21025.after the 2nd updown cycle it was sold for rs484416.what was the original price of the jewellery?(rs 525625)

Let the intial amount = 10000a

New amount after inc = 10000a + 100xa
when dec, 10000a + 100xa - (100ax+ax^2) = 10000a-ax^2.

total dec = ax^2 = 21025............1

Again inc., 10000a - ax^2 + 100xa - ax^3/100
Again dec.., 10000a - ax^2 + 100xa - ax^3/100 - 100xa + ax^3/100 - ax^2 + ax^4/10000
= 10000a -2ax^2+ax^4/10000 = 484416..........2

Solving 1 and 2, x=20

There substituting back we get a=52.5625
Therefore total = 525625

[masoom]

Originally Posted by manjushaa

hey how do i do these?
1.the amt of work in a leather factory is increased by 50%.By what % is it neccessary to increase the no of workers to complete the new amt of work in previously planned time,if the productivity of the new labour is 25% more(40%)
2.a shopkeeper first raises the price of jewellery by x% then he decreases the new price by x %.after 1 such up down cycle,the price of a jewellery is decreased by rs21025.after the 2nd updown cycle it was sold for rs484416.what was the original price of the jewellery?(rs 525625)

Yups for the first one.. i am also getting 20 % increase

[janvats]

The integers 1, 2, …, 40 are written on a blackboard. The following operation is then repeated 39 times: In each repetition, any two numbers, say a and b, currently on the blackboard are erased and a new number a + b – 1 is written. What will be the number left on the board at the end?

1.820
2.821
3.781
4.819
5.780

[subu_222412]

Originally Posted by masoom

is the answer is b) 294840...

Yes..
Can you please elaborate..

[first_timer]

Originally Posted by janvats

The integers 1, 2, …, 40 are written on a blackboard. The following operation is then repeated 39 times: In each repetition, any two numbers, say a and b, currently on the blackboard are erased and a new number a + b – 1 is written. What will be the number left on the board at the end?

1.820
2.821
3.781
4.819
5.780

Sum of the numbers on the board = (40*41)/2 = 820
Each operation reduces 1 from this sum.
After 39 operations, the number remaining on the board will be (820-39) = 781

[manjushaa]

Originally Posted by Decent_flirt

Leth the initial work = 1
let the number of workers be 1
Let the number of days in which it is completed = 1.
Therefor efficiency = 1.

Now work = 1.5
efficiency = 1.25
Therefore 1.25*x=1.5

x=1.2
Therefore increase = 20% and not 40%

hey but the answer given is 40%

[janvats]

What are the last two digits of 7^2008?

[first_timer]

Originally Posted by janvats

What are the last two digits of 72008?

didn't get ur question

[first_timer]

Originally Posted by janvats

What are the last two digits of 7^2008?

7^2008 = 2401^502 (7^4 = 2401)

So the second last digit is 0*2 = 0
and the last digit is 1.

So last two digits = 01

Approach using euler : last two digits can be obtained by finding the remainder of 7^2008/100

Euler of 100 = 40

So Rem[7^2008/100] = rem [ 7^8/100 ] = rem [2401^2/100] = Rem[01^2/100] = 01--> last two digits...

[MaskedMenace]

Originally Posted by janvats

What are the last two digits of 7^2008?

7^2008

2008 mod 4 = 0

so, 7^2008 = 7^4 = 2401