Posts and Comments

I am not getting what is actually being asked?

@Surbhih
2

Posted 06 Feb '12

plz help-
Two traders, Chetan and Michael, were involved in the buying and selling of MCS shares over five trading
days. At the beginning of the first day, the MCS share was priced at Rs 100, while at the end of the fifth
day it was priced at Rs 110. At the end of each day, the MCS share pr...

plz help-

Two traders, Chetan and Michael, were involved in the buying and selling of MCS shares over five trading

days. At the beginning of the first day, the MCS share was priced at Rs 100, while at the end of the fifth

day it was priced at Rs 110. At the end of each day, the MCS share price either went up by Rs 10, or else, it

came down by Rs 10. Both Chetan and Michael took buying and selling decisions at the end of each

trading day. The beginning price of MCS share on a given day was the same as the ending price of the

previous day. Chetan and Michael started with the same number of shares and amount of cash, and had

enough of both. Below are some additional facts about how Chetan and Michael traded over the five

trading days.

- Each day if the price went up, Chetan sold 10 shares of MCS at the closing price. On the other hand,

each day if the price went down, he bought 10 shares at the closing price.

- If on any day, the closing price was above Rs 110, then Michael sold 10 shares of MCS, while if it was

below Rs 90, he bought 10 shares, all at the closing price.

4. If Chetan sold 10 shares of MCS on three consecutive days, while Michael sold 10 shares only once

during the five days, what was the price of MCS at the end of day 3?

(1) Rs 90 (2) Rs 100

(3) Rs 110 (4) Rs 120

(5) Rs 130

Two traders, Chetan and Michael, were involved in the buying and selling of MCS shares over five trading

days. At the beginning of the first day, the MCS share was priced at Rs 100, while at the end of the fifth

day it was priced at Rs 110. At the end of each day, the MCS share price either went up by Rs 10, or else, it

came down by Rs 10. Both Chetan and Michael took buying and selling decisions at the end of each

trading day. The beginning price of MCS share on a given day was the same as the ending price of the

previous day. Chetan and Michael started with the same number of shares and amount of cash, and had

enough of both. Below are some additional facts about how Chetan and Michael traded over the five

trading days.

- Each day if the price went up, Chetan sold 10 shares of MCS at the closing price. On the other hand,

each day if the price went down, he bought 10 shares at the closing price.

- If on any day, the closing price was above Rs 110, then Michael sold 10 shares of MCS, while if it was

below Rs 90, he bought 10 shares, all at the closing price.

4. If Chetan sold 10 shares of MCS on three consecutive days, while Michael sold 10 shares only once

during the five days, what was the price of MCS at the end of day 3?

(1) Rs 90 (2) Rs 100

(3) Rs 110 (4) Rs 120

(5) Rs 130

@Surbhih
2

Posted 05 Feb '12

> I think this should be 14*5!
First, we can select 5 positions out of 6 on that circle. As it is circular arrangement, we need to fix a position. Say, we fix it and name it position 1.
Then, we have 2-3-4-5-6 as other places of circle with 2 and 6 next to position 1. All we need to do ...

I think this should be 14*5!

First, we can select 5 positions out of 6 on that circle. As it is circular arrangement, we need to fix a position. Say, we fix it and name it position 1.

Then, we have 2-3-4-5-6 as other places of circle with 2 and 6 next to position 1. All we need to do is select 4 more positions. This is like making 4 digit numbers using digits (2, 3, 4, 5 and 6) such that no two consecutive digits are adjecent; and numbers do not start with 2 or 6.

Starting with 3: 3 _ _ _

Second position can be taken by 5 or 6. When taken by 5; we can have 14 or 26 in the end.

Second position when taken by 6; we can have 24, 42 or 25 as last two digits.

=> Total 5 numbers for 3 at position 2.

=> Total 5 numbers for 5 at position 2 as we have symmetry.

Starting with 4:

We can have seocnd position as 2 or 6. With 2; we can have last two digits as 53 or 63. Two cases and two more symmetrical cases for with 6.

=> Total 4 cases

So, in all we have (5+5+4) i.e. 14 ways of selecting 5 positions.

And then, we can place 5 events in 5! positions for each of selected combination.

So, in all 14*5! ways.

but y answer nhi h

@Surbhih
2

Posted 05 Feb '12

plz tell-
Q) two line segments PQ and RS intersects at X in sucha way XP=XR. if PSX=RQX, then one must have-
1\. PR=QS
2\. PS=RQ
3\. XSQ=XRP
4\. ar(triangle PXR)=ar(triangle QXS)

plz tell-

Q) two line segments PQ and RS intersects at X in sucha way XP=XR. if PSX=RQX, then one must have-

1. PR=QS

2. PS=RQ

3. XSQ=XRP

4. ar(triangle PXR)=ar(triangle QXS)

Q) two line segments PQ and RS intersects at X in sucha way XP=XR. if PSX=RQX, then one must have-

1. PR=QS

2. PS=RQ

3. XSQ=XRP

4. ar(triangle PXR)=ar(triangle QXS)

@Surbhih
2

Posted 04 Feb '12

> I think this should be 14*5!
First, we can select 5 positions out of 6 on that circle. As it is circular arrangement, we need to fix a position. Say, we fix it and name it position 1.
Then, we have 2-3-4-5-6 as other places of circle with 2 and 6 next to position 1. All we need to do ...

I think this should be 14*5!

First, we can select 5 positions out of 6 on that circle. As it is circular arrangement, we need to fix a position. Say, we fix it and name it position 1.

Then, we have 2-3-4-5-6 as other places of circle with 2 and 6 next to position 1. All we need to do is select 4 more positions. This is like making 4 digit numbers using digits (2, 3, 4, 5 and 6) such that no two consecutive digits are adjecent; and numbers do not start with 2 or 6.

Starting with 3: 3 _ _ _

Second position can be taken by 5 or 6. When taken by 5; we can have 14 or 26 in the end.

Second position when taken by 6; we can have 24, 42 or 25 as last two digits.

=> Total 5 numbers for 3 at position 2.

=> Total 5 numbers for 5 at position 2 as we have symmetry.

Starting with 4:

We can have seocnd position as 2 or 6. With 2; we can have last two digits as 53 or 63. Two cases and two more symmetrical cases for with 6.

=> Total 4 cases

So, in all we have (5+5+4) i.e. 14 ways of selecting 5 positions.

And then, we can place 5 events in 5! positions for each of selected combination.

So, in all 14*5! ways.

its not like that.. its wrong

@Surbhih
2

Posted 04 Feb '12

> total arrangements will be 6P5.....now we will take scenario where we will consider two point as one...so it will be 5C5 x 2!...
Ans- 6P5 - 5C5 x 2!
ps: correct me if iam making some mistake in this question..
this is not d answer

total arrangements will be 6P5.....now we will take scenario where we will consider two point as one...so it will be 5C5 x 2!...

Ans- 6P5 - 5C5 x 2!

ps: correct me if iam making some mistake in this question..

this is not d answer

@Surbhih
2

Posted 03 Feb '12

> I think this should be 14*5!
First, we can select 5 positions out of 6 on that circle. As it is circular arrangement, we need to fix a position. Say, we fix it and name it position 1.
Then, we have 2-3-4-5-6 as other places of circle with 2 and 6 next to position 1. All we need to do ...

I think this should be 14*5!

First, we can select 5 positions out of 6 on that circle. As it is circular arrangement, we need to fix a position. Say, we fix it and name it position 1.

Then, we have 2-3-4-5-6 as other places of circle with 2 and 6 next to position 1. All we need to do is select 4 more positions. This is like making 4 digit number

THIS IS NOT THE ANSWER:|

@Surbhih
2

Posted 03 Feb '12

>
*According to question i assumed since they wants to travel together ,then they wants to sit together as well !! But i guess this is not the case !!*
*Then it shud be :
C(7,5)*C(2,2)*12!*8! + C(7,1)*12!*8!
=> 12! * 8! *[ 21 + 7]
=> 12! * 8! * [28]
Explanation :...

According to question i assumed since they wants to travel together ,then they wants to sit together as well !! But i guess this is not the case !!Then it shud be :

C(7,5)*C(2,2)*12!*8! + C(7,1)*12!*8!

=> 12! * 8! *[ 21 + 7]

=> 12! * 8! * [28]

Explanation :

When 7 sits in upper Deck :C(7,5)*C(2,2)*12!*8!

--> First make sit people who wants to sit together.

7 in upper Deck and 6 in lower Deck.

--> Now we left with 7 seats ( 5 upper and 2 lower ) and 7 people with no restriction

Hence Choose 5 out of 7 people and make them sit in upper deck and remaining in lower deck.

--> Now arrange whole upper and lower deck in 12! and 8! ways

Similarly for other case when 7 sits in lower deck.

BUT U R GETTING THE ANSWRS ODR WAY TOO [IMG]http://www.pagalguy.net/pagalguy/smilies/mgk.gif[/IMG]

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