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Hi puys, CAT 2011 season is finally over :: :: Let start a new season of Quants thread for CAT 2012.. :: :: Here are few links of Quants thread for CAT 2011 season... Part-8 http://www.pagalguy.com/forum/quanti...at-...

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@akalola
19

Q.) If denotes the greatest integer less than or equal to a, find the number of possible values of x (x is an integer, 0+ + = 31x/30a.) 31...........b.) 32............c.) 33............d.) 34..............e.) None of these

My take is 33

Above equation is satisfied if and only if x is multiple of 30 30x ( 0

so total possible value is 33

Correct me if i am wrong

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Jai Mehra
@jaikishan.n
1,520

[QUOTE=varun.tyagi;3235251**Q.) Mr. Khan usually leaves home at 8:00 am and reaches office at 8:30 am, travelling by his car at a constant speed. One day, he traveled half way to the office at his normal speed, when he realized that he had forgotten something at home. He sped home to collect it and went to office, all at a higher speed than normal. He reached at 8:45 am.If on the next day, he started at 8:00 am as usual but traveled at the higher speed throughout. At what time did he reach office? a.)8:15 am..........b.)8:18 am...........c.)8:20 am..........d.)8:24 am..............e.)8:25 am**

my take

On first day he reaches half way by 8:15

now he travels with a higher speed say 'S' and covers a distance of 1.5D in 30 mins

S= 1.5D/30 = D/20

Next day he starts by 8 and goes by speed S

time taken = D*20/D = 20 mins

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Start Downloadingplease post your approach as well.....you are missing something here :)

jain bhai check karo 503 hoga

galti to hoti rehti hai bhai.....is thread pe no sorry .....vaise check karo 1 se to bahut zyaada hogi values

:thumbsup: Bingo.....bang on target :d

q.) mr. Khan usually leaves home at 8:00 am and reaches office at 8:30 am, travelling by his car at a constant speed. One day, he traveled half way to the office at his normal speed, when he realized that he had forgotten something at home. He sped home to collect it and went to office, all at a higher speed than normal. He reached at 8:45 am.if on the next day, he started at 8:00 am as usual but traveled at the higher speed throughout. At what time did he reach office?a.) 8:15 am..........b.) 8:18 am...........c.) 8:20 am..........d.) 8:24 am..............e.) 8:25 am

since lhs is an integer...rhs must also be an integer.

Thus rhs=30+x/30

every multiple of 30 will satisfy this...hence till 1000 we have 1000/30=33 values.what am i missing sir???:|

ANS TO THE SECOND 1 IS 8:20 AM

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Abhishek Jain
@jain4444
40,426

Q.) Mr. Khan usually leaves home at 8:00 am and reaches office at 8:30 am, travelling by his car at a constant speed. One day, he traveled half way to the office at his normal speed, when he realized that he had forgotten something at home. He sped home to collect it and went to office, all at a higher speed than normal. He reached at 8:45 am.

If on the next day, he started at 8:00 am as usual but traveled at the higher speed throughout. At what time did he reach office?

a.)8:15 am..........b.)8:18 am...........c.)8:20 am..........d.)8:24 am..............e.)8:25 am

normal speed = 400 kmph

distance = 200 km

abnormal speed = (100+200)*2 = 600 kmph

200/600 = 1/3 = 20 mins

so ,

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Varun Tyagi
@varun.tyagi
14,220

pratyasha1990 Saysis the oa=33

Please post your approach as well.....You are missing something here :)

3 1 2 1 3 1 2 1.........3 1 2 1 3 1 2

cyclicity of four so , 2010 = 502*4 + 2

so ,502 times will 3 appear...doubt lag raha hai varun bhai

Jain bhai check karo 503 hoga

as LHS is an integer RHS should also be an integer.

30+x/30

so x should be a multiple of 30 say 30k

substituting the value

30k/2 + 30k/3 +30k/5 =15k+10k+6K

31k=30+30k/30

so k=1

x=30

so only one value

kuchh galti hui ho to maaf karna maalik

Galti to hoti rehti hai bhai.....Is thread pe no sorry .....Vaise check karo 1 se to bahut zyaada hogi values

502 +1 =503 hoga

position of 3 is

1st , 5th, 9th and so on till 2009th making a total of 503

:thumbsup: Bingo.....bang on target :D

If on the next day, he started at 8:00 am as usual but traveled at the higher speed throughout. At what time did he reach office?

a.)8:15 am..........b.)8:18 am...........c.)8:20 am..........d.)8:24 am..............e.)8:25 am

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q.) 2010 coins, each of value 1, 2 or 3 are arranged in a row such that between any two coins of value 1 there is at least one coin, between any two of value 2 there are at least two coins, and between any two of value 3 there are at least three coins. The maximum number of coins of value 3 that can be in the row?a.) 502............b.) 670............c.) 503.............d.) 669..............e.) 501

is oa 10=502

to have max no of 3 coins let us tk this arrangement

3_ _ _3 and fill it up lyk

3 1 2 1 3 1 2 1 ....

Hence 1 out of 4 coins is 3

2010=502*4 +2

hence 502 3 coins( the rem 2 are 1 &2 coins)

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@vinay_sharma
1,053

3 1 2 1 3 1 2 1.........3 1 2 1 3 1 2

cyclicity of four so , 2010 = 502*4 + 2

so ,502 times will 3 appear...doubt lag raha hai varun bhai

502 +1 =503 hoga

position of 3 is

1st , 5th, 9th and so on till 2009th making a total of 503

- 2 Likes

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Sumeet Nayak
@sumeet1489
3,111

Q.) If denotes the greatest integer less than or equal to a, find the number of possible values of x (x is an integer, 0+ + = 31x/30a.) 31...........b.) 32............c.) 33............d.) 34..............e.) None of these

as LHS is an integer RHS should also be an integer.

30+x/30

so x should be a multiple of 30 say 30k

substituting the value

30k/2 + 30k/3 +30k/5 =15k+10k+6K

31k=30+30k/30

so k=1,2,3.......33

so 33 values

kuchh galti hui ho to maaf karna maalik

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Varun Tyagi
@varun.tyagi
14,220

The probability of a missile hitting a bridge is 1/2 and it takes 3 missiles to destroy a bridge.what is the min. No of missiles to be fired so that the probability of the bridge being destroyed exceeds .99?

1.7

2.8

3.14

4.15

Probability of destroying a bridge = Probability of hitting 3 missiles > 99/100

=> Probability of not destroying a bridge = Probability of hitting 0,1,2 missiles

=> C(n,0) + C(n,1) + C(n,2)*(1/2)^n => *(1/2)^n

=> (n^2 + n + 2)/2^(n+1)

=> n^2 + n + 2

n = 7

=> 58 Not possible

=> 212 Possible

So,

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@akalola
19

60b + c - 60a - b = 60c + a.

=> a = 59(b-c)/61.

So only possible is when b = c, ie, a = 0. Otherwise it would be next day.

Yes you are right so there is no possible value of a cause a=0 is not possible .

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Abhishek Jain
@jain4444
40,426

Q.) 2010 coins, each of value 1, 2 or 3 are arranged in a row such that between any two coins of value 1 there is at least one coin, between any two of value 2 there are at least two coins, and between any two of value 3 there are at least three coins. The maximum number of coins of value 3 that can be in the row?

a.)502............b.)670............c.)503.............d.)669..............e.)501

3 1 2 1 3 1 2 1.........3 1 2 1 3 1 2

cyclicity of four so , 2010 = 502*4 + 2

so ,

skashwin89 Saysplease justify why 4-4-3 only?

coz we have 11 members and 3 cars and each car has maximum seating capacity of 4

read this line again again and think about why 4-4-3 only by this you will get it for life time

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My take is 20160

Number of people in car can be (4, 4, 3) only.

Now, we have 4 males for 3 cars. So, it has to be (2, 1, 1).

Now, we get two cases: 2 men in a car of 3 and 2 men in a car of 4.

Case 1: Two men in a car of 3

These two men can be selected in 4C2 ways. Then, they can pick a car in 3 ways. Remaining two men can pick cars in 2*1 ways.

Now, we have 7 women and (1, 3, 3) psostions in car.

One lady can be chosen in 7 ways. Second car can have ladies in 6C3 ways while other car will have remaining ladies.

So, total = 4C2 * 3 * 2*1 * 7 * 6C3 ways = 5040 ways

Case 2: Two men in a car of 4

These two men can be selected in 4C2 ways and they can pick a car in 3 ways. Remaining men can pick cars in 2*1 ways.

Now, we have 7 women and (2, 3, 2) posistions in cars of capacity (4, 4, 3) resp.

So, first car can have women in 7C2 ways. Second car can have them in 5C3 ways and remaining two will go to third car.

One more point here that the cars which have 1 man each, have capacity of 4 and 3. So, the car with capacity of 4 can be chosen in 2 ways. Other will have capacity of 1.

So, total = 4C2 * 3 * 2*1 * 7C2 * 5C3 * 2*1 ways = 15120 ways

In all, we have (5040+15120) ways = 20160 ways

please justify why 4-4-3 only?

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@akalola
19

pratyasha1990 Says2 values-11011 and 12221

I didnt get the answer .

How did you find that 2 values ?

Could you explain it in details ?

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q.) if denotes the greatest integer less than or equal to a, find the number of possible values of x (x is an integer, 0+ + = 31x/30a.) 31...........b.) 32............c.) 33............d.) 34..............e.) none of these

is the oa=33

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Varun Tyagi
@varun.tyagi
14,220

a.)502............b.)670............c.)503.............d.)669..............e.)501

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