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Please continue here with all the quant queries and their discussions. link to previous thread :- http://www.pagalguy.com/forum/quantitative-questions-and-answers/72542-official-quant-thread-cat-2011-a.html :cheerio: :cheerio: :c...

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Nachiket Deshmukh
@ThinkAce
11,961

And 2^1008 + 1 = (2^336)^3 + 1 => Divisible by 2^336 + 1Also, 2^1010 + 1 = (2^2)^505 + 1 => Divisible by 2^2 + 1How did u get that if u divide by this it will be divided by 2^336 + 1Ok let's value is 2^2012+1 then????1) ye to upar se gaya

(a^n + b^n) is divisible by (a+b) if n is odd.

So, I just adjusted some odd power other than 1 for the numbers.

2^2012 + 1 = (2^4)^503 + 1 .... 2012 = 4*503

Now Submarine and warship both are underwater..we need to find height of the warship over the submarine..

tan30 = height / 50

=> height = 50 * 1/root3

Now after 30 mins the distance is (50-x)

height will remain same..

=> tan60 = 50/(root3) /

=> 50-x = 50 / 3

=> x = 100/3 = 33.33m ?

could anyone make me undestand what is distance between them from the point of observation is 50 km? i am getting it ? you could explain by digram..what is observation point.

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(3) None of them is prime.

3 leaves remainder by 2 as -1.

So, 2^1009 + 1 and 2^1011 + 1 will leave remainder by 3 as (-1+1) = 0

And 2^1008 + 1 = (2^336)^3 + 1 => Divisible by 2^336 + 1

Also, 2^1010 + 1 = (2^2)^505 + 1 => Divisible by 2^2 + 1

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Take them to be a cyclic trapezium.Of the 4 triangles formed by the diagonals two are equal and the other two are similar.

APB and DPC are equal in area

ADP and BPC are similar

In triangle ABD,

BP:PD = 3:2 [ given area(APD)/area(BPC) = 9/4..sq root will be sides ]

given Area (APD) = 12

Area (ABP):12 = 3:2

=>Area (ABP) = 18

could you clearly explain why APD and BPC are similar?

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Nachiket Deshmukh
@ThinkAce
11,961

1. In an examination having 2 subjects, a student gets 90iles in each subject. What can be the overall maximum and minimum %ile of the student?2. How many Squares formed by Joining 4 points such that co-ordinate of each point is Intgral (both X And Y). And both X and Y co-ordinate is lies in range .3. 2^1008+1,2^1009+1,2^1010+1,2^1011+1which one are prime number?

(1) Minimum percentile will be 80 and maximum will be 100.

Minimum: He has least percentile of 90.

Say he got 10 marks out of 100 in bothe the subject which gives him 90 percentile.

Now say those 10% students ahead of him in each subject score more than 50 in resp. subject. And they are different 10% people.

So, totally 20% people surely have more than 50 marks.

So, he can get 80 percentile. Less than 80 is not possible as students ahead of him must do better than him in at least one section. So, at most 20% such people can be there.

Maximum: He can score 100

Say, he scores 90 and 90 in two subjects. People who are behind him score less than 10 in each. And there are different set of people ahead of him in two sections. So, maximim possible marks of other students are (100+10) = 110. So, he is way ahead of all.

(2) Will try this one after some time

(3) None of them is prime.

3 leaves remainder by 2 as -1.

So, 2^1009 + 1 and 2^1011 + 1 will leave remainder by 3 as (-1+1) = 0

And 2^1008 + 1 = (2^336)^3 + 1 => Divisible by 2^336 + 1

Also, 2^1010 + 1 = (2^2)^505 + 1 => Divisible by 2^2 + 1

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Take them to be a cyclic trapezium.Of the 4 triangles formed by the diagonals two are equal and the other two are similar.

APB and DPC are equal in area

ADP and BPC are similar

In triangle ABD,

BP:PD = 3:2 [ given area(APD)/area(BPC) = 9/4..sq root will be sides ]

given Area (APD) = 12

Area (ABP):12 = 3:2

=>Area (ABP) = 18

could you clearly explain why APD and BPC are similar?

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@gagan220988
57

3. 2^1008+1,2^1009+1,2^1010+1,2^1011+1

which one are prime number?

2^1010+1 ends with 5 so cant be prime

2^1009+1 2^1011 are divisible by 3

2)..20????

1)..80 and 100 waise hi tukka....;)

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Jai Mehra
@jaikishan.n
1,520

3. 2^1008+1,2^1009+1,2^1010+1,2^1011+1which one are prime number?

P.S :

- 1 Like

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which one are prime number?

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Jai Mehra
@jaikishan.n
1,520

a five digit no divisible by 3 is to be formed using numerical 0,1,2,3,4,5 without repetition . the toal no of ways this can be done is:

1. 122

2. 210

3. 216

4. 217

case 1: no zero

the digits must be 1,2,3,4,5

these can be arranged in 5! = 120 ways

case 2: when we use zero

you can only use 1,2,4,5

number of ways

for zer0 = 4 ways

rest 4! = 24

total = 120+4*24216 ways

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