@ashish100 said: @Stoicalme for ist one:denominator is (a-b)(a+b)now max value of a+b can be till 10 only as 11 won't divide 8!so when a+b=10 then a-b can be 2,4,6,8out of which following sol can be possible : (6,4)(7,3)(8,2)when a+b=9 then a-b can be 1,3,5,7here 0 possible solutions as max power of 3 can be two onlywhen a+b=8 a-b can be 2,4,6so (5,3)(6,2)possible solutionswhen a+b=7 and a-b can be 1,3,5possible sol are :(4,3)(5,2)when a+b=6 and a-b can be 2,4so sol can be :(4,2)when a+b =5 and a-b can be 1,3so sol can be 3,2 when a+b=4 then a-b can be 2so sol can be nothingIn all getting 9 solutions....kahan miss kar raha hoon bhai log

@sowmyanarayanan said: CAT 1998 You can collect Rubies and Emralds as many as you can. Each Ruby is worth 4 Crores and each Emrald is worth 5 Crores. Each Ruby weighs 0.3kg and each emrald weighs 0.4kg. Your bag can carry at most 12kg. What you should collect to get maximum wealth? a) 20R and 15E b) 40R c) 28R and 9E d) None of these Can someone explain the above with solution? Since the answer cjoice has 'None of these' i was not able to pick one from the top three. please help.

@[576959:Stoicalme]

Option C

@Stoicalme said: Q) Two ships sail in a fog towards each other with the same speed.When they are 4 kms apart,the captains decelerate the engines for 4 minutes with a deceleration rate of 0.1m/s^2,and then the ships continue sailing with the speeds attained.For what range of values of the initial speed V0(subscript) will the ships avoid collision?A) 0<10m/sB)0<20m/sC)0<30m/sD)N.O.TE)CBD

@sravan1989 said: Any one knows the sum of first n terms of Harmonic progression ???

Q) Two ships sail in a fog towards each other with the same speed.When they are 4 kms apart,the captains decelerate the engines for 4 minutes with a deceleration rate of 0.1m/s^2,and then the ships continue sailing with the speeds attained.For what range of values of the initial speed V0(subscript) will the ships avoid collision?

@[525438:taurean11]:

@htomar said: FoolNFinal Bhai, i need one clarification in first part. In first part, we have assumed the boxes to be distinct and balls to be identical, and used the formula to get the answer. Which is nothing but 8!/(4!*4!). Now,I will get the same answer, if i assume that there are 4 identical balls placed in a row and i put four identical bars in the gaps to divide the balls into 5 groups. Then the situation will be same as arraning 8 objects out of which 4 are of one type and 4 another type. This will come out same as 8!/(4!*4!) Question is, for your part you assumed boxes to be distinct and balls to be identical whereas in the above approach i assumed balls to be identical but nothing about boxes. Is it correct approach?

A square intersects a circle at eight distinct points such that the area of the circle outside the square is equal to the area of the square outside the circle. If the radius of the circle is ‘r’, then the area of the square is

a) root(2) pi r^2

b) 2 pi r^2

c) 3/2 r^2

d)none f dese

@Stoicalme said: 1) a and b are natural numbers such that a > b > 1. If 8! is divisible by a^2 — b^2, then how many such sets (a, b) are possible? a 5 b 6 c 7 d 8 2) The lines y = 3x and x + y = 40 and the x axis bound a triangular area. Find the total number of points on or inside this triangle with integral coordinates. a 600 b 641 c 655 d 662 e 625 _____/\_____

@[576959:Stoicalme]

CAT 1998
You can collect Rubies and Emralds as many as you can. Each Ruby is worth 4 Crores and each Emrald is worth 5 Crores. Each Ruby weighs 0.3kg and each emrald weighs 0.4kg. Your bag can carry at most 12kg. What you should collect to get maximum wealth?
a) 20R and 15E
b) 40R
c) 28R and 9E
d) None of these

Can someone explain the above with solution?
Since the answer cjoice has 'None of these' i was not able to pick one from the top three. please help.

@[470280:gs4890] @[271527:ashish100] @[525438:taurean11] @[509517:kpcat]

@Aizen said: sn = 1+5+14+28+47 + ....+tn sn = 1+5 +14+28+ ... +tn-1 +tn [Shift it by 1] Subtract.. 0 = 1+(4+9+14+19+... (n-1) terms) - tn [See in brackets the terms are in AP] => tn =1 + (n-1)/2*(2*1 +(n-2)*5) Now multiply and get tn in form of n. Just do sigma over 1 to 10. p.s: __/\__ @gs4890 bhai , @sumeet1489 bhai aur sabhi quant maharathi ...

@rachit_28 said: I have the same doubt as Raghav. Can you please tell the logic of taking the form 2^x * 3^y or could you redirect to the post you are mentioning?

@mani0303 said: Try this: the sum of the series - 1,5,14,28,47... until 10th term

Any one knows the sum of first n terms of Harmonic progression ???

@mani0303 said: Try this: the sum of the series - 1,5,14,28,47... until 10th term

@Stoicalme said: @gs4890 @sumeet1489 As @taurean11 pointed out i checked out the question again and it was indeed the multiplication of both. Extremely sorry for your loss of time and energy bhailogo.

@Debasis070 said: Q)Shyam went from Delhi to Shimla via Chandigarh by car. The distance from Delhi to Chandigarh is 3/4 times thedistance from Chandigarh to Shimla. The average speed from Delhi to Chandigarh was half as much again as that fromChandigarh to Shimla. If the average speed for the entire journey was 49 kmph. What was the average speed fromChandigarh to Shimla? a)39.2 kmphb)63 kmphc)42 kmphd)70 kmph Bhai log good evening

@rachit_28 said: i am unable to appreciate the multiplication of 2 qirh 12C2 - 8C2.Can you please tell why it is being multiplied?

@[573636:Debasis070]

@[573636:Debasis070]

@gs4890 said: bhai all my possible sets are satisfying the condition ...plz recheck the OA or point out any of the set not satisfying the condition

@[470280:gs4890] gs bhai (8,24) bhi satisfy kar raha hai

@Debasis070 said: Q)Shyam went from Delhi to Shimla via Chandigarh by car. The distance from Delhi to Chandigarh is 3/4 times thedistance from Chandigarh to Shimla. The average speed from Delhi to Chandigarh was half as much again as that fromChandigarh to Shimla. If the average speed for the entire journey was 49 kmph. What was the average speed fromChandigarh to Shimla? a)39.2 kmphb)63 kmphc)42 kmphd)70 kmph Bhai log good evening

@RaghavMittal said: If n is a positive integer such that 2n has 28 positive divisors and 3n has 30 positive divisors, then how many positive divisors does 6n have ?? A. 32B. 34C. 35D. 36

@Stoicalme said: @gs4890 The answer is 7, bhai.

@sowmyanarayanan said: i guess it is D. 104maximum number of points a circle & a circle can intersect is 2, same is the case for circle & staright line. But two straight lines can intersect at one point only.So the answer is 2 x 12C2 - 8C2

@[470280:gs4890] The answer is 7, bhai.

@Stoicalme said: 1) a and b are natural numbers such that a > b > 1. If 8! is divisible by a^2 — b^2, then how many such sets (a, b) are possible? a 5 b 6 c 7 d 8 2

@arpitrkothari said: please post the approach to find the sum of series whose difference is in ap??

@Stoicalme said: 1) a and b are natural numbers such that a > b > 1. If 8! is divisible by a^2 — b^2, then how many such sets (a, b) are possible? a 5 b 6 c 7 d 8

m getting 9 possible sets

2 ,4

2 ,6

3 ,6

3 ,9

4 ,8

4 ,12

6 ,12

6 ,18

8 ,16

??

@mani0303 said: @rachit_28 Yeah,rightdamn my imaginationThanks though!

please post the approach to find the sum of series whose difference is in ap??

@Enceladus said: 2^x*3^y.Factors = (x+1)*(y+1).It says (x+2)*(y+1) = 28.And (x+1)*(y+2) = 30.=> x = 5 and y = 3.N = 2^5*3^3.2N = 2^6*3^3, Factors = 28.3N = 2^5*3^4, Factors = 30.6N = 2^6*3^4, Factors = 35.solved it a few posts back

@[594243:rachit_28] Yeah,right:)damn my imagination;)

@mani0303 said: @culdip how could 69 and 96 are not confusable?

@Faruq said: ‎1) : if 1 of the roots of the eqn x^2 – px +24= 0 is 4 and the eqn x^2 – px +q = 0 has real and equal roots then find value of q.

@Stoicalme said: 1) a and b are natural numbers such that a > b > 1. If 8! is divisible by a^2 — b^2, then how many such sets (a, b) are possible? a 5 b 6 c 7 d 8 2) The lines y = 3x and x + y = 40 and the x axis bound a triangular area. Find the total number of points on or inside this triangle with integral coordinates. a 600 b 641 c 655 d 662 e 625 _____/\_____

@[482349:sowmyanarayanan]

@culdip said: 10 01 06 60 08 80 09 90 16 61 18 81 19 91 86 68 89 98 66 99 are confusable (00 11 88 69 96 are not confusable) so 80 will be ans

@Stoicalme said: 2) The lines y = 3x and x + y = 40 and the x axis bound a triangular area. Find the total number of points on or inside this triangle with integral coordinates. a 600 b 641 c 655 d 662 e 625 _____/\_____

please tell me how to these kind of questions?

@sumeet1489 said: 2nd question => 641

@Stoicalme said: 1) a and b are natural numbers such that a > b > 1. If 8! is divisible by a^2 — b^2, then how many such sets (a, b) are possible? a 5 b 6 c 7 d 8 2) The lines y = 3x and x + y = 40 and the x axis bound a triangular area. Find the total number of points on or inside this triangle with integral coordinates. a 600 b 641 c 655 d 662 e 625 _____/\_____

@sravan1989 said: an intelligence agency decided on a code of two digits selected from 0, 1,2, ......9 but the slip on which the code is hand written allows confusion between top and bottom, because those are indistinguishable. thus , for example , the code 91 could be confused with 16. how many codes are there such that there is no possibility of any confusion?? a)25 b)75 c)80 d)None of the above

@Stoicalme said: 1) a and b are natural numbers such that a > b > 1. If 8! is divisible by a^2 — b^2, then how many such sets (a, b) are possible? a 5 b 6 c 7 d 8 2) The lines y = 3x and x + y = 40 and the x axis bound a triangular area. Find the total number of points on or inside this triangle with integral coordinates. a 600 b 641 c 655 d 662 e 625 _____/\_____

@gs4890 said: enceladus bhai , shouldn't it be - => 16[(n*(n+1)*(2n+1)/6 - n*(n+1)/2] + 3*26 ??

@sravan1989 said: A square is drawn by joining the mid points of the side of a given square. A third square is drawn inside the second square in the same way and this process id continued indefinitely. If a side of the first square is 8 cm, the sum of the areas of all the squares(in sq. cm) is, ??? a)128 b)120 c)96 d)None of the above

@YouMadFellow said: I am not sure about it, but I think there is a theorem which says N^(p-1)/p will give a remainder of 1, where N and p are relatively prime. Using this, I am getting the remainder as 3.

@Enceladus said: Summation 16 (n^2 - n) + 3. => 16[(n*(n+1)*(2n+1)/6 - n*(n+1)/2] + 3.Put n = 26. I hope it is clear.

@rachit_28 said: Last term will be T26. Could you please how to do its summation?

@shivs said: @ajayarora116 3^1001= 3. 3^1000 = 3.(3^10)^100= 3.(59049)^100= 3.(1001.59-10)^100 which when divided by 1001 will give = 3.(10)^100= 30.(10^3)^33 = 30 (1000)^33= 30.(1001-1)^33 which when divided by 1001 will give -30 as the remainder. Hence the remainder 1001-30 = 971

@targeted said: The greatest possible number of points of intersection of 8 straight lines and 4 circles isa.32 b.64 c.76 d.104P.S.: I dont have the ans to this question.....plz explain ur approach in details.....it wil help me....

@[594243:rachit_28]

@Enceladus said: Tn = 16(n^2 - n)+3. Now calculate the sum.