rv_blitz

As the student requires to take atleast one history, one science, and one arts class , he can do it in 6c1*4c1*3c1 ways. As they can be arranged in 3! ways, number of schedules = 6c1*4c1*3c1*3! = 720. Is this correct ?

What I was thinking is (First throw being 4) or (second being 4) or .. (n'th throw being 4) > 0.25. Replacing this by probablity p(First throw being 4)+P(second being 4)+..+p(n'th throw being 4)>0.25. Based on calculations ... after the P(second being 4), the probablity becomes greater than 0.25. Please correct me if I am wrong.

Probablity that we get a 4 in the first throw = 1/6 = 0.166 Probablity that we get 4 in throws is = (Probablity that 4 in the first throw or Probablity that the first is not 4 and second is 4) = 1/6+5/6*1/6 = 11/36 > 0.25 / S0 , 2 throws are required. Is this correct ?

For them to be all of same color, they can be all blue or all green.As they are drawn with replacement, P(all 3 same color) = (3/9*3/9*3/9)+(4/9*4/9*4/9) = (27+64)/729 = 91/729

If the total number of students be 10, then First year = 6 Second year = 4. First year with fin. assistance = 3 First year, NO fin. assistance = 3 Second year with Fin. Assistance = 1 Second year wiith NO fin assitance = 3. Hence, Total with Fin assistance = 4, those of first year = 3, %age = 75 %.

This is possible only if none of the numbers rolled is even,i.e, all of them are odd. i.e 1/2*1/2*1/2 = 1/8. Is this correct ?

Is this 0.4 ??

Is it 75 % ?

Borrowing from dsaqwert, it can be written as 49*2401^1998 = 49 *(2400+1)^1998. NOw, as only the last 3 digits are relevent to us .. 2400^2 and greater powers do not add value as 2400^2, itself has the last 4 digits are zero. So, for the required 3 digits, we need only the last 2 terms of the expansion i.e 49 [1998c1*2400+1] = 49*1998*2400+49 = ..800+49=..849.

7^7994 = 49 ^3997 = (50-1)^3997. Now the last 3 digits are siginificant ... Start working from 3997c3994 * (50^3) *(-1^3994)+3997c3995*(50^2) *(-1^3995)+3997c3996(50^1) *(-1^3996)+3997c3997*(-1^3997). Considering only the last 3 terms of each .. (..000)-(...500)+(..850)-1 = (...349) But I am not sure about this .. Someone needs to check this.

Is it 349 ?

Great explanation , GMAT0805. THanx

Ans : 6 Let n be the number of unique colors n+nc2 >=20 n+[n(n-1)]/2 >=20 (2n+n^2-n)>=40. n^2+n>=40. The smallest value to satisfy this is 6.

Ans E. As the area is 2401, each side =49 feet. x-x-x-x-x-x-x-x If "-" represents the spacing between the centers of the trees, then number of trees in each row = 8. No. of trees = 64. So, the amount gained = 64*35 = 2240. Cost of raising = $896. Profit = 2240-896= 1344. % of total cost = (1344/896)*100 = 150 %

Let P = $1000= initial sum; r = rate of interest. 0.19P = P*(1+r/100)^3 -P; 1.19=(1+r/100)^3. At the end of 6 yrs , the interest is I = 1000*(1+r/100)^6-1000=1000*(1.19)^2 - 1000 = 416. %age of accrued int = (416.1/1000)*100=41.6 % ~42 %