@TootaHuaDil said:125*1 ...125*8 and 125*9 kyu nhi?? samjha do
4 digit ho jata na bhai woh...
for three positive integers a,b and c
a+b+c+ab+bc+ca+abc=1000.
what is the value of a+b+c ?
Certain quantities of 2 distinct varieties of rice V1 and V2 are mixed and d mixture is sold at rs.X per kg ,giving 25% profit.If rice of variety V1 is sold at Rs.X per kg,a loss of (100/11)% would be incurred.If the ratio of the quantities of rice of variety V2 and variety V1 in d mixture is 8:3,then what is the %age of profit obtained when rice of variety V2 is sold at rs.X per kg?
@The_Loser said:A _ _ _ _ _ _ _ _ BFfix 8 station bw A & B.Now we can select 4 station out of the remaining 9 positions. So 9c4 = 126
hw ?? we can also select the station frm these 9 stations consecutively..
bhai solve it @vijay_chandola
bhai solve it @vijay_chandola
@vishcat said:for three positive integers a,b and c
a+b+c+ab+bc+ca+abc=1000.
what is the value of a+b+c ?
(a+1)(b+1)(c+1)=1001=7*11*13
a+b+c=6+10+12=28
@Logrhythm said:4 digit ho jata na bhai woh...
bhai OA me 7 values possible bole hain....5^3*1=125....ko bhi consider kia hai...ye sahi hai kya?
@TootaHuaDil said:n! has x number of zeroes at the end and (n+1)! has x+3 zeroes at the end. Find the number of possible values of n if n is a three digit number.
my ans is coming 7 ..
@joyjitpal said:100 to 199 --- we have 20 numbers having 2 as a digit in them. 200 to 299 --- 120 numbers having 2 as a digit in them. 300 to 499 --- 20 . . . 700 to 799 --- 20 20 * 6 + 120 = 240
bhai we are counting the numbers having a 2 one of its digit.. not the number of 2's... doing same mistake as i did.. 
@joyjitpal said:There are 12 intermediate stations between two places A and B. In how many ways can a trainbe made to stop at 4 of these 12 intermediate stations provided no two of them areconsecutive?a. 45 b. 126 c. 84 d. 168
See here:
A_x1_x2_x3_x4_B
x1,x2,x3 and x4 are the four stops. The first and last '_' can be 0. other '_' has to be >=1.
So, a+b+c+d+e=8 (a,e>=0,b,c,d>=1)
So 9C4 solutions.....
@TootaHuaDil said:bhai OA me 7 values possible bole hain....5^3*1=125....ko bhi consider kia hai...ye sahi hai kya?
bhai 125*(1,2,3,4,6,7) hi toh hoga...
aur konsa??
@CrookDinu - how 7??
@TootaHuaDil said:n! has x number of zeroes at the end and (n+1)! has x+3 zeroes at the end. Find the number of possible values of n if n is a three digit number.
@CrookDinu said:my ans is coming 7 ..
1st value= 124,125
2nd value= 249,250
3rd value=374,375
4th value=499 ,500
5th value=749 ,750
6th value=874,875
so ans is 6
what is the correct ans ??
@CrookDinu said:1st value= 124,1252nd value= 249,2503rd value=374,3754th value=499 ,5005th value=749 ,7506th value=874,875so ans is 6what is the correct ans ??
@Logrhythm ..bhai wo ek jyada count ho gyi thi..i realized that later..ans is 6
@Logrhythm said:bhai 125*(1,2,3,4,6,7) hi toh hoga...aur konsa?? @CrookDinu - how 7??
125*8=1000...this is n+1 na...n to abhi bhi 3 digit hoga...to isko count kar sakte hain na??? Not sure...
@TootaHuaDil said:125*8=1000...this is n+1 na...n to abhi bhi 3 digit hoga...to isko count kar sakte hain na??? Not sure...
true...it would be 7... 
A square is inscribed in a circle and the circle is inscribed in a regular octagon. Find the ratio of the area of the square to that of the octagon.
@bs0409 said:A square is inscribed in a circle and the circle is inscribed in a regular octagon. Find the ratio of the area of the square to that of the octagon.
is it (rt2+1):4
Find the number of possible values of n = 125 * m if n is less than or equal to 1000 and m is not divisible by 5
@bs0409 said:A square is inscribed in a circle and the circle is inscribed in a regular octagon. Find the ratio of the area of the square to that of the octagon.
(rt2 + 1)/4 ?
side of big square = x
so side of octagon = a + rt2*a + a = x-----> a = x/(2 + rt2)
area of octagon = x^2 - 4*1/2 *a^2 = x ^2 - 2*x^2 (1/(2 + rt2))^2
area of smaller square = x^2/2
ratio = x^/2 / x^2(1 - 2 (1/(2 + rt2))^2
solving this eqn...will get
(rt2 + 1)/4
The quotient obtained by dividing n by 121 and the number 11 are co-prime to each other. Find the number of possible values of n if n is less than or equal to 10000.