@rahulgaur666 said:what is wrong in my solution /
A is appearing twice therefore divide by 2
PS : sorry for not replying due to office work :)
@Subhashdec2 said:end with04->512->416->4 20 -> 524->432->436->4 40->552->456->460->548??
its 52 check again
If n is the largest possible value of points in a plane that any three of them can form a right triangle then what is the value of n ?
@shattereddream
4? for values greater than 5 you can create a non-right triangle is why i thought ans might be 4
4? for values greater than 5 you can create a non-right triangle is why i thought ans might be 4
@anantn boss OA not available right now for me its too 4 only cxonsidered the case where the answer can be 5 but thats not possible
@shattereddream said:If n is the largest possible value of points in a plane that any three of them can form a right triangle then what is the value of n ?
5??
a square and the point of intersection of diagnols??
@Subhashdec2 said:
a square and the point of intersection of diagnols??
will they form a right traingle ....think this
A boat is to be rowed by 10 men; 5 on the left side and 5 on the right side. Of the 10 men, available two cannot row on the right side and three cannot row on the left side. In how many ways can the 10 men be arranged?
@pavimai said:A boat is to be rowed by 10 men; 5 on the left side and 5 on the right side. Of the 10 men, available two cannot row on the right side and three cannot row on the left side. In how many ways can the 10 men be arranged?
5c3 *5!*5!??
@pavimai said:A boat is to be rowed by 10 men; 5 on the left side and 5 on the right side. Of the 10 men, available two cannot row on the right side and three cannot row on the left side. In how many ways can the 10 men be arranged?
5c3*5!*5! ?
Given that f(x) = ax^2 + bx + c and f(4) = 100. If a, b, c are distinct positive integers, then the maximum possible value of a + b + c is
a 79
b 87
c 122
d 82
e Data Insufficient
a 79
b 87
c 122
d 82
e Data Insufficient
@pavimai said:A boat is to be rowed by 10 men; 5 on the left side and 5 on the right side. Of the 10 men, available two cannot row on the right side and three cannot row on the left side. In how many ways can the 10 men be arranged?
5C2 *5!^2
@amresh_maverick said:Given that f(x) = ax^2 + bx + c and f(4) = 100. If a, b, c are distinct positive integers, then the maximum possible value of a + b + c is a 79 b 87 c 122 d 82 e Data Insufficient
79??
Number of common terms in the 2 sequences
17,21,25,.......417
16,21,26........466??
Ans= 20
My approach : 21-17=4, 21-16=5
lcm(4,5)=20
So, on adding corresponding differences after every 20th no we'll get common terms
eg 37 .... 1st series => 37+4=41
36...2nd series => 36+5=41
so in all 20 common terms
is this approach correct????/
@amresh_maverick said:Given that f(x) = ax^2 + bx + c and f(4) = 100. If a, b, c are distinct positive integers, then the maximum possible value of a + b + c is a 79 b 87 c 122 d 82 e Data Insufficient
a=1,b=2,c=76.so a+b+c=79