@htomar said:is it 8?
nopes..wats ur fraction ??? is a
@nick_baba said:there u go man.....ur take on a and b??
1/a + 1/b =2/7
on solving
(2a-7)(2b-7)=49
solutions: a=7 & b= 7
a=4,b=28 or a=28, b=4
but as given in ques(a
then a=4 b =28
a+b=32
on solving
(2a-7)(2b-7)=49
solutions: a=7 & b= 7
a=4,b=28 or a=28, b=4
but as given in ques(a
then a=4 b =28
a+b=32
Now got it..@19rsb said:1/a + 1/b =2/7on solving(2a-7)(2b-7)=49 solutions: a=7 & b= 7a=4,b=28 or a=28, b=4but athen a=4 b =28a+b=32
perfect..
..but i did it this way..
..but i did it this way..one fraction is 1/a and other is 2/7-1/a
which gives b=7a/(2a-7)...now a
therefore..a
therefore a
and b= 7a/(2a-7)
for a=4 we get integer value of b=28
therefore a+b=32...
The remainder when F(x) is divided by (x-19) and (x-99) , is 99 and 19 respectively.Find remainder when F(x) is divided by(x-19)(x-99).
Let Xn denote the n-th element of the sequence {1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, .....}, where n is a positive integer. How many of the following statements are then true?
Statement I :
Xn is the largest integer less than ½ + sqrt(2n + ¼)
Statement II :
Xn is the largest integer not greater than ½ + Sqrt(2(n - 1) + ¼)
Statement III :
Xn is the smallest integer greater than -1/2 + sqrt(2n + ¼)
Case 1: Consider only the positive values for the square roots in the above statements.
Statement I :
Xn is the largest integer less than ½ + sqrt(2n + ¼)
Statement II :
Xn is the largest integer not greater than ½ + Sqrt(2(n - 1) + ¼)
Statement III :
Xn is the smallest integer greater than -1/2 + sqrt(2n + ¼)
Case 1: Consider only the positive values for the square roots in the above statements.
Case 2: Consider both positive and negative values of square roots in the above statements.
@19rsb said:The remainder when F(x) is divided by (x-19) and (x-99) , is 99 and 19 respectively.Find remainder when F(x) is divided by(x-19)(x-99).
118-x??
@nick_baba said:Let Xn denote the n-th element of the sequence {1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, .....}, where n is a positive integer. How many of the following statements are then true?Statement I : Xn is the largest integer less than ½ + sqrt(2n + ¼)Statement II : Xn is the largest integer not greater than ½ + Sqrt(2(n - 1) + ¼)Statement III :Xn is the smallest integer greater than -1/2 + sqrt(2n + ¼)Case 1: Consider only the positive values for the square roots in the above statements.Case 2: Consider both positive and negative values of square roots in the above statements.
case1:TTF
case2:FFF????
case2:FFF????
@Ashmukh said:kaise aaya??
Let F(x) = A(x-19)(x-99) + ax + b
F(19)=19a+b=99
F(99)=99a+b=19
From this.. a=-1 and b=118
When F(x) = A(x-19)(x-99) + 118-x is divided by (x-19)(x-99) remainder is 118-x
@ScareCrow28 said:Let F(x) = A(x-19)(x-99) + ax + bF(19)=19a+b=99F(99)=99a+b=19From this.. a=-1 and b=118When F(x) = A(x-19)(x-99) + 118-x is divided by (x-19)(x-99) remainder is 118-x
perfect
@19rsb
f(x) when div by (x-19) leaves a remainder of 99
=> f(19) = 99 -------- (i)
f(x) when div by (x-99) leaves a remainder of 19
=> f(99) = 19 -------- (ii)
let,
f(x) = (x-19)(x-99)*k + ax + b
when this is divided by (x-19)*(x-99), it will leave a remainder of ax +b
Using, equations (i) & (ii),
19a + b = 99
99a + b = 19
Solving for a &b;,
a = -1, b = 118
Thus, the remainder is 118-x
Two ladies POONAM PANDEY (PP) and SHERLYN CHOPRA(SC) are counting numbers at same speed, but their counting patterns are different.
=>PP counts normally starting from 1 and adding 1 to the previous number like,1,2,3,4,5....
=>SC counts from 1 to 10 and then back to 1,and then counts upto 20 and then back to 1,and then counts upto 30 and then back to 1,and then so on.Also while counting she is careful that any two consecutive numbers counted by her are not the same.
If both started counting at the same time,then at what number is SC, if PP is at the number 10101?
1)243
2)153
3)268
4)none of these
=>PP counts normally starting from 1 and adding 1 to the previous number like,1,2,3,4,5....
=>SC counts from 1 to 10 and then back to 1,and then counts upto 20 and then back to 1,and then counts upto 30 and then back to 1,and then so on.Also while counting she is careful that any two consecutive numbers counted by her are not the same.
If both started counting at the same time,then at what number is SC, if PP is at the number 10101?
1)243
2)153
3)268
4)none of these
Let {an} be an arithmetic sequence that is not constant and {bn} a geometric sequence that is not constant. Assume that a40 = b40 > 0 and a60 = b60 > 0. Then:
(1) a50 = b50 (2) a50 b50 (3) a50 > b50 (4) a40 = b60 (5) a60 = b40
(1) a50 = b50 (2) a50 b50 (3) a50 > b50 (4) a40 = b60 (5) a60 = b40
P.S.: i don have an OA for this...hence discussions, explanations,expert comments welcomed..
@nick_baba said:Let {an} be an arithmetic sequence that is not constant and {bn} a geometric sequence that is not constant. Assume that a40 = b40 > 0 and a60 = b60 > 0. Then:(1) a50 = b50 (2) a50 b50 (4) a40 = b60 (5) a60 = b40P.S.: i don have an OA for this...hence discussions, explanations,expert comments welcomed..
a60 = a40 + 20d
b60 = b40*r^20
Equate
a40 + 20d = b40*r^20
b40 + 20d = b40*r^20
20d = b40*(r^20 - 1)
a50 = a40 + 10d
= b40 + b40*(r^20 - 1)/2
=b40(r^20+1)/2
b50 = b40*r^10
a50/b50 = (r^20 + 1)/(2r^10)
= 1/2 * (r^10 + 1/r^10) > 1
a50 > b50
b60 = b40*r^20
Equate
a40 + 20d = b40*r^20
b40 + 20d = b40*r^20
20d = b40*(r^20 - 1)
a50 = a40 + 10d
= b40 + b40*(r^20 - 1)/2
=b40(r^20+1)/2
b50 = b40*r^10
a50/b50 = (r^20 + 1)/(2r^10)
= 1/2 * (r^10 + 1/r^10) > 1
a50 > b50
@19rsb said:Also while counting she is careful that any two consecutive numbers counted by her are not the same.
I didn't understand this part ^^^
@grkkrg said:I didn't understand this part ^^^
She counts up and down right !!! ... What that statement simply means is ... She doesn't count 1 twice consecutively
...5,4,3,2,1,2,3... Same for any other number: 10,20,30,etc.
...5,4,3,2,1,2,3... Same for any other number: 10,20,30,etc.