@gudda1122 said:correct method post karna mera wala kafi bada hai
Bro only 2 lines
See Trian ODE = 1/4 Train BOC (similar trian)
Also train BOC = 1/3 Train ABC ( Median Prop)
So Train Ode= 1/12 ABC 
@gudda1122 said:A wheel rotates 3.5 times in one second. What time in seconds does the wheel take torotate 55 radian of angle?1. 1.5 b. 2.5 c 3.5 d 4.5
2.5
@gudda1122 said:A wheel rotates 3.5 times in one second. What time in seconds does the wheel take torotate 55 radian of angle?1. 1.5 b. 2.5 c 3.5 d 4.5
2.5.....
@gudda1122 said:A wheel rotates 3.5 times in one second. What time in seconds does the wheel take torotate 55 radian of angle?1. 1.5 b. 2.5 c 3.5 d 4.5
2.5????
@JS-M said:2.5.....
@sandeepv882 said:2.5
correct
yaar yeh radian se angle ka funda batana ya koi link dena
kafi time phele padha tha 
@gudda1122 said:A wheel rotates 3.5 times in one second. What time in seconds does the wheel take torotate 55 radian of angle?1. 1.5 b. 2.5 c 3.5 d 4.5
b)2.5
@gudda1122 said:correctyaar yeh radian se angle ka funda batana ya koi link denakafi time phele padha tha
2pi radian=360 degree
@gudda1122 said:correctyaar yeh radian se angle ka funda batana ya koi link denakafi time phele padha tha
2pi rad=360deg
@JS-M said:What is the area (in square units) of the quadrilateral ABCD formed by the points A(0, 0), B(6, 0), C(8, 4) and D(2, 8) in the x-y plane?(a) 40 (b) 32 (c) 56 (d) 48
40
@JS-M said:What is the area (in square units) of the quadrilateral ABCD formed by the points A(0, 0), B(6, 0), C(8, 4) and D(2, 8) in the x-y plane?(a) 40 (b) 32 (c) 56 (d) 48
40
In ∆ABC, D is the midpoint of BC. E is a point on AC such that AE : EC = 2 : 1 and F is a point on AB such that AF : FB = 3 : 1. Line segments AD and FE intersect at point O. What is the ratio of the area of ∆DOF to the area of ∆DOE?
(a) 8 : 9 (b) 9 : 8 (c) 3 : 4 (d) 4 : 3
divisibility rules
Divisibility by 2,4,8,16,….
A number is divisible by 2,4,8,16,…., when the number formed by the last one, two, three, four, …., n digits is divisible by 2,4,8,16,…., respectively.
Divisibility by 3
A number is divisible by three if the sum of its digits is divisible by 3.
Divisibility by 5
A number is divisible by 5 if the digit at unit's place is 0 or 5.
Divisibility test of 7
Whenever we have to check whether a number is divisible by 7 or not follow the below procedure:
1).Double the last digit (digit at the rightmost place) and subtract it from the number left (excluding the last digit). If this number is divisible with 7 then the original number is divisible by 7.
This procedure can be followed as many times as required (until the number is reduced to 2 digit number). Then the number so obtained can be checked whether it is divisible by 7 or not. If the number so obtained is divisible by 7 then the original number is divisible by 7 and if not then original number is not divisible by 7.
E.g.-
Consider the number 1057.
Now the last digit is 7. On doubling it we get 14.
On subtracting it from 105 we get 91.
Now it can be seen that 91 is divisible by 7 so the original number is divisible by 7.
(It can further be simplified by doubling 1 and subtracting it from 9 and thus we get 7 which is divisible by7.)
Divisibility test of 8
To check whether a number is divisible by 8 or not we consider the last three digits of the number. If the last three digits are divisible by 8 then the whole number is divisible by 8 otherwise not.
E.g.-7645892
Now consider the last three digits 892. Now check whether the three digits are divisible by 8. On checking we get that the number is not divisible by 8 so the original number is not divisible by 8.
Divisibility by 9
A number is divisible by three if the sum of its digits is divisible by 9.
Divisibility test of 10
If last digit of a number is 0 then the given number is divisible by 10 otherwise not.
E.g.- 9860 is divisible by 10 as the last digit is 0.
divisibility test for 11
If the difference between sum of digits in the odd places and the sum of the digits in the even places is either 0 or is divisible by 11, then the number is divisible by 11.
Divisibility test of 13
To check whether a number is divisible by 13 we follow the procedure as follows:
1). Multiply the last digit with 4 and add it to the number left (after removing the last digit).
2). Follow this method again and again and reduce the number to 2-digit number form.
3). Now check whether the number is divisible by 13 or not.
If the 2-digit number so obtained is divisible by 13 then the original number is divisible by 13 otherwise not.
E.g.-
Let us consider the number 195.
Now the last digit is 5 and on multiplying it with 4 we get 20.
Now on adding this with the remaining number (i.e. 19) we get 39. Now as 39 is divisible by 13 therefore the original number id divisible by 13.
Divisibility test of 17
To check whether a number is divisible by 17 we follow the procedure as follows:
1). Multiply the last digit with 5 and subtract it from the number left (after removing the last digit).
2). Follow this method again and again and reduce the number to 2-digit number form.
3). Now check whether the number is divisible by 17 or not.
If the 2-digit number so obtained is divisible by 17 then the original number is divisible by 17 otherwise not.
E.g.-
Let us consider the number 221.
Now the last digit is 1 and on multiplying it with 5 we get 5.
Now on subtracting 5 from the remaining number (i.e. 22) we get 17. Now as 17 is divisible by 17 therefore the original number id divisible by 17.
Divisibility test of 19
To check whether a number is divisible by 19 we follow the procedure as follows:
1). Multiply the last digit with 2 and add it to the number left (after removing the last digit).
2). Follow this method again and again and reduce the number to 2-digit number form.
3). Now check whether the number is divisible by 19 or not.
If the 2-digit number so obtained is divisible by 19 then the original number is divisible by 19 otherwise not.
E.g.-
Let us consider the number 209.
Now the last digit is 9 and on multiplying it with 2 we get 18.
Now on adding th18 with the remaining number(i.e. 20) we get 38. Now as 38 is divisible by 19 therefore the original number id divisible by 19.
Divisibility by 7, 11, 13
Consider any number abcdefghij.
Starting from the right towards left, we make groups of 3 digit numbers successively and continue till the end. It is not necessary that the leftmost group has three digits.
Now we have
a bcd efg hij
Now we number these groups starting from the left.
So, we have
Group 1= hij
Group 2= efg
Group 3= bcd
Group 4= a
And if the number is bigger this number can be continued.
Now, we add the alternate groups i.e. 1,3,5,… and 2,4,6,…
So we get two sums
Sum1= group(1+3+5+….)
And
Sum2=group(2+4+6+….)
Now we find the difference between the two sums obtained i.e. sum1 and sum2.
Let the difference be D.
D= sum1 - sum2
Now if D is-
i). divisible by 7, then the original number is divisible by 7.
ii). divisible by 11, then the original number is divisible by 11.
iii). divisible by 13, then the original number is divisible by 13.
Corollary: Any 6-digit or 12-digit ,or any such number with number of digits equal to multiple of 6,is divisible by each of 7,11 and 13 if all of its digits are same. E.g. 666666, 888888888888 etc. are divisible by 7,11 and 13.
Divisibility rules for composite numbers like 6,12,14,18,….
Whenever we have to check the divisibility of a number N by a composite number C, then the number N should be divisible by all the prime factors (the highest power of every prime factor) present in c.
Now the divisibility rules for various composite numbers will be-
1). Divisibility by 6
The given number should be divisible by both 2 and 3.
2). Divisibility by 12
The given number should be divisible by both 4 and 3.
3).Divisibility by 14
The given number should be divisible by both 2 and 7.
4). Divisibility by 15
The given number should be divisible by both 3 and 5.
5). Divisibility by 18
@skepticeye said: @sampras 20kb photo wali problem solve ho gayi ?
ha bhai ho gayi ... finally went to studio and got it done through adobe photoshop... thanks for concern yaar...
@sampras said:ha bhai ho gayi ... finally went to studio and got it done through adobe photoshop... thanks for concern yaar...
bhai hua kya tha?
@rahulshaitan said:divisibility rules
div by 19 .... interesting one... actually had posted qn related to it in previous link... but was sent into oblivion... hardly anyone responded to it except truly bhai...
@rahulshaitan
Hey shaitan bhai chk my method for ur MAXIMA wala ques
1/(sin^2 y + 3 sin y cos y + 5 cos^2 y)
= 1/ (sin^2 y + 4cos^2 y - 4sin y cos y + 7sin y cos y+ cos^2 y)
= 1/ (sin y 창€“ 2cos y)^2 + cos^2 y +7 Sin y cos y
For this to be max. denominator should be minimum.
so sin y = 2 cos y
Tany = 2
Or sin y = 2/rt5
Cos y = 1/rt5
on putting these
1/( 1/5 + 7*2/5)
= 1/3
Ans
@sampras said:div by 19 .... interesting one... actually had posted qn related to it in previous link... but was sent into oblivion... hardly anyone responded to it except truly bhai...
matlab mere notes kuch to kaam aa he rahe hai
@JS-M said:In ∆ABC, D is the midpoint of BC. E is a point on AC such that AE : EC = 2 : 1 and F is a point on AB such that AF : FB = 3 : 1. Line segments AD and FE intersect at point O. What is the ratio of the area of ∆DOF to the area of ∆DOE?(a) 8 : 9 (b) 9 : 8 (c) 3 : 4 (d) 4 : 3
d???/