Total number of pairs is NC2^{N}C_2NC2. Number of pairs standing next to each other = N. Therefore, number of pairs in question = NC2^{N}C_2NC2 – N = 28/2 = 14. If N = 7,
7C2 – 7 = 21 – 7 = 14….
N =7
105, 85, 60, 30, 0, – 45, – 90
85
Let the number be x.
So x-2 =15/x, or
x^2–2x-15=0, or
(x-5)(x+3)=0
Hence x = 5 or -3.
Check: When x=5, 5–2=3 which is 15/5=3.
When x=-3, -3–2=-5 which is 15/-3=-5.
The number is 5 or -3.
let square be x (squares are 4 sides)
i.e., X+X+X+X=4X
4X+3=1460
4X=1460-3
4X=1457
X=364.25
31
I solve it as the 280m train being stationary… and the 120m train moving at (42+30) = 72kph…. or 20mps
The front of the train will have to go 280m to pass the stationary train plus another 120m for the backs to clear… that is 400m in all.
The time to travel 400m at 20mps is 20secs
126
cos if one of the factor is given for hcf and lcm and multiply hcf and lcm and them divide it by the given factor
so 18*3780/540=18*7=126
2* pi* r ( h + r )
r=3cm
h=7cm
2 * 22/7 *7 ( 10 ) = 440 sq cm
Mm
1 1 2+4
2 2 1+4
3 1+2 4
4 4 1+2
5 4+1 2
6 4+2 1
7 4+2+1 0
17:3
c
Lets call the 5 litre jug as jug A and 3 litre jug as jug B. Now, follow the steps:
Fill jug A completely. Now it contains 5 litres.
Slowly pour the water from jug A to jug B until jug B is completely filled. Now, jug A contains 2 litres and jug B contains 3 litres.
Throw away the water in jug B so that it is completely empty. Now, jug A contains 2 litres and jug B is empty.
Transfer the water from jug A to jug B. Now, jug A is empty and jug B contains 2 litres.
Fill jug A completely. Now, jug A contains 5 litres and jug B contains 2 litres.
Transfer water from jug A to jug B until jug B is completely filled. Now, jug A contains 4 litres and jug B contains 3 litres.
Now you have 4 litres of water in jug A.
A