1/2:1/3:1/4 = 6:4:3
Ram = 6/13 * 3250 = 1500,
Shyam = 4/13 * 3250 = 1000,
Mohan = 3/13 * 3250 = 750
390
3800
1^1,2^2,3^3,4:^4,5^5,6^6
1,4,27,256,3125,46656
6+3.50= 2.50
let no of boys in group is = x
then total sum = 30 * x
after joining one more boys with a weight of 35 kg the total sum is = 30x + 35
after joining the new student the weight will increase 1 kg
so total sum
30x + 35 = 31(x+1)
x = 4
8, 13, 21, 32, 47, 63, 83
47 is answer by consecutive adding of
8+5=13
13+8=21
21+11=32
32+14=46
46+17=63
Answer: z and u
Explanation:
Y is to the right of U and exactly in front of V. Therefore,
U Y
V
Z is behind W and W and X are at extreme ends. So, W has to be to the right of Y. The final arrangement is as follows.
U Y W
X V Z Therefore, Z and U are at extreme ends is true.
– While the train is moving, the jogger will also be running in the same direction.
– for the head(engine) of the train to get to the current position of the jogger 240m away, it will take:
45km/hr => 12.5m/s => 240/12.5 = 19.2 seconds.
– But in the same period of time, the jogger will still be running and will have moved to a new location by: 9km/hr => 2.5m/s => 2.5 * 19.2 = 48m
To get to the new location at the speed of 12.5m/s will take the train:
48/12.5 = 3.84sec
In this additional time, the jogger will move forward by:
3.84 * 2.5 = 9.6m
at a speed of 12.5m/s, it will take the train less than a second to cover the additional 9.6m
If we add the distance the jogger will cover in 1 second to 9.6, it is still less than what the train can cover per second. let us see (9.6 + 2.5 = 12.1)
Therefore, the head of the train will pass the runner at approximately: 19.2 + 3.84 + 1 => 24.04 seconds.
For the train to completely pass the runner, it will need its whole length of 120m to be in front of the runner.
This will take an additional (9.6 + 2) seconds.
Therefore for the length of the train to be ahead of the runner it will take approx. 35.65 (24.04 + 9.6 + 2) seconds
8/15 left.
1/15 * 4 + 1/20 * 4 = 7/15
1 – 7/15 = 8/15
30/70*100=42.87
Explanation:
The number of ways of selecting three men, two women and three children is:
= ⁴C₃ * ⁶C₂ * ⁵C₃
= (4 * 3 * 2)/(3 * 2 * 1) * (6 * 5)/(2 * 1) * (5 * 4 * 3)/(3 * 2 * 1)
= 4 * 15 * 10
= 600 ways.
1
6=1x2x3
24=2x3x4
60=3x4x5
120=4x5x6
210=5x6x7
336=6x7x8
answer
R-R*d/100-5(1-d/100)
x and y can be equal to 1..
If so,then he gets 1/1 of Rs.10 which is equal to 10 and
again 1/1 of Rs.10=Rs.10..so,he gets total of 20..and
returns 20..so,no gain and no loss..
If x=1,y=2,he gets, 5+20=25..and returns 20..so he may not
lose..
So,whatever be the values of x and y,only these two answers
are possible..
so its a)He never loses..
130
520 = 26 * 20 = 2 * 13 * 22 * 5 = 23 * 13 * 5
Required smallest number = 2 * 13 * 5 = 130
130 is the smallest number which should be multiplied with 520 to make it a perfect square.
145.2