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
D. Rs. 187500
21 days
600
In order to enhance my career, I asked my colleagues about my strength and weaknesses, the areas where I need to improve more, and acquired more knowledge from my superiors. I worked on my weaknesses to convert into a positive response.
Ans alphabet = Q alphabet – 1
CHRONRD
1!=1
2!=2×1=2
3!=3x2x1=6
4!=4x3x2x1=24
5!=5x4x3x2x1=120
6!=6x5x4x3x2x1=720
The answer is 120.
Let the distance between each pole be x m. Then, distance up to 12th pole = 11x m
∴ Speed = 11x22m/s
∴ Time to cover total distance up to 20th pole
= 19x×2411x
= 41.45 s
The last person covered 120.71 meters.
It is given that the platoon and the last person moved with
uniform speed. Also, they both moved for the identical
amount of time. Hence, the ratio of the distance they
covered – while person moving forward and backword – are
equal.
Let’s assume that when the last person reached the first
person, the platoon moved X meters forward.
Thus, while moving forward the last person moved (50+X)
meters whereas the platoon moved X meters.
Similarly, while moving back the last person moved [50-(50-
X)] X meters whereas the platoon moved (50-X) meters.
Now, as the ratios are equal,
(50+X)/X = X/(50-X)
(50+X)*(50-X) = X*X
Solving, X=35.355 meters
Thus, total distance covered by the last person
= (50+X) + X
= 2*X + 50
= 2*(35.355) + 50
= 120.71 meters
Note that at first glance, one might think that the total
distance covered by the last person is 100 meters, as he
ran the total lenght of the platoon (50 meters) twice.
TRUE, but that’s the relative distance covered by the last
person i.e. assuming that the platoon is stationary.
Balls- B1, B2, B3, B4, B5, B6, B7, B8, B9.
Group1 – (B1, B2, B3), Group2 – (B4, B5, B6), Group3 – (B7, B8, B9)
Now weigh any two groups. Group1 on left side of the scale and Group2 on the right side.
When weighing scale tilts left – Group1 has a heavy ball or right – Group2 has a heavy ball or balanced – Group3 has a heavy ball.
Lets assume Group 1 has a heavy ball.
Now weigh any two balls from Group1. B1 on left side of the scale and B2 on right side.
When weighing scale tilts left – B1 is the heavy or tilts right – B2 is the heavy or balanced – B3 is the heavy.
1>=y>x => y belongs to (-infinity,1]
x belongs to (-infinity,y)
if both x and y are negitive z can be greater than zero and
can be greater than y(1 and 4 are true)
if y equals 1 then x can be equal to z(2 is true)
therefore y=z is not true for any value of x and y
9000
5:6::7:10::6:5
=((5/6)/(7/10)/(6/5))
=((5/6)*(6/5)/(7/10))
=(1/(7/10))
=10/7
13
sorry this question is irrelevant.
45
2:1