From the first statement
Speed=distance/time
=>240/24=10m/s
so speed=10m/s
From the second statement
distance=length of the platform+length of the train=650+240=890m
Time=distance/speed
here, the speed of the train is already calculated
=>890/10=89s
So the time taken by the train to cross the platform is 89 seconds
22.5
340
let the average of 25 person be x
total age of a person=average×no.of person
total age of 25 person=X×25
=25X
when a new person of 46kg come average decrease by 5kg(X-5)
Total age of 26 person =25x+46
Average age of 26 person=X-5
A/Q
.25X+46=26(X-5)
25X+46=26X-130
26X-25X=130+46
X =176
hence the av. of 25 person be 176kg
Niece
1.5
The ratio flat vs hill is unknown so you can’t calculate this normally. But since the average for uphill/downhill is also 4 kmph ((1.5h)/6km)), the calculation is: 6 hours * 4kmph = 24km.
114
area is doubled.
actual area = 1/2 bh
after increase area = 1/2 *4b* h /2
=bh
1/2 *bh *2 = bh
therefore area is doubled
A perfect square is a square number of a digit. eg 64 is a perfect number, a square of 8
Now digits AB9 is a square number of a number.
AB9 Can also be written as A multiply by B multiply by 9
Get the Square root of AB9
Assumption, A=1, B=1
1*1=1=A, 1*1=1=B Therefore,
Square root of A = A, B = B and 9=3
Therefore
An odd number is a number indivisible by 2.
for example 1,3,5,7…….
Therefore Squares A*B*9= AB9
Where a=1, b=1, 3 as digits.
Conclusion
A=1 is an odd number
Speed Ratio = 1:7/6 = 6:7
Time Ratio = 7:6
1 ——– 7
4 ——— ? 28 m
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..
How will you know the odd is in lighter one or heavier one from only one weighing. It will require 2 weighing to find the odd set and one weighing for odd coin in that set i.e total 3 weighings.
15km/hr