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1×2×…100=100!
Number of zeros in product of n numbers =[5n]+[52n]+[53n]+…
Number of zeros in product of 100 numbers =[5100]+[52100]+[53100]
where [.] is greatest integer function
=[20]+[4]+[0.8]=20+4=24
5c1 / 9c1
60 more men
Nice
Chambal
Reduction of 40% or 4/10th in price of bananas will lead to an increase of 4/(10 – 4) = 2/3rd part in quantity if expenditure is constant.
If original quantity is q, then :
2q/3 = 64
=> q = 64 x 3/2 = 96 = 8 dozens.
=> Original price per dozen = 40/8 = Rs 5
1 : 2
A certain sum amounts to Rs. 1725 in 3 years
and amounts to Rs.1875 in 5 years
so interest of 2 years = 1875 -1725
= 150
so interest of 1 year = 75
so interest of 3 years = 75 × 3 =225 rs
so , Principal = Amount – SI
= 1725 – 225
= 1500 rs
now ,
S.I. = P × N × R /100
75 = 1500 × 1 × R /100
R = 75 / 15
R = 5%
2, 5, 10, 50, 500, 5000
5*2=10
10*5=50
50*10=500
500*50=25000
So odd one is 5000
1%=50
12%=600
A
1/x
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
3000(4x/100) + 5000(2x/100)=10
a