The numbers that lie between 100 and 1000 which are divisible by 14 are 112, 126,140 …,994
a = 112; l = 994, d = 14
n= (l−a)/d+1
= (994-112)/14+1
= 64
Sn=n/2(l+a)
= 64/2(994+112)
= 32*1106
= 35392
( A ) 124
19, 26, 33, 46, 59, 74, 91
Answer 33
Adding 7,9.13,15,17
Correct is 35
Can someone confirm that it is : 5/9 * 4/8 * 3/7 = 11.9% that you choose 3 good bulbs in a row
Matches played: 60.
Matches won: 30% of 60 => (60*(30/100)) = 18 matches.
Iterative approach:
On adding 1 to matches played and matches won, on every iteration until the win percentage gets to 50. So
19 / 61 = 0.3114754098360656
20 / 62 = 0.3225806451612903
21 / 63 = 0.3333333333333333
22 / 64 = 0.34375
…
…
…
…
Similarly,
41 / 83 = 0.4939759036144578
42 / 84 = 0.5
So, after 60th match 24 more matches has to be played and won to get 50% average winning rate.
132
perimeter = pi(r) + 2r
= 19.782 + (12.6)
= 32.382 cm
#1: N = 1, f(N) = 1
#2: N = 199981, f(N) = 199981
#3: N = 199982, f(N) = 199982
#4: N = 199983, f(N) = 199983
#5: N = 199984, f(N) = 199984
#6: N = 199985, f(N) = 199985
#7: N = 199986, f(N) = 199986
#8: N = 199987, f(N) = 199987
#9: N = 199988, f(N) = 199988
#10: N = 199989, f(N) = 199989
#11: N = 199990, f(N) = 199990
#12: N = 200000, f(N) = 200000
#13: N = 200001, f(N) = 200001
#14: N = 1599981, f(N) = 1599981
#15: N = 1599982, f(N) = 1599982
pass English =80%
fail=100-80=20%
pass Math = 70%
fail in Math=100-70=30%
fail in both=10%
total fail students= fail in Eng+fail in Math-common
= 20+30-10=40%
if 40% fail then 60% will pass
let total students=x
hence
60% 0f (total students)=144
60/100 of (x)= 144
x=(144×100)/60
X=240
total students=240
3×8=24=2xt ===> t=12hours
divide 100 by 7, and you get 14.28. (Obviously you aren’t talking about decimals here) and so 14 numbers can be divisible by 7 up to 100.
The answer is 14.