twenty members
2637
4815
OR
2817
4635
3×3=9
5 litre jug contains 4 litres of salt water solution
with 15% of salt, that means it has 4000*.15 = 600 ml salt
in it.
If 1.5 litres solution is spills out, remaining solution
is 2.5 litres, then the salf content is 2500*.15 = 375ml.
Now, the jug is filled to full capacity with water i.e.,
now the jug has 5 litre solution in it.
Now, the salt content is 375/5000 = 7.5%
1)Port
2)Trop
They’ll never meet unless some of them stop and other continue to run
Ramji sir
time (dist covrd by husband) (dist covrd by wife)
0.5hr 20 miles 0 miles
1 hr 40 miles 25 miles
1.5 hr 60 miles 50 miles
2 hrs 80 miles 75 miles
2.5 hrs 100 miles 100 miles
Therefore wife catches up after 2.5 hrs..
In the example first we have to find the area of the field
we have given the following values ,
cost of fencing per meter = Rs 1.50
Diameter of circular field = 28
We have to find the area , a = ?
We know that,
Area = 2 π r
= 2 x 22/7 x 14
= 2 x 22 x 2
= 44 x 2
= 88 sq. m
So, area of circle is 88 sq.m
Now just multiply it with 1.50
cost of fencing = 88 x 1.50
= Rs. 132
So, the cost of fencing the circular field is Rs. 132
#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
original cost is 100^2 / (100^2 – p^2) Rs
Total number of pairs is NC2^{N}C_2NC2. Number of pairs standing next to each other = N. Therefore, number of pairs in question = NC2^{N}C_2NC2 – N = 28/2 = 14. If N = 7,
7C2 – 7 = 21 – 7 = 14….
N =7
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