area of a square = a² sq cm
length of the diagonal = cm
area of equilateral triangle with side
=
required ratio =
When we divide 1234 by 8, remainder is 2
When we divide 1235 by 8, remainder is 3
When we divide 1237 by 8, remainder is 5
---> 2 x 3 x 5 = 30
As 30 will not be the remainder because it is greater than 8,
when 30 divided by 8, remainder = 6.
Let Speed of boat in still water = b
Let Speed of still water = w
Then we know that,
Speed of Upstream = U = boat - water
Speed of Downstream = D = boat + water
Given, U + D = 82
b - w + b + w = 82
2b = 82
=> b = 41 kmph
From the given data,
41 - w = 105/3 = 35
w = 6 kmph
Now,
b + w = 126/t
=> 41 + 6 = 126/t
=> t = 126/47 = 2.68 hrs.
As we know that a week has 7 days,
k weeks k days = (7k + k) days = 8k days.
Number of runs scored more to increse the ratio by 1 is 26 - 14 = 12
To raise the average by one (from 14 to 15), he scored 12 more than the existing average.
Therefore, to raise the average by five (from 14 to 19), he should score 12 x 5 = 60 more than the existing average. Thus he should score 14 + 60 = 74.
Let the average expenditure per head be Rs. p
Now, the expenditure of the mess for old students is Rs. 44p
After joining of 15 more students, the average expenditure per head is decreased by Rs. 3 => p-3
Here, given the expenditure of the mess for (44+15 = 59) students is increased by Rs. 33
Therefore, 59(p-3) = 44p + 33
59p - 177 = 44p + 33
15p = 210
=> p = 14
Thus, the expenditure of the mess for old students is Rs. 44p = 44 x 14 = Rs. 616.
Average speed = (2xy) /(x + y) km/hr
= (2 * 50 * 30) / (50 + 30) km/hr.
37.5 km/hr.
Speed in still water = Average of Speed in Upstream and speed in Downstream
= 1/2 (12 + 6) kmph = 9 kmph.
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