Since we know that the difference b/w the age of any two persons remains always constant, while the ratio of their ages gets changed as the time changes.
so, if the age of his child be x (presently)
Then the age of wife be x + 26 (presently)
Thus the total age = x + ( x + 26) = 32 [ 252-220 =32]
=> x = 3
Therefore, The age of her child is 3 years and her self is 29 years. Hence her age at the time of the birth of her child was 26 years.
also ( M - 11111 = 11111 - S)
=>A = 11111
Manager's monthly salary
= Rs. (1900 x 25 - 1500 x 24) = Rs. 11,500
Let 'K' be the total number of sweets.
Given total number of students = 112
If sweets are distributed among 112 children,
Let number of sweets each student gets = 'L'
=> K/112 = L ....(1)
But on that day students absent = 32 => remaining = 112 - 32 = 80
Then, each student gets '6' sweets extra.
=> K/80 = L + 6 ....(2)
from (1) K = 112L substitute in (2), we get
112L = 80L + 480
32L = 480
L = 15
Therefore, 15 sweets were each student originally supposed to get.
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.
Let the average bill paid by twenty members = 'x'
But 19 men paid each = Rs. 70
20th man paid Rs. 90.25 more than the avg bill of 20 = x + 90.25
20x = 19(70) + x + 90.25
19x = 1330 + 90.25
19x = 1420.25
x = 1420.25/19 = Rs. 74.75
But the total bill = 20 x 74.75 = Rs. 1495.
Average = (11 + 22 + 33 + 44 + 55 + 66 + 77 + 88 + 99) / 9
=( (11 + 99) + (22 + 88) + (33 + 77) + (44 + 66) + 55) / 9
= (4 * 110 + 55)/9 = 495 / 9 = 55.
Average of 26,29,35 and 43 is 33.25 . Also the average of 26 , 29, n, 35 and 43 lies between 25 and 35 i.e,
=> 125 < 26+29+n+35+43 < 175
=> 125 < 133 + n < 175
=> n < 42
Since the value of n is an integer and greater than 33.25 then 33 < n < 42 for every integer n.
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.
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