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.
Manager's monthly salary
= Rs. (1900 x 25 - 1500 x 24) = Rs. 11,500
Sum of the present ages of husband, wife and child = (27 x 3 + 3 x 3) years = 90 years.
Sum of the present ages of wife and child (20 x 2 + 5 x 2) years = 50 years.
Husband's present age = (90 - 50) years = 40 years.
also ( M - 11111 = 11111 - S)
=>A = 11111
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.
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.
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.
Copyright ©CuriousTab. All rights reserved.