Out of the three annual examinations, each with a total of 500 marks, a student secured average marks of 45% and 55% in the first and second annual examinations. To have an overall average of 60%, how many marks does the student need to secure in the third annual examination?

Total age of 5 sister = 20 x 5 = 100 yr.
4 yr ago, total sum of ages
= 100 - ( 5 x 4) = 100 - 20 = 80 yr
But at that time (4 yr ago), there were 4 sister in the group.
? Average age at that time(4 yr ago) = 80/4 = 20 yr.

Let ages of Rakesh, Mohan and Ramesh be R, M and r, respectively .
Then, R + M = 15 x 2 = 30.......(i)
M + r = 12 x 2 = 24 ........(ii)
r + R = 13 x 2 = 26 ..... (iii)
Adding Eqs. (i), (ii) and (iii), we get
2(R + M + r) = 30 + 24 + 26 = 80
? R + M + r = 40
Subtracting Eq. (iii) from Eq. (iv), we get
M = 40 - 26 = 14

The average number of pens and pencils = 180
? Totol number of pens and pencils = 180 x 2 = 360
? Total number of pencils in the shop = 360 x (2/5) = 144
? 30% of 144 = (30/100) x 144 = 43.2 = 43
= 43 (approx)

According to the question,
Average age of man and his twin sons = 30
? Total age = 30 x 3 = 90 yr.

According to the question,
Ratio of father and one son = 5 : 2
? Ratio of father and both the sons = 5 : 2 : 2
(As sons are born on the same day.)

? 5k + 2k + 2k = 90 or 9k = 90
? k = 10
? Age of father = 5 x 10 = 50 yr.

Total actual age of 54 girls = (54 x 14) - 13 + 10 .5
= 753.5 yr

? Require average age = 753.5/54
= 13.95 yr
= 14 yr(approx).

Let the number of women workers be Y.

Then, 22y + 16 x 8 = 15 (16+y)
? 22y + 128 = 240 + 15y
? 22y - 15y= 240 - 128
? y = 16

? Unmarried women workers = 16 - 10 = 6

Let ages of Dinesh, Mahesh and Rahul be D, M and R, respectively.
Then,
D + M = 2 x 16 = 32 ...(i)
M + R = 2 x 13 = 26 ....(ii)
D + R = 2 x14 = 28 ....(iii)

On adding Eqs. (i), (ii) and (iii), we get
2(D + M + R) =86
? D + M + R = 43 ....(iv)

On subtracting Eqs. (iii) from Eq. (iv), we get
M = 43 - 28
? M = 15 Yr

Let the number of failed student = Y
? Number of passed student = 80 - Y

sum of marks of all students = sum of marks of passed students + sum of marks of failed students
80 x 40 = (80 - Y) x 60 + 35Y
? 3200 =4800 - 60Y + 35Y
? 25Y = 1600
? Y = 64
? Number of students who failed = 64

Required average
= Old average - Sold average
= ( 250 ) - ( 10 ) = 240

Let the present ages of father be 3x and daughter be x .
so the 4 year ago father's age and daughter's age was
( 3x - 4 ) and ( x - 4 )
? ( 3x - 4 / x -4 ) = 4 / 1
? x = 12 year and 3x = 36 year