The average age of a man and his two twin sons born on the same day is 30 yr. The ratio of the ages of father and one son is 5 : 2, What is the father's age?

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).

According to the question,
a + b + c = 11 x 3 = 33 ......(i)
c + d + e = 17 x 3 = 51 ...........(ii)
e + f = 22 x 2 = 44 ........... (iii)
e + c = 17 x 2 = 34 ..........(iv)
From Eqs. (ii) and (iv), we get
34 + d = 51
or d = 17 ....(v)

Now, by adding Eqs. (i), (iii) and (v), we get
a + b + c + d + e + f = 33 + 44 + 17 = 94

? Average of a, b, c, d, e and f = 94/6
= 47 / 3
= 15²/₃

Number of oranges bought by the person
= 36/1 + 36/1.50 + 36/1.80 + 36/2 + 36/2.25
= 36 + 24 + 20 + 18 + 16
= 114

Total expenditure = 36 x 5 = ? 180
Thus, average price of each orange
= ? 180/114
= ? 1.58

3 yr ago, total age of 5 members of the family = 17 x 5 = 85 yr
Present age of all the members of the family = 85 + 3 x 5 = 100 yr
Let the age of the child = N yr
Present avrage age of family = (100 + N) / 6
[At present total number of members = 5 + 1 = 6]
17 = 100N / 6
? 102 = 100 + N
? N = 2 yr

Total age of all members of the family (at present) = 20 x 4 = 80
4yr ago, total age of all members = 80 - ( 4 x 4) = 80- 16 = 64 yr.

As 4 yr ago, there were only 3 members in the family

? Average age of the family at the time of the birth of the youngest member
= 64/3
= 21¹/₃ yr

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)

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

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.

Total marks for three examinations = 3x 500 = 1500

Total required marks in three examinations = 60% of 1500
=( 3 x 500 x 60 ) / 100
= 900

Marks secured in first examination = 45 % of 500
=( 500 x 45 ) / 100
=225

Marks secured in third examination = 55 % of 500
=( 500 x 55 ) / 100
=275

Thus, the required marks in third examination
=900 - ( 225 + 275 )
=900 - 500
= 400

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