The average age of 54 girls in a class was calculated as 14 yr. It was later relized that the actual age of one of the girls in the class was 10.5 yr but it was calculated as 13 yr. What is the actual averge age (round off) of the the girls in the class? (Find the approximate answer.)

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

Here n₁ = 20, n₂ = 30, a₁ = 80%, a₂ = 70%
? Total average = (n₁a₁ + n₂a₂) / (n₁ + n₂)
= (20 x 80 + 30 x 70) / (20 + 30)
= (1600 + 2100) / 50
= 74%

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.

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