Let the total number of students be 100.
Also, let the mean score = N
According to the question,
20 x 80 + 25 x 31 + (100 - 20 - 25) x N = 100 x 52
? 1600 + 775 + (55) x N = 5200
? 55N = 5200 - 1600 - 775
? N = 2825/55 = 51.36 = 51.4%
So, mean score of remaining 55% is 51.4%
Let the required runs be N.
According to the question,
(80 x 99 + N) / 80 = 100
? 7920 + N = 8000
? N = 80
Let money of C be ? N.
According to the question
Total money of B = N + N x 50%
= N + 50N/100 = 3N/2
Total money of A = 2 x 3N/2 = 3N
Average money of three persons = 12000
? Total money of three = 12000 x 3
? 3N + (3N/2) + N = 12000 x 3
? (6N + 3N + 2N)/2 = 36000
? N = (36000 x 2)/11 = 72000/11
Now, money of A = 3N = (3 x 72000)/11
= ? 216000/11
Let the ages of mother, father and boys be M, F, B1 and B2, respectively.
The total age of four member
= 19 x 4
= 76 yr.
Given (B1 + B1)/2 = 11/2
? B1 + B2 = 11
M + F + B1 + B2 = 76 - 11
? M + F = 65 .......(i)
According to the question,
(B1 + B2 + F) / 3 = (B1 + B2 + M) / 3 + 3
? B1 + B2 + F) = B1 + B2 + M + 9
? F = M + 9
? F - M = 9 ............. (ii)
From Eqs (i) and (ii), we get
F = 37 yr and M = 28 yr
Let x be the number of wickets taken till the last match
According to the question,
(12.4 x + 26) = 12(x + 5)
? 12.4 x + 26 = 12x + 60
? 0.4 x = 34
? x = 34/0.4 = 85
Let average after 10th innings = x.
According to the question.
10x + 216 = 11(x +12)
? x = 216 - 132 = 84
? New average = 84 + 12 = 96
Sum of the weight of A, B, C and D = 67 x 4 = 268 kg
and average weight of A, B , C , D and E = 67 - 2 = 65kg
? Sum of the weight of A, B, C, D and E = 65 x 5 = 325 kg
? Weight of E = 325 - 268 = 57 kg
? Weight of F = 57 + 4 = 61 kg
Now, average weight of F, B, C, D and E = 64 kg
? Sum of the weight of F, B, C, D and E = 64 x 5 = 320 Kg
? Sum of the weights of B, C and D = 320 - 57 - 61 = 202 kg
? Weight of A = 268 - 202 = 66 kg
Total age of Rashmi and Surbhi (5 yr ago) 20 x 2 = 40 yr.
Total age of Rashmi and Surbhi (at present) = 40 + 5 + 5 = 50 yr.
Total age of Rashmi, Surbhi and Meeta (at present) = 30 x 3 = 90 yr.
? Present age of Meeta = 90 - 50 = 40 yr.
? Age of meeta after 15yr. = 40 + 15 = 55 yr.
Let number of students in the sections
A, B, C and D be a, b, c and d respectively.
Then, total weight of students of sections A = 45a
Total weight of students of section B = 50b
Total weight of students of section C = 72c
Total weight of students of section D = 80d
According to the question,
Average weight of students of sections of A and B = 48 kg
? (45a + 50b)/(a + b) = 48
? 45a +50b = 48a + 48b
? 3a = 2b
? 15a = 10b
And average weight of students of sections B and C = 60 kg
? 50b + 72c = 60(b + c)
? 10b = 12c
Now, average weight of students of A, B, C and D = 60 kg
? 45a + 50b + 72c + 80d = 60(a + b + c + d)
? 15a + 10b - 12c - 20d = 0
? 15a = 20d
? a : d = 4 : 3
125.009 + 69.999 + 104.989 = ?
Lets assume, each value is approximated to nearest whole number
? ? ? 125 + 70 + 105
? ? ? 300
? = 16.003 x 27.998 - 209.010
? ? 16 x 28 - 210 = 448 - 210
? = 238 ? 240
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