Total age of all three boys
= 3 x 15 = 45 yr
Ratio of ages = 3 : 5 : 7
? Age of the oldest boy
= [7/(3 + 5 + 7)] x 45 = (7/15) x 45 = 21 yr
Total age of 5 memeber (3 yr ago) = 27 x 5 = 135 yr
Total age of 5 members (at present) = 135 + 5 x 3 = 150 yr.
Let the persent age of child = x
? 27 = (150 + x)/6
? x = 162 - 150 = 12 yr.
Let average of a inning = x
Total runs = 9x
According to the question,
(9x + 100) /10 = x + 8
? 9x + 100 = 10x + 80
x = 20
Average after 10 inning = 20 + 8 = 28
Let one integer be x.
? Other ingeger x + 15
According to the question,
29 + x + x + 15+ 108 = 4 x 73.5
? 2x + 152 = 294
? 2x = 294 - 152 = 142
? x = 142/2 = 71
If two equal distance are covered at different speeds at A km/h and B km/h respectively, then
Average speed during the whole journey = 2AB/(A + B) km/h
? Average speed of the car = (2 x 42 x 48) / (42 + 48)
= (2 x 42 x 48)/90
= 44.8 km/h
Here n = 120, a = 35, x = 39, y = 15
Number of students passed = n (a - y)/(x - y)
= 120(35 - 15)/(39 -15)
= 100 students passed
Let average expenditure of 9 persons = x.
According to the question,
12 x 8 + (x + 16) = 9x
? 8x = 112
? x = 112/8 = 14
? Total money spent = 9x = 9 x 14 = ? 126
Let the number of papers = x.
According to the question,
64x + 18 + 4 = 66x
? 2x = 22
? x = 22/2 = 11
Let number of passed students = N
Total marks = 120 x 30
According to the question,
40N + (120 - N) x 10 = 120 x 30
? 3600 = 40N + 1200 - 10N
? 30N = 2400
? N = 2400/30 = 80
? 25% of 80 = 80/4 = 20
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%
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
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