Increase in sum of age during 3 years = 3 x 5 years
= 15 years
So , the difference between ages of old and new member is 15 years
M + T + W = 3 x 37 = 111° C .....(1)
T + W + Th = 3 x 34 = 102 ° C ......(ii)
Let M = y
Then, Th = 4y/5
subtraction (ii) from (i) we get,
M - Th = 9° C
? y - 4y / 5 = 9° C
? y = 45° C
? Temperature on thursday = (4 x 45) / 5° C
= 36° C
Let the age of youngest child be y.
Then y + ( y + 2 ) = ( 24 x 6 ) - ( 24 x 4 + 4 x 10)
= 144 - 136
? 2y + 2 = 8
? y =3 years
Age of child = sum of age of husband , wife and child - sum of age of husband and wife.
= [(20 x 3) - (23 x 2 + 5 x 2)] years
= 4 years.
Let the original expenditure be Rs. y per day ,
Then, y / 35 - [( y + 42 ) / 42 ] = 1
? 42y - 35 ( y + 42 ) = 35 x 42
? 7y = 35 x 42 + 35 x 42
? y = ( 2 x 35 x 42 ) / 7
= 420
Hence , the original expenditure is Rs . 420
Let the number of wickets taken before the last match = y
Bowler's overall average = total run scored / total wicket taken by bowler
Then, (12.4y + 26 ) / (y + 5) = 12
? x = 85
Decrease in sum of age = 11 x 2 months
= 1 year 10 months
Total age of reserves = (17 + 20 ) years - (1 year 10 month)
= 35 years 2 months
Average age of the reserves = (35 years 2 months ) / 2
= 17 years 7 months
Total age of Ram and shyam 5 years ago
= 2 x 20 = 40 years
? Total age of Ram and Shyam at present
= 40 + 5 + 5 = 50 years
Mohan's age now =90 - 50 years = 40 years
Mohan's age 10 years hence = 40 + 10 years
= 50 years
Let a, b, c are the ages of A, B and C respectively
a + b = 2 X 20 = 40 ...(1)
b + c = 2 x 19 = 38 ...(2)
c + a = 2 x 21 = 42 ...(3)
Adding all the 3 equalities
2a + 2b + 2c = 120
a + b + c = 60 --- (4)
From eq. (2) and (4)
a = 22
From eq. (1) and (4)
b = 18
From eq. (4)
c= 20
? Age of A = 22 years
Age of B = 18 years
Age of C = 20 years
Even numbers between 11 to 63
= 12,14,16,18,20,..........,62.
Clearly, this is a series of consecutive even number
According to the formula
Average of consecutive even numbers = (First number + Last number)/2
= (12 + 62)/2
= 74/2
= 37
As per the formula,
Average of the cubes os 1st 'n' natural numbers= n(n+1)2/4
where, n= 5.
? Required average = 5(5+1)2/4
= (5 x 36)/4 = 5 x 9= 45
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