Let the number be 'x'.
= 11x
= 22x
=> x = 22.
One way to deal with fractions is to convert them all to decimals.
In this case all you would need to do is to see which is greater than 0.5.
Otherwise to see which is greater than ½, double the numerator and see if the result is greater than the denominator. In B doubling the numerator gives us 8, which is bigger than 7.
Required average = (67 * 2 + 35 * 2 + 6 * 3) / (2 + 2 + 3)
= (134 + 70 + 18) / 7 = 222 / 7 = 31(5/7) years.
Total number of students = 54 x 30
When arranged in rows of 45, number of rows formed are,
= 36.
Number of runs scored more to increse the ratio by 1 is 26 - 14 = 12
To raise the average by one (from 14 to 15), he scored 12 more than the existing average.
Therefore, to raise the average by five (from 14 to 19), he should score 12 x 5 = 60 more than the existing average. Thus he should score 14 + 60 = 74.
Average of 26,29,35 and 43 is 33.25 . Also the average of 26 , 29, n, 35 and 43 lies between 25 and 35 i.e,
=> 125 < 26+29+n+35+43 < 175
=> 125 < 133 + n < 175
=> n < 42
Since the value of n is an integer and greater than 33.25 then 33 < n < 42 for every integer n.
Amithab's total expenditure for Jan - June = 4200 x 6 = 25200
Expenditure for Feb - June = 25200-1200 = 24000
Expenditure for the months of Feb - July = 24000 + 1500 =25500
The average expenditure = 25500/6 = 4250
3X + 4y = 240
by substitute through options 48 is correct
Assuming Sripad has scored the least marks in subject other than science,
Then the marks he could secure in other two are 58 each.
Since the average mark of all the 3 subject is 65.
i.e (58+58+x)/3 = 65
116 + x = 195
x = 79 marks.
Therefore, the maximum marks he can score in maths is 79.
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