1 2 3 4 5 6(Shankar) 7(Nitu) 8(Althaf) 9 10 11 12 13 14
Here, Althaf is 8th from front, Shankar is 9th from rear end and Nitu is between them
So minimum no. of boys standing in the queue = 14
The least common multiple of 12 and 15 is
12 = 2 x 2 x 3
15 = 3 x 5
LCM of 12 and 15 is 2 x 2 x 3 x 5 = 60.
Assume first child (the youngest) get = Rs. x
According to the question ;
each son having Rs. 30 more than the younger one
Second child will get = Rs. x + 30
Third child will get = Rs. x + 30 + 30 = x + 60
Forth child will get = Rs. x + 30 + 30 + 30 = x + 90
Fifth child will get = Rs. x + 30 + 30 + 30 + 30 = x + 120
Total amount they got = Rs. 2000
x + (x+30) + (x+60) + (x+90) + (x+120) = 2000
5x + 300 = 2000
5x = 1700
x = Rs. 340
So the youngest child will get Rs. 340.
Let the required no of hectares be x. Then
More men, More hectares (Direct proportion)
More days, More hectares (Direct proportion)
=> ( 8 * 24 * x) = (36 * 30 * 80)
=> x= 450
But at x=-1/3, log x is not defined.
The only admissible value of x is 1.
Let the required number of revolutions made by larger wheel be x.
Then, More cogs, Less revolutions (Indirect Proportion)
14 : 6 :: 21 : x => 14 x x = 6 x 21
x=6*21/14 => x=9
After 10 days : 150 men had food for 35 days.
Suppose 125 men had food for x days.
Now, Less men, More days (Indirect Proportion)
125 : 150 :: 35 : x 125 x x = 150 x 35
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