P = 0.6 Q
Q= 0.25 R
So P = 0.6*0.25 R
R = 0.4 S
So, P = 0.6*0.5 * 0.4 S
S = 0.45*T
Again, P = 0.6* 0.5 * 0.4 * 0.45*T = 54/1000T
So P/T = 27/500
Option 8)
Isha’s grandfather was 5 times older to her 4 years ago. He would be three times of her age 8 years from now. 7 years ago, what was the ratio of Isha’s age to that of her grandfather’s age?
A father said to his son, "I was as old as you are at the present at the time of your birth". If the father's age is 55 years now, what would be the son's age 4.5 years later?
Let the sons current age be x
So fathers age x years back when son was born => 55-x = x
Solving we get x = 27.5
4.5 Years later son's age = 27.5 + 4.5 = 32 years
Jack enters a lift on the 25th floor, which is going up @ 56 floors/min speed. At the same time Jill enters another lift on the 70th floor which is going down @ 64 floors/min speed. At what floor will
Chintu works at a soft drink company and was arranging some particular number of cylindrical aluminium soft drink cans in a square box but the box became full and there were 8 cans remaining to be put in. Then, he started arranging all Cans in a rectangular box where he could arrange 6 cans more along the length than the breadth after putting all the cans in the rectangular box. He found out that there was still space left for another 11 cans. What is the number of cans Chintu had if it is known that the number of cans were more than 15 and no can was put on top of another can?
In a square box the number of cans arranged along the length and breadth would be same.
In case of rectangular box, lets assume that number of cans along breadth is x then number of boxes along the length would be x+6
Total number of cans the rectangular box can take = x*(x+6)
Total number of cans Chintu was able to put in the rectangular box (as some empty space was remaining ) = x*(x+6) - 6 ---- (i)
Given that the number of cans are more than 15, hence 2 & 3 cans along the length are ruled out
Lets try with 4 cans along length of square box, total cans in box would be 16
Total number of cans which Chintu has = 16+8 = 24 ---- (ii)
(i = (ii)
x*(x+6) - 11 = 24
=> x^2 + 6x = 35
Trying to solve we realize that there is no such whole number which x can take
Lets try with 5 cans along length of square box, total cans in box would be 25
Total number of cans which Chintu has = 25+8 = 33 ---- (ii)
(i = (ii)
x*(x+6) - 11 = 33
=> x^2 + 6x = 44
Trying to solve we realize that there is no such whole number which x can take
Lets try with 6 cans along length of square box, total cans in box would be 36
Total number of cans which Chintu has = 36+8 = 44 ---- (ii)
(i = (ii)
x*(x+6) - 11 = 44
=> x^2 + 6x = 55
Solving we get x = 5
So total number of cans which Chintu had = x*(x+6) - 11 = 5(5+6) - 11 = 5*11 -11 = 44
CP of 1 toffees = 5850/1300 = Rs. 4.5
SP of 1 toffees for 56% profit = 4.5*1.56 = Rs. 7.02
So in 1053 Rupees number of toffees to be sold = 1053/7.02 = 150
Any factor of this number should be of the form 37^a * 19^b * 11^c * 5^d * 3^e * 31^f
All the factors of this number would be an odd number as there is no even factor
a can take values 0, 1, 2
b can take values 0, 1,
c can take values 0, 1, 2, 3
d can take values 0, 1, 2, 3
e can take values 0, 1, 2, 3
and f can take value 0, 1