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
A mixture contains sugar solution and colored water in the ratio 3 : 5. If 16 liters of colored water is added to the mixture, the ratio becomes 4 : 7. Find the initial quantity of colored water in the given mixture (in liters).
The initial ratio is 3 : 5
Let â€˜kâ€™ be the common ratio.
=> Initial quantity of sugar solution = 3 k
=> Initial quantity of colored water = 5 k
=> Final quantity of sugar solution = 3 k
=> Final quantity of colored water = 5 k + 16
Final ratio = 3 k : 5 k + 16 = 4 : 7
=> 21 k = 20k + 64
=> k = 64
Therefore, initial quantity of colored water in the given mixture (in liters) = 5 * 64 = 320 liter
When A covers 500m B Covers 470m
When A Covers 500m C covers 480
A : B = 500 : 470
A : C = 500 : 480
C/B = (C/A*A/B) = (480/470) = 480:470
When C covers 480 m, B covers 470 m.
When C covers 1200 m, B covers (470/480)*1200)m = 1175m
Therefore, C beats B by (1200 - 1175) m = 25 m.
A letter is randomly picked from the following longest word in the dictionary: pneumonoultramicroscopicsilicovolcanoconiosis
what is the probability that it is a vowel?