2x + 3y = 12067y is always oddLeast value of y = 1Largest value of y = 4021So, y can take {(4021 - 1)/2} + 1 = 2011 valuesYeah for n = 12066, we have 2012 solutions (didn't check for 12066)
Yes Sir! And I didn't check for 12067 Thanks a lot sir :)
there are two quantities of milk-amul and sudha having different prices per litre,their volumes being 130 litres and 180 litres respectively.After equal amounts of milk was removed from both,the milk removed from amul was added to sudha and vice versa.The resulting two types of the milk now have the same price.Find the amount of milk drawn out from each type of milk. 1)58.66 2)75.48 3)81.23 4)none of these.
there are two quantities of milk-amul and sudha having different prices per litre,their volumes being 130 litres and 180 litres respectively.After equal amounts of milk was removed from both,the milk removed from amul was added to sudha and vice versa.The resulting two types of the milk now have the same price.Find the amount of milk drawn out from each type of milk.1)58.662)75.483)81.234)none of these.
there are two quantities of milk-amul and sudha having different prices per litre,their volumes being 130 litres and 180 litres respectively.After equal amounts of milk was removed from both,the milk removed from amul was added to sudha and vice versa.The resulting two types of the milk now have the same price.Find the amount of milk drawn out from each type of milk.1)58.662)75.483)81.234)none of these.
let x and y per litre be the rate of 130l and 180l resp also let L litre of exchange is required then ,as condition given in the question 130x-Lx+Ly/130=180y-Ly+Lx/180 on solving,L=75.48 litres
@19rsb can u please explain Why did u divide LY on l.h.s by 130 and LX on right hand side by 180 ?
arey bhai.... the expression is lyk this (130x-Lx+Ly)/130=(180y-Ly+Lx)/180...........in order to equate the new rates after exchange of L litres(since rate= total money/volume)
In a factory where toys are manufactured, machines A, B and C produce 25%. 35% and 40% of the total toys, respectively. Of their output, 5%, 4% and 2% respectively, are defective toys. If a toy drawn at random in found to be defective, What is the probability that it is manufactured on machine B?
In a factory where toys are manufactured, machines A, B and C produce 25%. 35% and 40% of the total toys, respectively. Of their output, 5%, 4% and 2% respectively, are defective toys. If a toy drawn at random in found to be defective, What is the probability that it is manufactured on machine B?solution plz explain
In a factory where toys are manufactured, machines A, B and C produce 25%. 35% and 40% of the total toys, respectively. Of their output, 5%, 4% and 2% respectively, are defective toys. If a toy drawn at random in found to be defective, What is the probability that it is manufactured on machine B?solution plz explain
@Shray14 total amount of work =16*32= 512 man days work done in 16 days=16*16=256 man days Let x be the extra number of days required. so 512=256 + 8*x x=32 so total days=32+16=48 days
And I didn't check for 12067