8 litres are drawn from a cask full of wine and is then filled with water.This operation is performed three more times.The ratio of the quantity of wine now left in cask to that of water is 16:81.How much wine did the cask hold originally ?
8 litres are drawn from a cask full of wine and is then filled with water.This operation is performed three more times.The ratio of the quantity of wine now left in cask to that of water is 16:81.How much wine did the cask hold originally ?
8 litres are drawn from a cask full of wine and is then filled with water.This operation is performed three more times.The ratio of the quantity of wine now left in cask to that of water is 16:81.How much wine did the cask hold originally ?
Wine to water? or wine to total? doubt lag raha hai mujhe...
this formula is (ratio of solution not replaced)/(total volume) = initial ratio * (vol. after removal/vol. after replacing)^n 16/97 = 1 * [(V - 8)/V]^4
A shipping clerk has five boxes of different but unkown weights each weighing less than 100 kg,the clerk weighs the boxes in pairs.The weights obtained are 110,112,113,114,115,116,117,118,120 and 121 kgs.What is the weight,in kgs of the heaviest box ?a) 60b) 62c) 64d) can't be determined.
My approach : solve it using options ..
trying option a ie. 60 ..so take 121 ..we need a weight of 61 kg ..this violates the condition as 60 kg is supposed to be the heaviest ..so A option is rejected ..
trying option B.. assume heaviest weight is 62 kg.. to make 118 ..other weight has to be 59 kg... similarly to make 120(62+58),,, so our third weight is 58kg... similarly solving we get fourth and fifth weights as 56 and 54 kg respectively ..
Hence our weight s are 62,59,58,56 and 54 kg .. This option B satisfies..
One from my side: Three cubes of edge lengths 8cm, 6m and 2cm are placed face to face to form a solid object. Find the difference between the maximum and minimum possible surface area of the solid object formed.Team BV - Kamal Lohia
A shipping clerk has five boxes of different but unkown weights each weighing less than 100 kg,the clerk weighs the boxes in pairs.The weights obtained are 110,112,113,114,115,116,117,118,120 and 121 kgs.What is the weight,in kgs of the heaviest box ?a) 60b) 62c) 64d) can't be determined.
4(w1+w2+w3+w4+w5) = 1156 , so w1+w2+w3+w4+w5 = 289
since , w1+w2 =121 and w4+w5 = 110 ,we have w3 = 58
No we can find w2 as w2+w3 = 117 is a possible case , so w2 = 59