A mixture of milk and water in the ratio of 8:7. Another mixture of milk and water in the ratio of 5:9. In what ratio should we mix these two, if you need a mixture of milk and water in the ratio of 1:1?Copied from CGL Thread
A mixture of milk and water in the ratio of 8:7. Another mixture of milk and water in the ratio of 5:9. In what ratio should we mix these two, if you need a mixture of milk and water in the ratio of 1:1?Copied from CGL Thread
A dice is rolled six times. One, two, three, four, five and six appears on consecutive throws of dices. How many ways are possible of having 1 before 6?1) 1202) 3603) 2404) 280
OA: 360 No. of ways to place 1 and 6= 6c2 No. of arrangements of the rest of the numbers= 4! Answer= 6c2. 4! =300
A mixture of milk and water in the ratio of 8:7. Another mixture of milk and water in the ratio of 5:9. In what ratio should we mix these two, if you need a mixture of milk and water in the ratio of 1:1?Copied from CGL Thread
A mixture of milk and water in the ratio of 8:7. Another mixture of milk and water in the ratio of 5:9. In what ratio should we mix these two, if you need a mixture of milk and water in the ratio of 1:1?Copied from CGL Thread
A dice is rolled six times. One, two, three, four, five and six appears on consecutive throws of dices. How many ways are possible of having 1 before 6?1) 1202) 3603) 2404) 280
@ScareCrow28Bhai I did using Alligation.Most fundamental rule in Alligation is that it could only and only be applied on a single quantity at a time. Either take milk or water but not both.Milk in 1st mix=8:15Milk in 2nd mix=5:14Milk in final mix=1:28/15.........5/14........1/21/7...........1/30Ratio 1/7 : 1/30 =30:7Will get same answer if we take only water.Here I think you are taking both quantities at same time that's why you got a wrong answer though fundamentally this approach looks right.
If you look at the question first then it seems wrong, but what if we look at the answer first?
We can see that it , eventually, is satisfying the problem, isn't it? :rolleyes:
I still don't know where and how am I wrong. @scrabbler Please help on this.
(50-a)+(50-b)+(50-c)+(100-d)= 150,a+b+c+d = 100 = > 103c 3 none of the above?The answer is not yet over this will contain values such as (51,x,y,z)= invalid. we have to exclude those cases.so consider when(51+a)+b+c+d= 100 or a+b+c+d = 49 or 52 c3 cases for a so in total , 3 * 52 C3 cases we have to exclude. a and b or c cannot be simultaneously greAter than 50. and d alone cannot be greater than 100. so no other exclusions. Hence OA should be 103C3 - 3*52C3
yup i was about to write all this thanks u save me from so much typing :)
it will be 103c3 - 3*52c3
it is because in 103c3 we will have all the cases when a or b or c is greater than 50 which is unfavourable
OA: 30:7 That's right. Tell me where am I wrong ( Must be fundamentally wrong somewhere )Mix-1: Milk = 8ml, Water = 7 mlMix-2: Milk = 5ml, Water = 9mlRatio: k:18k + 5 / (5k + 9) = 1from here, k = 4Hence ratio = 4:1You can also check it by putting the value of "k" Please tell me where am I going wrong.
A mixture of milk and water in the ratio of 8:7. Another mixture of milk and water in the ratio of 5:9. In what ratio should we mix these two, if you need a mixture of milk and water in the ratio of 1:1?Copied from CGL Thread
First one is divided into 8 + 7 = 15 parts, second in 5 + 9 = 14 parts. So take 210 (LCM of 14 and 15) parts. Now milk in first solution is 14 x 8 = 112, second is 15 x 5 = 75 and we need 1:1 i.e. 105 out of 210 parts milk. So see the difference of 112 and 75 from 105 - they are 7 and 30 so reverse this and voila, 30:7 is the ratio required. regards scrabbler
First one is divided into 8 + 7 = 15 parts, second in 5 + 9 = 14 parts. So take 210 (LCM of 14 and 15) parts. Now milk in first solution is 14 x 8 = 112, second is 15 x 5 = 75 and we need 1:1 i.e. 105 out of 210 parts milk. So see the difference of 112 and 75 from 105 - they are 7 and 30 so reverse this and voila, 30:7 is the ratio required. regardsscrabbler
)
Nyce explanation 
