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Bunuel
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assume water as 0 % milk

initial milk is 80 % final milk is 60 %

apply allegations and mixture concept

x/45 = ( 0.8 - 0.6) / ( 0.6 - 0)
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PrabhatKC
How to do this?
Milk = 4 * 45 / 5 = 36 now this ratio of milk to the water 36 / (40-9) we need to change it in 3/2 so for converting 36 / 9 into we will need 15 more ltr of water therefore if we add 15 into 9 now water will become 24 and ratio will be 36/24 = 3/2
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Purnank
Milk = 4 * 45 / 5 = 36 now this ratio of milk to the water 36 / (40-9) we need to change it in 3/2 so for converting 36 / 9 into we will need 15 more ltr of water therefore if we add 15 into 9 now water will become 24 and ratio will be 36/24 = 3/2
sorry miss type "36 / (45-36) = 36 / 9 ratio we need to change it in 3/2"
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prashi1503
initial milk is 80 % final milk is 60 %
Can you explain your 150 IQ approach?
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Simple go with ratio only
Milk:water =36/9
36/(9+x)=3/2
Ok solving x will get 15
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4:1 of 45 litres would mean 36 l of milk and 9 l of water

Now we want for 36 l of milk, we need 24 of water to get 3:2 ratio, means we need 24-9=15l
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Why isn’t the answer 12?
My logic: There are 12 centers and each center is represented by one or two colors.
To minimize the no of color used, we take 1 color for each centers,
So the answer has to be 12.
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Bunuel
Check here: a-company-th ... 68427.html
If n is the number of colors (say).
For n = 4, total possible combinstions = 4C1 +4C2 = 10
For n = 5, total possible combinstions = 5C1 +5C2 = 15

So the minimumn possible value of n is 5 since 10 <= 12 <= 15.
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ani12345678
first is c as each of them arent indivdually sufficient but both of them combined are sufficient ,next question its a option is a+b=ab and a=b doesnt necessarily mean ab=a+b (using a=3 b=3 example)
Hey I have a ques what if the value of x is 100/3, that would mean value of m can be 100 and the same ambiguity may come even if we combine with statement 1

could someone please helpme with this?
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Pranjal committed a mistake in finding the LCM of three distinct positive integers greater than 1 namely A, B and C, and found it to be 840, which is a common multiple of A, B and C all, but is not the lowest. The HCF of A, B and C is 1. Find the maximum possible value of A + B + C.

1) 631
2) 613
3) 563
4) 257

I have doubt in options
Can someone help ?
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What are your steps to solving?
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If the HCF is one the 3 numbers have to be primes.
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840=3x5x7x8
We can take a b c as 3 5 7 or 8 7 5
So max will will 8 7 5
So a+b+c =8+7+5=>20

my ans is 20
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and the given answer?
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No ans given
It was from test

I had skipped this qn during test as my ans was beyond options
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