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When processing flower-nectar into honeybees' extract, a considerable amount of water gets reduced. How much flower-nectar must be processed to yield 1kg of honey, if nectar contains $50%$ water, and the honey obtained from this nectar contains $15%$ water?


1.5 kgs


1.7 kgs


3.33 kgs


None of these

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Option(B) is correct

Flower-nectar contains $50\%$ of non-water part. 

In honey this non-water part constitutes $85\% (100-15)$.

Therefore $0.5X$ Amount of flower-nectar = $0.85X$ Amount of honey = $0.85\times 1$ kg 

Therefore amount of flower-nectar needed

\(=\left(\dfrac{0.85}{0.5}\right)\times 1\)

\(=1.7\) kg

(6) Comment(s)



Extraction process--> Nector(+50%water) --> Honey(+15%water)--> Honey (pure).

End result is 1 Kg honey.

That means if we go one step up, before its extraction it will contain 15% water as well. So now total quantity:

X-15% of X = 1kg honey.

==> .85X = 1

X= 1/.85 ~ 1.176 Kg (Honey plus water)

Now this mixture was extracted from Nector-water mixture.

let us assume it is extracted from X kg of flower nector.


X- 50% of X = 1.176

.5X = 1.176

=> X = 2.352 Kg


i also got this answer but it is correct or not?


Are there any other different method to solve this question ?


As of now no other substitute becoz it is already simple. Let's make it simpler with some verbal concepts.

1. Honey in Nectar = Honey in honeybee = Honey we got. i.e no loss of honey anywhere only the proportion of water kept changing at different stage.

2. The value of honey at all stages is equal . so calculate the value at all three stages at keep them equal.

3. First stage - Honey is 50% of something.say, Value= 0.5X

4. Second stage - Honey is 85% of something. say, Value =0.85X

5. Last stage - Honey is 1 Kg. (Given)

so, 0.5X=0.85X=1kg. X=1.7 Kg (Pretty Obvious will me more than the honey)

Hope it helps.


this is a good place for the preparation of aptitude tests.Tongue


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