solve these for me..plz...with detailed solutions

[yahooboy]

Solve these:

1) 10 cu.cm of ink can write ,on an average,340 words ,how many can one write with 0.568 litres of ink?
(a)4828 (b)8484 (c) 19312 (d) 24467

2) Fresh grapes contain 84% water while raisins contain 20% water.how many Kg of raisins can be made from
80 Kg of grapes?
(a)16 kg (b) 18 kg (c) 20 kg (d) 22 kg

3) A shop keeper mixes 3 varieties of rice costing Rs 10,12,17 per Kg.In what ratio should they be mixed so that the
resulting mixture can be sold at Rs 15.60 per kg making 20% profit?
(a) 9:14:36 (b) 11:14:25 (c) 14:36:43 (d) 2:6:3

[thereisno2ndtime]

Originally Posted by yahooboy

Solve these:

1) 10 cu.cm of ink can write ,on an average,340 words ,how many can one write with 0.568 litres of ink?
(a)4828 (b)8484 (c) 19312 (d) 24467

ans: 568*34 = 19312


2) Fresh grapes contain 84% water while raisins contain 20% water.how many Kg of raisins can be made from
80 Kg of grapes?
(a)16 kg (b) 18 kg (c) 20 kg (d) 22 kg

80 Kg Grapes==>84% water = 67.2Kg, grapes content = 12.8Kg
12.8Kg==> 80% in a certain quantity of raisin, which is = (12.8/80)*100 = 16Kg.

3) A shop keeper mixes 3 varieties of rice costing Rs 10,12,17 per Kg.In what ratio should they be mixed so that the
resulting mixture can be sold at Rs 15.60 per kg making 20% profit?
(a) 9:14:36 (b) 11:14:25 (c) 14:36:43 (d) 2:6:3

i see the ratio as 1:1:1

if u mix 1Kg of each, total cp of the mixture = Rs.39
now, SP = 3*15.6 = Rs.46.8, this will be a clear 20% profit on Rs. 39...
so... the reatio should be 1:1:1... correct?

[yahooboy]

"thereisno2ndtime" i think u got the 1st wrong....
the answers are:
(1)a
(2)a
(3)d
I dont have aclue as to how (1) is a and (3) is d..but this is what the answers given...In fact these are from TIME materials SM100501 "averages-mixtures-alligation"..could someone send a solution for these....[/spoiler]

[THUSSU]

no. 3 also works with D (wonder of wonders!)

where the cost of 11 kg = 2x10+6x12+3x17 = 143
cost of 1 kg = 13, which is the cp before a 20% profit on 15.6!

woner!

[ImpossibleGuy]

The first ans has to be (c).
Is there any quick method to solve third que barring the elimination technique?

[ImpossibleGuy]

The first ans has to be (c).
Is there any quick method to solve third que barring the elimination technique?

[peter]

Originally Posted by yahooboy

Solve these:

1) 10 cu.cm of ink can write ,on an average,340 words ,how many can one write with 0.568 litres of ink?
(a)4828 (b)8484 (c) 19312 (d) 24467

2) Fresh grapes contain 84% water while raisins contain 20% water.how many Kg of raisins can be made from
80 Kg of grapes?
(a)16 kg (b) 18 kg (c) 20 kg (d) 22 kg

3) A shop keeper mixes 3 varieties of rice costing Rs 10,12,17 per Kg.In what ratio should they be mixed so that the
resulting mixture can be sold at Rs 15.60 per kg making 20% profit?
(a) 9:14:36 (b) 11:14:25 (c) 14:36:43 (d) 2:6:3

1) 0.568 lt = 568 cu cm. and no. of words/cu cm = 34.
=> No. of words = 568 X 34 = 19312.
(c)
2) 80 Kg grapes contain 16% pulp = 12.8 Kg.
Raisins contain 80% pulp = 12.8 Kg.
=> Raisins = 12.8/0.8 = 16 Kg.
(a)
3) 20% profit => CP of mixture = 15.6 / 1.2 = 13 = Cost of mixture per Kg.
We can mix the 3 in more than 1 combinations 2 possibilities are 1:1:1, 2:6:3
(d)

[cat_cracker1]

There is a question
There are four pens each a different colour and four caps each a different colour but,same as that of one of the pens.In how many ways can these caps be put on the puns so that a cap does not go to the pen of same colour.

These prob. Can be done by considering all the cases for c1 going in p2,p3 and p4 and so on.In which the answer is 9.But if the above quantities are increased would it be feasible to go as in the above way.Is there any shortcut method to solve these type of ques.

[mizzou]

Originally Posted by cat_cracker1

There is a question
There are four pens each a different colour and four caps each a different colour but,same as that of one of the pens.In how many ways can these caps be put on the puns so that a cap does not go to the pen of same colour.

These prob. Can be done by considering all the cases for c1 going in p2,p3 and p4 and so on.In which the answer is 9.But if the above quantities are increased would it be feasible to go as in the above way.Is there any shortcut method to solve these type of ques.

c1 can go to wrong pens in 3 ways
c2 can go to wrong pens in 3 ways
c3 can go to wrong pens in 3 ways
After abov 3, c4 will get remaining pen.

Since these events are mutually exclusive, total no. of such events = 3+3+3=9

[whocarez]

Originally Posted by cat_cracker1

There is a question
There are four pens each a different colour and four caps each a different colour but,same as that of one of the pens.In how many ways can these caps be put on the puns so that a cap does not go to the pen of same colour.

These prob. Can be done by considering all the cases for c1 going in p2,p3 and p4 and so on.In which the answer is 9.But if the above quantities are increased would it be feasible to go as in the above way.Is there any shortcut method to solve these type of ques.

Yes there is a formula for this type of questions. The type mentioned above is a classic case of dearrangement!!

It should be 4! [1 - 1/1! + 1/2! - 1/3! + 1/4!] which equals 24-24+12-4+1=9