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Solution:
Profit on 1st part 8% Profit on 2nd part 18%
Mean Profit 14%
4 6
Ratio of 1st and 2nd parts =4:6 =2:3.
Quantity of 2nd ind =3/5*1000Kg =600 Kg.
6.A jar full of whiskey contains 40% alcohol. A part of
this whiskey is replaced by another containing 19% alcohol
and now the percentage of alcohol was found to be 26%.
The quantity of whiskey replaced is?
Solution:
Strength of first jar 40% Strength of 2nd jar 19%
Mean Strength 26%
7 14
So,ratio of 1st and 2nd quantities =7:14 =1:2
Therefore required quantity replaced =2/3.
7.A container contains 40lit of milk. From this container
4 lit of milk was taken out and replaced by water.
This process was repeated further two times.
How much milk is now contained by the container?
Solution:Amount of milk left after 3 operations = 40(1-4/40)3lit
=(40*9/10*9/10*9/10)
= 29.16 lit
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<b>Complex Problems</b>
1.Tea worth Rs 126 per Kg are mixed with a third variety in
the ratio 1:1:2. If the mixture is worth Rs 153 per Kg ,
the price of the third variety per Kg will be?
Solution:
Since First and second varieties are mixed in equal proportions
so their average price =Rs (126+135)/2 = 130.50.
So the mixture is formed by mixing two varieties ,one at
Rs 130.50 per Kg and the other at say Rs x per Kg in the
ratio 2:2 i e,1:1 we have to find x.
Costof 1Kg tea of 1st kind RS 130.50 Costof 1Kg tea of 2n d
kind Rs x.
Mean Price Rs 153
x-153 22.50
(x=153)/22.5 = 1 =>x-153 = 22.5
x = 175.50.
Price of the third variety =Rs 175.50 per Kg.
2.The milk and water in two vessels A and B are in the ratio 4:3
and 2:3 respectively. In what ratio the liquids in both the
vessels be mixed to obtain a new mixture in vessel c
consisting half milk and half water?
Solution:Let the C.P of milk be Re 1 per liter.
Milk in 1 liter mixture of A = 4/7 liter.
Milk in 1 liter mixture of B = 2/5 liter.
Milk in 1 liter mixture of C = 1/2 liter.
C.P of 1 liter mixture in A=Re 4/7
C.P of 1 liter mixture in B=Re 2/5.
Mean Price = Re 1/2.
By rule of allegation we have:
C.P of 1 liter mixture in A C.P of 1 liter mixture in B
4/7 2/5
Mean Price ½
1/10 1/14
Required ratio = 1/10 : 1/14 = 7:5.
3.How many Kg s of wheat costing him Rs 1.20,Rs 1.44
and Rs 1.74 per Kg so that the mixture may be worth
Rs 1.41 per Kg?
Solution:
Step1:Mix wheat of first and third kind to get a mixture
worth Rs 1.41 per Kg.
C.P of 1 Kg wheat of 1st kind 120p C.P of 1 Kg wheat of
3rd kind 174p
Mean Price 141p
33 21
They must be mixed in the ratio =33:21 = 11:7
Step2:Mix wheats of 1st and 2n d kind to obtain a mixture
worth of 1.41.per Kg.
C.P of 1 Kg wheat of 1st kind 120p C.P of 1 Kg wheat of 2n d
kind 144p
Mean Price 141p
3 21
They must be mixed in the ratio = 3:21=1:7.
Thus,Quantity of 2n d kind of wheat / Quantity of
3rd kind of wheat = 7/1*11/7= 11/1
Quantities of wheat of 1st :2n d:3rd = 11:77:7.
4.Two vessels A and B contain spirit and water mixed in
the ratio 5:2 and 7:6 respectively. Find the ratio n which
these mixture be mixed to obtain a new mixture in vessel
c containing spirit and water in the ratio 8:5?
Solution:Let the C.P of spirit be Re 1 per liter.
Spirit in 1 liter mix of A = 5/7 liter.
C.P of 1 liter mix in A =5/7.
Spirit in 1 liter mix of B = 7/13 liter.
C.P of 1 liter mix in B =7/13.
Spirit in 1 liter mix of C = 8/13 liter.
C.P of 1 liter mix in C =8/13.
C.P of 1 liter mixture in A 5/7 C.P of 1 liter mixture
in B 7/13
Mean Price 8/13
1/13 9/91
Therefore required ratio = 1/13 : 9/91 = 7:9.
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5.A milk vendor has 2 cans of milk .The first contains 5% water
and the rest milk. The second contains 50% water. How much milk
should he mix from each of the container so as to get 12 liters
of milk such that the ratio of water to milk is 3:5?
Solution:Let cost of 1 liter milk be Re 1.
Milk in 1 liter mixture in 1st can = 3/4 lit.
C.P of 1 liter mixture in 1st can =Re 3/4
Milk in 1 liter mixture in 2n d can = 1/2 lit.
C.P of 1 liter mixture in 2n d can =Re 1/2
Milk in 1 liter final mixture = 5/8 lit.
Mean Price = Re 5/8.
C.P of 1 lt mix in 1st Re3/4 C.P of 1 lt mix in 2nd Re1/2
Mean Price 5/8
1/8 1/8
There ratio of two mixtures =1/8 :1/8 = 1:1.
So,quantity of mixture taken from each can=1/2*12
= 6 liters.
6.One quantity of wheat at Rs 9.30 p
er Kg are mixed
with another quality at a certain rate in the ratio 8:7.
If the mixture so formed be worth Rs 10 per Kg ,what is
the rate per Kg of the second quality of wheat?
Solution:Let the rate of second quality be Rs x per Kg.
C.P of 1Kg wheat of 1st 980p C.P of 1 Kg wheat of 2nd 100x p
Mean Price 1000p
100x-1000p 70 p
(100x-1000) / 70 = 8/7
700x -7000 = 560
700x = 7560 =>x = Rs 10.80.
Therefore the rate of second quality is Rs10.80
7.8lit are drawn from a 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 the water
is 16:81. How much wine did the cask hold originally?
Solution:
Let the quantity of the wine in the cask originally be
x liters.
Then quantity of wine left in cask after
4 operations = x(1- 8/x)4lit.
Therefore x((1-(8/x))4)/x = 16/81.
(1- 8/x)4=(2/3) 4
(x- 8)/x=2/3
3x-24 =2x
x=24.
8.A can contains a mixture of two liquids A and B in the
ratio 7:5 when 9 liters of mixture are drawn off and the
can is filled with B,the ratio of A and B becomes 7:9.
How many liters of liquid A was contained by the can initially?
Solution:
Suppose the can initially contains 7x and 5x liters
of mixtures A and B respectively .
Quantity of A in mixture left = (7x- (7/12)*9 )lit
= 7x - (21/4) liters.
Quantity of B in mixture left = 5x - 5/12*9
= 5x - (15/4) liters
Therefore (7x – 21/4)/ (5x – 15/4+9)=7/9
(28x-21)/(20x +21)= 7/9
(252x -189)= 140x +147
112x = 336
=> x=3.
So the can contains 21 liters of A.
9.A vessel is filled with liquid,3 parts of which are water
and 5 parts syrup. How much of the mixture must be drawn off
and replaced with water so that the mixture may be
half water and half syrup?
Solution:
Suppose the vessal initially contains 8 liters of liquid.
Let x liters of this liquid be replaced with water
then quantity of water in new mixture
= 3-(3x/8)+x liters.
Quantity of syrup in new mixture = 5 - 5x/8 liters.
Therefore 3 - 3x/8 +x = 5 - 5x/8
5x+24 = 40-5x
10x = 16.
x= 8/5.
So part of the mixture replaced = 8/5*1/8 =1/5.
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