Complex Problems:
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.
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.
