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SOLVED PROBLEMS Simple problems: 1.In what ratio must rice at Rs 9.30 per Kg be mixed with rice at Rs 10.80 per Kg so that the mixture be worth Rs 10 per Kg? Solution:
C.P of 1 Kg rice of 1st kind 930 p C.P of 1 Kg rice of 2n d kind 1080p
Mean Price 1000p
80 70
Required ratio=80:70 = 8:7 2.How much water must be added to 60 liters of milk at 11/2 liters for Rs 20 so as to have a mixture worth Rs 10 2/3 a liter? Solution:C.P of 1 lit of milk = 20*2/3 = 40/3
C.P of 1 lit of water 0 C.P of 1 lit of milk 40/3
Mean Price 32/3
8/3 32/3
Ratio of water and milk =8/3 : 32/3 = 1:4 Quantity of water to be added to 60 lit of milk =1/4*60=15 liters. 3.In what ratio must water to be mixed with milk to gain 20% by selling the mixture at cost price? Solution:Let the C.P of milk be Re 1 per liter Then S.P of 1 liter of mixture = Re.1 Gain obtained =20%. Therefore C.P of 1 liter mixture = Rs(100/120*1) =5/6
C.P of 1 liter of water 0 C.P of 1 liter of milk1
Mean Price 5/6
1/6 5/6
Ratio of water and milk =1/6 : 5/6 = 1:5. 4.In what ratio must a grocer mix two varieties of pulses costing Rs 15 and Rs 20 per Kg respectively so as to get a mixture worth Rs 16.50 per Kg? Solution:
Cost of 1 Kg pulses of 1 kind 15 Cost of 1 Kg pulses of 2nd kind 20
Mean Price Rs 16.50
3.50 1.50
Required ratio =3.50 : 1.50 = 35:15 = 7:3. 5. 4Kg s of rice at Rs 5 per Kg is mixed with 8 Kg of rice at Rs 6 per Kg .Find the average price of the mixture? Solution:
rice of 5 Rs per Kg rice of 6 Rs per Kg
Average price Aw
6-Aw Aw-5
(6-Aw)/(Aw-5) = 4/8 =1/2 12-2Aw = Aw-5 3Aw = 17 Aw = 5.66 per Kg. 6.5Kg of rice at Rs 6 per Kg is mixed with 4 Kg of rice to get a mixture costing Rs 7 per Kg. Find the price of the costlier rice? Solution:Using the cross method:
rice at Rs 6 per Kg rice at Rs x per Kg
Mean price Rs 7 per Kg
5 4
x-7:1=5:4 4x-28 = 5 4x=33=>x=Rs 8.25. Therefore price of costlier rice is Rs 8.25 per Kg BACK |