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anusha

Plz help me solve these questions

hi frnz,Can anyone help me solve these three problems.1)What is the maximum number of half-pint bottles of cream that can be filled with a 4-gallon can of cream(2 pt.=1 qt. and 4 qt.=1 gal)A.16 B.24 C.30 D.64 (Ans.D)2)A coffee shop blends 2 kinds of coffee,putting in 2 parts of a 33p. a gm. grade to 1 part of a 24p. a gm.If the mixture is changed to 1 part of the 33p. a gm. to 2 parts of the less expensive grade,how much will the shop save in blending 100 gms.A.Rs.90 B.Rs.1.00 C.Rs.3.00 D.Rs.8.00 (Ans.C)3)In june a baseball team that played 60 games had won 30% of its game played. After a phenomenal winning streak this team raised its average to 50% .How many games must the team have won in a row to attain this average?A. 12 B. 20 C. 24 D. 30 (Ans. C)4)M men agree to purchase a gift for Rs. D. If three men drop out how much more will each have to contribute towards the purchase of the gift/ A. D/(M-3) B. MD/3 C. M/(D-3) D. 3D/(M2-3M) (Ans. D)

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hi frnz,Can anyone help me solve these three problems.1)What is the maximum number of half-pint bottles of cream that can be filled with a 4-gallon can of cream(2 pt.=1 qt. and 4 qt.=1 gal)A.16 B.24 C.30 D.64 (Ans.D) 2 pt = 1 qt 8 pt = 4 qt 8 pt = 1 gal 8 * 4 pt = 1* 4 gal 36 pt = 4 gal so we reguire 64 half bottles to fill a 4-gal can. 2)A coffee shop blends 2 kinds of coffee,putting in 2 parts of a 33p. a gm. grade to 1 part of a 24p. a gm.If the mixture is changed to 1 part of the 33p. a gm. to 2 parts of the less expensive grade,how much will the shop save in blending 100 gms.A.Rs.90 B.Rs.1.00 C.Rs.3.00 D.Rs.8.00 (Ans.C)3)In june a baseball team that played 60 games had won 30% of its game played. After a phenomenal winning streak this team raised its average to 50% .How many games must the team have won in a row to attain this average?A. 12 B. 20 C. 24 D. 30 (Ans. C)4)M men agree to purchase a gift for Rs. D. If three men drop out how much more will each have to contribute towards the purchase of the gift/ A. D/(M-3) B. MD/3 C. M/(D-3) D. 3D/(M2-3M) (Ans. D)

1 M = D/M 3 M = 3D/M M-3 =3D/(M-3)M

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