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11. In two alloys A and B the ratio of zic to tin is 5:2 and 3:4 respectively. 7kg of alloy A and 21kg of the alloy B are mixed together to form a new alloy. What will be the ratio of zinc and tin in the new alloy ?
Solution:-

Method 1:In alloy A ⟹ Z T ⟹ 7 kg5 : 25 kg2 kg1In alloy B ⟹ Z T ⟹ 21 kg3 : 43912× 3× 3In new alloy ⟹ Z T ⟹ Z : T5+92+12:1 : 1Ans

method (2):

A BZ TZ T5 23 4Z Z7 21131311In new alloy zinc : alloy = 1:2zinc : tin = 1:1Ans
alternately:Tin Tin7 211 3××1311

∴ in new alloy tin : alloy = 1 : 2
zinc : tin = 1 : 1    Ans

method (3):zinc Tin CapacityAB5 2 7 kg3 4 21 kg1 1 = 1= 3A + B9125 + 9 = 142 + 12 = 14 28 kg1 : 1Ans

12. Three vessels whose capacities are in the ratio 3:2:1 are completely filled with milk mixed with water. The ratio of milk and water in the mixture of vessels are 5:2, 4:1 and 4:1 respectively. Taking \(\frac{1}{3}\) of first, \(\frac{1}{2}\) of second and \(\frac{1}{7}\) of the third mixtures, a new mixture kept in a new vessel is prepared. Find the % of water in the new mixture ?
Solution:-

method (1): Let the capacity of vessels be 3 litres, 2litres and 1 litre respectivelty.

vessel I:m w capacity5 2 31taking of the firstvessel II:m w capacity4 1 21taking of the second::vessel III:m w capacity4 1 1taking of the third:

when we mix these three vessels then in new vessel
milk = \(\frac{5}{7} + \frac{4}{5} + \frac{4}{35} = \frac{57}{35}\)
water = \(\frac{2}{7} + \frac{1}{5} + \frac{1}{35} = \frac{18}{35}\)
total new mixture = \(\frac{57}{35} + \frac{18}{35} = \frac{75}{35} = \frac{15}{7}\)
∴ % of water = \(\frac{\frac{18}{35}}{\frac{15}{7}} \times 100 = 24\% \)     Ans

method (2):milk water capacityV5 = 752 = 30VV4 = 564 = 281 = 141 = 7755now taking the quantities of the mixture as given milk water VVV25 1028 74 1∴ required % = Ans

13. 60 kg of alloy A is mixed with 100 kg of alloy B. If alloy A has lead and copper in the ratio 3:2 and alloy B has copper and tin in the ratio 1:4, then find the amount of copper in the new alloy.

solution:A Bcopper copper60 100(60+100)16013 : 51111×4 = 44 kg1435method (1):
method (2):In alloy A lead = copper = In alloy B copper = tin = ∴ In new alloy copper = 24 + 20 = 44 kg Ans

14. Two variety of tea costs Rs. 35 per kg and Rs. 40 per kg respectively are mixed in the ratio 2:3 by weight. If one-fifth of the mixture is sold at Rs. 46 per kg and remaining at the rate of Rs. 55 per kg, then find the profit percent.
Solution:

method(1):
Let 5 kg of mixture be prepared.
∴ CP of 5 kg of mixture = 2×35 + 3×40 = 190
total SP of 5 kg of mixture = 46 + 4×55 = 266
∴ profit % = \(\frac{266 – 190}{190} \times 100\) = 40%    Ans

method (2):3835 4023CP of mixture40 - 35 = 52346 55 55 - 46 = 914155 - 1 4SP of mixture

∴ profit % = \(\frac{53.2 – 38}{38} \times 100\) = 40%       Ans

15. 30 litres of a mixture contains milk & water in the ratio 4:1. Then find the amount of milk to be added to the mixture so as to have milk and water in the ratio 6:1.
Solution:

30mw6LAns

16. A man purchased two books in Rs. 1000. He sells the first book at \(\frac{4}{5}\) of its cost price and the second book at \(\frac{5}{4}\) of its cost price. If during the whole transaction he earns a profit of Rs. 100 then find the cost price of cheaper book.
Solution:-

SPCPi.e. Loss of 20%SPCPi.e. profit of 25%overall profit % = I II-20% 25%15 3010%1 : 2310001price of cheaper book = 1 unit = Ans

17. Silver is 19 times as heavy as water and copper is 10 times as heavy as water. In what ratio these be mixed to get an alloy 13 times as heavy as water ?
Solution:-

Silver Copper19 103 6131 : 2Ans

18. Ankit buys 8 pens and 4 pencils for Rs. 2400. He sells pencils at a profit of 20% and pen at a loss of 10%. If his overall profit is 240 then find the cost price of each pencil and each pen.
Solution:-

overall profit % = \(\frac{240}{2400} \times 100\) = 10%

Pen Pencil-10%20%102012:CP88×8×4+2400⟹ 1 unit150∴ CP of pen = 150& CP of pencil = 300Wrong10%(1+2)24001800∴ CP of pen = & CP of pencil = AnsAns

19. Ratio of land and water on earth is 1:2 and ratio of land and water in northern hemisphere is 2:3 then find the ratio of land and water in southern hemisphere.
Solution:-

on earth \(\frac{water}{water + land} = \frac{2}{3}\)
on N.H. \(\frac{water}{water + land} = \frac{3}{5}\)
Let on S.H. \(\frac{water}{water + land} = x\)

N.H. S.H.:ratio of (land + water) on N.H : S.H.Earth

⟹ \(x – \frac{2}{3} = \frac{1}{15}\) ⟹ \(x = \frac{11}{15}\)
∴ on S.H.  \(\frac{water}{water + land} = \frac{11}{15}\implies \frac{water}{land} = \frac{11}{4}\)
∴ required ratio = 11:4   Ans

20. In what proportion must water be mixed with spirit to gain 40% by selling it at cost price.
Solution:-

Let C.P. of spirit = x
∴ S.P. of mixture = y

Let C.P. of mixture = y
∴ \(\frac{x – y}{y} = \frac{40}{100} = \frac{2}{5}\)
5x – 5y = 2y ⟹ y = \(\frac{5}{7}x\)

Spirit Water5 : 2AnsC.P.C.P. of mixture

required ratio of water and spirit = 2:5    Ans

⬤ In these type of questions covert % into ratio, that  will be the answer               Ans