1.The proportion of acid and water in three samples is 2 : 1 ,3 : 2 and 5 : 3. A mixture containing equal quantities of all three samples is made. The ratio of water and acid in the mixture is :
(a) 120 : 133
(b) 227 : 133
(c) 227 : 120
(d) 133 : 227
2.40 litres of a mixture of milk and water contains 10% of water, the water to be added, to make the water content 20% in the new mixture is :
(a) 6 litres
(b) 6.5 litres
(c) 5.5 litres
(d) 5 litres
3.Zinc and copper are in the ratio 5 : 3 in 400 gm of an alloy. How much of copper (in grams) should be added to make the ratio 5 : 4 ?
(a) 50
(b) 66
(c) 72
(d) 200
4.Two vessels A and B contain milk and water mixed in the ratio 8 : 5 and 5 : 2 respectively. The ratio in which these two mixtures be mixed to get a new mixture containing 69 3/13% milk is:
(a) 3 : 5
(b) 5 : 2
(c) 5 : 7
(d) 2 : 7
5.A barrel contains a mixture of wine and water in the ratio 3 : 1. How much fraction of the mixture must be drawn off and substituted by water so that the ratio of wine and water in the resultant mixture in the resultant mixture in the barrel becomes 1 : 1 ?
(a) 1/4
(b) 1/3
(c) 3/4
(d) 2/3
6.A jar contained a mixture of two liquids A and B in the ratio 4 : 1 when 10 litres of the mixture was taken out and 10 litres of liquid B was poured into the jar, this ratio become 2 : 3. The quantity of liquid A contained in the jar initially was
(a) 4litres
(b) 8litres
(c) 16litres
(d) 40litres
7.A and B are tow alloys of gold and copper prepared by mixing metals in the ratio 5 : 3 and 5 : 11 respectively. Equal quantities of these alloys are melted to form a third alloy C. The ratio of gold and copper in the alloy C is
(a) 25 : 33
(b) 33 : 25
(c) 15 : 17
(d) 17 : 15
8.A mixture contains wine and water in the ratio 3 : 2 and another mixture contains them in the ratio 4 : 5. How many litres of the latter must be mixed with 3 litres of the former so that the resulting mixture may contain equal quantities of wine and water?
(a) 5 2/5
(b) 5 2/3
(c) 4 1/2
(d) 3 3/4
9.An alloy contains zinc, copper and tin in the ratio 2 : 3 : 1 and another contains copper, tin and lead in the ratio 5 : 4 : 3. If equal weights of both alloys are melted together to form a third alloy, then the weight of lead per kg in the new alloy will be
(a) 1/2 kg
(b) 1/8 kg
(c) 3/14 kg
(d) 7/9 kg
10.An alloy contains copper, zinc and nickel in the ratio of 5 : 3 : 2. The quantity of nickel in kg that must be added to 100 kg of this alloy to have the new ratio 5 : 3 : 3 is
(a) 8
(b) 10
(c) 12
(d) 15
ANSWERS AND SOLUTION
1.(b) Required ratio
= (2/3+3/5+5/8) :(1/3+2/5+3/8)
= ((80+72+75 )/120) : ((40+48+45 )/120)
= 227 : 133
2.(a) In 400 gm of alloy,
Zinc = 5/8 * 400 = 250 gm.
Copper = 3/8 * 40 = 150 gm.
If x gm of copper be mixed, then 250/(150 + x) = 5/4
=> 750 + 5x = 1000
=> 5x = 1000 – 750 = 250
x = 50 gm.
3.(d) Water content in 40 litres of mixture = 40 * 10/100 = 4 litres
so Milk content = 40 – 4 = 36 litres
Let x litres of water is mixed.
so (4+x )/(40+x )=20/100 ? (4+x )/(40+x )=1/5
=> 20 + 5x = 40 + x
4x = 20
X = 5 litres
4.(d)
5.(b) Let the barrel contain 4 liters of mixture.
so Wine = 3 litres
Water = 1 litre
Let x litre mixture is taken out.
so Wine in (4 – x) litres mixtures = 3/4(4 –x)
On adding x litres water, water in mixture
= (4 – x) * 1/4 + x
= 1 – x/4 + x
= (4 – x+ 4x)/4 = (4 + 3x)/4
? 3/4 (4-x)= (4+3x )/4
=> 3 – 3x/4 = 1 + 3x/4
=> 2 = 6x/4
x = 2 * 4/6 = 4/3
Required answer
= 4/3//4 = 1/3
6.(d) Let the initial quantity of liquids A and B in the jar be 4x and x litres respectively.
After taking out 10 litres of the mixture.
Liquied A = 4x – 4/5 * 10 = (4x – 8) litres
Liquied B = 4x – 1/5 * 10 = (4x -2) litres
After pouring 10 litres of liquied B, (4x-8 )/(4x-2+10 )= 2/3
12x – 24 = 8x + 16
4x = 40
X = 40/4 = 10
so quantity of liquid A = 4x
= 4 x 10 = 40 litres
7.(c) Let 1 kg of each of the alloys A and B be mixed together.
In alloy A,
Quantity of gold = 5/8 kg.
Quantity of copper = 3/8 kg.
In alloy B,
Quantity of gold = 5/16 kg.
Quantity of copper = 11/16 kg.
so Required ratio = (5/8+5/16) :(3/8+11/16)
= 15/16 : 17/16 = 15 : 17
8.(a)
9.(b) Quantity of glass in 1 kg of first alloy = zero
Quantity of glass in 1 kg.of second alloy = 3/12 = 1/4 kg
so Quantity of glass in 1 kg of new alloy = 1/8 kg
10.(b) Let x kg of nickel be mixed.
so (20+x)/(100+x)=3/11
=> 220 + 11x = 300 + 3x
=> 11x – 3x = 300 – 220
=> 8x = 80
=> X = 10 kg.
1.(b) Required ratio
= (2/3+3/5+5/8) :(1/3+2/5+3/8)
= ((80+72+75 )/120) : ((40+48+45 )/120)
= 227 : 133
2.(a) In 400 gm of alloy,
Zinc = 5/8 * 400 = 250 gm.
Copper = 3/8 * 40 = 150 gm.
If x gm of copper be mixed, then 250/(150 + x) = 5/4
=> 750 + 5x = 1000
=> 5x = 1000 – 750 = 250
x = 50 gm.
3.(d) Water content in 40 litres of mixture = 40 * 10/100 = 4 litres
so Milk content = 40 – 4 = 36 litres
Let x litres of water is mixed.
so (4+x )/(40+x )=20/100 ? (4+x )/(40+x )=1/5
=> 20 + 5x = 40 + x
4x = 20
X = 5 litres
4.(d)
5.(b) Let the barrel contain 4 liters of mixture.
so Wine = 3 litres
Water = 1 litre
Let x litre mixture is taken out.
so Wine in (4 – x) litres mixtures = 3/4(4 –x)
On adding x litres water, water in mixture
= (4 – x) * 1/4 + x
= 1 – x/4 + x
= (4 – x+ 4x)/4 = (4 + 3x)/4
? 3/4 (4-x)= (4+3x )/4
=> 3 – 3x/4 = 1 + 3x/4
=> 2 = 6x/4
x = 2 * 4/6 = 4/3
Required answer
= 4/3//4 = 1/3
6.(d) Let the initial quantity of liquids A and B in the jar be 4x and x litres respectively.
After taking out 10 litres of the mixture.
Liquied A = 4x – 4/5 * 10 = (4x – 8) litres
Liquied B = 4x – 1/5 * 10 = (4x -2) litres
After pouring 10 litres of liquied B, (4x-8 )/(4x-2+10 )= 2/3
12x – 24 = 8x + 16
4x = 40
X = 40/4 = 10
so quantity of liquid A = 4x
= 4 x 10 = 40 litres
7.(c) Let 1 kg of each of the alloys A and B be mixed together.
In alloy A,
Quantity of gold = 5/8 kg.
Quantity of copper = 3/8 kg.
In alloy B,
Quantity of gold = 5/16 kg.
Quantity of copper = 11/16 kg.
so Required ratio = (5/8+5/16) :(3/8+11/16)
= 15/16 : 17/16 = 15 : 17
8.(a)
9.(b) Quantity of glass in 1 kg of first alloy = zero
Quantity of glass in 1 kg.of second alloy = 3/12 = 1/4 kg
so Quantity of glass in 1 kg of new alloy = 1/8 kg
10.(b) Let x kg of nickel be mixed.
so (20+x)/(100+x)=3/11
=> 220 + 11x = 300 + 3x
=> 11x – 3x = 300 – 220
=> 8x = 80
=> X = 10 kg.
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