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**Quiz Description** :

**Name: **Mixture and Alligation Question answer : Aptitude Test – 1

**Subject: **Aptitude

**Topic: **Mixture and Alligation

**Questions: **10 Objective Type

**Time Allowed: **15 minutes

**Important for: ** IBPS PO, IBPS Clerk, SBI PO, SBI Clerk, RBI, LIC AAO, SSC CGL, SSC CHSL, State Police, Indian Army, CRPF, State PCS, Engineering Job Interview etc.

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- Question 1 of 10
##### 1. Question

1 pointsIn what proportion must rice at Rs. 3.10 per kg be mixed with rice at Rs. 3.60 per kg, so that the mixture be worth Rs. 3.25 a kg?

CorrectC.P of 1 Kg. cheaper rice C.P of 1 Kg. dearer rice 310 Paise 360 Paise Mean Price

(325 Paise)35 15 By applying Alligation rule:

(Quality of cheaper Rice)/(Quality of dearer Rice)= 35/15= 7/3

Hence they must be mixed in the ratio 7 : 3.

IncorrectC.P of 1 Kg. cheaper rice C.P of 1 Kg. dearer rice 310 Paise 360 Paise Mean Price

(325 Paise)35 15 By applying Alligation rule:

(Quality of cheaper Rice)/(Quality of dearer Rice)= 35/15= 7/3

Hence they must be mixed in the ratio 7 : 3.

- Question 2 of 10
##### 2. Question

1 pointsWhat amount of salt at 56 per Kg. must a salesman mix with 36 Kg. of salt at 28 P per Kg. so that salesman may, on selling the mixture at 48 P per Kg., gain 20% on the outlay?

CorrectThe cost price of mixture = 48 × 100/120 P=40 P per Kg (By the rule of fraction)

56 28 40 12 16 3 4 ∴ Ratio = 3 : 4

Thus for every 4 Kg. of salt at 28 P, 3 Kg. of salt at 56 P is used.

∴ The required no. of Kg. = 36 × 3/4 = 27 KgIncorrectThe cost price of mixture = 48 × 100/120 P=40 P per Kg (By the rule of fraction)

56 28 40 12 16 3 4 ∴ Ratio = 3 : 4

Thus for every 4 Kg. of salt at 28 P, 3 Kg. of salt at 56 P is used.

∴ The required no. of Kg. = 36 × 3/4 = 27 Kg - Question 3 of 10
##### 3. Question

1 pointsA milkman mixes of a certain quantity of milk with 16 liters of water is worth 90 P per liter. If pure milk be worth Rs. 1.08 per liter how much milk is there in the mixture?

CorrectThe mean value is 90 P and the price of water is 0 P.

Milk Water 108 0 90 90 18 By Applying Alligation Rule;

Milk and Water are in the ratio of 5 : 1.

∴ Quantity of milk in the mixture = 5 × 16 = 80 litresIncorrectThe mean value is 90 P and the price of water is 0 P.

Milk Water 108 0 90 90 18 By Applying Alligation Rule;

Milk and Water are in the ratio of 5 : 1.

∴ Quantity of milk in the mixture = 5 × 16 = 80 litres - Question 4 of 10
##### 4. Question

1 pointsA butler stole wine from a butt of sherry which contained 40% of spirit. He replaced what he had stolen by wine containing only 16% spirit. The butt was then of 24% strength only. How much of the butt did he steal?

CorrectExplanation:

Wine Containing Wine Containing 40% Spirit 16% Spirit Water Containing

(24% Spirit)8 16 ∴ By allegation rule:

(Wine with 40% spirit)/(Wine with 16% spirit)= 8/16= 1/2

i.e., they must be mixed in the ratio (1 : 2). Thus 1/3 of the butt of sherry was left and hence the butler drew out 2/3 of the butt.IncorrectExplanation:

Wine Containing Wine Containing 40% Spirit 16% Spirit Water Containing

(24% Spirit)8 16 ∴ By allegation rule:

(Wine with 40% spirit)/(Wine with 16% spirit)= 8/16= 1/2

i.e., they must be mixed in the ratio (1 : 2). Thus 1/3 of the butt of sherry was left and hence the butler drew out 2/3 of the butt. - Question 5 of 10
##### 5. Question

1 pointsIn what proportion may three kinds of wheat at Rs. 1.27, Rs. 1.29 and Rs. 1.32 per Kg. be mixed to produce mixture worth Rs. 1.30 per Kg?

CorrectExplanation:

**Method 1:**1 ^{st}Wheat2 ^{nd}Wheat3 ^{rd}WheatMean Price 127 P 129 P 132 P 130 P Here the first two prices are less and the third price is greater than the mean price.

We first find the proportion in which wheat at 127 P and 132 P must be mixed to produce a mixture at 130 P.

(i)1 ^{st}Wheat3 ^{rd}Wheat127 132 130 2 3 The proportion is 2 : 3.

We next find the proportion in which wheat at 129 P and 132 P must be mixed to produce a mixture at 130 P.2 ^{nd}Wheat3 ^{rd}Wheat129 132 130 2 1 The proportion is 2 : 1.

∴ The required proportion is = 2 : 2 : 4 = 1 : 1 : 2

IncorrectExplanation:

**Method 1:**1 ^{st}Wheat2 ^{nd}Wheat3 ^{rd}WheatMean Price 127 P 129 P 132 P 130 P Here the first two prices are less and the third price is greater than the mean price.

We first find the proportion in which wheat at 127 P and 132 P must be mixed to produce a mixture at 130 P.

(i)1 ^{st}Wheat3 ^{rd}Wheat127 132 130 2 3 The proportion is 2 : 3.

We next find the proportion in which wheat at 129 P and 132 P must be mixed to produce a mixture at 130 P.2 ^{nd}Wheat3 ^{rd}Wheat129 132 130 2 1 The proportion is 2 : 1.

∴ The required proportion is = 2 : 2 : 4 = 1 : 1 : 2

- Question 6 of 10
##### 6. Question

1 pointsMilk and water are mixed in a vessel A in the proportion 5 : 2, and in vessel B in the proportion 8 : 5. In what proportion should quantities be taken from the two vessels so as to from a mixture in which milk and water will be in the proportion of 9 : 4?

CorrectExplanation:

In vessel A, milk = 5/7 of the weight of mixture

In vessel B, milk = 8/13 of the weight of mixture. Now we want to from a mixture in which milk will be 9/13 of the weight of this mixture.Applying allegation rule:

5/7 8/13 9/13 1/13 2/91 ∴ Required proportion is = 1/13 : 2/91 = 7 : 2

IncorrectExplanation:

In vessel A, milk = 5/7 of the weight of mixture

In vessel B, milk = 8/13 of the weight of mixture. Now we want to from a mixture in which milk will be 9/13 of the weight of this mixture.Applying allegation rule:

5/7 8/13 9/13 1/13 2/91 ∴ Required proportion is = 1/13 : 2/91 = 7 : 2

- Question 7 of 10
##### 7. Question

1 points300 gm of sugar solution has 40% sugar in it. How much sugar should be added to make it 50% in the solution?

Correct**Method 1:**The existing solution has 40% sugar. And sugar is to mixed; so the other solution has 100% sugar. So by allegation method:

40% 100% 50% 50% 10% ∴ The ratio mixture should be added in the ratio 5 : 1.

Therefore, required sugar = 300/5 ×1 = 60 gms.**Method 2:**Quantity of sugar added = (Solution (Required % value – Present % value))/((100-required % value)) = ([300 ×(50 – 40)])/((100 – 50) ) = 60 gms.

Incorrect**Method 1:**The existing solution has 40% sugar. And sugar is to mixed; so the other solution has 100% sugar. So by allegation method:

40% 100% 50% 50% 10% ∴ The ratio mixture should be added in the ratio 5 : 1.

Therefore, required sugar = 300/5 ×1 = 60 gms.**Method 2:**Quantity of sugar added = (Solution (Required % value – Present % value))/((100-required % value)) = ([300 ×(50 – 40)])/((100 – 50) ) = 60 gms.

- Question 8 of 10
##### 8. Question

1 pointsA person travels 285 Km in 6 hrs in two stages. In the first part of the journey, he travels by bus at the speed of 40 Km per hr. In the second part of the journey, he travels by train at the speed of 55 km per hr. How much distance did he travel by train?

CorrectExplanation:

By applying the allegation method on the speed.

Speed of Bus Speed of Train 40 55 Average Speed

285/645/6 45/6 ∴ Time spent in bus : Time spent in train = 45/6 : 45/6 = 1 : 1

∴ Distance travelled by Train = 285/2 = 142.5 km.IncorrectExplanation:

By applying the allegation method on the speed.

Speed of Bus Speed of Train 40 55 Average Speed

285/645/6 45/6 ∴ Time spent in bus : Time spent in train = 45/6 : 45/6 = 1 : 1

∴ Distance travelled by Train = 285/2 = 142.5 km. - Question 9 of 10
##### 9. Question

1 pointsMira’s expenditure and saving are in the ratio 3 : 2. Her income increase by 10%. Her expenditure also increase by 12%. By how many % does her saving increase?

CorrectExplanation:

Expenditure Saving 12 x % Increase in Expenditure % Increase in Saving 10 % Increase in income 3 2 ∴ Required % increase = 7%

IncorrectExplanation:

Expenditure Saving 12 x % Increase in Expenditure % Increase in Saving 10 % Increase in income 3 2 ∴ Required % increase = 7%

- Question 10 of 10
##### 10. Question

1 pointsA container contained 80 Kg of milk. From this container 8 Kg of milk was taken out and replaced by water. This process was further repeated two times. How much milk is now contained by the container?

CorrectAmount of liquid left after n operation, when the container originally contains x units of liquid form which y units is taken out each time is

=x((x – y)/x)^n units

Thus, in the above case, amount of milk left = =80((80 – 8)/80)^3 Kg. = 58.32 Kg.IncorrectAmount of liquid left after n operation, when the container originally contains x units of liquid form which y units is taken out each time is

=x((x – y)/x)^n units

Thus, in the above case, amount of milk left = =80((80 – 8)/80)^3 Kg. = 58.32 Kg.