Explore Exams
Login Signup
Sign in Open App
All Exams  >   SSC MTS / SSC GD  >   Numerical Aptitude for SSC Exams  >   All Questions

All questions of Mixtures & Alligations for SSC MTS / SSC GD Exam

Clear all your doubts with EduRev

A Jar contains 200 liters of Milk a thief stole ‘X’ liters of Milk and replaced it with water. Next, he stole 40 liters of Milk and replaced it with water. Again he stole 50 liters of Milk and replaced with water. If the quantity of water in the final mixture is 92 liters. Then what is the value of X?
  • a)
    10 Liter
  • b)
    15 Liter
  • c)
    20 Liter
  • d)
    30 Liter
  • e)
    50 Liter
Correct answer is option 'C'. Can you explain this answer?

Aarav Sharma answered
50 liters of milk from the jar. How much milk is left in the jar?

There are two ways to approach this problem:

Method 1: Subtraction
To find the amount of milk left in the jar, we can simply subtract the amount stolen from the total amount:
200 - 50 = 150 liters
Therefore, there are 150 liters of milk left in the jar.

Method 2: Proportion
We can also use proportions to find the amount of milk left in the jar. Let x be the amount of milk left in the jar. Then we can set up a proportion:
200/100 = x/50
This proportion states that the initial amount of milk (200 liters) is 100% of the jar, and the stolen amount (50 liters) is 50% of the jar. We can solve for x by cross-multiplying:
200 * 50 = 100 * x
10,000 = 100x
x = 100 liters
Therefore, there are 100 liters of milk left in the jar.

How many kgs of rice of variety-1 costing Rs.42/kg should a shopkeeper mix with 25 kgs of rice of variety-2 costing Rs.24 per kg so that he makes a profit of 25% on selling the mixture at Rs.40/kg?
  • a)
    15
  • b)
    20
  • c)
    25
  • d)
    30
  • e)
    None of these
Correct answer is option 'B'. Can you explain this answer?

Bank Exams India answered
Answer – B.20 Explanation : 25% profit by selling the mixture at Rs.40/kg, his cost per kg of the mixture = Rs.32/kg. (x×42)+(25×24)=32(x+25) 42x+600=32x+800 10x=200 x=20kgs.
Shortcut: 125 = 40
100 = 32
42……………….24
………32…………..
8…………………….10
10 = 25
8 = 20

 A vessel is filled with liquid, 4 parts of which are water and 5 parts syrup. How much of the mixture must be drawn off and replaced with water so that the mixture may be half water and half syrup?
  • a)
    1/9
  • b)
    1/10
  • c)
    1/5
  • d)
    1/12
  • e)
    1/7
Correct answer is option 'B'. Can you explain this answer?

Divey Sethi answered
Answer – B.1/10 Explanation : Suppose the vessel initially contains 9litres of liquid.
Let x litres of this liquid be replaced with water. (4-(4x/9) +x) = (5-(5x/9)) X = 9/10
So, part of the mixture replaced = 9/10 * 1/9 = 1/10 Shortcut: 4:5—->9
1:1—->2
8:10
9:9
1/10

A mixture of wheat is sold at Rs.3 per Kg. This mixture is formed by mixing the Wheat of Rs.2.10 per kg and Rs.2.52 per kg. What is the ratio of price of cheaper to the costlier quality in the mixture if the profit of 25% is being earned
  • a)
    2:5
  • b)
    2:1
  • c)
    2:3
  • d)
    2:7
  • e)
    2:9
Correct answer is option 'A'. Can you explain this answer?

Alok Verma answered
Solution:
  1. Calculate the cost price of the mixture:
    • Since there's a 25% profit on the selling price of Rs. 3/kg, the cost price = 3 / (1 + 25/100) = Rs. 2.4/kg.
  2. Apply the alligation rule:
    • The alligation rule states that the ratio of quantities of two ingredients in a mixture is inversely proportional to the difference of their prices from the mean price.
    • Mean price (cost price of the mixture) = Rs. 2.4/kg
    • Price of cheaper wheat = Rs. 2.10/kg
    • Price of costlier wheat = Rs. 2.52/kg
    • Difference between cheaper wheat and mean price = 2.40 - 2.10 = Rs. 0.30/kg
    • Difference between costlier wheat and mean price = 2.52 - 2.40 = Rs. 0.12/kg
    • Ratio of cheaper to costlier wheat = 0.12 : 0.30 = 2 : 5
Therefore, the ratio of the price of cheaper to the costlier quality in the mixture is 2:5.
So, the correct answer is option A: 2:5.

Vikram covered 180 km distance in 10 hours. The first part of his journey he covered by Car, then he hired a Rickshaw. The speed of the car and rickshaw is 25 kmph and 15 kmph respectively. The ratio of the distances covered by the car and the rickshaw is
  • a)
    7:9
  • b)
    7:3
  • c)
    7:5
  • d)
    7:2
  • e)
    7:4
Correct answer is option 'C'. Can you explain this answer?

Харе Кришна answered
let the time taken by car = x 
then time taken by rickshaw would be = 10-x
Now,
distance = speed * time
180 = 25x + 15(10-x)
x=3 = time taken by the car.
7 = time taken by rickshaw.
Ratio of distance covered by Car/ rickshaw = (25*3)/(15*7) = 5/7 = 5:7
But there is no such option. 

When one litre of water is added to a mixture of milk and water, the new mixture contains 25% of milk. When one litre of milk is added to the new mixture, then the resulting mixture contains 40% milk. What is the percentage of milk in the original mixture?
  • a)
    100/6 %
  • b)
    50/6 %
  • c)
    100/3 %
  • d)
    50/3 %
  • e)
    None of the Above
Correct answer is option 'C'. Can you explain this answer?

Ravi Singh answered
Answer – C. 100/3 % Explanation : Original Mixture = x L In (x + 1) Mixture, quantity of milk = (x + 1)* (25/100) = (x + 1)/4 one litre of milk is added to the new mixture [((x + 1)/4 )+ 1 ]/ x + 2 = 40% x = 3 ; quantity of milk = (3 + 1)/4 = 1L percentage of milk in the original mixture = 1/3 * 100 = 100/3 %

 A vessel which contains a mixture of acid and water in ratio 13:4. 25.5 litres of mixture is taken out from the vessel and 2.5 litres of pure water and 5 litres of acid is added to the mixture. If resultant mixture contains 25% water, what was the initial quantity of mixture in the vessel before the replacement in litres?
  • a)
    58 litre
  • b)
    68 litre
  • c)
    78 litre
  • d)
    48 litre
  • e)
    None of the Above
Correct answer is option 'B'. Can you explain this answer?

Kavya Saxena answered
Answer – B. 68 litre Explanation : Quantity of Acid = 13x Quantity of water = 4x Total = 17x Resultant Mixture = 17x – 25.5 + 2.5 + 5 = 17x – 18 Resultant water = 4x – 25.5 * (4/17) + 2.5 = 4x – 3.5 Resultant mixture contains 25% water (17x – 18)*25/100 = 4x – 3.5 x = 4 Initial quantity = 17*4 = 68

18 litres of Petrol was added to a vessel containing 80 litres of Kerosene. 49 litres of the resultant mixture was taken out and some more quantity of petrol and kerosene was added to the vessel in the ratio 2:1. If the respective ratio of kerosene and petrol in the vessel was 4:1, what was the quantity of kerosene added in the vessel?
  • a)
    1 litre
  • b)
    2 litre
  • c)
    5 litre
  • d)
    3 litre
  • e)
    None of the Above
Correct answer is option 'E'. Can you explain this answer?

Aarav Sharma answered
Given:
Initial quantity of petrol = 18 litres
Initial quantity of kerosene = 80 litres
Quantity of mixture taken out = 49 litres

Let's solve this problem step by step.

Step 1: Finding the quantity of petrol and kerosene in the vessel after adding 18 litres of petrol
Total quantity of petrol = Initial quantity of petrol + Added petrol = 18 litres + 18 litres = 36 litres
Total quantity of kerosene = Initial quantity of kerosene = 80 litres

Step 2: Finding the quantity of petrol and kerosene in the mixture taken out
Let's assume x litres of petrol and y litres of kerosene were present in the mixture taken out.
According to the given information, the ratio of kerosene and petrol in the mixture taken out is 4:1.
So, we have the equation:
y/x = 4/1

Also, the total quantity of mixture taken out is 49 litres.
So, we have another equation:
x + y = 49

Solving these two equations, we get:
x = 7 litres
y = 28 litres

Step 3: Finding the quantity of petrol and kerosene in the vessel after adding more petrol and kerosene in the ratio 2:1
Let's assume a litres of petrol and b litres of kerosene were added to the vessel.
According to the given information, the ratio of kerosene and petrol added is 2:1.
So, we have the equation:
b/a = 2/1

Also, the total quantity of petrol and kerosene in the vessel after adding is:
Total quantity of petrol = Initial quantity of petrol + Added petrol = 36 litres + a litres
Total quantity of kerosene = Initial quantity of kerosene + Added kerosene = 80 litres + b litres

Step 4: Finding the quantity of kerosene added in the vessel
Now, we need to find the value of b (quantity of kerosene added to the vessel).
To find b, we can use the equation y/x = 4/1 from Step 2.
Substituting the values of y and x from Step 2, we get:
28/a = 4/1
Cross-multiplying, we get:
4a = 28
a = 7

Substituting the value of a in the equation b/a = 2/1 from Step 3, we get:
b/7 = 2/1
Cross-multiplying, we get:
b = 14

Therefore, the quantity of kerosene added in the vessel is 14 litres.

Hence, the correct answer is option E) None of the above.

A jar was full with Milk. A person used to draw out 20% of the Milk from the jar and replaced it with water. He has repeated the same process 4 times and thus there was only 512 gm of milk left in the jar, the rest part of the jar was filled with the water. The initial amount of milk in the jar was: 
  • a)
    1.50 kg
  • b)
    1.30 kg
  • c)
    1.40 kg
  • d)
    1.25 kg
  • e)
    1.75 kg
Correct answer is option 'D'. Can you explain this answer?

Aarav Sharma answered
Problem: A jar was full with Milk. A person used to draw out 20% of the Milk from the jar and replaced it with water. He has repeated the same process 4 times and thus there was only 512 gm of milk left in the jar, the rest part of the jar was filled with the water. The initial amount of milk in the jar was:

Solution:
Let's assume the initial amount of milk in the jar was x kg.

After the first process, the amount of milk left in the jar will be:
x - (20% of x) + (20% of x) of water = 0.8x + 0.2(0.8x)

After the second process, the amount of milk left in the jar will be:
0.8x + 0.2(0.8x) - (20% of 0.8x) + (20% of 0.8x) of water = 0.64x + 0.36(0.8x)

Similarly, after the third process, the amount of milk left in the jar will be:
0.64x + 0.36(0.8x) - (20% of 0.64x) + (20% of 0.64x) of water = 0.512x + 0.488(0.64x)

After the fourth process, the amount of milk left in the jar will be:
0.512x + 0.488(0.64x) - (20% of 0.512x) + (20% of 0.512x) of water = 0.4096x + 0.5904(0.512x)

We know that there was only 512 gm of milk left in the jar after the fourth process, so we can equate the above equation to 0.512 kg and solve for x.

0.4096x + 0.5904(0.512x) = 0.512
0.4096x + 0.3022x = 0.512
0.7118x = 0.512
x = 0.72 kg or 720 gm

Therefore, the initial amount of milk in the jar was 1.25 kg (since 1 kg = 1000 gm). Answer: (d)

Chapter doubts & questions for Mixtures & Alligations - Numerical Aptitude for SSC Exams 2025 is part of SSC MTS / SSC GD exam preparation. The chapters have been prepared according to the SSC MTS / SSC GD exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for SSC MTS / SSC GD 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.

Chapter doubts & questions of Mixtures & Alligations - Numerical Aptitude for SSC Exams in English & Hindi are available as part of SSC MTS / SSC GD exam. Download more important topics, notes, lectures and mock test series for SSC MTS / SSC GD Exam by signing up for free.

Numerical Aptitude for SSC Exams

172 videos|87 docs|127 tests

Top Courses SSC MTS / SSC GD

© EduRev
Education Revolution
Signup to see your scores go up
within 7 days!
Takes less than 10 seconds to signup