SSC CGL level Solution Set 5, Arithmetic Ratio & Proportion

Fifth SSC CGL level Solution Set, topic Arithmetic Ratio & Proportion

SSC CGL Solution set5 arithmetic ratio and proportion

This is the fifth solution set of 10 practice problem exercise for SSC CGL exam on topic Arithmetic Ratio & Proportion. Students must complete the corresponding question set in prescribed time first and then only refer to the solution set.

It is emphasized here that answering in MCQ test is not at all the same as answering in a school test where you need to derive the solution in perfectly elaborated steps.

In MCQ test instead, you need basically to deduce the answer in shortest possible time and select the right choice. None will ask you about what steps you followed.

Based on our analysis and experience we have seen that, for accurate and quick answering, the student

  • must have complete understanding of the basic concepts of the topics
  • is adequately fast in mental math calculation
  • should try to solve each problem using the most basic concepts in the specific topic area and
  • does most of the deductive reasoning and calculation in his head rather than on paper.

Actual problem solving happens in items 3 and 4 above. But how to do that?

You need to use your problem solving abilities only. There is no other recourse. Your problem solving skill can be significantly increased by intelligent and systematic practice.

If you have not taken the test yet, refer to SSC CGL level Question Set 5, Arithmetic Ratio & Proportion 2 and then come back to go through the solution.

Fifth solution set- 10 problems for SSC CGL exam: topic Arithmetic Ratio & Proportion - time 12 mins

Q1. Ratio of ages of A and B is 2 : 3 and between B and C is 4 : 5, while the total age of C and A is 46 years. Age of B in years is,

  1. 16
  2. 48
  3. 24
  4. 54

Solution: By analyzing the problem goal we observe that to get B, we need to know the value of either A or C, as in the given ratios B is related to one of A or C.

This can be done if we get a second relation between A and C, one relation, A + C = 46 being already available.

We know, two linear equations in two variables would be sufficient to get values of the two unknown variables.

So the primary requirement boils down to finding a new relation between A and C.

This can be done by joining the two given ratios and as B is common to both the ratios, we transform the ratios in such a way that B's term values in each ratio equals LCM of its two values, that is, 12. So,

$\displaystyle\frac{A}{B} = \displaystyle\frac{2}{3} = \displaystyle\frac{8}{12}$, and

$\displaystyle\frac{B}{C} = \displaystyle\frac{4}{5} = \displaystyle\frac{12}{15}$.

From these two relations we get the three quantity ratio as,

$ A : B : C = 8 : 12 : 15$, and between A and C as,

$\displaystyle\frac{A}{C} = \displaystyle\frac{8}{15}$

Or,$\displaystyle\frac{A + C}{C} = \displaystyle\frac{23}{15}$

Or, $C = \displaystyle\frac{46\times{15}}{23} = 30$, and $B = 30\times{\displaystyle\frac{4}{5}} = 24$.

Answer: Option c: 24.

Key concepts used: Ratio joining technique, Joining two ratios by base equalization -- using new relation between A and C obtained by joining two ratios and the given relation A + C = 46 -- solving for either of the term values finally gets us to the value of B using appropriate given ratio relation.

Q2. Instead of dividing 117 among A, B, and C in the ratio of $\displaystyle\frac{1}{2} : \displaystyle\frac{1}{3} : \displaystyle\frac{1}{4}$, it was divided by mistake in the ratio of 2 : 3 : 4. Who gains most and by how much because of the mistake?

  1. B, 3
  2. A, 28
  3. C, 20
  4. C, 25

Solution: Change from lowest fractional share in the first case to highest whole integer share in second case would result in maximum gain for C.

Assuming $x$ to be the common 1 portion of each share, in the first case we get the total expressed as,

$\frac{1}{2}x + \frac{1}{3}x + \frac{1}{4}x = \frac{x}{12}(6 + 4 + 3) = \frac{13}{12}x = 117$, or $x=108$. So C got $\frac{1}{4}x = 27$ in this case.

In the mistaken scenario, assuming $y$ to be one common portion simlarly we have the total as,

$2y + 3y + 4y = 9y = 117$, or, $y=13$.

So, C = 52, and thus gain of $C = 52 - 27 = 25.$

Answer: Option d : C, 25 .

Key concepts used: Identifying the pair of values with highest difference -- finding the before mistake and after mistake values by using cancelled out HCF reintroduction technique in both cases.

Q3. After replacing an old member by a new member it was found that the average age of five members of a club remained same as it was three years back. The difference between the ages of two members involved in replacement is,

  1. 10 years
  2. 2 years
  3. 15 years
  4. 3 years

Solution: Three years ago average age of the members was three years less, as each member was younger by 3 years. That means for five members, going back by a total of 15 years. As this average and total is same as the average and total now after replacement, the replaced member must have compenseated for this 15 years excess by being younger than the replaced member by the same number of years.

Answer: Option c: 15 years.

Key concepts used: Clear problem definition and problem modeling -- average age of a group of persons increases by $n$ number of years after $n$ years, and thus total age increases by $n$ times number of persons in the group -- Basic average concept -- Average compensation for shortfall concept.

Q4. A container has in it a mixture of two liquids A and B in the ratio 7 : 5. When 9 litres of mixture is replaced by 9 litres of liquid B, the ratio of A and B changes to 7 : 9. How many litres of liquid A was there to start with?

  1. 25
  2. 10
  3. 20
  4. 21

Solution: First case,

In 1 litre of mixture, A was $\frac{7}{12}$ litres

So in $x$ litres of mixture, A was $\frac{7x}{12}$ litres, assuming total volume of the mixture as $x$ litres.

And in 9 litres of mixture, A was $\frac{63}{12}$ litres.

Similarly in the second case after replacement,

In 1 litre of mixture, A is $\frac{7}{16}$ litre and in $x$ litre of mixture, $\frac{7x}{16}$ litre.

The reduction in A by the replacement of the mixture by only B wasn't made up in any way by the replacement, and so,

$\frac{7x}{12} - \frac{63}{12} = \frac{7x}{16}$,

Or,$\frac{7x}{12} - \frac{7x}{16} = \frac{63}{12}$,

Or, $\frac{7x}{48} = \frac{63}{12}$,

Or, $x = 36$.

So volume of A before replacement is $A = \frac{7x}{12} = 21$ litres.

Answer: Option d: 21.

Key concepts used: Basic ratio concepts -- Concentration in mixtures technique -- Use of the ratio concept of Proportion of a component and the total of all components selecting the component as liquid A. This approach eliminates the complexity in dealing with other components and focuses on only the component of interest -- unitary method and assumption of $x$ as the total in both cases -- the basic concept of proportion of component A in the 9 litre volume mixture that was replaced -- context awareness, that is, exactly what happened to the quantities because of the activities.

Q5. Five members in a team were weighed one by one and after each weighing the average weight increased by 1 kg. How much is the difference in weights in kg of the first and last player?

  1. 5
  2. 8
  3. 4
  4. 20

Solution: Let $x$ kg be the weight of the first member. So after weighing the first member, the total is $x$ and average is $x$.

In second weighing, average is $x + 1$ and number of players 2, So total is $2x + 2$ and weight of second player, $x + 2$. Notice how we start with finding the average first (this increases at each step by 1 as specified and thus is known) and then deduce all other values to get finally the weight of the player added at this step.

In third weighing, average is $x + 2$, number of players 3, total weight $3x + 6$ and weight of the player added in this third step is $x + 4$.

The weights of the players then follow an AP series, $x$, $x + 2$, $x + 4$, $x+6$, and $x+8$. The fifth term of this series will then be $x+8$. So the difference in weight between the first and the last player will be, $(x+8) -x=8$ kg.

Answer: Option b: 8.

Key concept used: Charting out exactly what happens with the entities at each step and detecting the progressive pattern in the entity of interest, that is, the weight of the player introduced at each step -- detecting AP series -- finding 5th term of series.

Q6. The students in three classes are in the ratio of 4 : 6 : 9. If 12 students are increased in each class, the ratio changes to 7 : 9 : 12. Then the total number of students in the three classes before the increase was,

  1. 100
  2. 76
  3. 95
  4. 114

Solution: The first ratio represents 4 portions of students of class A : 6 portions of class B : 9 portions of class C. The number of students in the three classes are such that this relationship holds. The portion here represents the HCF between the three numbers that was canceled out in forming the minimized ratio terms. Let's assume $x$ is equal to one portion in this case, that is $x$ is the HCF of the three numbers. This is HCF reintroduction technique.

Likewise let's assume $y$ to be equal to 1 portion of the changed total number of students that has been canceled out between the terms of the second ratio.

As $4x + 12 = 7y$ and $6x + 12 = 9y$, subtracting one from the other we get, $2x = 2y$, or, $x = y$. In other words, in both the ratios, 1 portion equals $x$.

From the first relation, then we have $3x = 12$, or, $x = 4$, or, required total number before increase = $4x + 6x + 9x = 19x = 76$.

Answer: Option b : 76.

Key concepts used: Basic ratio concepts -- HCF between the terms of a ratio canceled out represents one common portion -- HCF reintroduction technique, by introducing the HCF in the proceedings, we can get to actual numbers.

Q7. Two vessels contain milk and water in the ratio 3 : 2 and 7 : 3. Find the ratio in which the contents of the two vessels have to be mixed to get a new mixture in which the ratio of milk and water is 2 : 1.

  1. 1 : 2
  2. 2 : 1
  3. 1 : 4
  4. 4 : 1

Solution: In first vessel, 1 litre of mix contains, $\frac{3}{5}$ litres of milk.

So assuming $x$ litres of this diluted milk is used for the desired mixing,

$x$ litres of first diluted milk contains $\frac{3x}{5}$ litres of milk.

Assuming that $y$ litres of second diluted milk has been used in desired mixture, similarly we have,

$y$ litres of second diluted milk contains, $\frac{7y}{10}$ litres of milk.

So $x + y$ litres of new third type of diluted milk contains, $\frac {3x}{5} + \frac{7y}{10}$ litres of milk.

But by the problem statement, 1 litre of this third type of diluted milk contains $\frac{2}{3}$ litres of milk, that is, $x + y$ litres contains, $\frac{2}{3}(x + y)$ litre of milk.

Equating the two we get,

$\frac{3x}{5} + \frac{7y}{10} = \frac{2}{3}(x + y)$

Or, $\frac{9x}{5} + \frac{21y}{10} = 2x + 2y$

Or, $\frac{1}{5}x = \frac{1}{10}y$

Or, $x : y = 1 : 2$

Answer: Option a: 1 : 2.

Key concepts used: Using milk to total mix as a relationship simplifies the process -- milk in 1 lite of mixture is considered, Concentration in mixtures concept -- unitary method.

Q8. A person distributes his pens among four friends A, B, C and D in the ratio, $\frac{1}{3} : \frac{1}{4} : \frac{1}{5} : \frac{1}{6}$. What is the minimum number of pens that the person should have?

  1. 65
  2. 57
  3. 45
  4. 75

Solution: The terms of a ratio series usually are expressed in integer form for ease of manipulation and understanding. To transform the given ratio to integer form, we need to multiply each term by the LCM of the denominators of the fractions which is 60. The transformed ratio is then,

$20 : 15 : 12 : 10$.

Assuming $x$ as the HCF that has been canceled out, we get the total number of pens as, $20x + 15x + 12x + 10x=57x$. The minimum value of this total would be 57 when $x=1$.

Answer: Option b: 57.

Key concepts used: Basic ratio concepts -- HCF reintroduction technique, reintroduction of canceled out HCF as a factor of each term -- total of ratio terms -- minimization concept.

Q9. Sum of three numbers is 68. If the ratio of the first to the second is 2 : 3 and the second to the third is 5 : 3, then the second number is,

  1. 58
  2. 30
  3. 20
  4. 48

Solution: A relation between the three numbers is given as a total of the numbers. This urges us the get a second relation between the three by joining the two ratios. To join the two ratios, we need to equalize the common base term values of the second number. Transforming the two ratios thus we get,

2 : 3 = 10 : 15 and 5 : 3 = 15 : 9, giving the ratio of the three numbers as 10 : 15 : 9.

Let's assume the common factor canceled out between the term values as $x$. Thus,

$10x + 15x + 9x = 34x = 68$

Or, $x = 2$

Or, the second number = $15x = 30$.

Answer: Option b: 30.

Key concepts used: Ratio joining technique, Joining of two ratios by base equalization -- HCF reintroduction technique -- basic ratio total concept.

Q10. A milk seller wants to make a profit of $5{\%}$ by selling his milk diluted with water at the cost price of milk he purchased. What should be the ratio in which he need to mix water to his purchased milk?

  1. 1 : 2
  2. 2 : 5
  3. 1 : 20
  4. 1 : 4

Solution: To make a profit of $5{\%}$, the milk seller must sell at Rs. 105 milk that he purchased at Rs. 100. Let's assume that the amount of milk he purchased at Rs. 100 is $100x$ litre. To keep the selling price officially equal to the cost price, he has to sell $105x$ litres of diluted milk at Rs. 105.

As this $105x$ litre of diluted milk contains $100x$ litres of milk, he need to mix $5x$ litres of water to each $100x$ litres of milk, that is, in a ratio of 1 : 20.

Answer: Option c : 1 : 20.

Key concepts used: Basic percentage concepts -- basic mixing concepts.


Resources that should be useful for you

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