## 24th SSC CGL level Solution Set, 3rd on topic Arithmetic Ratio & Proportion

This is the 24th solution set of 10 practice problem exercise for SSC CGL exam and 3rd on topic Arithmetic Ratio & Proportion. Students should complete the corresponding question set in prescribed time first and then only refer to this solution set.

We will repeat here the method of taking the test if you have not gone through it already.

### Method for taking the test and get the best results from the test set:

**Before start,**go through the tutorials**Basic concepts on fractions and decimals part 1**,**, Basic concepts on Ratio and Proportion**,**How to solve problems on mixing liquids and based on ages**or any other short but good material to refresh your concepts if you so require.**Basic and Rich Percentage Concepts for solving difficult SSC CGL problems****Answer the questions**in an undisturbed environment with no interruption, full concentration and alarm set at 12 minutes.**When the time limit of 15 minutes is over,**mark up to which you have answered,**but go on to complete the set.****At the end,**refer to the answers given at the end to mark your score at 15 minutes. For every correct answer add 1 and for every incorrect answer deduct 0.25 (or whatever is the scoring pattern in the coming test). Write your score on top of the answer sheet with date and time.**Identify and analyze**the problems that**you couldn't do**to learn how to solve those problems.**Identify and analyze**the problems that**you solved incorrectly**. Identify the reasons behind the errors. If it is because of**your shortcoming in topic knowledge**improve it by referring to**only that part of concept**from the best source you get hold of. You might google it. If it is because of**your method of answering,**analyze and improve those aspects specifically.**Identify and analyze**the**problems that posed difficulties for you and delayed you**. Analyze and learn how to solve the problems using basic concepts and relevant problem solving strategies and techniques.**Give a gap**before you take a 10 problem practice test again.

Important:bothandpractice testsmust be timed, analyzed, improving actions taken and then repeated. With intelligent method, it is possible to reach highest excellence level in performance.mock tests

If you have not taken the test yet, refer to * SSC CGL level Question Set 24, Arithmetic Ratio & Proportion 3* and then come back to go through the solution.

### 24th solution set- 10 problems for SSC CGL exam: topic Arithmetic Ratio & Proportion - Answering time 15 mins

**Problem 1.**

Divide Rs. 4067 among $A$, $B$ and $C$ in such a way that 4 times the share of $A$, 5 times the share of $B$ and 7 times the share of $C$ are all equal. What is then the share of $A$ in Rs.?

- 1614
- 1715
- 1825
- 1516

**Solution 1 - Problem analysis**

Let us first jot down the given facts.

$A + B + C = 4067$, and

$4A = 5B = 7C$.

So we have three equations and three variables. Solution is possible.

#### Solution 1 - Problem solving execution

The * standard technique of variable elimination is to substitute values of variable from simpler expression to more complex expression.* In this case, from two share equalities we will substitute the values of $B$ and $C$ in terms of $A$ into the equation 1.

$4A = 5B$,

Or. $B = \displaystyle\frac{4}{5}A$.

$4A = 7C$,

Or, $C = \displaystyle\frac{4}{7}A$.

Substituting in first equation,

$A + \displaystyle\frac{4}{5}A + \displaystyle\frac{4}{7}A = 4067$,

Or, $A(35 + 28 + 20) = 35\times{4067}$,

Or, $A = \displaystyle\frac{35\times{4067}}{83}$

Or, $A = 35\times{49} = 1715$

**Answer:** Option b: 1715.

**Key concepts used:** Following best practices of variable elimination from linear equations -- simplification -- delayed evaluation technique -- mostly algebraic.

**Problem 2.**

The ratio of present age of a gentleman and his wife is 5 : 4 and after 4 years it will be 11 : 9. When they were married the ratio was 7 : 5. How long ago were they married?

- 10 years
- 8 years
- 12 years
- 6 years

**Solution 2 - Problem analysis:**

It is a classic ratio problem and we need to introduce the common factor $x$ eliminated between the two present ages in the proper fraction form of the ratio, so that the age of the husband and wife are, $5x$ and $4x$ respectively.

After 4 years age of both will increase by 4 years and so we have,

$\displaystyle\frac{11}{9} = \frac{5x + 4}{4x + 4}$.

Before cross-multiplication, we subtract 1 from both sides to simplify,

$\displaystyle\frac{2}{9} = \frac{x}{4x + 4}$

Or, $9x = 8x + 8$,

So, $x = 8$.

Let us assume they were married $y$ years ago. So,

$\displaystyle\frac{5x - y}{4x - y} = \frac{7}{5}$

Or, $\displaystyle\frac{40 - y}{32 - y} = \frac{7}{5}$

Again subtracting 1 from both sides to simplify,

$\displaystyle\frac{8}{32 - y} = \frac{2}{5}$

Or, $40 = 64 - 2y$,

Or $y =12$.

They were married 12 years ago.

**Answer:** Option c : 12 years.

**Key concepts used:** Reintroducing the canceled common factor $x$ for the ratio of present ages -- expressing in terms of ratio of ages 4 years later to get value of $x$ and the present ages -- again assuming unknown as the desired number of years and going back $y$ years to form the ratio and get the value of $y$.

**Problem 3.**

1% of tea is wasted at the time of mixing of two kinds of tea priced at Rs.600 per kg and Rs. 800 per kg. In what ratio the two kinds of tea are to be mixed so that there will be a profit of 10% on selling mixed tea at the rate of Rs. 700 per kg?

- 3 : 19
- 17 : 3
- 5 : 17
- 17 : 5

**Solution 3 - Problem analysis**

It is obvious that two problems are combined into one. We won't even think of the 1% wastage problem at first. We will put our head down to figure out the familiar mixing problem first.

#### Solution 3 - Problem solving execution

Assume $x$ kg of first type and $1-x$ kg of second type are mixed to form 1 kg of mixed tea.

* Note:* Instead of two variables $x$ and $y$ for two amounts we recognize clearly,

Only one variable is enough and sufficient.

This is because we are dealing with ratios and not actual amounts and in ratios, we can always assume the total as 1, and the mixing amounts of two materials, $x$ and $1-x$. This approach simplifies solving of mixing problems greatly.

From the 10% profit requirement we have,

$SP = CP + 10\text{% of }CP $

$\hspace{6mm}= CP + 0.1CP$

$\hspace{6mm}= 1.1CP$

$\hspace{6mm}=700$,

Or, $CP = \displaystyle\frac{700}{1.1}$.

Again from the costs of the tea mixed we have the * cost price of 1 kg without considering wastage* as,

$CP =600 x + (1 - x)800$.

Now let us try to understand how the 1% wastage affects the cost price of 1 kg of mixed tea.

When we went for mixing 1 kg, because of the wastage we actually got in our hands $0.99$ kg at the cost of $CP =600 x + (1 - x)800$.

Using unitary method then we have,

0.99 kg costs $600x + (1 - x)800$

So, 1 kg will be costing $\displaystyle\frac{600x + (1 - x)800}{0.99}$

This is our actual CP for actual 1 kg for mixed tea arrived at by actual mixing. We will equate this cost now with the cost price of 1 kg arrived at from the 10% profit requirement.

So the relation obtained from the profit requirement turns to,

$\displaystyle\frac{600x + (1 - x)800}{0.99} = \displaystyle\frac{700}{1.1}$,

Or, $\displaystyle\frac{6x + (1 - x)8}{9} = 0.7$,

Or, $2x = 8 - 6.3 = 1.7$,

Or, $x = 0.85$.

So the ratio will be,

$85 : 15 = 17 : 3$.

**Answer:** Option b: $17 : 3$.

**Key concepts used:** By * problem breakdown technique*, without bothering about the wastage of 1% issue, we went on to derive the cost price of 1 kg of mixed tea equating it with the cost price obtained from 10% profit requirement -- getting the cost price for 10% profit requirement is based on

*, but while forming the cost price of 1 kg of mixed tea*

**basic profit and loss concept***, as the problem involves ratios and we are interested in portions of 1 only, this approach simplifies deductions greatly -- now only we visualize what happens due to the wastage, we lose .01 kg, that is, get physically 0.99 kg at the cost of formally arrived 1 kg -- this is an ideal place to apply unitary method to get the actual cost of actual 1 kg of mixed tea -- once we get it we equate it with the cost price obtained from the profit requirement and simplify the equation as quickly and simply as possible.*

**we took care to define only one variable and not two****Problem 4.**

Three persons start a business with capitals in the ratio 3 : 8 : 5. If the first man gets Rs. 600 less than the third as his profit what was the total profit?

- 2400
- 5400
- 4000
- 4800

**Solution 4 - Problem analysis**

The first conclusion: the three persons should get their share of profits in the same ratio as the capitals they have invested, that is, in the ratio of 3 : 8 : 5.

#### Solution 4 - Problem solving execution

In ratios the first action we take is to reintroduce the canceled out HCF $x$ between the terms of the ratio expressing it as $3x : 8x : 5x$ thus converting the ratio terms to actual amounts even if these are unknown at this moment.

Now we consider the given problem statement,

$5x - 3x = 600$,

Or, $x = 300$.

So * total profit* $=3x + 8x + 5x = 16x = 4800$.

**Answer:** Option d: 4800.

**Key concepts used:** Recognizing the basic concept that capitals invested and profits accrued should be in proportion, that is, their ratios should be equal-- the ratio of capitals is thus transformed by this concept to ratio of profits -- reintroduction of canceled out HCF to the ratio terms -- forming a simple linear equation from problem statement to find out the unknown HCF introduced -- final result is only a simple step away.

**Problem 5.**

If a milkman sells his milk at Rs. 30 per litre he incurs losses but if he sells at Rs. 34 per litre he makes a profit. If the loss to gain ratio is 1: 3, his cost price per litre (in Rs.) was,

- 31.5
- 31
- 31.75
- 32

**Solution 5 - Problem analysis**

We understand from the problem statements that we would get two equations from two transactions, but considering that the amount of milk purchased is unknown and the actual amount of gain or loss is also unknown, while we need to find out a third unknown, the cost price $CP$, the problem seems a little dicey at first.

To get over the problem, thinking deeper, we discover a rich concept in the area of profit and loss that we thought will deliver our solution.

#### Rate of profit or loss - rich concept on profit and loss topic area

Usually we deal with total profit or loss by the formulas,

$Profit = SP - CP$, and

$Loss = CP - SP$.

In both these relations, all three components, $SP$, $CP$ and Profit or Loss * can be either in terms of per unit or total.*This concept will hold good in all cases when total profit or loss will be directly proportional to the per unit buying cost and selling cost.

In the competitive test problems that we encounter, this situation holds good.

**Solution 5 - Problem solving execution**

With the awareness of this concept, let us reintroduce the canceled HCF in ratio of loss to gain as $y$ so that actual total loss and gain are $y$ and $3y$ respectively.

Let us also assume that the milkman had purchased $x$ litres of milk.

With these assumptions for the first lossy transaction we have the expression,

Total loss $= y = xCP - 30x$, where $CP$ is per litre buying cost of milk

and by second statement,

Total gain $=3y = 34x - xCP$.

**Note:** It might seem that we could have used $x$ as HCF of loss and gain, but think over, can we move along that path with full confidence?

We will now take the ratio of the two equations eliminating $x$ and $y$ both in the process,

$\displaystyle\frac{1}{3} = \frac{CP -30}{34 - CP}$,

Or, $34 - CP = 3CP - 90$,

Or, $4CP = 124$,

Or $CP = 31$.

**Answer:** Option b: 31.

**Key concept used:** Discovered a * rich concept of per unit profit, loss, CP and SP proportionality *with the intention of using this concept as the breakthrough idea -- while forming the transaction expressions though, noted that being a ratio we can use its

*, but*

**missing HCF concept**

**found***Thus both sides of each equation had two unknowns aside from $CP$.*

**relating the ratio items with number of litres purchased is not required .**But again, taking a ratio eliminated both the unknowns at one stroke leaving only $CP$ in one simple linear equation.

**Note:** In ratio problems this type of cancellation can always be expected. So you need to go ahead and explore. By the way, the $x$ and $y$ values have to be same.

**Problem 6.**

A shopkeeper sells shoes at Rs. 1134 each after giving a discount on the marked price where discount to discounted price ratio was 19 : 81. Had he not given the discount, he would have earned a profit of 40% on the cost price. What is the cost price of each shoe in Rs.?

- 900
- 1100
- 1200
- 1000

**Solution 6 - Problem analysis**

We will use the basic concepts in this topic area of profit and loss,

- Profit percentage is on base of Cost Price, CP
- Discount percentage is on base of marked price, MP
- Sale Price(SP) = Marked Price(MP) - Discount.

#### Solution 6 - Problem solving execution

The very first problem we need to resolve is to get the discount percentage on the marked price, because in profit and loss problems we work with percentages at the base level and not with ratios.

As the ratio of discount to discounted price is 19 : 81, and discount plus discounted price = marked price which is (19 + 81 = 100%), we get the percentage of discount on marked price as the Discount ratio term itself, that is, 19% only.

By the first statement now,

$1134 = \text{Sale price} = MP - 0.19MP = 0.81MP$,

Or, $MP = \displaystyle\frac{1134}{0.81}= \frac{1260}{0.9} = 1400$.

Had he not given the discount the SP would have been the MP itself, that is, Rs. 1400. By the second statement then,

$SP = CP + 0.4CP = 1.4CP =1400$,

So, $CP = Rs. 1000$.

**Answer:** Option d : 1000.

**Key concepts used:** Basic profit, loss and discount concepts.

**Problem 7.**

Diluted milk from three vessels with milk to water ratios 2 : 3, 3 : 5 and 1: 2 are mixed together taking equal volume from each mixture. What will be the milk to water ratio in the final mixture?

- 113 : 247
- 133 : 227
- 147 : 213
- 101 : 259

** Solution 7 - Problem analysis**

As equal volumes from each mixture need to be mixed, we will assume the convenient volume of 1 litre to be taken from each mixture. With this assumption, we will calculate the exact amount of milk in each mixture, add the three values and divide by 3 to nullify the effect of 3 litres of total volume after addition. We will consider only one liquid milk for the deduction because amount of milk in total 1 litre will automatically give use the water in total 1 litre of final mixture.

**Solution 7 - Problem solving execution**

Milk in 1 litre of first mixture - $\displaystyle\frac{2}{5}$ litre.

Milk in 1 litre of second mixture - $\displaystyle\frac{3}{8}$ litre.

Milk in 1 litre of third mixture - $\displaystyle\frac{1}{3}$ litre.

Adding the three fraction volumes, Milk in 3 litres of final mixture of three mixtures,

$\displaystyle\frac{2}{5} + \displaystyle\frac{3}{8} + \displaystyle\frac{1}{3}$

$=\displaystyle\frac{48+45+40}{120}$

$= \displaystyle\frac{133}{120}$.

So milk in 1 litre of new mixture is,

$= \displaystyle\frac{133}{360}$ litre.

The water in 1 litre mixture is then

$= \displaystyle\frac{360 - 133}{360} = \frac{227}{360}$ litres.

So the desired milk to water ratio in the new mixture is,

$=\displaystyle\frac{133}{227}$.

**Answer:** Option b: 133 : 227.

**Key concepts used:** Using milk to total mix as a relationship simplifies the process -- proportion of one liquid per litre of mixture for adding mixtures simplifies the whole proceedings.

**Problem 8.**

The number of students in three classes are in the ratio 2 : 3 : 4. If 12 students are increased in each class, the ratio changes to 8 : 11 : 14. the total number of students in three classes in the beginning was,

- 108
- 54
- 96
- 162

** Solution 8 - Problem analysis**

This is a classic ratio problem and to get hold of the actual number of students, even if in unknown form, we reintroduce the canceled out HCF $x$ (let's assume) between the terms and express the ratio of number students at present as, $2x : 3x : 4x$.

When 12 students joined in each class then, we have,

$(2x + 12) : (3x + 12) : (4x + 12) = 8 : 11 : 14$.

We get two equations from this expression, but we require only one as we have only one unknown variable. Taking the first expression we have,

$(2x + 12) : (3x + 12) = 8 : 11$,

Inverting and subtracting 1 from both sides we simplify,

$\displaystyle\frac{3x + 12}{2x + 12} - 1 = \displaystyle\frac{11}{8} - 1$,

Or, $\displaystyle\frac{x}{2x + 12} = \displaystyle\frac{3}{8}$,

Or, $8x = 6x + 36$,

Or, $x = 18$,

So total number of students in the beginning was,

$2x + 3x + 4x = 9x = 162$.

**Answer:** Option d: 162.

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

**Problem 9.**

In two bags $a$ and $b$ the ratio of number of balls is 2 : 3. Five balls are taken out from bag $b$ and dropped into bag $a$ making the two numbers equal. The number is,

- 45
- 30
- 25
- 20

**Solution 9. **

Assuming the number of balls in the two bags to be $2x$ and $3x$ where $x$ is the canceled out factor, we get,

$3x - 5 = 2x + 5$,

Or, $x = 10$.

So the equal number after exchanging 5 balls is, 25. 20 balls in first bag increased by 5 to 25 and 30 balls from bag 2 decreased by 5 to 25 to become equal to each other.

**Answer:** Option c: 25.

**Key concepts used:** Basic ratio concepts.

** Problem 10.**

Three numbers are in the ratio 2 : 3 : 4. If sum of their squares is 1856, then the numbers are,

- 16, 24, 32
- 8, 12, 16
- 12, 18, 24
- none of these

**Solution 10**

Reintroducing the canceled out common HCF to the ratio terms, we have the actual numbers, $2x$, $3x$ and $4x$.

So by the problem statement we have,

$4x^2 + 9x^2 + 16x^2 = 1856$

Or, $29x^2 = 1856$.

Or, $x^2 = 64$.

So $x=8$, and the numbers are, 16, 24, 32.

**Answer:** Option a : 16, 24, 32.

**Key concepts used:** Basic ratio concepts.

**Resources that should be useful for you**

* 7 steps for sure success in SSC CGL tier 1 and tier 2 competitive tests* or

*to access all the valuable student resources that we have created specifically for SSC CGL, but*

**section on SSC CGL****generally for any hard MCQ test.**

**Concept Tutorials on related topics**

**Basic concepts on fractions and decimals part 1**

**Basic concepts on Ratio and Proportion**

**Componendo dividendo applied on number system and ratio proportion problems**

**How to solve problems on mixing liquids and based on ages**

**Basic and Rich Percentage Concepts for solving difficult SSC CGL problems**

#### Efficient solution techniques on related topics

**How to solve SSC CGL level arithmetic mixture problems in a few simple steps 1**

**How to solve SSC CGL level arithmetic mixture problems in a few simple steps 2**

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**SSC CGL level solved question set 85 on mixture or alligation 2**

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**SSC CHSL level Solved Question set 10 on Mixture or Alligation 2**

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**SSC CGL Tier II level solved question set 25 on mixture or alligation 2**

**SSC CGL Tier II level question and solution sets on Ratio and Proportion**

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