SSC CGL level Solution Set 4, Arithmetic Ratio & Proportion

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

SSC CGL Solution set4 arithmetic ratio and proportion

This is the Fourth solution set of 10 practice problem exercise for SSC CGL exam and first 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 related test yet, you may refer to SSC CGL level Question Set 4, Arithmetic Ratio & Proportion, take the test and come back for this solution for maximum learning.

Fourth solution set- 10 problems for SSC CGL exam: topic Arithmetic Ratio & Proportion - time 20 mins

Q1. Ratio of incomes of A and B as well as B and C are 3 : 2. If one-third of A's income exceeds one-fourth of C's income by Rs.1000, then B's income in Rs. is,

  1. 2500
  2. 3000
  3. 4000
  4. 3500

Solution: Let's put down the last statement first, in the form of an expression,

$\frac{1}{3}A - \frac{1}{4}C = 1000$

As we need to explore the relation between A and C, let's transform the two ratios to equalize base B to its LCM value 6, giving, $A : B = 9 : 6$ and $B : C = 6 : 4$, so that, joining the two ratios, we have $A : B : C = 9 : 6 : 4$. Eliminating B, $A : C = 9 : 4$ giving, $4A = 9C$.

Now we take up the target expression again,

$\frac{1}{3}A - \frac{1}{4}C = 1000$, or, $4A - 3C = 12000$, Or, $6C = 12000$ or, $C=2000$.

As $B : C = 3 : 2$, $B = 3000$.

Answer: Option b: 3000.

Key concepts used: Ratio joining, Joining two ratios of by base equalization, it is a Rich ratio concept -- Basic ratio concepts -- picking up useful relation between the two terms that are used in the last given expression -- solving for either of the term values finally gets us to the value of B using appropriate given ratio relation.

Q2. If the cost of pins reduces by Rs. 4 per dozen, 12 more pins can be purchased for Rs. 48. After reduction, the cost of pins per dozen is,

  1. Rs. 12
  2. Rs. 8
  3. Rs. 20
  4. Rs. 16

Solution: As all rates and purchases are by the dozens, let's assume dozen to be a new unit (12 may complicate things). 12 more pins purchased after reduction of price is also a dozen.

So we restate the problem as: due to a reduction in price by Rs. 4 per unit we could purchase 1 more unit with Rs. 48. Note that in both cases, before or after reduction of prices, the total amount being used for purchases has been a fixed amount of Rs. 48.

Whatever be the price per unit, as we are getting whole number of units, the price per unit must be a factor of 48. By this logic, choice value 20 is eliminated.

Testing further with the choice value of starting price as 12, we find it buys 4 units. The ending price would then be 8 buying 6 units - a violation of given condition (1 more unit not 2).

Starting price 16 and ending price 12 leads us to the desired answer quickly.

A sense that the starting price, a factor of 48, should be large enough so that a reduction of Rs. 4 results in only 1 more unit's worth, helps here. Smaller the value of starting price, say 8, the difference in number of purchased units would be larger (12 - 6 = 6).

Here we have used the choice values as a useful problem solving resource.

A more mathematical way to the solution is to assume $x$ as the items purchased now. So by the problem statement we get the expression of prices as,

$\frac{48}{x} - 4 = \frac{48}{x + 1}$

Or, $\frac{12}{x} - 1 = \frac{12}{x + 1}$,

Or, $(12 - x)(x + 1) = 12x$

Or, $x^2 + x - 12 = (x + 4)(x - 3) = 0$, giving positive root $x = 3$, old price Rs. 16 and new price Rs. 12.

Answer: Option a : 12 .

Key concepts used: Simplifying problem statement -- simple problem modelling using context details accurately and unitary method concept.

Q3. A dishonest grocer sells rice at a profit of 10%, but he uses weights 20% less than the marked. His total gain will be,

  1. 30.5%
  2. 37.5%
  3. 40%
  4. 35%

Solution: Officially the dishonest grocer earns a profit of 10%. It means, if Rs. 100 per kg were the cost price, he sells 1 kg rice at Rs. 110. But as he uses spurious weight 20% less than marked weight, he actually sells 0.8 kg rice at the price of Rs. 110.

Effectively thus he sells a quantity of rice purchased at a cost of Rs. 80 at Rs. 110. The total gain he accrues is then Rs. 30 on Rs. 80, his cost price. The % gain is,

$\frac{3}{8}100$% = $37.5$%

Answer: Option b: 37.5%.

Key concepts used: Clear problem definition and problem modeling -- assuming the cost price as Rs. 100 on which the first profit of 10% is mentioned -- all subsequent actions can then be kept track of with respect to this reference starting point -- context awareness about weight reduction using spurious weight.

Q4. The smallest integer which when subtracted from both the terms of 6 : 7 results in a ratio less than 16 : 21 is,

  1. 5
  2. 4
  3. 3
  4. 2

Solution: Before anything we must get an idea of how much the first given fraction 6 : 7 is to reduce to be less than the second fraction 16 : 21. Transforming 6 : 7 by equalizing the base we get, 18 : 21. That is, it needs to be reduced by a very small amount (in comparison to its own value now) to push it below the valued of 16 : 21.

Now we look at the problem definition and find it wants the smallest integer needed to reduce both the numerator and numerator so that the desired state is reached with the changed fraction.

Here we should take recourse to our knowledge of Principle of relative increment (or decrement). It says, in case of fractions, if the two terms are reduced by same amount, the fraction itself will be reduced and will be reduced more as the term reduction is made larger if, the numerator is less than the denominator. Considering this, we decide that if we use choice value 5 we would go far below 16 : 21 (it will be 1 : 2 or 21 : 42 in comparison to 32 : 42).

In other words, this concept urges us to use the lowest choice value 2 to get,

$\frac{6}{7} \Rightarrow \frac{(6 - 2)}{7 - 2}= \frac{4}{5}= \frac{16}{20}$ still larger than $\frac{16}{21}$ (numerator equalization technique).

Taking the next choice value 3,

$\frac{6}{7} \Rightarrow \frac{(6 - 3)}{7 - 3}= \frac{3}{4}= \frac{63}{84}$, whereas $\frac{16}{21} = \frac{64}{84}$. We have just been able to reduce 6 : 7 to a value lesser than 16 : 21.

By our earlier knowledge we know that if we reduce the two terms further, the fraction will be reduced further. So we stop here with target value 3.

Answer: Option c: 3.

Key concepts used: Use of principle of relative increment -- fraction comparison -- Principle of free resource use, use of choice values as problem solving resource.

Q5. In a vessel A, milk and water are mixed in ratio 8 : 5 and in a second vessel B, in a ratio 5 : 2. In what ratio these two mixtures are to be mixed to get a mixture containing $69\frac{3}{13}{\%}$ milk?

  1. 5 : 3
  2. 2 : 7
  3. 3 : 5
  4. 5 : 7

Solution: Let us first convert the target percentage of milk to ratio of milk.

$69\frac{3}{13}{\%} = \frac{69\times{13} + 3}{13\times{100}} = \frac{900}{13\times{100}}=\frac{9}{13}$

This represents milk to total mixture volume ratio as $\frac{9}{13}$.

Let's assume $x$ litres of mixture in vessel A is mixed with $y$ litres of mixture in vessel B to get a mixture C with this target ratio 9 : 13 of milk to total mixture.

Now we will consider only proportion of milk in total mixture without thinking of water at all because, water proportion will be included or embedded in milk to total mixture proportion and by this approach we will simplify the whole process of reasoning significantly.

In 1 litre of mixture A milk is $\frac{8}{13}$ litre (as milk : water is 8 : 5, total is 13 and milk : total becomes 8 : 13).

In $x$ litre of mixture A milk is $\frac{8x}{13}$ litre.

Similarly in $y$ litre of mixture B, milk is $\frac{5y}{7}$ litre.

In $x + y$ litre of the new mixture C then, milk is $\frac{8x}{13} + \frac{5y}{7}$ litre.

Taking up the new mixture C with milk to total mixture volume ratio as 9 : 13 we get,

1 litre C contains $\frac{9}{13}$ litres of milk.

$x + y$ litres of C then contains $\frac{9}{13}(x + y)$ litres of milk.

So equating the two results,

$\frac{8x}{13} + \frac{5y}{7} = \frac{9}{13}(x + y) = \frac{9x}{13} + \frac{9y}{13}$

Or, $\frac{x}{13} = y(\frac{5}{7} - \frac{9}{13})$

Or, $\frac{x}{y} = 13(\frac{5}{7} - \frac{9}{13})=\frac{65}{7} - 9= \frac{2}{7}$.

Answer: Option b: 2 : 7.

Key concepts used: Use of very basic concepts of ratio -- identifying the target ratio as x : y -- considering only milk in total mixture thus simplifying deduction -- Concentration in mixtures concept, Considering amount of milk in each 1 litre of mixture -- unitary method.

Advantage of this approach of using only the very basic concepts of ratio instead of any formula is faster and surer way reach to the solution with greater accuracy.

Less memory load of remembering advanced formula or method relieves the mind from chance of forgetting the formula as well as enables the problem solver to reach the solution quickly because of use of more basic and frequently used simpler relations. This is called Less facts more procedures approach.

Q6. Rs. 180 are to be divided among 66 men and women. Men and women receive total amount in the ratio 5 : 4, but individually men and women receive money in the ratio 3 : 2. The number of women is,

  1. 42
  2. 36
  3. 30
  4. 46

Solution: As before working from the very basic concepts, amount recieved by men is $5\times{\frac{180}{9}}= Rs.100$ (5 portions of total 9 portions, total being 180) and total for women, Rs.80.

If $x$ be number of women, ratio of amount that each man receives to the amount each woman receives is, $\frac{100}{66 - x}$ : $\frac{80}{x}$ = $ 3 : 2$

Or, $\frac{100x}{80(66 - x)} = \frac{3}{2}$

Or, $200x = 240(66 - x)$,

Or, $440x = 240\times{66}$, or, $11x = 6\times{66}$ or, $x=36$.

Answer: Option b : 36.

Key concepts used: Basic ratio concepts to determine actual total amounts received by men and women -- using single unknown as number of women and known value of total number of men and women we get the target ratio 3 : 2 as a ratio in expressions involving only unknown number of women -- simplification.

Q7. Water is $1\frac{2}{5}$ times as heavy as a liquid L1 and another liquid L2 is $1\frac{3}{7}$ times as heavy as water. The amount of liquid L2 that must be added to 7 litres of liquid L1 so that the mixture may weigh as much as an equal volume of water will be,

  1. 5 litres
  2. 7 litres
  3. $4\frac{2}{3}$ litres
  4. $5\frac{1}{6}$ litres

Solution: Ultimately, weight of a volume of water equals weight of equal volume of mixture.

From first statement, $W = \frac{7}{5}L_{1}$, where $W$ represents weight of 1 litre water and $L_{1}$ represents weight of 1 litre of liquid L1. From this we get, 7 litres of liquid L1 is equivalent in weight to 5 litres of water.

Similarly from the second given weight relation we get, $L_{2} = \frac{10}{7}W$.

With this we know, if we add $x$ litres of liquid L2 that is heavier than water, it will compensate for the 2 litres weight shortfall plus weight of $x$ litres water (we need to have weight of 7 litres of water after all, and with 7 litres of liquid L1 we have weight of only 5 litres of water).

So, finally for the mix,

Weight of $(2 + x)$ litres of water = Weight of $x$ litres of L2 = Weight of $\frac{10}{7}x$ litres of water

Or, $14 + 7x = 10x$,

Or, $x=\frac{14}{3}=4\frac{2}{3}$ litres.

Answer: Option c: $4\frac{2}{3}$ litres.

Key concepts used: Identifying the target term in which all else to be converted using the given relations -- formation of crucial expression using shortfall concept.

Q8. A cricketer with bowling average of 24.85 runs per wicket takes 5 wickets for 52 runs and consequently decreases his average by 0.85. Number of wickets taken by him till last match was,

  1. 64
  2. 96
  3. 72
  4. 80

Solution: For $n$ wickets average was 24.85 runs per wicket. Taking 5 more wickets makes his total number of wickets $n + 5$. We have then,

Total runs now, $24.85n + 52 = 24(n + 5)$

Or, $0.85n = 120 - 52 = 68$

Or, $n = \frac{100\times{68}}{85}=80$.

Answer: Option d: 80.

Key concepts used: Basic Average and changing total concept.

Q9. Sum of the two terms of a fraction is 11. If 2 is added to both of its terms, the new fraction exceeds the old by $\frac{1}{24}$. Then one term is larger than the other by, 

  1. 1
  2. 3
  3. 9
  4. 5

Solution: By the concept of relative change in fraction terms, we know that if by adding same value to both terms increases the fraction, the numerator must be smaller than the denominator. In other words, addition of same number to both terms contributes more proportional increase of the numerator (being less than the denominator) resulting an increase of the fraction.

With this knowledge and the fact that the total of numerator and denominator is 11, we can enumerate the possible fractions as,

$\displaystyle\frac{1}{10}$, $\displaystyle\frac{2}{9}$, $\displaystyle\frac{3}{8}$, $\displaystyle\frac{4}{7}$, $\displaystyle\frac{5}{6}$.

From the concept of relative change in fraction terms, we also know that smaller is the difference between the numerator and denominator, smaller will be the value increase of the fraction by adding the same number to both the terms. In this case the increase $\frac{1}{24}$ is quite small compared to the fractions being in consideration.

This urges us to start testing from the largest of the possible solutions, that is, $\frac{5}{6}$. Thus we get, $\frac{5}{6}\Rightarrow\frac{5 + 2}{6 + 2}=\frac{7}{8}$ and $\frac{7}{8} - \frac{5}{6} = \frac{21 - 20}{24} = \frac{1}{24}$, a success at the very first attempt.

This shows the power of using a basic problem solving principle and its related concepts.

Answer: Option a: 1.

Key concepts used: Principle of relative increment (decrement) of fractions -- enumeration of possible solutions -- guided trials.

Q10. The incomes of A, B and C are in the ratio of 7 : 9 : 12 and their spendings are in the ratio of 8 : 9 : 15. If A saves $\displaystyle\frac{1}{4}$th of his income, then the savings of A, B and C are in the ratio of,

  1. 69 : 56 : 99
  2. 99 : 56 : 69
  3. 56 : 99 : 69
  4. 99 : 69 : 56

Solution: The first basic concept of ratios says, when forming the ratio of two quantities in a minimized fraction form, HCF of the compared quantities have been eliminated between the numerator and denominator to get the minimized fraction. Thus to get back the original quantities we multiply the two terms of a ratio by their HCF.

Thus from the first relation we have the actual incomes as,

$7x : 9x : 12x$, where $x$ is the HCF that was canceled out to form the minimized ratio terms. This is use of cancelled out HCF reintroduction technique.

Similarly from the second relation we get the actual expenditures, $8y : 9y : 15y$, where $y$ is the HCF that was eliminated while forming the minimized ratio terms.

For A the saving is, $7x - 8y = \frac{7}{4}x$, or, $x = \frac{32}{21}y$. We will use this value in savings expressions of A, B and C to transform them in terms of $y$. Accordingly,

Savings of A in terms of $y$ is, $7x - 8y = \frac{32}{3}y - 8y = \frac{8}{3}y$

Savings of B in terms of $y$ is, $9x - 9y = \frac{96}{7}y - 9y = \frac{33}{7}y$

Savings of C in terms of $y$ is, $12x - 15y = \frac{128}{7}y - 15y = \frac{23}{7}y$

So the ratios of savings of A, B and C is,

$\frac{8}{3}y : \frac{33}{7}y : \frac{23}{7}y$

In this expression $y$ cancels out and we multiply each term by $3\times{7}=21$, the LCM of the denominators of the fractions. This process is normalization towards an expression in integers.

The target ratio of savings is then, $56 : 99 : 69$.

Answer: Option c : 56 : 99 : 69.

Key concepts used: Basic concept of HCF elimination (and introduction when required to get the actual quantity) between terms of a ratio -- transforming the ratio terms to actual incomes and actual savings respectively -- HCF reintroduction technique -- from given savings of A getting $x$ in terms of $y$ -- transforming each savings as actual income minus actual expenditure -- transforming each savings thus obtained to expression in terms of $y$ only -- taking the ratio of three savings -- normalizing the fraction terms by multiplying each term by LCM of their denominators and cancelling common $y$.


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