SSC CGL level Solution Set 83, Ratio proportion 7 | SureSolv

SSC CGL level Solution Set 83, Ratio proportion 7

83rd SSC CGL level Solution Set, 7th on topic Ratio and Proportion

ssc cgl level solution set 83 ratio proportion 7

This is the 83rd solution set of 10 practice problem exercise for SSC CGL exam and 7th on topic Ratio and Proportion. Students should complete the corresponding question set in prescribed time first and then only refer to this solution set for extracting maximum benefits from this resource.

If you have not taken the test yet, you can refer to SSC CGL level Question Set 83, Ratio proportion 7, and then after taking the test come back to this solution.


83rd solution set—10 problems for SSC CGL exam: topic Ratio and Proportion 7—Answering time 12 mins

Problem 1.

10 women can do a work in 6 days, 6 men can do the same piece of work in 5 days and 8 children can do it in 10 days. What is the ratio of efficiency of a woman, a man and a child respectively?

  1. 4 : 8 : 3
  2. 4 : 5 : 3
  3. 4 : 6 : 3
  4. 2 : 4 : 3

Solution 1: Problem analysis and solving in mind by worker-days as amount of work, efficiency as inverse of worker-days and basic ratio concepts

Worker-days, a product of number of workers and the number of days the worker takes to finish the job, for a woman, a man and for a boy are respectively,

$10\times{6}=60$,

$6\times{5}={30}$, and,

$8\times{10}=80$.

Worker-days, a measure of the quantum of work in terms of how many days a worker takes to finish the job, when inversed gives the fraction of the work the worker does in a day. That incidentally is also the efficiency of the worker.

So the ratio of the efficiencies of a woman, a man and a boy is,

$\displaystyle\frac{1}{60} : \displaystyle\frac{1}{30} : \displaystyle\frac{1}{80}=\displaystyle\frac{1}{6}:\displaystyle\frac{1}{3}:\displaystyle\frac{1}{8}$.

Multiplying the ratio by the LCM 24 of 6, 3 and 8 to eliminate the fractions, the desired ratio is,

$4:8:3$.

Answer: Option a: 4 : 8 : 3.

Key concepts used: Basic ratio concepts -- Worker-days concept -- Efficiency as inverse of worker-days -- Fraction ratio values to be multplied by LCM of denominators of fraction ratio values -- Solving in mind.

With conceptual clarity, the problem can easily be solved wholly in mind.

Problem 2.

The railway fares of AC 2 tier sleeper and Sleeper classes are in the ratio 4 : 1. The number of passengers travelled by AC 2 tier sleeper and Sleeper classes were in the ratio 3 : 25. If the total fare collection was Rs.37000, how much did AC 2 tier sleeper class passengers pay? 

  1. Rs.12000
  2. Rs.15000
  3. Rs.10000
  4. Rs.16000

Solution 2: Problem analysis and solving in mind by product of ratios concept

By introducing the cancelled out HCF in the first and second ratio as $x$ and $y$, the ticket fares and number of passengers travelled for AC 2 tier and Sleeper claases are,

$4x$, $x$ and $3y$, $25y$.

Total revenue would then be,

$12xy+25xy=37000$,

Or, $xy=1000$.

The product of two HCF's $xy$ can be treated as a single variable. The revenue amount as product of two ratio values and resulting $xy$ as product of two cancelled out HCFs we call as product of ratios concept.

This is possible because the desired revenue from AC 2 tier passengers would also be in terms of $xy$ as

$12xy=12000$.

Answer: Option a: Rs.12000.

Key concepts used: Basic ratio concepts -- HCF reintroduction technique -- Revenue as product of two ratio values -- Product of ratios -- Solving in mind.

Problem 3.

The working rates of A and B are in the ratio 2 : 3. The number of days taken by each to finish the work are in the ratio,

  1. 2 : 3
  2. 4 : 9
  3. 9 : 4
  4. 3 : 2

Solution 3: Problem analysis and solving in mind by the concept of days to complete as inverse of work rate concept

Work rate is portion of work finished by a worker in a day. The number of days the worker would take to finish the work is inverse of his work rate. So the ratio of number of days the two workers would take to finish the job when ratio of their work rates are 2 : 3 would be inverse of 2 : 3, that is, 3 : 2.

As the same work is being done by the two workers, the second ratio can be directly derived as the inverse of the first ratio.

Answer: Option d: 3 : 2.

Key concepts used: Basic ratio concept -- Work rate concept -- Work rate as the inverse of time to finish the work concept -- Solving in mind.

Problem 4.

Two numbers are in the ratio 3 : 5. If 9 be subtracted from each, their ratio becomes 12 : 23. Find the numbers.

  1. 15, 28
  2. 60, 69
  3. 36, 115
  4. 33, 55

Solution 4: Problem analysis and solving in mind

If $x$ be the cancelled out HCF of the ratio of the original numbers, the ratio after subtraction of 9 from each can be expressed as,

$\displaystyle\frac{3x-9}{5x-9}=\frac{12}{23}$.

We will use the multiple of ratio value tachnique on the prime ratio value of 23 for faster solution. Basically we have to try for multiples of 23 and add 9 to check whether it is a multiple of 5. If so we will check the same for the numerator ratio value of 12.

23 fails, but with $2\times{23}=46$, we get $x=5\times{11}$. The value of $x$ as 11 also satisfies the numerator ratio variable and its value relationship-$3\times{11}-9=2\times{12}$, with same factor of 2 as in denominator trial value of 46. 

With cancelled out HCF $x$ as 11, we have the original numbers as, 

$3x=33$, and $5x=55$.

Answer: Option d: 33, 55.

Key concepts used: Basic ratio concept -- Change in ratio because of change in ratio variable values -- Multiple of ratio value technique -- Solving in mind.

Problem 5.

If Rs.782 is divided into three parts in the ratio, $\displaystyle\frac{1}{2}:\displaystyle\frac{2}{3}:\displaystyle\frac{3}{4}$, the first part is,

  1. Rs.204
  2. Rs.196
  3. Rs.182
  4. Rs.190

Solution 5: Problem analysis and solving in mind using HCF reintroduction technique and fraction ratio value conversion concept

If in a ratio, the ratio values are fractions, the values are to be converted to integers by multiplying with their LCM for ease of understanding of total portions that make up the whole as well as ease of manipulation of the ratio values.

In this case, to get rid of the denominators of the fractions, we multiply the ratio values by the LCM of the denominators, 2, 3 and 4, that is, by 12.

The ratio is then transformed to, 6 : 8 : 9.

The total number portions is,

$6+8+9=23$.

And value of each portion is,

$\displaystyle\frac{782}{23}=34$.

The value of first part will then be,

$6\times{34}=204$.

Answer: Option a: Rs.204.

Key concept used: Faction ratio conversion concept --- Portion of a ratio concept, total of portions multiplied by value of each portion makes up the whole -- Solving in mind.

Problem 6.

Three numbers are in the ratio 5 : 7 : 12. If the sum of the first and third numbers is greater than the second number by 50, the sum of the three numbers is,

  1. 95
  2. 125 
  3. 120
  4. 85

Solution 6: Problem analysis and solving in mind by HCF reintroduction technique

Introducing the calncelled out HCF as $x$, the three numbers are, $5x$, $7x$ and $12x$. The difference between the sum of the first and the third number and the second number is then,

$(5x+12x)-7x=10x=50$.

So, $x=5$, and sum of the three numbers, $24x=120$.

Answer: Option c: 120.

Key concepts used: Basic ratio concepts -- HCF reintroduction technique -- Solving in mind.

Problem 7.

The numbers to be added to each of the numbers 7, 16, 43 and 79 to make the numbers in proportion is,

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

Solution 7: Problem analysis and solving in mind by proportion concept

If four numbers are to be in proportion, the ratio of the first two numbers would be equal to the ratio of the third and fourth numbers afrer adding up say, $x$ to all the four numbers. This can then be expressed as,

$\displaystyle\frac{7+x}{16+x}=\frac{43+x}{79+x}$.

We will use factors multiples concept on numerator values, $7+x$ and $43+x$.

Only 5 satisfies the factor multiple relation between $7+5=12$ and $43+5=48=4\times{12}$.

Addition of 5 also satisfies the denominator ratio variable and its value relationship,

$16+5=21$ and $79+5=84=4\times{21}$.

Answer: Option d: 5.

Key concepts used: Basic ratio concepts --Proportion concept -- Factors and multiples concept -- Solving in mind.

Problem 8.

What number should be added to or subtracted from each term of the ratio 17 : 24 so that it becomes equal to 1 : 2?

  1. 5 is subtracted
  2. 10 is added
  3. 10 is subtracted
  4. 7 is added

Solution 8: Problem solving in mind by number sense

We just subtract 10 to get the denominator double the numerator. It is by a mix of trial by choice values and number sense.

If you are not satisfied, you can also use the reasoning that the number to be added or subtracted must be even for the denominator ratio value to be even. Subtracting 10 will be just a small step away.

The step,

$\displaystyle\frac{17-10}{24-10}=\frac{7}{14}=\frac{1}{2}$.

Answer: Option c: 10 is subtracted.

Key concepts used: Basic ratio concepts -- Mathematical reasoning -- Solving in mind.

Problem 9.

Three numbers are in the ratio 3 : 4 : 5. The sum of the largest and the smallest equals the sum of the second and 52. The smallest number is,

  1. 52
  2. 27
  3. 20
  4. 39

Solution 9: Problem solving in mind by HCF reintroduction technique

If the cancelled out HCF be $x$, the relation described can be expressed as,

$3x+5x=4x+52$,

Or, $4x=52$,

Or, $x=13$.

The smallest number is,

$3x=39$.

Answer: Option d: 39.

Key concepts used: Basic ratio concepts -- HCF reintroduction technique -- Solving in mind.

Problem 10.

Three numbers are in the ratio 1 : 2 : 3 and the sum of their cubes is 4500. The smallest number is,

  1. 10
  2. 6
  3. 5
  4. 4

Solution 10: Problem analysis and solving in mind by HCF reintroduction technique

If the cancelled out HCF be $x$, the numbers are, $x$, $2x$ and $3x$ and sum of their cubes is,

$x^3+8x^3+27x^3=36x^3=4500$.

So, $x^3=125$, and $x=5$.

The smallest number is just 5.

Answer: Option c: 5.

Key concepts used: Basic ratio concept -- HCF reintroduction technique -- Solving in mind.

Overall, all of the ten problems could be solved in mind using varieties of concepts and techniques.

For solving these 10 problems, the concepts and techniques used were—basic and rich ratio concepts, HCF reintroduction technique, Multiple of ratio value technique, factors multiples concept, product of ratios concept, fraction ratio conversion, portions concept, change in ratio variable values, proportion concept, revenue as product of two ratio values, mandays concept, work rate as inverse of time to finish the work concept, worker efficiency as inverse of worker-days concept and mathematical reasoning. The problems needed an wide array of basic and rich concepts for quick solution.

Just remember, understanding and applying basic and rich concepts should enable you to solve any such problem easily under a minute with no dependence on varieties of formulas.


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