SSC CGL Tier II level Solution Set 23, Ratio proportion 2 | SureSolv

SSC CGL Tier II level Solution Set 23, Ratio proportion 2

23rd SSC CGL Tier II level Solution Set, 2nd on topic Ratio and Proportion

ssc cgl tier II level solution set 23 ratio proportion 2

This is the 23rd solution set of 10 practice problem exercise for SSC CGL Tier II exam and 2nd 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 Tier II level Question Set 23, Ratio proportion 2, and then after taking the test come back to this solution.


23rd solution set—10 problems for SSC CGL Tier II exam: topic Ratio and Proportion 2—Answering time 12 mins

Problem 1.

Ratio of number of boys to the number of girls in a school of 432 students is 5 : 4. When some new boys and girls are admitted, the number of boys increased by 12 and the ratio of boys to girls changed to 7 : 6. The number of new girls admitted is,

  1. 12
  2. 14
  3. 20
  4. 24

Solution 1: Problem analysis and solving in mind by multiple of ratio value technique

By introducing the cancelled out HCF in original ratio of boys to girls 5 : 4 as $x$, the total number of students originally was,

$5x+4x=9x=432$,

Or, $x=48$.

The original number of boys and girls were then respectively,

$5x=240$, and

$4x=192$.

After the new admission, with additional 12 new boys number boys changed to, $240+12=252$.

Assuming the number of new girls to be $a$, the new ratio is expressed as,

$\displaystyle\frac{252}{192+a}=\frac{7}{6}$.

Applying multiple of ratio value technique we find that LHS numerator ratio value, $252=7\times{36}$, where 7 is the RHS numerator ratio value. The cancelled out HCF in this new ratio must then be 36.

The LHS denominator will then be,

$6\times{36}=216$.

And number of new girls,

$216-192=24$.

Answer: Option d: 24.

Key concepts used: Basic ratio concepts -- Rich ratio concepts -- HCF reintroduction technique -- Multiple of ratio value technique -- Basic factors multiples concept -- Solving in mind.

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

Problem 2.

If by increasing the price of a ticket in the ratio 8 : 11 the number of tickets sold falls in the ratio 23 : 21 then what is the increase in revenue if revenue before increase in price of ticket were Rs.36800?

  1. Rs.7850
  2. Rs.12850
  3. Rs.9400
  4. Rs.21250

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 price of ticket and number of tickets sold before price rise were,

$8x$ and $23y$.

And the revenue before the price rise was,

$8\times{23}xy=36800$.

This is a product of corresponding ratio terms and the product of two HCF's $xy$ can be treated as a single variable. We name it as product of ratios concept.

Simplifying,

$xy=200$.

In the same way price and number of tickets sold after the price rise were respectively,

$11x$ and $21y$.

So the revenue after the price rise became,

$11\times{21}xy=231\times{200}=46200$.

The revenue increase is then,

$46200-36800=9400$.

Answer: Option c: Rs.9400.

Key concepts used: Basic ratio concepts -- HCF reintroduction technique -- New variable revenue as product of corresponding terms of two ratios so that product of two introduced HCFs $xy$ is used as a single variable -- Product of ratios -- Solving in mind.

Problem 3.

Rs.2420 were divided among A, B and C so that A : B = 5 : 4 and B : C = 9 : 10. Then C gets,

  1. Rs.900
  2. Rs.680
  3. Rs.950
  4. Rs.800

Solution 3: Problem analysis and solving in mind by ratio joining

Share of each would be proportional to the corresponding ratio value in the joint three-party ratio. So we need to join the two ratios to get the three-party joint ratio of shares of A, B and C.

To join the two given ratios, the ratio value corresponding to the middle variable B is to equalized in the two ratios to the LCM of the two values 4 and 9, that is, 36.

The first ratio A : B = 5 : 4 is thus converted to A : B = 45 : 36 by multiplying the numerator and denominator by 9.

Similarly the second ratio B : C = 9 : 10 is converted to B : C = 36 : 40, by multiplying numerator and denominator by 4.

The joined ratio is then, A : B : C = 45 : 36 : 40.

Value of total number of 121 portions = Rs.2420.

Value of each portion is then,

$\displaystyle\frac{2420}{121}=20$.

And share of C,

$20\times{40}=800$.

Answer: Option d: Rs.800.

Key concepts used: Basic ratio concept -- Fund share in proportion to corresponding ratio value -- Ratio joining -- Portions concept -- Share of each of three parties can be determined only if the three share variables are expressed in a single ratio which gives the total fund being shared also -- Dividing in ratios -- Solving in mind.

Problem 4.

In a regiment the ratio between the number of officers to soldiers was 3 : 31 before a battle. In the battle 6 officers and 22 soldiers were killed and the ratio became 1 : 13. The number of officers before the battle was,

  1. 28
  2. 21
  3. 38
  4. 31

Solution 4: Problem analysis and solving in mind

If $x$ be the cancelled out HCF of the ratio of officers and soldiers before the battle, the ratio after the battle can be expressed as,

$\displaystyle\frac{3x-6}{31x-22}=\frac{1}{13}$.

Cross-multiplying,

$31x-22=39x-78$,

Or, $8x=56$

Or, $x=7$.

So the number of officers before the battle was,

$3x=21$.

Answer: Option b: 21.

Key concepts used: Basic ratio concept -- Change in ratio because of change in ratio variable values -- Context awareness -- Solving in mind.

Problem 5.

If the three numbers in the ratio 3 : 2 : 5 be such that the sum of their squares is equal to 1862, then which is the middle number?

  1. 13
  2. 14
  3. 15
  4. 16

Solution 5: Problem analysis and solving in mind using HCF reintroduction technique

Introducing cancelled out HCF as $x$, the sum of squares of the three number can be expressed as,

$x^2(9+4+25)=38x^2=1862$

Or, $x^2=49$.

Or, $x=7$.

The middle number is then,

$2x=14$.

Answer: Option b: 14.

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

Problem 6.

In a college union there are 48 students. The ratio of number of boys to number of girls is 5 : 3. The number of girls to be added to the union so that the ratio of boys to girls becomes 6 : 5 is,

  1. 7
  2. 17
  3. 12

Solution 6: Problem analysis and solving in mind

48 students total up to $(5+3)=8$ portions, so each portion value is 6, the number of boys is 30 and the number of girls is 18.

If $y$ be the number of girls to be added to make the ratio of boys to girls to 6 : 5,

$\displaystyle\frac{30}{18+y}=\frac{6}{5}$.

As the LHS denominator needs to have the value of 25 to make the ratio 6 : 5,

$y=25-18=7$.

Answer: Option a: 7.

Key concepts used: Basic ratio concepts -- Portions concept -- HCF reintroduction technique -- Factors multiples concept -- Change in ratio variable values -- Solving in mind.

Problem 7.

A and B start an enterprise together with A as active partner. A invests Rs.4000 and Rs.2000 more after 8 months. B invests Rs.5000 and withdraws Rs.2000 after 9 months. Being the active partner, A takes Rs.100 as allowance from the profit. What is the share of B if the profit for the year is Rs.6700?

  1. Rs.3250
  2. Rs.3350
  3. Rs.2800
  4. Rs.2700

Solution 7: Problem analysis and solving in mind by use of investment-month concept

Each individual profit share will be proportional to the figure of total investment-month. This will be the sum over a year, of the product of the amount of capital invested during a period and the period duration in months.

Total investment-month for A will be,

$4000\times{12}+2000\times{4}=56000$,

And for B it will be,

$5000\times{12}-2000\times{3}=54000$.

The ratio in which profit will be shared between A and B is then,

$\text{Profit}_A:\text{Profit}_B=56000 : 54000=28:27$, a total of 55 portions.

As A takes special allowance of Rs.100 per month for the whole year from the profit, the profit to be shared reduces to,

$6700-1200=5500$.

As this total profit of Rs.5500 is equivalent to 55 portions to be shared between A and B in the ratio of 28 : 27, each portion value will be Rs.100 and B will get Rs.2700.

Answer: Option d: 2700.

Key concepts used: Basic ratio concepts -- Investment-month concept in capital investment -- Profit sharing -- Portions concept -- Solving in mind.

Problem 8.

Annual incomes of Abhi and Riju are in the ratio 3 : 2, while the ratio of their expenditures is 5 : 3. If at the end of the year each saves Rs.1000, the annual income of Abhi is,

  1. Rs.9000
  2. Rs.8000
  3. Rs.6000
  4. Rs.7000

Solution 8: Problem analysis and solving in mind using Income expenditure ratios concept, HCF reintroduction and linear equation solution

In the two given ratios, if the cancelled out HCF were $x$ and $y$ respectively then for income and savings of Abhi, we have the relation,

$3x-5y=1000$, and for Riju,

$2x-3y=1000$.

Subtracting, $x=2y$,

Or, $y=\frac{1}{2}x$.

Substituting this value of $y$ in the first equation,

$3x-\frac{5}{2}x=1000$,

Or, $x=2000$.

So annual income of Abhi is,

$3x=6000$.

Answer: Option c: 6000.

Key concepts used: Basic ratio concepts -- Rich ratio concept -- Income expenditure ratios concept -- HCF reintroduction technique -- Linear equations -- Solving in mind.

Problem 9.

The ratio of sum of salaries of A and B to the difference of their salaries is 11 : 1 and the ratio of sum of the salaries of B and C to the difference of their salaries is also 11 : 1. If A's salary is the highest and C's the lowest, then what is B's salary, given total of their salaries to be Rs.182000?

  1. Rs.50000
  2. Rs.86400
  3. Rs.60000
  4. Rs.72000

Solution 9: Problem analysis and solving in mind by componendo dividendo and ratio joining

If the salaries of A, B and C were $A$, $B$ and $C$, by the first ratio statement,

$(A+B) : (A-B) = 11 : 1$.

Expressing it to a fraction,

$\displaystyle\frac{A+B}{A-B}=\frac{11}{1}$

Applying three-step componendo dividendo,

$\displaystyle\frac{A}{B}=\frac{12}{10}=\frac{6}{5}$.

Similarly by the second ratio statement,

$\displaystyle\frac{B+C}{B-C}=\frac{11}{1}$,

Again applying componendo dividendo,

$\displaystyle\frac{B}{C}=\frac{6}{5}$.

To join the two ratios, the ratio values corresponding to the middle variable B are to be equalized to the LCM of the two values, 6 and 5, that is 30.

Accordingly the first ratio is transformed to,

$\displaystyle\frac{A}{B}=\frac{36}{30}$.

And the second ratio to,

$\displaystyle\frac{B}{C}=\frac{30}{25}$.

Joining the two ratios,

$A : B : C=36 : 30 : 25$ with total number of portions as,

$36x+30x+25x=91x=182000$,

So portion value is,

$x=2000$.

And B's salary is,

$30x=60000$.

Answer: Option c: Rs.60000.

Key concepts used: Basic ratio concepts -- Rich ratio concepts -- Componendo dividendo -- HCF reintroduction technique -- Portions concept -- Ratio joining -- Solving in mind.

Being aware of the three step method of componendo dividendo, each of the two operations is executed directly without elaboration.

Problem 10.

What is the fourth proportional to 189, 273 and 153?

  1. 117
  2. 187
  3. 299
  4. 221

Solution 10: Problem analysis and solving in mind by fourth proportional concept

The proportional concept is another way of identifying the variables in an equal ratio of four variables,

$A:B=C:D$.

Variable $D$ is the fourth proportional in this ratio of four variablT%he values of the first three variables as given are,

$A=189$,

$B=273$, and

$C=153$.

Thus,

$\displaystyle\frac{189}{273}=\frac{153}{D}$,

So,

$D=\displaystyle\frac{153}{189}\times{273}$

$=\displaystyle\frac{153}{63}\times{91}$

$=17\times{13}$

$=221$.

Answer: Option d: 221.

Key concepts used: Basic ratio concept -- Fourth proportional concept  -- Solving in mind.

Overall, all of the eight 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, dividing in ratios, ratio joining, portions concept, change in ratio variable values, Investment-month concept, profit sharing, context awareness, componendo dividendo and fourth proportional concept. 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.


Resources that should be useful for you

7 steps for sure success in SSC CGL tier 1 and tier 2 competitive tests or section on SSC CGL to access all the valuable student resources that we have created specifically for SSC CGL, but 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

How to solve SSC CGL level number and age ratio problems lightning fast

How to solve SSC CGL level tricky age ratio problem lightning fast

SSC CGL level solved question sets on mixture or alligation

SSC CGL level solved questions sets 78 on mixture or alligation 1

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

SSC CGL Tier II level solved question sets on mixture or alligation

SSC CGL Tier II level solved question set 24 on mixture or alligation 1

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

SSC CGL Tier II level Solution Set 23 Ratio proportion 2

SSC CGL Tier II level Question Set 23 Ratio proportion 2

SSC CGL Tier II level Solution Set 22 Ratio proportion 1

SSC CGL Tier II level Question Set 22 Ratio proportion 1

SSC CGL question and solution sets on Ratio and Proportion and Percentage

SSC CGL level Solution Set 84, Ratio proportion 8

SSC CGL level Question Set 84, Ratio proportion 8

SSC CGL level Solution Set 83, Ratio Proportion 7

SSC CGL level Question Set 83, Ratio Proportion 7

SSC CGL level Solution Set 76, Percentage 4

SSC CGL level Question Set 76, Percentage 4 

SSC CGL level Solution Set 69, Percentage 3

SSC CGL level Question Set 69, Percentage 3

SSC CGL level Solution Set 68, Ratio Proportion 6

SSC CGL level Question Set 68, Ratio proportion 6

SSC CGL level Solution Set 31, Ratio Proportion 5

SSC CGL level Question Set 31, Ratio and Proportion 5

SSC CGL Level Solution Set 25, Percentage, Ratio and Proportion 4

SSC CGL level Question Set 25, Percentage, Ratio and Proportion 4

SSC CGL level Solution Set 24, Arithmetic Ratio & Proportion 3

SSC CGL level Question Set 24, Arithmetic Ratio & Proportion 3

SSC CGL level Solution Set 5, Arithmetic Ratio & Proportion 2

SSC CGL level Question Set 5, Arithmetic Ratio & Proportion 2

SSC CGL level Solution Set 4, Arithmetic Ratio & Proportion 1

SSC CGL level Question Set 4, Arithmetic Ratio & Proportion 1

If you like, you may subscribe to get latest content from this place.