SSC CGL Tier II level Solution Set 22, Ratio proportion 1 | SureSolv

SSC CGL Tier II level Solution Set 22, Ratio proportion 1

22nd SSC CGL Tier II level Solution Set, 1st on topic Ratio and Proportion

ssc cgl tier II level solution set 22 ratio proportion 1

This is the 22nd solution set of 10 practice problem exercise for SSC CGL Tier II exam and 1st 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. The set of 10 problems being a bit more difficult than usual, 15 minutes answering time was given.

If you have not taken the test yet, you can refer to SSC CGL Tier II level Question Set 22, Ratio proportion 1, and then after taking the test come back to this solution.


22nd solution set—10 problems for SSC CGL Tier II exam: topic Ratio and Proportion 1—Answering time 15 mins

Problem 1.

When the son was born, the ratio of ages of his father and mother was 11 : 10. When the son will be twice his present age the ratio of the ages of father and mother will be 19 : 18. What is the ratio of ages of father and mother at present?

  1. 16 : 5
  2. 17 : 16
  3. 15 : 14
  4. 14 : 13

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

Assuming $F$ and $M$ to be the actual ages of father and mother when their son was born, and $x$ to be the cancelled out HCF for their ratio, $F=11x$ and $M=10x$. This is by applying HCF reintroduction technique.

If $a$ be the present age of the son, when he will be twice his present age, that is, $2a$ years old, the ages of the father and the mother would also be increased by $2a$ years and their ratio will be,

$\displaystyle\frac{11x+2a}{10x+2a}=\frac{19}{18}$.

This is the crucial ratio relation that we have to solve.

We have explained in details two solutions of this problem in our article, How to solve a SSC CGL level tricky age ratio problem lightning fast.

Here we won't explain the conventional solution and will explain only the elegant and lightning fast solution by the use of multiple of ratio value technique based on factors multiples concept.

In this conceptual technique, we will use trial values of multiples of RHS numerator 19 (we could have used denominator value 18 as well) and for each trial value, we would try combination of values of $x$ and $a$ in the LHS numerator so that the result becomes equal to the trial value of 19 chosen. This is the trial phase.

For trial value, $2\times{19}=38$, it is easy to see that $x=2$ and $a=8$ results in the same value of 38 for the LHS numerator. The trial phase is then successful.

In the second phase of verification we would use the same values of $x=2$ and $a=8$ in the LHS denominator to see whether the resulting denominator is a multiple of RHS denominator 18 with same factor 2 as in the trial value 38 of the numerator. This would ensure unchanged ratio which is a stringent condition. 

Indeed, with $x=2$ and $a=8$, the LHS denominator turns out to be $2\times{18}=36$, satisfying the RHS ratio relation. We have got our correct values. Now with these values of $x$ and $a$ we will find the value of the desired ratio of present ages of father and mother as, 

$\displaystyle\frac{11x+a}{10x+a}=\frac{30}{28}=\frac{15}{14}$.

So the desired ratio is 15 : 14.

Answer: Option c: 15 : 14.

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

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

Problem 2.

A man divides his property so that his son's share to his wife's and wife's share to his daughter's both are as in the ratio 3 : 1. If the daughter gets Rs.10000 less than the son, value of the whole property is,

  1. Rs.16250
  2. Rs.16000
  3. Rs.17000
  4. Rs.18250

Solution 2: Problem analysis and solving in mind by ratio joining

If $S$, $W$ and $D$ are the shares that the son, the wife and the daughter gets, by the first ratio statement,

$S:W=3:1$,

By the second ratio statement,

$W : D=3:1$,

For joining the two ratios, we need to equalize the middle ratio value of $W$ in both ratios. So as a first step, value of $W$ is changed to 3 in the first ratio by converting it to,

$S:W=9:3$.

Now we can join the two ratios,

$S:W:D=9:3:1$.

Total number of portions is,

$9+3+1=13$, 

Share value of son is nine times that of daughter from the joined ratio $S:W:D=9:3:1$,

$S=9D$.

Daughter's share value, which is also the unit portion value of each of the 13 portions, is less by Rs.10000 from the son's. So,

$9D-D=8D=10000$,

Or, portion value, $D=1250$,

And total property value for 13 portions,

$13D=13\times{1250}=16250$.

Answer: Option a: Rs.16250.

Key concepts used: Basic ratio concepts -- Ratio joining -- Portions concept -- Solving in mind.

The problem could have been solved without using the ratio joining as well by taking, $S=3W=9D$, or, $9D-D=8D=10000$ and so on.

Problem 3.

In a partnership business, B's capital was half of A's. If after 8 months, B withdrew half his capital and after 2 months more A withdrew one-fourth of his capital, then the profit ratio of A and B at the end of the year will be,

  1. 2 : 5
  2. 23 : 10
  3. 10 : 23
  4. 5 : 2

Solution 3: Problem analysis and solving by using investment-month concept

The ratio of the profits of A and B will be in the ratio of the amounts they have invested over the periods of months. The product of amount invested and the period of investment in months gives the measure of amount of investment over period invested. This is the single figure of investment-month summed up for each of A and B for the year that gives the share of profit for each of them.

Assuming amount of capital invested by A and B at the beginning to be $C_A$ and $C_B$,

$C_A=2C_B$.

For the first 8 months investment-month for A and B were respectively, $8C_A$ and $8C_B$.

At the end of 8 months B withdrew half his capital to reduce his capital to $0.5C_B$.

At the end of 10th month A withdrew one-fourth his capital reducing his capital to, $0.75C_A$.

For the two month period after 8th month then investment-months for A and B were respectively, $2C_A$ and $2\times{0.5C_B}=C_B$.

For the last two months similarly, the investment-month figures for A and B were respectively, $2\times{0.75C_A}=1.5C_A$ and $C_B$.

Over the year the total investment-month figures for A will then be,

$C_A(8+2+1.5)=11.5C_A$ and for B will be,

$C_B(8+1+1)=10C_B$.

Ratio of profits of A and B will be in the ratio of their cumulative investment-months, that is,

$\displaystyle\frac{\text{Profit}_A}{\text{Profit}_B}$

$=\displaystyle\frac{11.5C_A}{10C_B}$

$=\displaystyle\frac{23C_B}{10C_B}$, as $C_A=2C_B$

$=\displaystyle\frac{23}{10}$.

Answer: Option b: 23 : 10.

Key concepts used: Basic ratio concept -- Capital investment and profit concept -- Investment-month concept -- Investment to profit proportionality concept.

Problem 4.

A's income is Rs.140 more than B's income and C's income is Rs.80 more than D's income. If the ratio of A's and C's incomes is 2 : 3 and the ratio of B's and D's incomes is 1 : 2, then the incomes of A, B, C and D (in Rs.) are respectively,

  1. 320, 180, 480, and 360
  2. 400, 260, 600, and 520
  3. 300, 160, 600, and 520
  4. 260, 120, 320, and 240

Solution 4: Problem analysis and solving in mind

Let us assume incomes of A, B, C and D are respectively, $A$, $B$, $C$ and $D$. So,

$\displaystyle\frac{A}{C}=\frac{B+140}{D+80}=\frac{2}{3}$, whereas,

$\displaystyle\frac{B}{D}=\frac{1}{2}$,

Or, $D=2B$.

Substituting,

$\displaystyle\frac{B+140}{2B+80}=\frac{2}{3}$.

Simplifying,

$B=420-160=260$.

$D=2B=520$.

$C=D+80=600$.

$A=B+140=400$.

Instead of dealing with solving linear equations, use of the expressions in ratio form has clarity and solution is reached quickly.

Answer: Option b: 400, 260, 600 and 520.

Key concepts used: Basic ratio concept -- Careful tracking of ratio variables -- Context awareness -- Efficient simplification by judicious substitution -- Solving in mind.

Problem 5.

The ratio of amount of work done by $(x+1)$ workers in $(x+2)$ days and that done by $(x-1)$ workers in $(x+1)$ days is 6 : 5. Then the value of $x$ is,

  1. 16
  2. 17
  3. 15
  4. 14

Solution 5: Problem analysis and solving in mind using mandays concept

Using the concept of work amount as a product of number of workers and number of days which is the well known concept of mandays, we have,

$\displaystyle\frac{(x+1)(x+2)}{(x-1)(x+1)}=\frac{6}{5}$.

Cross-multiplying,

$5x^2+15x + 10=6x^2-6$,

Or, $x^2-15x-16=0$,

Or, $(x+1)(x-16)=0$.

As $x$ can't be negative, $x=16$.

Answer: Option a: 16.

Key concept used: Basic ratio concept --- Basic time and work concept -- Mandays technique -- Solving quadratic equation -- Solving in mind.

Problem 6.

If $x$ runs are scored by A, $y$ runs are scored by B and $z$ runs are scored by C, then $x:y=y:z=3:2$. If total number of runs scored by A, B and C is 342, the runs scored by each would respectively be,

  1. 189, 126, 84
  2. 162, 108, 72 
  3. 144, 96, 64
  4. 180, 120, 80

Solution 6: Problem analysis and solving in mind by ratio joining

By the given information, $x$, $y$ and $z$ are in a continued equation,

$\displaystyle\frac{x}{y}=\frac{y}{z}=\frac{3}{2}$.

We split up the continued equation to its component two part equations,

$\displaystyle\frac{x}{y}=\frac{3}{2}$, and

$\displaystyle\frac{y}{z}=\frac{3}{2}$.

To join the two ratios, we equalize the ratio value of the middle variable to the LCM of its two values 2 and 3, that is 6.

The first equation is thus converted to,

$\displaystyle\frac{x}{y}=\frac{9}{6}$.

And the second equation to,

$\displaystyle\frac{y}{z}=\frac{6}{4}$.

Joining the two ratios we have,

$x : y : z=9 : 6 : 4=9a : 6a : 4a$, introducing the cancelled out HCF as $a$.

Total number of runs is then,

$a(9+6+4)=19a=342$,

Or, $a=18$.

So, $x=9a=162$, $y=6a=108$, and $z=4a=72$.

Answer: Option b: 162, 108, 72.

Key concepts used: Splitting up the continued equation into two ratios -- Ratio joining by equalizing the middle term ratio value -- HCF reintroduction technique -- Portions concept -- Solving in mind.

Problem 7.

If 25 less are selected while 50 less had applied, the ratio of selected to unselected would have been 9 : 4. So, how many candidates had applied if the ratio of selected to unselected were 2 : 1?

  1. 500
  2. 125
  3. 250
  4. 375

Solution 7: Problem analysis and solution by careful problem analysis and context awareness

The problem involves two selection events. The ratio of selected to unselected of the first event is given at the end of problem statement, while the ratio result of the second selection event in terms of the first was stated in the beginning. With careful reading and analysis of the problem, its meaning should be understood very clearly. This is problem analysis and problem definition phase.

Assuming $A$ and $B$ to be the number of selected and unselected candidates during first selection, by the problem statement, 

$\displaystyle\frac{A}{B}=\frac{2}{1}$,

Or, $A=2B$.

Number of candidates applied then is,

$A+B=3B$.

In the second selection event where 25 less were selected and 50 less had applied (take careful note of this, it is not the figure of unselected), the number of unselected in this case would be,

$(3B-50)-(A-25)=3B-A-25=B-25$.

The ratio of selected to unselected would then be by definition, 

$\displaystyle\frac{A-25}{B-25}=\frac{9}{4}$,

Or, $\displaystyle\frac{2B-25}{B-25}=\frac{9}{4}$.

Cross-multiplying,

$8B-100=9B-225$,

Or, $B=125$, and total number candidates that had applied,

$3B=375$.

Answer: Option d: 375.

Key concepts used: Basic ratio concepts -- Rich ratio concepts -- Problem analysis -- Problem definition -- Context awareness.

With careful tracking of events and variables, the problem has turned out to be simple enough.

Problem 8.

Before a war the ratio of planes to tanks in an army was 3 : 5. During the war 800 planes and 1000 tanks were destroyed. The ratio of planes to tanks became 1 : 2. What is the number of tanks after the war?

  1. 2000
  2. 4000
  3. 1000
  4. 3000

Solution 8: Problem analysis and solving in mind using substitution

If $P$ and $T$ were the number of planes and number of tanks before the war,

$\displaystyle\frac{P}{T}=\frac{3}{5}$,

Or, $5P=3T$,

Or, $P=\displaystyle\frac{3}{5}T$.

After the war the ratio is changed to,

$\displaystyle\frac{P-800}{T-1000}=\frac{1}{2}$.

Substituting the value of $P$,

$\displaystyle\frac{\frac{3}{5}T-800}{T-1000}=\frac{1}{2}$,

Or, $\displaystyle\frac{3T-4000}{5T-5000}=\frac{1}{2}$.

Simplifying,

$5T-5000=6T-8000$,

Or, $T=3000$.

So the number of tanks after the war is, $3000-1000=2000$.

Answer: Option a: 2000.

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

Problem 9.

If $A: B=\displaystyle\frac{1}{2}:\displaystyle\frac{1}{3}$, $B:C=\displaystyle\frac{1}{5}:\displaystyle\frac{1}{3}$, then $(A+B):(B+C)$ is,

  1. 5 : 8
  2. 6 : 15
  3. 9 : 10
  4. 15 : 16

Solution 9: Problem analysis and solving in mind by modified form of adapted componendo dividendo

Eliminating the fraction in the first ratio by multiplying both numerator and denominator of the RHS of the ratio by the LCM of 2 and 3, that is, 6, we get,

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

Or, $\displaystyle\frac{(A+B)}{B}=\frac{5}{2}$, by adding 1 to both sides.

This is the first step of modified or adapted componendo dividendo.

Similarly eliminating the fractions in the second ratio we get,

$\displaystyle\frac{B}{C}=\frac{3}{5}$.

To keep the denominator $B$, same as the first result, we first invert the second ratio,

$\displaystyle\frac{C}{B}=\frac{5}{3}$.

Now by modified or adapted componendo dividendo, taking the second step, we add 1 to both sides of this ratio equation to get,

$\displaystyle\frac{(B+C)}{B}=\frac{8}{3}$.

Taking the third step of modified or adapted componendo dividendo, and dividing the first result by the second to eliminate the LHS denominator $B$,

$\displaystyle\frac{(A+B)}{(B+C)}=\frac{15}{16}$.

Answer: Option d: 15:16.

Key concepts used: Basic ratio concepts -- Fraction equivalent of ratios -- Fraction elimination in ratio values -- Modified  or adapted componendo dividendo -- Algebra concepts -- Solving in mind.

Being aware of the three step modified form of adapted componendo dividendo, the problem could be solved easily and quickly in mind.

Problem 10.

The average of 11 numbers is 36, whereas average of 9 of them is 34. If the remaining two numbers are in the ratio of 2 : 3, find the value of the smaller number between the remaining two.

  1. 45
  2. 54
  3. 36
  4. 48

Solution 10: Problem analysis and solving in mind by average concept and portions concept

With average of 36 for 11 numbers, total of 11 numbers is, $11\times{36}=396$.

Again with average of 34 for 9 numbers, total of 9 numbers is, $9\times{34}=306$.

Total of the remaining two numbers is then, $396-306=90$.

As these two remaining numbers are in ratio $2:3$, total number of portions the two numbers is, $2+3=5$, and value of each of the portions is, 

$\displaystyle\frac{90}{5}=18$.

The smaller number with 2 portions is then,

$2\times{18}=36$.

Answer: Option c: 36.

Key concepts used: Basic ratio concept -- Rich ratio concept -- Average concept -- Portions concept -- Solving in mind.

Overall, for careful tracking of events and variable values, the problems 3 and 7 needed a few steps in writing. Rest of the eight problems could be solved easily 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, ratio joining, portions concept, Investment-month concept, context awareness, mandays concept of time and work, careful problem analysis and problem definition, modified form of componendo dividendo that we call as adapted componendo dividendo and average 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.


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