SSC CGL level Solution Set 31, Ratio & Proportion 5

31st SSC CGL level Solution Set, 5th on topic Ratio & Proportion

SSC CGL Solution Set 31 Ratio and Proportion 5

This is the 31st solution set of 10 practice problem exercise for SSC CGL exam and 5th on topic Ratio & Proportion. Students must complete the corresponding question set in prescribed time first and then only refer to the solution set for extracting maximum benefits from this resource.

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Actual problem solving is done in the fourth layer. You need to use your problem solving abilities to gain an edge in competition.

If you have not taken the corresponding test yet, refer to SSC CGL level Question Set 31 on Ratio Proportion 5 for taking the test before coming back to this solution set.


31st solution set- 10 problems for SSC CGL exam: 5th on topic Ratio & Proportion- time 15 mins

Problem 1.

A certain amount of money is divided among A, B and C. If A receives 25% more than B and B receives 25% less than C, then A : B : C is,

  1. 12 : 10 : 11
  2. 10 : 9 : 12
  3. !5 : 12 : 16
  4. 14 : 12 : 13

Solution 1 : Problem analysis:

We need to combine two binary (two-term) ratios between A and B, and B and C to a three term ratio A : B : C.

Solution 1 : Problem execution

By first statement of relation between A and B,

$A = 1.25B$,

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

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

By second stement of relation between B and C,

$B = 0.75C$,

Or, $B = \displaystyle\frac{3}{4}C$,

Or, $\displaystyle\frac{B}{C} = \frac{3}{4}$.

To combine these two ratios, the ratio term value of the common term B must be made equal to the LCM of its corresponding two values 3 and 4 which is 12 here.

Thus transforming the two ratios we get,

$\displaystyle\frac{A}{B} = \frac{5}{4} = \frac{15}{12}$, and

$\displaystyle\frac{B}{C} = \frac{3}{4} = \frac{12}{16}$.

Combining the two binary ratios to a three term ratio by the common term B,

$A : B : C = 15: 12 : 16$.

Answer: c: 15 : 12 : 16.

Key concepts used: Two term ratio formation -- binary ratio to three term ratio conversion by common equal LCM value term.

Problem 2.

Two numbers are in the ratio 17 : 45. One third of the smaller is less than one fifth of the larger by 15. The smaller number is,

  1. $25\displaystyle\frac{1}{2}$
  2. $86\displaystyle\frac{1}{2}$
  3. $67\displaystyle\frac{1}{2}$
  4. $76\displaystyle\frac{1}{2}$

Solution 2 : Problem analysis and execution:

From first statement,

$\displaystyle\frac{S}{L} = \frac{17}{45}$, where $S$ is the smaller and $L$ is the larger number.

From second statement,

$\displaystyle\frac{1}{5}L - \displaystyle\frac{1}{3}S = 15$,

From first relation,

$S = 17x$ and $L=45x$, where $x$ is the HCF of S and L

Putting these in last relation,

$9x - \displaystyle\frac{17}{3}x = 15$,

Or, $\displaystyle\frac{10}{3}x = 15$,

Or, $x = \displaystyle\frac{9}{2}$

So,

$S = 17x = \displaystyle\frac{153}{2} = 76\displaystyle\frac{1}{2}$.

Answer: Option d : $76\displaystyle\frac{1}{2}$ .

Key concepts used: Ratio concepts -- transformation of ratio terms to actual values by introduction of canceled out HCF $x$ -- formulation of relation between smaller and larger number from problem statement -- substitution to get value of $x$ and then the desired value of the smaller number.

Problem 3.

A policeman starts to chase a thief and moves 8 steps when the thief takes 10 steps, but 7 steps of the thief equals 5 steps of the policeman. The ratio of the speeds of the policeman and the thief is,

  1. 25 : 28
  2. 26 : 25
  3. 25 : 26
  4. 28 : 25

Solution 3 : Problem analysis

The policeman takes smaller number of steps than the thief in same amount of time. If the length of steps of the two were same, the policeman was sure to have a speed smaller than the thief. But though the policeman takes smaller number of steps, his steps are longer than the thief. Combining the two results only will indicate who is faster and by how much.

Solution 3 : Problem execution

By first statement in same length of time ratio of the distance covered by the two, which is the ratio of speeds of the two is,

$\displaystyle\frac{8P}{10T}=\frac{4P}{5T}$, where $P$ is the step length of the policeman and $T$ is the step length of the thief.

Again by the second statement,

$7T = 5P$,

Or, $\displaystyle\frac{P}{T} = \frac{7}{5}$.

Putting this in speed ratio relation we have the ratio of speeds as,

$\displaystyle\frac{4P}{5T} = \frac{28}{25}$.

Answer: Option d: 28 : 25.

Key concepts used: Basic concepts of speed time distance to form the ratio of speeds as ratio of number of steps covered in same length of time -- qualifying this ratio with the ratio of step lengths gives us the ratio of speeds.

Problem 4.

Two numbers are in ratio of 3 : 5 and their LCM is 225. The smaller number is,

  1. 75
  2. 45
  3. 60
  4. 90

Solution 4 : Problem analysis and execution

Introducing the canceled out HCF as $x$ in the ratio we have the transformed ratio as $3x : 5x$ where the actual numbers are $3x$ and $5x$.

Now applying the method of forming the LCM by multiplying the two numbers and excluding one occurrence of HCF from the product, we have,

$LCM = 15x = 225$.

So $x = 15$

Thus the smaller number is 45.

Answer: b: 45.

Key concepts used: Concept that introduction of HCF as factor in both the ratio terms generate the two actual numbers -- basic LCM formation concept.

Problem 5.

Two numbers are in the ratio of $1\displaystyle\frac{1}{2} : 2\displaystyle\frac{2}{3}$. When each of these is increased by 15, the ratio changes to $1\displaystyle\frac{2}{3} : 2\displaystyle\frac{1}{2}$. The larger of the numbers is,

  1. 48
  2. 27
  3. 36
  4. 64

Solution 5 : Problem analysis

When a ratio is given in terms of mixed fraction, it needs to be transformed to the form involving proper fraction only. After this first step the canceled out HCF introduction as $x$ and addition of 15 to two terms will give us a linear equation in $x$.

Solution 5 : Problem execution

Transforming the first ratio to proper fraction form we get ratio of the two numbers before term increase by 15 as,

$1\displaystyle\frac{1}{2} : 2\displaystyle\frac{2}{3} = \displaystyle\frac{3}{2} : \displaystyle\frac{8}{3} = 9 : 16=9x : 16x$, where $x$ is the canceled out HCF of the two numbers and $9x$ and $16x$ are the actual numbers.

Similarly the transformed ratio after term increase by 15 is,

$1\displaystyle\frac{2}{3} : 2\displaystyle\frac{1}{2} = \displaystyle\frac{5}{3} : \displaystyle\frac{5}{2} = 2 : 3$.

Thus by the statement of term increase by 15 we have,

$\displaystyle\frac{9x + 15}{16x + 15} = \frac{2}{3}$,

Or, $27x + 45 = 32x + 30$,

Or, $5x = 15$,

Or, $x=3$,

The larger number is then,

$16x = 48$.

Answer: Option a: 48.

Key concepts used: Ratio transformation to proper fraction form -- introduction of canceled out HCF to the two terms to get the actual terms -- forming the linear equation in a single variable and solution.

Problem 6.

The ratio of savings to expenditure of a person is 2 : 3. If his savings increases by 6% while his income increases by 15% then by how much percentage did his expenditure increase?

  1. 21%
  2. 24%
  3. 12%
  4. 25%

Solution 6 : Problem analysis

Income (I) = Expenditure (E) + Savings (S).

Using this relation, ratio between any two of these three variables can be converted to the ratio between any other two of these three variables.

Solution 6 : Problem execution

As the increases are in Income and Savings we need to deal first with the ratio of Savings to Income.

Given, $\displaystyle\frac{S}{E} = \frac{2}{3}$,

Or, $\displaystyle\frac{E}{S} = \frac{3}{2}$,

Adding 1 to both sides,

$\displaystyle\frac{S + E}{S} = \frac{3 + 2}{2}$,

Or, $\displaystyle\frac{I}{S} = \frac{5}{2}$. This is the present situation.

By problem statement, savings increases by 6% while his income increases by 15%. So increased new savings becomes, $S_2=1.06S$ and increased new income, $I_2=1.15I$.

Thus ratio of new income to new savings is,

$\displaystyle\frac{I_2}{S_2} =\frac{1.15I}{1.06S} =  \frac{1.15}{1.06}\times{\frac{5}{2}} = \frac{5.75}{2.12}$,

Subtracting 1 from both sides,

$\displaystyle\frac{I_2 - S_2}{S_2} = \frac{5.75 - 2.12}{2.12}$,

Or, $\displaystyle\frac{E_2}{S_2} = \frac{3.63}{2.12}$, where $E_2$ is the new increased expenditure.

It means, while savings has increased by 6% to its new portion of 2.12, original being 2 portion ($2 + 0.06\times{2}=2.12$), the expenditure has increased from its 3 portions to 3.63, that is, an increase of 63% for 3 portions and 21% per portion.

Answer: Option a : 21%.

Key concepts used: Income, savings expenditure relationship -- percentage increase and effect of the increase on the ratio.

Problem 7.

A trader sells 20 kg of sugar at Rs. 400 and agrees to the request of a customer to give him a discount of 20%. But being a dishonest trader he uses a spurious weight which is less by 4% per kg of weight. The effective discount that the customer then gets,

  1. 15.5%
  2. 16%
  3. 16.66%
  4. 19.6%

Solution 7 : Problem analysis

Because of using a spurious weight less by 4% per kg, the trader sells less sugar by that amount per kg and takes the agreed to discounted price from the customer. Effectively the customer gets less than 1 kg at the price of 1 kg and so his price for 1 kg is not the agreed to discounted price but more than that and to be normalized by the shortfall in 1 kg.

Solution 7 : Problem execution

The discounted price is,

$\displaystyle\frac{400}{20} - 20$% of $\displaystyle\frac{400}{20} = 20 - 4 = 16$ per kg.

As each 1 kg is actually 1 kg less 4%, that is, 0.96kg, the effective price per kg to the customer is,

$\displaystyle\frac{16}{0.96} = \frac{100}{6}$ per kg.

So the effective discount is,

$20 - \displaystyle\frac{100}{6} = \displaystyle\frac{20}{6}$, and percentage discount on marked price of Rs 20 as,

$\displaystyle\frac{20}{6}\times{\displaystyle\frac{100}{20}} = \displaystyle\frac{100}{6} = 16.66$%, not the agreed to 20%.

Answer: Option c: 16.66%.

Key concepts used: Discount, percentage and ratio concepts.

Problem 8.

When A, B and C do a task together the ratio of work that A and B do together with respect to what C does is 7 : 3 and the ratio of work that B and C together do to the work A does is 1 : 1. Which worker among the three is most efficient?

  1. B
  2. A
  3. C
  4. Cannot be answered with the given information

Solution 8 : Problem analysis

In the statements here, when the word work is mentioned, it is always the portion of the whole work in same time period that may assumed to be the unit time, say a day.

Solution 8 : Problem execution

Let $A$, $B$ and $C$ be the portion of work that A, B and C do per unit time when they work together.

By the first statement then,

$\displaystyle\frac{A + B}{C} = \frac{7}{3}$,

and by the second statement,

$B + C = A$,

By adding 1 to both sides of the first equation we have,

$\displaystyle\frac{A + B + C}{C} = \displaystyle\frac{2A}{C} = \frac{10}{3}$,

Or, $\displaystyle\frac{A}{C} = \frac{5}{3}$,

Or $C$ in terms of $A$,

$C =\displaystyle\frac{3}{5}A$,

Putting this in the second equation we get,

$B = \displaystyle\frac{2}{5}A$,

Or, $\displaystyle\frac{B}{A} = \frac{2}{5}$.

Combining the two ratios we get,

$B : A : C = 2 : 5 : 3$.

So A is the most efficient worker among the three, doing 5 portions of work when C does 3 portions and B does 2 portions.

Answer: Option b: A.

Key concepts used: Work rate per unit time concept allowed formation of two ratio expressions transforming and joining of which gave us the relative portions of work done by the three workers. The method is quite general and can be applied in diverse situations.

Problem 9.

In three containers of equal capacity, the ratio of milk to water is 3: 2, 7 : 3 and 11 : 4 respectively. If the liquids in the three containers are mixed together, what will be the milk to water ratio in the new mixture?

  1. 61 : 28
  2. 60 : 29
  3. 59 : 29
  4. 61 : 29

Solution 9 : Problem analysis

For each ratio we can say with certainty how much milk 1 litre of a particular mixture contains. If we add these three 1 litre mixtures we get 3 litres of mixture and also the total amount of milk in this 3 litres of mixture with certainty. Dividing this milk amount by 3 will give us the milk portion in 1 litre of new mixture and consequently the milk to water ratio.

Dealing with milk to total mixture ratio simplifies this type of problem considerably.

Solution 9 : Problem exceution

Milk to water ratio in the first mixture is, $\displaystyle\frac{3}{2}$,

Or, milk to total mixture ratio in the first mixture is, $\displaystyle\frac{3}{5}$, that is,

1 litre of the first mixture contains, $\displaystyle\frac{3}{5}$ litres of milk.

Similarly for the second mixture,

1 litre of the mixture contains, $\displaystyle\frac{7}{10}$ litres of milk, and

for the third mixture 1 litre of the mixture contains $\displaystyle\frac{11}{15}$ litres of milk.

Adding up we get 3 litres of new mixture containing,

$\displaystyle\frac{3}{5} + \displaystyle\frac{7}{10} + \displaystyle\frac{11}{15} = \displaystyle\frac{61}{30}$ litres of milk.

So 1 litre of new mixture will contain, $\displaystyle\frac{61}{90}$ litres of milk and hence, $1 - \displaystyle\frac{61}{90} = \displaystyle\frac{29}{90}$ litres of water.

So the desired milk to water ratio in the new mixture is, 61 : 29.

Answer: Option d: 61 : 29.

Key concepts used: Mixture concepts -- ratio and proportion concepts.

Problem 10.

Vinny got twice as many marks in English as in Science. Her total marks in English, Science and Maths is 180. If the ratio of her marks in Maths and English is 3 : 2, what is her marks in Science?

  1. 30
  2. 72
  3. 60
  4. 90

Solution 10 : Problem analysis

Three linear relations between three variables are given and it is then easily possible to find the value of each unknown. The main task is to form the relations between the variables.

Solution 10 : Problem execution

Let $E$, $S$, and $M$ be the marks Vinny got in English, Science and Maths.

From first statement then,

$E = 2S$,

Or, $S = \displaystyle\frac{1}{2}E$.

From third statement,

$\displaystyle\frac{M}{E} = \frac{3}{2}$,

Or, $M = \displaystyle\frac{3}{2}E$,

We have converted all other variables in terms of the single variable $E$. When we substitute all these values to the third equation in terms of the three variables we will get a single linear equation in terms of a single variable.

Thus,

$E + M + S = 180$,

Or, $E + \displaystyle\frac{3}{2}E + \displaystyle\frac{1}{2}E = 3E = 180$,

Or $E = 60$,

and, marks in Science, $S = \displaystyle\frac{1}{2}E = 30$.

Answer: Option a: 30.

Key concepts used: Relation formation and linear equation concepts.


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