SSC CGL level Solution Set 68, Ratio Proportion 6

68th SSC CGL level Solution Set, 6th on topic Ratio Proportion

SSC CGL Solution Set 68 Ratio Proportion 6

This is the 68th solution set of 10 practice problem exercise for SSC CGL exam and 6th 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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If you have not taken the corresponding test yet, refer to SSC CGL level Question Set 68 on Ratio Proportion 6 for taking the test before coming back to this solution set.


68th solution set - 10 problems for SSC CGL exam: 6th on topic Ratio Proportion- time 12 mins

Problem 1.

The ratio of number of boys to girls in a school was 3 : 4 with total number of students as 1554. If the ratio of girls to boys changed to 6 : 7 when 30 more girls joined the school and a few boys left, how many boys left the school?

  1. 86
  2. 76
  3. 84
  4. 74

Solution 1 : Problem analysis and solving

Reintroducing the canceled out HCF in the first ratio, we can express it as,

$3x:4x$, where $x$ is the canceled out HCF so that actual number of girls and boys were $3x$ and $4x$.

By the given information then, the total number of students was,

$3x+4x=7x=1554$,

Or, $x=222$,

Number of girls,

$3x=666$,

Number of boys,

$4x=888$.

In the second case then we get the changed ratio as $\displaystyle\frac{6}{7}$ with the relation,

$\displaystyle\frac{666+30}{888-n} = \frac{6}{7}$, where $n$ is assumed to be the number of boys who left,

Or, $\displaystyle\frac{116}{888-n} = \frac{1}{7}$, cancelling common factor 6 in the numerators,

Or, $812=888-n$,

Or, $n=76$.

So, 76 boys left.

Answer: b: 76.

Key concepts used: Basic ratio concepts -- HCF reintroduction technique, reintroduction of canceled out HCF to form actual values -- Sequencing of events.

Problem 2.

The ratio of milk and water in mixtures of four containers A, B, C, and D are respectively, 5 : 3, 2 : 1, 3 : 2 and 7 : 4. Which container then has the minimum quantity of milk relative to water?

  1. C
  2. D
  3. A
  4. B

Solution 2 : Problem analysis and execution

The total number of portions in the four containers are, $5+3=8$, $2:1=3$, $3+2=5$, and $7+4=11$ respectively.

On the other hand the milk quantities are, 5, 2, 3 and 7 portions respectively.

So for containers A, B, C, and D, the milk quantity with respect to total volume are respectively,

$5\text{ out of }8=\displaystyle\frac{5}{8}$,

$2\text{ out of }3=\displaystyle\frac{2}{3}$,

$3\text{ out of }5=\displaystyle\frac{3}{5}$,

$7\text{ out of }11=\displaystyle\frac{7}{11}$.

The problem is now transformed to a fraction comparison problem.

With the fractions showing no usable pattern for quick comparison, we resort to pair by pair comparison starting with the first pair of fractions, then comparing the smaller of the two with the third fraction and so on.

This is the most basic method of comparison of a series of values to determine which one is the smallest value.

Comparison of two fractions by denominator equalization and numerator comparison pair by pair

Here, for the pair of fractions being compared, we just form the two numerator components by cross multiplication to see which one is larger (without bothering about the LCM of the denominators, thus eliminating one action).

For the first two fractions, $\displaystyle\frac{5}{8}$ and $\displaystyle\frac{2}{3}$, numerator components are,

$15 \lt 16$, so the first fraction $\displaystyle\frac{5}{8}$ is smaller.

For the second comparison between $\displaystyle\frac{5}{8}$ and $\displaystyle\frac{3}{5}$, numerator components are,

$25 \gt 24$, so third fraction $\displaystyle\frac{3}{5}$ is smaller.

Similarly, for the third comparison between $\displaystyle\frac{3}{5}$ and $\displaystyle\frac{7}{11}$,

$33 \lt 35$, so third fraction $\displaystyle\frac{3}{5}$ corresponding to vessel C is the smallest.

Answer: Option a : C.

Key concepts used: Portion use technique -- Basic ratio concepts -- Problem transformation -- Fraction comparison -- Numerator component comparison by cross-multiplication.

Problem 3.

Two blends of tea costing Rs.40 per kg and Rs.35 per kg are mixed in a ratio of 3 : 2 by weight. If one-fifth of the mixture is sold at Rs.46 per kg and the rest at Rs.55 per kg, the profit percentage would be,

  1. 20
  2. 50
  3. 30
  4. 40

Solution 3 : Problem analysis and execution

In the mixing operation, 3 portions of tea costing Rs.40 per kg are mixed with 2 portions of tea costing Rs.35 per kg. Thus 5 portions of mixture is formed with a cost price of,

$\text{CP}=3\times{40}+2\times{35}=190$.

Mark that effectively we have mixed 3 kgs of the first type of tea with 2 kgs of second type of tea, making portion size as 1 kg and the total price of Rs.190 as price of 5 kgs of mixed tea.

Similarly in the selling operation, the total number of portions in the mixture is 5 and one fifth and four-fifths are sold at different prices. It means 1 portion of the mixture is sold at Rs.46 per kg and the rest 4 portions at Rs.55 per kg.

The sale price of the five portions, that is, 5 kgs would then be,

$\text{SP}=46+4\times{55}=266$. Here again 1 portion is 1 kg.

Percent profit is then,

$\text{Profit}=\displaystyle\frac{266-190}{190}=\displaystyle\frac{4}{10}=40$%.

Note that the mixing and selling operations used portions that were in synchronization without breaking down the portion size throughout. This enabled us to avoid calculating the cost price or sale price per kg as the profit required is a percentage and the total weight of 5 portions, that is, 5 kgs cancelled out.

Answer: Option d: 40.

Key concepts used: Portion use technique -- Basic ratio concepts -- Basic mixing concepts -- Basic profit and loss concepts.

Problem 4.

When two numbers, in ratio 5 : 7, are reduced each by 40, the ratio changes to 17 : 27. The difference of the numbers is,

  1. 18
  2. 137
  3. 50
  4. 52

Solution 4 : Problem analysis and execution

Introducing the canceled out HCF as $x$ in the ratio, let us form the two numbers as $a=5x$, and $b=7x$.

After reduction of each of the numbers by 40 then,

$\displaystyle\frac{5x-40}{7x-40}=\frac{17}{27}$.

Inverting,

$\displaystyle\frac{7x-40}{5x-40}=\frac{27}{17}$, we are targetting to get the difference, $2x$

Or, $\displaystyle\frac{2x}{5x-40}=\frac{10}{17}$, subtracting 1 from each side.

Now we apply multiple of ratio value technique on the two numerators. By this concept, RHS numerator being 10, the LHS numerator $2x$ must also be a factor of 10.

Using now the free resource of the choice values, we find only the choice value 50 for $2x$ satisfies the condition.

So the difference of the two numbers is 50.

Answer: c: 50.

Key concepts used: Basic ratio concepts -- HCF reintroduction technique, reintroduction of cancelled out HCF, $x$ as a factor of both ratio terms, thus forming the actual values -- Target driven input transformation, forming target difference $2x$ in the numerator by inverting and subtracting 1 from both sides of the equation -- Multiple of ratio value technique -- Basic number domain concepts -- Principle of free resource use, free resource use of choice values -- Choice value set test.

Note: Try solving the problem by another method and compare methods.

Problem 5.

Two numbers $a$ and $b$ are increased by 1 to form the ratio of two numbers thus formed, $c : d$. When again the new numbers are increased by 1 each, the ratio becomes $1:2$. What is the sum of the original two numbers?

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

Solution 5 : Problem analysis and execution

As the first pair of increments don't have any value towards the solution, we focus directly on the second pair of increments that result in,

$\displaystyle\frac{a+2}{b+2}=\frac{1}{2}$.

We need the sum of the two numbers $a$ and $b$, that is, $(a+b)$. So we transform the relation above by adding 1 to both sides to get $(a+b)$ in the numerator,

$\displaystyle\frac{(a+b)+4}{b+2}=\frac{3}{2}$.

This is target driven input transformation.

Now we would apply the powerful Multiple of ratio value technique along with the free resource use of the choice value set test on the numerator.

By Basic number domain concepts, from the RHS and LHS numerators of the expression we can conclude that, The LHS numerator $(a+b)+4$ must be a multiple of 3.

Now we check the choice value set and find that only the choice value of $(a+b)=5$ resulting in $(a+b)+4=9$ results in an LHS that is a multiple of 3.

The sum of the numbers is then 5.

Answer: Option b: 5.

Key concepts used: Problem analysis to ignore the first pair of increments -- Target driven input transformation, forming target sum of $(a+b)$ in the numerator by adding 1 to both sides of the equation -- Multiple of ratio value technique -- Basic number domain concepts -- Principle of free resource use, free resource use of choice values -- Choice value set test.

Note: Usually, in a situation of incrementing numerator and denominator of a ratio, if you can set up the ground well, multiple of ratio value technique with choice value set test will reach you to the solution fastest.

Problem 6.

The expenditure in a hostel is partly variable, and directly proportional to the number of students, rest being a fixed amount. If the expenditures were Rs.8400 for 120 students and Rs.10000 for 150 students, what would be the expenditure for 330 students?

  1. Rs.22400
  2. Rs.19600
  3. Rs.18000
  4. Rs.21500

Solution 6 : Problem analysis and execution

Each of the two expenditures includes the fixed expenditure component along with the variable component that depends on the number of students. To eliminate the fixed expenditure component then we just subtract smaller expenditure from the larger.

Thus, expenditure for $(150-120)=30$ students is, $\text{Rs.}10000-\text{Rs.}8400=\text{Rs.}1600$.

To avoid calculation of the fixed expenditure, we now use the already given value of expenditure of Rs.10000 for 120 students that includes the fixed expenditure.

330 students is an additional number of $(330-150)=180$ students. This is 6 times of unit of 30 students and will increase the expenditure of Rs.10000 for 150 students by,

$\text{Rs.}(6\times{1600})=\text{Rs.}9600$.

The expenditure for 330 students will then be,

$\text{Rs.}10000+\text{Rs.}9600=\text{Rs.}19600$.

Answer: Option b : Rs.19600.

Key concepts used: Problem analysis -- Fixed to variable component relation -- Efficient simplification.

Problem 7.

Two sugar mixtures of concentrations 30% and 40% are mixed together to form a mixture of sugar concentration of 45%. What is the ratio in which the two mixtures are mixed?

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

Solution 7 : Problem analysis and execution

By the given conditions, in 1 kg of first mixture, sugar is 0.3 kg and in 1 kg of second mixture sugar is 0.4 kg (sugar being a soluble solid, it is to be measured by weight).

Let us assume that 1 kg mixture is formed by taking $x$ kg of first mixture and $(1-x)$ kg of second mixture.

Thus sugar quantity in 1 kg of final mixture is,

$0.45=0.3x+0.5(1-x)$,

Or, $0.2x=0.05$,

Or, $x=0.25$.

So the ratio of mixing the first and second mixtures is,

$x :(1-x)=0.25:0.75=1:3$.

Answer: Option a: 1 ; 3.

Key concepts used: Basic ratio concepts -- Basic mixing concepts - Concentration in mixtures technique, which considers one component amount in 1 kg (or 1 litre if all components are liquids) of mixture.

Problem 8.

Two numbers are in ratio of 3 : 4 and their LCM is 120. What is the sum of the numbers?

  1. 50
  2. 55
  3. 60
  4. 70

Solution 8 : Problem analysis and execution

By HCF reintroduction technique, the actual numbers are formed as $3x$ and $4x$ where $x$ is assumed to be the cancelled out HCF between the two numbers.

We know,

$\text{LCM}\times{HCF}=n_1\times{n_2}$, where $n_1$ and $n_2$ are the two numbers involved.

In this problem case then,

$120x=12x^2$,

Or, $12x=120$,

Or, $x=10$.

So sum of the two numbers is,

$3x+4x=7x=70$.

Answer: Option d: 70.

Key concepts used: HCF reintroduction technique -- Basic ratio concepts -- HCF LCM product relation.

Problem 9.

Three containers A, B, and C contain milk and water in the ratio of 5 : 2, 4 : 1, and 4 : 1 respectively. If the capacities of A, B, and C are in the ratio of 3 : 2 : 1 and a mixture is made taking one-third from the first container, one-half from the second and one-seventh from the third, the percentage of milk in the new mixture will be,

  1. 76
  2. 70
  3. 72
  4. 68

Solution 9 : Problem analysis and execution

Milk portion to total portion in the three mixtures respectively are,

5 out of 7,

4 out of 5, and

4 out of 5.

In 1 litre of the three mixtures each, milk amount will then be respectively, $\displaystyle\frac{5}{7}$, $\displaystyle\frac{4}{5}$, and $\displaystyle\frac{4}{5}$ litres.

To incorporate the volume ratio of three containers into the per litre milk amount, let us assume the container volumes as, 3 litres, 2 litres and 1 litre.

When a new mixture is formed by taking one-third from the first, one-half from the second and one-seventh from the third, the total volume of the new mixture will be,

$1+1+\displaystyle\frac{1}{7}=\displaystyle\frac{15}{7}$ litres.

And the milk amount in this volume of new mixture will be,

$\displaystyle\frac{5}{7}+\displaystyle\frac{4}{5}+\displaystyle\frac{4}{35}=\displaystyle\frac{57}{35}$ litres.

Finally then the percentage of milk in the new mixture will be,

$\displaystyle\frac{57}{35}\times{\displaystyle\frac{7}{15}}=\displaystyle\frac{19}{25}=76$%

Answer: Option a: 76%.

Key concepts used: Basic mixing concepts -- Basic ratio concepts -- Concentration technique, which considers amount of one component in 1 litre of mixture (for liquids).

Problem 10.

In a library, the ratio of science books to other books is 7 : 2, the total number of science books being 1512. When a new lot of science books were added to the library collection, the ratio of science books to other books changed to 15 : 4. What was the number of books in the new lot of science books?

  1. 97
  2. 205
  3. 108
  4. 100

Solution 10 : Problem analysis and execution

By using HCF reintroduction technique let's assume cancelled out HCF of the actual values of the ratio terms as $x$.

So number of science books is,

$7x=1512$,

Or, $x=216$.

And number of other books is,

$2x=432$.

When $n$, (say) number of new science books were added to the collection we get a new ratio of science books to other books,

$\displaystyle\frac{7\times{216}+n}{432}=\frac{15}{4}$,

Or, $\displaystyle\frac{7\times{216}+n}{108}=15$,

Or, $n=(15-14)\times{108}=108$.

Answer: Option c: 108.

Key concepts used: Basic ratio concepts -- HCF reintroduction technique -- Sequencing of events -- Efficient simplification.


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