Ratio and Proportion Questions with Solutions SSC CGL Set 84

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Ratio and proportion questions with solutions SSC CGL Set 84

Learn to solve 10 ratio and proportion questions for SSC CGL Set 84 in 12 minutes. All the the questions are solved by specially quick methods.

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SSC CGL level Question Set 84, Ratio proportion 8.

Solutions to 10 Ratio and proportion questions SSC CGL Set 84 - Answering time was 12 mins

Problem 1.

If $A:B:C=2:3:4$, then the ratio $\displaystyle\frac{A}{B}:\displaystyle\frac{B}{C}:\displaystyle\frac{C}{A}$ is equal to,

8 : 9 : 12
8 : 9 : 24
8 : 9 : 16
4 : 9 : 16

Solution 1: Problem analysis and solving in mind conceptually by mathematical reasoning

This is a ratio of ratios problem.

In a ratio, a ratio value represents the relative share of the whole for the corresponding ratio variable. For example, in our problem ratio value 2 for A means ratio variable A contributes to 2 portions or shares to the whole amount of $2+3+4=9$ portions. The 3 portions are contributed by B and the 4 by C. This is one fundamental way of looking at a ratio.

Descriptively stated, the ratio means: for each 2 of A, B is 3 and C is 4.

So when in this ratio we divide A by B, B by C and C by A, we can get corresponding ratio values by dividing 2 by 3, 3 by 4 and 4 by 2, the respective shares or portions of A, B and C by shares of B, C and A.

This makes the changed ratio to,

$\displaystyle\frac{A}{B}:\displaystyle\frac{B}{C}:\displaystyle\frac{C}{A}=\displaystyle\frac{2}{3}:\displaystyle\frac{3}{4}:\displaystyle\frac{4}{2}$.

Converting the fractions to integers by multiplying the values by the LCM of denominators 12 we get,

$\displaystyle\frac{A}{B}:\displaystyle\frac{B}{C}:\displaystyle\frac{C}{A}=8:9:24$.

Mentally this conceptual solution can be reached very quickly.

We will show you a more deductive solution to this interesting problem.

Deductive Solution to Problem 1.

$A:B:C=2:3:4$.

From this we get three ratios,

$A:B=2:3$

$B:C=3:4$, and

$C:A=4:2$.

Taking ratio of the first two,

$\displaystyle\frac{A}{B}:\displaystyle\frac{B}{C}=\frac{8}{9}$

And taking ratio of the second and the third,

$\displaystyle\frac{B}{C}:\displaystyle\frac{C}{A}=\frac{6}{16}=\frac{3}{8}$.

Joining the two by multiplying and dividing the second ratio values by 3, we get,

$\displaystyle\frac{A}{B}:\displaystyle\frac{B}{C}:\displaystyle\frac{C}{A}=8:9:24$.

Answer: Option b: 8 : 9 : 24.

Key concepts used: Basic ratio concepts -- Portions cioncepts -- Splitting of a ratio -- Joining a ratio -- Ratio of ratios -- Modifying a ratio term and its corresponding value -- Solving in mind.

Problem 2.

If $p:q=r:s=t:u=2:3$, then $(mp+nr+ot):(mq+ns+ou)$ is equal to,

1 : 3
3 : 2
1 : 2
2 : 3

Solution 2: Problem analysis and solving in mind by use of actual ratio values based on HCF reintroduced concept : Ratio merging

By the given ratio, $p$, $r$ and $t$ represents $2x$, $2y$ and $2z$ actual values while, $q$, $s$ and $u$ represents, $3x$, $3y$ and $3z$ actual values, where $x$, $y$ and $z$ are the three HCFs reintroduced for the three ratios respectively.

Substituting these values in the target ratio we have,

$(mp+nr+ot):(mq+ns+ou)$

$=(2xm+2yn+2zo):(3xm+3yn+3zo)$

$=2:3$.

Answer: Option d: 2 : 3.

Key concepts used: Basic ratio concepts -- HCF reintroduction technique -- Use of actual ratio values in a new ratio -- Ratio merging -- Solving in mind.

This is case of ratio merging (not joining). When equivalent ratio variables of three (or more) equal ratios are multiplied by same factors for numerator and denominators, the ratio of the sum of their products remains unchanged. Even if similar operations were performed for the numerator and denominator variables instead of all additions, the ratio would have remained unchanged. Though this is an advanced concept on ratios, it becomes believable by using the basic concept to build the advanced concept.

Problem 3.

Find the mean proportionals between 2 and 54.

6 and 18
6 and 9
12 and 18
6 and 12

Solution 3: Problem analysis and solving in mind by mean proportional concept and trial

Mean proportionals of two numbers $a$ and $d$ are such that, $a:b=c:d$.

By trial 6 and 18 satisfies this criterion,

$2:6=18:54=2:3$.

Answer: Option a: 6 and 18.

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

Problem 4.

The reciprocals of squares of the numbers $1\displaystyle\frac{1}{2}$ and $1\displaystyle\frac{1}{3}$ are in the ratio,

81 : 64
4 : 85
64 : 81
8 : 9

Solution 4: Problem analysis and solving in mind by ratio of two fractions concept

Converting the mixed fractions we get, $\displaystyle\frac{3}{2}$ and $\displaystyle\frac{4}{3}$.

Reciprocals of squares of the two are,

$\displaystyle\frac{4}{9}$ and $\displaystyle\frac{9}{16}$.

Ratio of the two will be,

$\displaystyle\frac{\displaystyle\frac{4}{9}}{\displaystyle\frac{9}{16}}=\frac{64}{81}$.

Answer: Option c: 64 : 81.

Key concepts used: Basic ratio concept -- Ratio of fractions by division -- Solving in mind.

Problem 5.

If 177 is divided into three parts in the ratio, $\displaystyle\frac{1}{2}:\displaystyle\frac{2}{3}:\displaystyle\frac{4}{5}$, then the second part is,

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

Converting the fraction ratio to integer ratio by multiplying by the LCM 30 of the denominators we get,

$15:20:24$.

Total number of portions is then,

$15+20+24=59$, and the value of each portion,

$\displaystyle\frac{177}{59}=3$.

So the value of the second part, that is, 20 portions is 60.

Answer: Option a: 60.

Key concept used: Basic ratio concept --- Amount division in a ratio -- Fraction ratio conversion -- Portions concept -- Solving in mind.

Problem 6.

If the ratio of two numbers is 1 : 5 and their product is 320, then the difference between squares of these two numbers is,

1536
1024
1435
1256

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

Assuming $x$ to be the cancelled out HCF, the two numbers are, $x$ and $5x$ respectively and their product,

$5x^2=320$,

Or, $x^2=64$

Difference of squares of the two numbers is,

$25x^2-x^2=24x^2=24\times{64}=1536$.

To speed up we didn't evaluate the numbers.

Answer: Option a: 1536.

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

Problem 7.

If a reduction in number of workers in a factory is in the ratio 15 : 11 and an increment in their wages in the ratio 22 : 25, then the ratio by which the total wage of the workers is decreased is,

3 : 7
3 : 5
5 : 6
6 : 5

Solution 7: Problem analysis and solving in mind by product of ratios concept

Total wage is the product of number of workers and wage per worker. Assuming $x$ and $y$ to be the cancelled out HCFs of the two ratios respectively, we get the ratio of original total wage to the changed total wage as,

$\displaystyle\frac{15xy\times{22}}{11xy\times{25}}=\frac{6}{5}$.

Before change, actual number of workers and wage per worker were $15x$ and $22y$, and after change these became, $11x$ and $25y$. Total wage in both cases was the product of the two. This is why we could take product of the ratios.

Answer: Option d: 6 : 5.

Key concepts used: Basic ratio concepts-- Product of ratios -- Total wage as a product of number of workers and wage per worker -- HCF reintroduction technique -- Solving in mind.

Problem 8.

Four numbers are in the ratio 1 : 2 : 3 : 4. Their sum is 16. The sum of the first and the fourth number is equal to,

Solution 8: Problem analysis and solving in mind portions concept

Total number of portions is 10 of value 16. So each portion value is 1.6 and the sum of the first and fourth numbers is, 5 portions of value, 8.

Answer: Option c: 8.

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

Problem 9.

A sum of Rs.730 is divided among A, B and C in such a way that if A gets Rs.3, then B gets Rs.4 and if B gets Rs.3.50 then C gets Rs.3. The share of B exceeds that of C by,

Rs.30
Rs.210
Rs.40
Rs.70

Solution 9: Problem analysis and solving in mind by fraction ratio conversion, ratio joining and portions concept

Assuming $a$, $b$ and $c$ to be the amount of shares of A, B and C,

$a:b=3:4$, and $b:c=3.50:3$.

Converting the second ratio to ratio of integers,

$b:c=7:6$.

We have to join the two ratios to the ratio of $a:b:c$ for which we have to equalize the middle ratio value of $b$ to the LCM 28.

Accordingly converting the first ratio we get,

$a:b=21:28$, and the second ratio to,

$b:c=28:24$.

Now joining the two ratios,

$a:b:c=21:28:24$.

Total number of portions is,

$21+28+24=73$.

As total amount is Rs.730, value of each portion is Rs.10.

Share of B exceeds that of C by 4 portions, that is by Rs.40.

Answer: Option c: Rs.40.

Key concepts used: Basic ratio concepts -- Fraction ratio conversion -- Ratio joining -- Portions concept -- Amount division in a ratio -- Solving in mind.

Problem 10.

If 378 coins consist of 1 rupee, 50 paise and 25 paise coins whose values are in the ratio of 13 : 11 : 7, the number of 50 paise coins will be,

Solution 10: Problem analysis and solving in mind by Ratio value unit conversion

The ratio of values of 1 rupee, 50 paise and 25 paise coins is,

$13:11:7$.

Actual values being product of each with cancelled out HCF, say, $x$, we can get the ratio of their numbers by multiplying each ratio value by the corresponding number of coins making up a rupee (the original unit), that is, by 1, 2 and 4, getting,

$13:22:28$.

Total portions is, $13+22+28=63$ and total portion value, 378. So value of each portion is,

$\displaystyle\frac{378}{63}=6$.

Number of 50 paise coins is then, $22\times{6}=132$.

Effectively we have changed the given ratio value unit of rupee to new ratio value unit of number of coins by multiplying each ratio value with number of coins equivalent to 1 unit of rupee.

Answer: Option b: 132.

Key concepts used: Basic ratio concept -- Money value ratio to number of coins ratio -- Ratio value unit conversion -- Solving in mind.

Overall, all the 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, ratio of fractions by division, division of ratios, fraction ratio conversion, total wage as a product of number of workers and the wage per worker, speed up techniques, product of ratios concept, splitting of ratios, ratio joining, portions concept, ratio merging, mean proportional, amount division in a ratio, money value ratio to number of coins ratio conversion, and ratio value unit conversion. 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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SSC CGL level Solution Set 84, Ratio proportion 8