SSC CGL level Solution Set 70, Fractions and Surds 6 | SureSolv

SSC CGL level Solution Set 70, Fractions and Surds 6

70th SSC CGL level Solution Set, 6th on topic fractions and surds

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This is the 70th solution set of 10 practice problem exercise for SSC CGL exam and 6th on topic fractions and surds. Students must complete the corresponding question set in prescribed time first and then only refer to this solution set for extracting maximum benefits from this resource.

You may refer to the related tutorials, question and solution sets listed at the end.

70th solution set- 10 problems for SSC CGL exam: 6th on topic Fractions and surds - time 15 mins

Problem 1.

The product of two fractions is $\displaystyle\frac{14}{15}$ and their quotient is $\displaystyle\frac{35}{24}$. The larger of the two fractions is,

  1. $\displaystyle\frac{7}{4}$
  2. $\displaystyle\frac{7}{6}$
  3. $\displaystyle\frac{7}{3}$
  4. $\displaystyle\frac{4}{5}$

Solution 1: Problem analysis and solving

Let's assume the first fraction as $a$ and the second as $b$. The product is,

$ab=\displaystyle\frac{14}{15}$.

Quotient, which is the result of dividing the larger fraction by the smaller is,

$\displaystyle\frac{a}{b}=\frac{35}{24}$.

Multiplying the two, $b$ cancels out leaving $a^2$,

$a^2=\displaystyle\frac{49}{36}$.

So the larger fraction is,

$\displaystyle\frac{7}{6}$.

Answer: Option b: $\displaystyle\frac{7}{6}$.

Key concepts used: Key pattern identification -- basic factor concept --mathematical reasoning.

This solution is fast and wholly in mind. The fractions are to be considered as individual numbers in this case, which poses a mild mental barrier.

Problem 2.

$\displaystyle\frac{4}{5}$, $\displaystyle\frac{7}{8}$, $\displaystyle\frac{6}{7}$, $\displaystyle\frac{5}{6}$ in ascending order is,

  1. $\displaystyle\frac{4}{5}$, $\displaystyle\frac{7}{8}$, $\displaystyle\frac{6}{7}$, $\displaystyle\frac{5}{6}$
  2. $\displaystyle\frac{7}{8}$, $\displaystyle\frac{6}{7}$, $\displaystyle\frac{5}{6}$, $\displaystyle\frac{4}{5}$
  3. $\displaystyle\frac{4}{5}$, $\displaystyle\frac{5}{6}$, $\displaystyle\frac{6}{7}$, $\displaystyle\frac{7}{8}$
  4. $\displaystyle\frac{5}{6}$, $\displaystyle\frac{6}{7}$, $\displaystyle\frac{7}{8}$, $\displaystyle\frac{4}{5}$

Solution 2: Problem analysis

The pattern we immediately discover in all four fractions as,

The difference between the denominator and the numerator is exactly 1. Under this condition, the fraction with larger numerator will be larger.

So the answer would be,

Answer. Option c: $\displaystyle\frac{4}{5}$, $\displaystyle\frac{5}{6}$, $\displaystyle\frac{6}{7}$, $\displaystyle\frac{7}{8}$.


Fraction comparison by equal denominator numerator difference

The principle is,

If the difference between the denominator and the numerator is exactly same for two fractions, then the fraction with larger numerator will be the larger fraction.

This is a powerful fraction comparison rule.

Proof of fraction comparison rule for equal denominator numerator difference

Let two fractions be,

$\displaystyle\frac{a}{b}$ and $\displaystyle\frac{c}{d}$, so that $b-a=d-c=p$.

Subtracting the second from the first,

$\displaystyle\frac{a}{b}-\displaystyle\frac{c}{d}$

$=\displaystyle\frac{ad-bc}{bd}$.

The numerator of the difference,

$ad-bc$

$=ad-bc +bd -bd$

$=-d(b-a)+b(d-c)=p(b-d)$.

So if denominators, $b \gt d$, or numerators $a \gt c$, the first fraction will be larger of the two, and vice versa.

With equal denominator numerator difference, whichever denominator or numerator is larger, the corresponding fraction will the greater one.


Problem 3.

A fraction having denominator 30 and lying between $\displaystyle\frac{5}{8}$ and $\displaystyle\frac{7}{11}$ is,

  1. $\displaystyle\frac{21}{30}$
  2. $\displaystyle\frac{18}{30}$
  3. $\displaystyle\frac{19}{30}$
  4. $\displaystyle\frac{20}{30}$

Solution 3: First solution to inclusion in a range problem

The problem involves fraction comparisons between the lower and the upper range boundaries with the choice values. Without a proper strategy, this process is time consuming.

We will employ Range comparison strategy alongwith easy fraction comparison possibilities.

While comparing the larger fraction of the range $\displaystyle\frac{7}{11}$ with the first choice $\displaystyle\frac{21}{30}$, we equalize the numerators by converting the first fraction to $\displaystyle\frac{21}{33}$.


Fraction comparison by numerator equalization

If one numerator out of the two fractions to be compared is a factor of the second numerator, multiply the first numerator (and also the first denominator) by a suitable factor to make it equal to the second numerator. Now with equal numerators, the fraction with lower denominator will be the larger.

For example, if we have to compare fractions $\displaystyle\frac{7}{9}$ with the fraction $\displaystyle\frac{35}{43}$, finding first numerator 7 as a factor of second numerator 35, we multiply the first numerator and first denominator both by $\displaystyle\frac{35}{7}=5$ to transform the fraction to,

$\displaystyle\frac{35}{45}$.

With the two numerator now equal, as the second denominator 43 is smaller than the first 45, the second fraction will be larger than the first (smaller number dividing same number produces larger result).

In general,

In two fractions,

$\displaystyle\frac{a}{b}$ and $\displaystyle\frac{a}{c}$, numerators are equal, and

if $c \lt b$, second fraction will be larger,

if $b \lt c$, first fraction will be larger.

This is fraction comparison by numerator equalization, which is infrequently used but is equally effective to the more frequently used fraction comparison by denominator equalization.

Fraction comparison by denominator equalization

If two fractions to be compared are $\displaystyle\frac{3}{4}$ and $\displaystyle\frac{19}{24}$, multiply both numerator and denominator of first fraction by $\displaystyle\frac{24}{4}=6$ to transform it to,

$\displaystyle\frac{18}{24}$.

With both denominators now 24, as the first numerator 18 is smaller than the second 19, the second fraction is larger.

In general, in two fractions,

$\displaystyle\frac{a}{b}$ and $\displaystyle\frac{c}{b}$, denominators are equal;

if $c \gt a$, second fraction will be larger,

if $a \gt c$, first fraction will be larger.

This is fraction comparison by denominator equalization.

Direct method of fraction comparison

When you can't use the faster methods of numerator or denominator equalization, you have to go by the conventional direct method.

When we are to compare the two fractions,

$\displaystyle\frac{13}{18}$ and $\displaystyle\frac{17}{24}$,

identify the unique factor in a denominator that is not present in the other denominator and cross-multiply the numerators with these unique factors. Whichever result is larger will result in the larger fraction.

In this case, the unique factor of 18 not present in 24 is 3, and unique factor of of 24 not present in 18 is 4.

As cross-multiplying the numerators with these unique fctors,

$13\times{4}=52$ is larger than $17\times{3}=51$,

the first fraction is larger.

This is a derived method from method of addition of subtraction of fractions where we find the LCM of the two denominators.


Coming back to our problem, in this case, the first choice is not only the largest of the four choices (and that's why to be compaed with the upper range boundary first), it can also be compared easily by numerator equalization.

With equal numerator and smaller denominator then the first choice is larger than the given range and so cannot be our choice.

Next we take up comparing the next smaller choice, $\displaystyle\frac{20}{30}$, or $\displaystyle\frac{2}{3}$ with $\displaystyle\frac{7}{11}$. This time by direct comparison we find the fourth choice also larger than the upper range boundary $\displaystyle\frac{7}{11}$.

Leaving the upper range boundary now we will take the lower range boundary $\displaystyle\frac{5}{8}$ and compare it with the lowest of the four choices, $\displaystyle\frac{18}{30}$, or $\displaystyle\frac{9}{15}$. This second choice being smaller than the lower range boundary, this is also out of consideration leaving the third choice $\displaystyle\frac{19}{30}$ as the only remaining choice possible and so the answer.

Answer: Option c: $\displaystyle\frac{19}{30}$.

Key concepts used: Range comparison strategy -- Fraction comparison by numerator equalization -- Fraction comparison by direct methos -- pattern identification -- efficient simplification -- strategic choice of comparison fraction pairs to minimize calculations -- Range comparison strategy - Analytical approach example.

Though all processing was done mentally, without the strategic selection of fraction pairs for comparison, the time taken would have been longer.


Range inclusion strategy

When upper and lower values of a range are given and we are to find which one of the four given choice values is within the range, we will try to find outside range choices by comparing the largest choice with the upper range value and the smallest choice with the lower range boundary.


Solution 3: Recommended quickest solution

First step: Sorted choice values

First thing we notice is the perfectly sorted four choice values. In this range inclusion type problem where we have to place one of four given values within a given range of two values, we must first mentally arrange the four choice values in sorted (ascending or descending) order. In this case the choice values are very conveniently sorted, and we identify the pattern.

Second step - first comparison: Compare the higher range boundary value with the second largest choice value (and not the largest choice value)

Effectively we halve the set of choice values to be compared with this modified step. This concept is taken from Binary search algorithm.

If result of comparison gives choice value as smaller, in this special case, this particular choice value becomes the answer, as in an MCQ problem there can't be more than one answer, that is, more than one choice value smaller than the higher range but larger than the lower range.

In reality, second largest choice value $\displaystyle\frac{20}{30}$ is larger than the higher range boundary $\displaystyle\frac{7}{11}$, and so is out of range. Only two possible choices are left.

At this step the equivalent action would have been to compare the second lowest choice value with the lower range boundary, but convenience of comparing $\displaystyle\frac{20}{30}$ or $\displaystyle\frac{2}{3}$ with $\displaystyle\frac{7}{11}$ is much more than comparing $\displaystyle\frac{19}{30}$ with $\displaystyle\frac{5}{8}$.

Convenience factor of action

When there are two possible courses of action, we take the action with higher convenience factor that results in quicker solution.

So we compare $\displaystyle\frac{20}{30}$ or $\displaystyle\frac{2}{3}$ with $\displaystyle\frac{7}{11}$ in the first comparison step.

Third step - second comparison

Now we can either compare the higher boundary value with the next lower choice value or use the lower boundary value in comparison. We take the second course of action as convenience factor of comparing $\displaystyle\frac{18}{30}$ or $\displaystyle\frac{9}{15}$ with $\displaystyle\frac{5}{8}$ is more.

This is a systematic process and will give you the result quickest in just maximum of two comparisons.

Problem 4.

If the difference between the reciprocal of a positive proper fraction and the fraction itself be $\displaystyle\frac{9}{20}$, then the fraction is,

  1. $\displaystyle\frac{5}{4}$
  2. $\displaystyle\frac{4}{5}$
  3. $\displaystyle\frac{3}{5}$
  4. $\displaystyle\frac{3}{10}$

Solution 4: Problem analysis and solving execution

Let the fraction be $\displaystyle\frac{a}{b}$. So by the given condition,

$\displaystyle\frac{b}{a}-\displaystyle\frac{a}{b}=\displaystyle\frac{b^2 - a^2}{ab}=\frac{9}{20}$

Only the second choice, $\displaystyle\frac{4}{5}$ satisfies this criterion.

Answer: Option b: $\displaystyle\frac{4}{5}$.

Key concepts used: Fraction arithmetic-- free resource use -- choice value test.

The main challenge in this problem is to recognize that by the problem statement, "the fraction is subtracted from the reciprocal of the fraction" and not the other way round. 

Problem 5.

The sum of three fractions is, $2\displaystyle\frac{11}{24}$. On dividing the largest fraction by the smallest result obtained is $\displaystyle\frac{7}{6}$ which is greater than the middle fraction by $\displaystyle\frac{1}{3}$. The smallest fraction then is,

  1. $\displaystyle\frac{5}{8}$
  2. $\displaystyle\frac{3}{7}$
  3. $\displaystyle\frac{3}{4}$
  4. $\displaystyle\frac{5}{6}$

Solution 5: Problem analysis and solving execution

Out of the choice values, the second choice has 7 in the denominator. By the sum LCM, it is not likely that any fraction has 7 as a denominator. The division value doesn't have 5 in the denominator, so it is not likely that the smallest fraction has 5 in its numerator. Most probable is then the third choice,

$\displaystyle\frac{3}{4}$.

We can make a quick verification check.

Middle fraction is, 

$\displaystyle\frac{7}{6}-\displaystyle\frac{1}{3}=\displaystyle\frac{5}{6}$.

Subtracting this from the sum of three we get, $\displaystyle\frac{39}{24}$ as the sum of the first and third fractions. Subtracting again the value of first we get the third fraction as, $\displaystyle\frac{7}{8}$.

Dividing this by the first produces $\displaystyle\frac{7}{6}$ as desired.

Answer: Option c: $\displaystyle\frac{3}{4}$.

Key concepts used: Pattern identification -- Fraction arithmetic concept -- Factor analysis of fractions -- Mathematical reasoning -- Free resource use of choice set -- Analytical approach example.

Factor analysis of fractions

Many times by analyzing the relationships between the given values, factors as numerator or denominator in the choice value set can be eliminated thus leading to quick solution.

In this problem we have eliminated 5 as numerator and 7 as denominator of the smallest fraction.

This is an example of analytical approach based on basic concepts.

Problem 6.

The value of $\left[\displaystyle\frac{\sqrt{3}+1}{\sqrt{3}-1}+\displaystyle\frac{\sqrt{2}+1}{\sqrt{2}-1}+\displaystyle\frac{\sqrt{3}-1}{\sqrt{3}+1}+\displaystyle\frac{\sqrt{2}-1}{\sqrt{2}+1}\right]$ is,

  1. $10$
  2. $12$
  3. $18$
  4. $14$

Solution 6: Problem analysis and solving execution

We will combine the promising first and third terms first,

$\displaystyle\frac{\sqrt{3}+1}{\sqrt{3}-1}+\displaystyle\frac{\sqrt{3}-1}{\sqrt{3}+1}$

$=\displaystyle\frac{(\sqrt{3}+1)^2+(\sqrt{3}-1)^2}{3-1}$

$=4$, the middle surd terms cancel out.

Similarly combining the second and fourth terms we get,

$\displaystyle\frac{\sqrt{2}+1}{\sqrt{2}-1}+\displaystyle\frac{\sqrt{2}-1}{\sqrt{2}+1}$

$=6$.

So the sum is, $4+6=10$

Answer: Option a: $10$.

Key concepts used: Pattern identification -- Promising term pairing -- Surd algebra.

Done easily in mind. Quick solution depended on the similar pattern identification and term pairing.

Problem 7.

The value of $\displaystyle\frac{1}{\sqrt{12-\sqrt{140}}}-\displaystyle\frac{1}{\sqrt{8-\sqrt{60}}}-\displaystyle\frac{2}{\sqrt{10+\sqrt{84}}}$ is

  1. 3
  2. 0
  3. 1
  4. 2

Solution 7: Problem analysis and solving execution

Whenever we meet a surd expression under a square root, we know the surd expression under the square root must be converted to a square of sum. This problem has three such expressions.

Taking these one by one,

$12 - \sqrt{140}=12-2\sqrt{35}=(\sqrt{7}-\sqrt{5})^2$,

$8-\sqrt{60}=8-2\sqrt{15}=(\sqrt{5}-\sqrt{3})^2$, and 

$10 +\sqrt{84}=10+2\sqrt{21}=(\sqrt{7}+\sqrt{3})^2$.

Surd sums in the brackets are then the three denominators (after taking out of the square roots). We rationalize to form the simplified expression,

$\displaystyle\frac{\sqrt{7}+\sqrt{5}}{2}-\displaystyle\frac{\sqrt{5}+\sqrt{3}}{2}-\displaystyle\frac{\sqrt{7}-\sqrt{3}}{2}$

$=0$, all terms cancel out.

Answer: Option b: 0.

Key concepts used: Conversion to square of sum surd expression-- Surd rationalization.

Problem 8.

If $\displaystyle\frac{4+3\sqrt{3}}{\sqrt{7+4\sqrt{3}}}=A + \sqrt{B}$ then $B-A$ is,

  1. $-13$
  2. $3\sqrt{3}-7$
  3. $13$
  4. $\sqrt{13}$

Solution 8: Problem analysis and solving execution

First we convert the denominator surd expression under square root to a square of sum,

$7 + 4\sqrt{3}=(2+\sqrt{3})^2$.

Rationalizing the transformed denominator $2+\sqrt{3}$, we get the given expession as,

$(4+3\sqrt{3})(2-\sqrt{3})=A + \sqrt{B}$,

Or, $-1+2\sqrt{3}=A+\sqrt{B}$.

As $\sqrt{B}$ is the surd term it must be equal to the surd term on the LHS (irrational and rational can't be added, though we represent an addition we can't arrive at any result by carrying out the addition).

This is what we call Coefficient equalization of similar variables that don't mix together. This is a fundamental algebraic principle.

Thus,

$A=-1$, and

$2\sqrt{3}=\sqrt{B}$,

Or, $B=12$, and

$B-A=13$.

Answer: Option c: 13.

Key concepts used: Surd square of sum conversion -- Surd rationalization -- Surd product -- Similar coefficient equalization.

Problem 9.

The value of $\left(3+\displaystyle\frac{1}{\sqrt{3}}+\displaystyle\frac{1}{3+\sqrt{3}}+\displaystyle\frac{1}{\sqrt{3}-3}\right)$ is,

  1. $1$
  2. $3$
  3. $3-\sqrt{3}$
  4. $3+\sqrt{3}$

Problem analysis and solving execution

Ignoring the first two terms we identify the convenient pattern of the last two terms and decide to combine them first,

$\displaystyle\frac{1}{3+\sqrt{3}}+\displaystyle\frac{1}{\sqrt{3}-3}$

$=-\displaystyle\frac{2\sqrt{3}}{6}$

$=-\displaystyle\frac{1}{\sqrt{3}}$.

This cancels out the second term leaving only 3 as result.

Answer. Option b: $3$.

Key concepts used: Pattern identification -- Convenient term pairing -- Surd factoring by taking $\sqrt{3}$ out of $6$.

Could be carried out in mind.

Problem 10.

If $\displaystyle\frac{\sqrt{a+2b}+\sqrt{a-2b}}{\sqrt{a+2b}-\sqrt{a-2b}}=\sqrt{3}$ then $a:b$ is,

  1. $4:\sqrt{3}$
  2. $\sqrt{3}:2$
  3. $\sqrt{3}:4$
  4. $2:\sqrt{3}$

Solution 10: Problem analysis and solving

We replace temporarily for convenience, $\sqrt{a+2b}=p$ and $\sqrt{a-2b}=q$. This is Component expression substitution and transforms the given expression to a simpler form,

$\displaystyle\frac{p+q}{p-q}=\sqrt{3}$.

Carrying out Componendo dividendo,

$\displaystyle\frac{\sqrt{a+2b}}{\sqrt{a-2b}}=\displaystyle\frac{\sqrt{3}+1}{\sqrt{3}-1}$

$=2 + \sqrt{3}$, rationalizing and simplifying.

Squaring,

$\displaystyle\frac{a+2b}{a-2b}=(2+\sqrt{3})^2=7+4\sqrt{3}$.

Carrying out Componendo dividendo again,

$a:b=2\left(\displaystyle\frac{8+4\sqrt{3}}{6+4\sqrt{3}}\right)$

$=4\left(\displaystyle\frac{2+\sqrt{3}}{3+2\sqrt{3}}\right)$

$=4:\sqrt{3}$, $3+2\sqrt{3}=\sqrt{3}(2+\sqrt{3})$ and cancels out the numerator, leaving the $\sqrt{3}$ in the denominator.

Answer: Option a: $4:\sqrt{3}$.

Key concepts used: Surd simplification -- Surd factorization.


Note: If you are an SSC CGL aspirant, you may refer to our useful article 7 steps for sure success in SSC CGL tier 1 and tier 2 tests in which we have included all our student resources on SSC CGL topics as links. Watch out for more student resources on SSC CGL in this article.

You may also like to go through the Fraction and surd related article directly by referring to the following valuable articles.


Concept tutorials on Fractions, Surds, decimals and related topics

Fractions and Surds concepts part 1

Base equalization technique is indispensable for solving fraction problems

How to solve surds part 1, Rationalization

How to solve surds part 2, Double square root surds and surd term factoring

How to solve surds part 3, Surd expression comparison and ranking


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