SSC CGL level Solution Set 17, Fractions Decimals and Surds

Seventeenth SSC CGL level Solution Set, topic Fraction Decimals and Surds

SSC CGL solution set17 fractions decimals surds

This is the seventeenth solution set of 10 practice problem exercise for SSC CGL exam on topic Fraction Decimals and Surds. Students should complete the corresponding question set in prescribed time first and then only refer to the solution set.

We found from our experience that this set of topics together pose some degree of difficulties to the students in a way similar to Algebra. Without going into the reasons behind the difficulties we will try to address them through detailed explanation of efficient problem solutions.

Before attempting this timed question set, you may like to go through our tutorial article on Fractions and Surds concepts part 1 in which this set of problems was given as an exercise.

Seventeenth solution set- 10 problems for SSC CGL exam: topic Fractions Decimals and Surds - time 12 mins

Problem 1.

How much more is $\sqrt{12} + \sqrt{18}$ compared to $\sqrt{3} + \sqrt{2}$?

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


If surd terms are not same you can't do any addition or subtraction between the terms. With this knowledge let us transform the higher valued surd terms because those should have square factors that can be taken out of the square roots.

$(\sqrt{12} + \sqrt{18}) - (\sqrt{3} + \sqrt{2})$

$=(2\sqrt{3} + 3\sqrt{2}) - (\sqrt{3} + \sqrt{2})$

$=\sqrt{3} + 2\sqrt{2}$

Answer: Option a: $\sqrt{3} + 2\sqrt{2}$.

Key concepts used: Surd arithmetic -- factorization.

Problem 2.

The value of $\displaystyle\frac{1}{30} + \displaystyle\frac{1}{42} + \displaystyle\frac{1}{56} + \displaystyle\frac{1}{72} + \displaystyle\frac{1}{90} + \displaystyle\frac{1}{110}$ is,

  1. $\displaystyle\frac{1}{9}$
  2. $\sqrt{2}\displaystyle\frac{2}{27}$
  3. $\displaystyle\frac{6}{55}$
  4. $\displaystyle\frac{5}{27}$


If you try to add together all the six fractions it usually turns out to be a lengthy time-consuming process. As it is expected that this long fraction sum may be completed in a short time it is highly probable that the denominators have a good amount of factor commonality.

With this reasoning as we examine the denominators closely we find the pattern,

$\displaystyle\frac{1}{30} + \displaystyle\frac{1}{42} + \displaystyle\frac{1}{56} + \displaystyle\frac{1}{72} + \displaystyle\frac{1}{90} + \displaystyle\frac{1}{110}$

$=\displaystyle\frac{1}{5\times{6}} + \displaystyle\frac{1}{6\times{7}} + \displaystyle\frac{1}{7\times{8}}$

$\hspace{5mm} + \displaystyle\frac{1}{8\times{9}} + \displaystyle\frac{1}{9\times{10}} + \displaystyle\frac{1}{10\times{11}}$

$=\displaystyle\frac{2}{35} + \displaystyle\frac{2}{63} + \displaystyle\frac{2}{99}$

$=\displaystyle\frac{2}{5\times{7}} + \displaystyle\frac{2}{7\times{9}} + \displaystyle\frac{2}{9\times{11}}$

$=\displaystyle\frac{4}{45} + \displaystyle\frac{2}{9\times{11}}$

$=\displaystyle\frac{4}{5\times{9}} + \displaystyle\frac{2}{9\times{11}}$

$=\displaystyle\frac{1}{9}\left( \displaystyle\frac{4}{5} + \displaystyle\frac{2}{11}\right)$



At the last step only we have shown the technique of factorng out the one common factor from the summation of a pair of fraction terms, making the summation process quick and easy.

Answer: Option a : $\displaystyle\frac{6}{55}$

Key concepts used: Pattern recognition -- Pairwise simplification technique -- factorization -- factoring out.

Problem 3.

$3.\overline{36} - 2.\overline{05} + 1.\overline{33}$ equals,

  1. $2.64$
  2. $2.\overline{64}$
  3. $2.60$
  4. $2.\overline{61}$


The first thought that comes to our mind, how can we convert the three overlined awkwards! But we remembered that the integers can always be separated from the decimals and the operations of addition or subtraction be carried on separately. That will simplify matters a bit. So we take the first step,

$3.\overline{36} - 2.\overline{05} + 1.\overline{33}$

$=(3 - 2 + 1) + (0.\overline{36} - 0.\overline{05} + 0.\overline{33})$

$=2 + (0.\overline{36} - 0.\overline{05} + 0.\overline{33})$

Now we discover the first rule by looking at the overlined decimals, all the three are two digit overlines. It means, these three pairs of digits will repeat indefinitely, but always will appear at exactly same places coinciding with each other.

Because of adherence to the second rule that the subtraction or addition does not create any carry or borrow, we can resort to direct normal arithmetic on the same length repeating decimals as if the overlines don't exist. We call this as Simple repeating decimal aritmetic technique. To be able to do this, the two conditions should be satisfied.

In other words, we can directly carry out the addition and subtraction operations on them with the overline function working like, say a bracket enclosing the three. This place value synchronization of all the digits of three overlines with each other occurs because of their equal length of two.

This technique is classified under the broader set of concepts, the basic concept of overline arithmetic or more formally, the basic concept of repeating non-terminating decimal arithmetic.

Naturally a question comes up, what happens if the number of repeating digits were all different? Think over. We will answer that later. But for now we will simply do our decimal arithmetic on the three decimals as if the overlines do not exist.

As we don't convert the repeating decimals to fractions, this simple technique takes up much smaller time.

Example, $0.\overline{6} - 0.\overline{3} = 0.\overline{3}$.

In our case,

$0.\overline{36} - 0.\overline{05} + 0.\overline{33} = 0.\overline{64}$

So our result is,


Answer: Option b: $2.\overline{64}$.

Key concepts used: Separating out the integers and decimals for arithmetic operations separately -- Simple repeating decimal arithmetic technique.

Problem 4.

$\displaystyle\frac{1\displaystyle\frac{1}{4} \div{1\displaystyle\frac{1}{2}}}{\displaystyle\frac{1}{15} + 1 -\displaystyle\frac{9}{10}}$ is equal to,

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


Target expression,

$E=\displaystyle\frac{1\displaystyle\frac{1}{4} \div{1\displaystyle\frac{1}{2}}}{\displaystyle\frac{1}{15} + 1 -\displaystyle\frac{9}{10}}$

$\hspace{5mm}=\displaystyle\frac{\displaystyle\frac{5}{4} \div{\displaystyle\frac{3}{2}}}{\displaystyle\frac{1}{15} + \displaystyle\frac{1}{10}}$.

We simplify bit by safe bit.





Answer: Option a: $5$.

Key concepts used: Compound fraction simplification -- use of simplification technique, simplifying a bit as soon as it is possible and simplifying in steps to ensure correctness.

Problem 5.

$\displaystyle\frac{1}{3 - \sqrt{8}} - \displaystyle\frac{1}{\sqrt{8} - \sqrt{7}} + \displaystyle\frac{1}{\sqrt{7} - \sqrt{6}}$

$\hspace{10mm}- \displaystyle\frac{1}{\sqrt{6} - \sqrt{5}} + \displaystyle\frac{1}{\sqrt{5} - 2} =$

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


The first step to take with any surd fraction is to rationalize them eliminating irrationals in the denominator. Furthermore when we examine the five denominators we find that on squaring the two terms in each, the difference will be 1.

Thus the transforming target expression we have,

$E=\displaystyle\frac{3 + \sqrt{8}}{3^2 - 8} - \displaystyle\frac{\sqrt{8} + \sqrt{7}}{8 - 7} + \displaystyle\frac{\sqrt{7} + \sqrt{6}}{7 - 6}$

$\hspace{10mm} - \displaystyle\frac{\sqrt{6} + \sqrt{5}}{6 - 5} + \displaystyle\frac{\sqrt{5} + 2}{5 -4}$

$\hspace{5mm}=(3 + \sqrt{8}) - (\sqrt{8} + \sqrt{7}) + (\sqrt{7}  + \sqrt{6}) $

$\hspace{10mm} - (\sqrt{6} + \sqrt{5}) + (\sqrt{5} + 2)$

$\hspace{5mm}=5$, all irrationals conveniently canceling out.

Answer: Option d: 5.

Key concepts used: Surd rationalization technique.

Problem 6.

$(159\times{21} + 53\times{87} + 25\times{106})$ equals

  1. 16000
  2. 1060
  3. 60100
  4. 10600


The only way to avoid large multiplications involved in the given expression is to look for common factors in a pair of products and taking the common factor out of the two terms. We call this as Pairwise simplification technique.

In a long fraction addition or subtraction or a long sum of products, this technique only can produce quick efficient results. Formally we state this rich simplification concept and technique as,

In a  long series of sum of products or fraction addition, simplify by taking up the operation on a pair of terms, simplifying it and then taking up the next pair of terms. In other words, simplify the terms pair by pair. The pairs are to be chosen suitably by common pattern identification.

In our problem applying this technique we get,

$(159\times{21} + 53\times{87} + 25\times{106})$

$=53(3\times{21} + 87) + 25\times{106}$

$=3\times{53}\times{(21 + 29)} + 25\times{106}$

$=159\times{50} + 50\times{53}$

$=50\times(159 + 53)$



The easy way to multiply a number by 50 is to multiply it by 100 and then divide by 2.

Answer: Option d : 10600.

Key concepts used: Pairwise simplification technique -- Factorization -- taking out common factor.

Problem 7.

$\displaystyle\frac{0.8\overline{3} \div{7.5}}{2.3\overline{21} - 0.0\overline{98}}$ equals

  1. 0.1
  2. 0.6
  3. 0.05
  4. 0.06


The first term in the numerator can simply be converted by inspection to a fraction as,

$0.8\overline{3} = \displaystyle\frac{1}{10}(8.\overline{3}) = \displaystyle\frac{1}{10}\left(8\frac{1}{3}\right)=\displaystyle\frac{25}{30}$

The numerator thus turns to,

$\displaystyle\frac{25}{30} \div{7.5} = \displaystyle\frac{1}{9}$.

Here we have used the technique of transforming the overlined decimal in such a way that the overlined digits appear just after the decimal point. This is done by taking out factors of suitable powers of 10. 

This is the form of pure overlined decimal with no other leading digit inside the decimal. By applying a simple algebraic technique we can then transform such a pure overlined decimal into a fraction easily. Note that unless we convert a repeating rational decimal into a fraction, simplification may not be possible.

Let us see how it works.

Conversion of repeating rational decimal to a fraction

First step: converting the number to a pure ovelined decimal with no leading digit before the repeating digits on their left

Let us take up the first term in the denominator to show how the method works,

$2.3\overline{21} = \displaystyle\frac{1}{10}(23 + 0.\overline{21})$.

We have isolated the pure overlined decimal as $0.\overline{21}$. Now we will transform it to a fraction.

Second step: transformation of repeating rational decimal to a fraction

Let's assume,

$x = 0.\overline{21}$,

Multiplying both sides by 100,

$100x = 21.\overline{21} = 21 + x$,

Or $x = \displaystyle\frac{21}{99}$.

Just note, instead of $21$ if the repeating digits were $98$, the same method would have produced instead,

$x = \displaystyle\frac{98}{99}$.

This is the advantage of applying a template of a technique. If you know how the technique works and how to use it, without the proving or deduction steps you can arrive at the results directly using the pattern.

Instead of two repeating digits if it were 3, work it out in your mind and you will find that it will follow the same simple pattern. For example,

$0.\overline{213} = \displaystyle\frac{213}{999}$, instead of 100 we multiply $x$ here by 1000 and then transpose the terms.

So the first term in the denominator is,

$\displaystyle\frac{1}{10}\left(23 + \displaystyle\frac{21}{99}\right)$.

In the same way the second term in the denominator is,

$\displaystyle\frac{1}{10}(0.\overline{98}) = \displaystyle\frac{1}{10}\left(\displaystyle\frac{98}{99}\right)$

Thus the target expression is,

$\displaystyle\frac{10}{9\left(23 + \displaystyle\frac{21}{99} - \displaystyle\frac{98}{99}\right)}$

$=\displaystyle\frac{10}{9\left(23 - \displaystyle\frac{77}{99}\right)}$

$=\displaystyle\frac{10}{9\left(23 - \displaystyle\frac{7}{9}\right)}$



Answer: Option c: 0.05.

Key concepts used: Conversion of a term with repeating non-terminating decimal to fraction -- simplification.

Note: If we know how to convert a two digit repeating non-terminating decimal to a fraction, we can easily replace the decimal with the fraction of regular pattern without going through the conversion process. But in any case, when dealing with decimals we must be careful about the negative powers of 10.

Problem 8.

The value of $(\sqrt{72} - \sqrt{18}) \div{\sqrt{12}}$ is,

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


Let us factorize the terms for canceling out,

$(\sqrt{72} - \sqrt{18}) \div{\sqrt{12}}$

$=(\sqrt{12} - \sqrt{3}) \div{\sqrt{2}}$, canceling out $\sqrt{6}$

$=(2\sqrt{3} - \sqrt{3}) \div{\sqrt{2}}$

$= \displaystyle\frac{\sqrt{6}}{2}$.

Answer: Option b: $\displaystyle\frac{\sqrt{6}}{2}$.

Key concepts used: Factorization and simplication of surd terms.

Problem 9.

The value of $\sqrt{900} + \sqrt{0.09}  - \sqrt{0.000009}$ is,

  1. 30.27
  2. 30.097
  3. 30.197
  4. 30.297


It is a straightforward case of simplification of terms under square roots,

$\sqrt{900} + \sqrt{0.09}  - \sqrt{0.000009}$

$=\sqrt{9\times{100}} + \sqrt{9\times{10^{-2}}} - \sqrt{9\times{10^{-6}}}$

$=30 + 0.3 - 0.003$


Answer: Option d: 30.297.

Key concepts used: Factorization -- Simplification of square root terms.

Problem 10.

The value of $(3 + \sqrt{8}) + \displaystyle\frac{1}{3 - \sqrt{8}} - (6 + 4\sqrt{2})$ is,

  1. $0$
  2. $1$
  3. $8$
  4. $\sqrt{2}$


Finding a surd denominator first we rationalize it and then take further simplifying steps,

$(3 + \sqrt{8}) + \displaystyle\frac{1}{3 - \sqrt{8}} - (6 + 4\sqrt{2})$

$=(3 + \sqrt{8}) + (3 + \sqrt{8}) - (6 + 4\sqrt{2})$

$=(6 + 4\sqrt{2}) - (6 + 4\sqrt{2})$


Answer: Option a: 0.

Key concepts used: Surd rationalization -- simplification.

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,

Tutorial: on Fractions and Surds concepts part 1.

Question sets and Solution Sets in Fractions and related topics

SSC CGL level Question Set 61 on fractions indices surds 5

SSC CGL level Solution Set 61 on fractions indices surds 5

SSC CGL level Question Set 60 on fractions indices surds 4

SSC CGL level Solution Set 60 on fractions indices surds 4

SSC CGL level Question Set 59 on fractions square roots and surds 3

SSC CGL level Solution Set 59 on fractions square roots and surds 3

SSC CGL level Question Set 47 on fractions decimals and surds 2

SSC CGL level Solution Set 47 on fractions decimals and surds 2

SSC CGL level Question Set 17 on fractions decimals and surds 1