SSC CGL level Solution Set 59, Fractions square roots and surds 3

59th SSC CGL level Solution Set, 3rd on topic fractions, square roots and surds

ssc cgl level solution set 59 fractions square roots surds 3

This is the 59th solution set of 10 practice problem exercise for SSC CGL exam and 3rd on topic Fractions, square roots and surds. 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.

In MCQ test, you need to deduce the answer in shortest possible time and select the right choice.

Based on our analysis and experience we have seen that, for accurate and quick answering, the student

  • must have complete understanding of the basic concepts in the topic area
  • is adequately fast in mental math calculation
  • should try to solve each problem using the basic and rich concepts in the specific topic area and
  • does most of the deductive reasoning and calculation in his or her head rather than on paper.

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 companion test yet, first take the test on SSC CGL level Question set 59 on fractions square roots surds 3 and then go through the solutions for gaining maximum benefit.

59th solution set- 10 problems for SSC CGL exam: 3rd on topic Fractions, square roots and surds - time 12 mins

Problem 1.

The value of $\sqrt{32}-\sqrt{128}+\sqrt{50}$ correct to 3 decimal places is,

  1. 1.441
  2. 1.732
  3. 1.414
  4. 1.141

Solution 1 : Problem analysis

To quickly evaluate the three term surd expression we must be able to factorize and simplify the target expression,





The problem reduces to evaluating the value of $\sqrt{2}$ correct to three decimal places.

We have a standard method for finding the square roots. Instead, we will arrive at the solution faster by using number system and algebraic concept based Square root estimation technique.

Solution 1 : Finding value of $\sqrt{2}$ using number system concepts

By the knowledge on squares of two digit numbers we know,

$(1.4)^2=1.96$, which is very near 2 and thus the Option value b of 1.732 and Option value d of 1.141 are eliminated.

We need now to have a dependable estimate of the second digit after the decimal to differentiate between the two options left, choice value a, that is, 1.441 and choice value c, that is, 1.414.

To compare the second digit after decimal for the first option 1.441, we will estimate how much the square of 1.44 is,



$\approxeq [1.96+0.112]$


We ignore the negligible third term of $(0.04)^2$. For such estimations where second digit effect is being considered, this assumption is always valid. The result exceeds 2 by a fair margin.

So the left out choice value 1.414 must be answer.

Answer: c: 1.414.

Key concepts used: Factorization -- principle of free resource use -- digit differentiation -- number system concepts -- algebraic square estimation -- mathematical reasoning -- approximation -- efficient simplification.

Note: The standard method of finding square root is procedural and should take more time.

To us the number system concept based approximation technique of finding square root of 2 is faster. The smallness of the third term of $(0.04)^2)$ makes it possible to ignore it and thus the concept based system becomes viable by using only one simple multiplication.

This technique though depends on use of free resource of the choice values and then differentiation between the two second digits (after decimal) of two choice values. We avoided considering the third digit altogether.

This approach is conceptual. We needed only to add $2\times{(1.4)}\times{(0.04)}$ to $(1.4)^2$.

Problem 2.

The square root of $\displaystyle\frac{\left(3\displaystyle\frac{1}{4}\right)^4-\left(4\displaystyle\frac{1}{3}\right)^4}{\left(3\displaystyle\frac{1}{4}\right)^2-\left(4\displaystyle\frac{1}{3}\right)^2}$ is,

  1. $1\displaystyle\frac{1}{12}$
  2. $1\displaystyle\frac{7}{12}$
  3. $7\displaystyle\frac{1}{12}$
  4. $5\displaystyle\frac{5}{12}$

Solution 2 : Problem analysis and execution

The target expression numerator being in the form of, $a^4-b^4=(a^2+b^2)(a^2-b^2)$, the presence of $a^2-b^2$ in the denominator leaves $a^2+b^2$ in the numerator for final evaluation.

So the target expression simplifies to,



Here we detect the pattern of 13 in the numerator of both the terms being squared and so we take out the square of 13, that is 169,





The square root of the target expression is then,



Answer: Option d : $5\displaystyle\frac{5}{12}$.

Key concepts used: Basic algebraic concepts -- Efficient simplification in the form of recognizing the common factor of $13^2$ in the numerator of both the terms.

Problem 3.

$\sqrt{11.981+7\sqrt{1.2996}}$ is closest to,

  1. 4.1
  2. 5.1
  3. 4.9
  4. 4.5

Solution 3 : Problem analysis and execution:

Instead of detailed calculation we will use approximation and mathematical reasoning based on number system concepts here.

The second term inside the square root, $1.2996$ is very close to $1.3$ and when square root is taken of the actual value $1.2996$, the value will be still closer to the square root of $1.3$. This deduction is by number system concepts.

The assumption is still more valid because the rounding off we are considering is at the position of fourth digit from the decimal. After two square roots, its effect, if at all, will extend up to the second digit after decimal, whereas our choice values are up to the first digit after decimal. This analysis is again based on mathematical reasoning.

Similarly the first term we assume to be approximately equal to 11.98, dropping the .001 effect of which should be negligible.

For estimating value of $\sqrt{1.3}$, we will assess the value to be between 1.1 and 1.2 as, $(1.1)^2=1.21$ and $(1.2)^2=1.44$. We are going to apply the square root range estimation technique.

Target square value 1.3 lying a little below the midpoint of the range 1.21 to 1.44, we will test first the square of 1.14 to see how close it is to 1.3.

Again applying the technique of adding $2\times{1.1}\times{0.04}=0.088$ to 1.21 we get 1.298 which is very close to 1.3.

So we assume, $\sqrt{13}=1.14$ and the target expression,





The second and third choice values of 5.1 and 4.9 are much further away (near 25 inside the sqare root) and so are rejected. In the same way square of 4.1 cannot be near 20 at all (it is less than 17), leaving 4.5 as the answer.

Answer: Option d: 4.5.

Key concepts used: Useful pattern identification, the rounding off at the fourth digit position and third digit position -- square root by number system concept -- approximation -- mathematical reasoning -- number system rich concepts.

Problem 4.

The digit at the unit's place in the square root of 15876 is,

  1. 2
  2. 4
  3. 6
  4. 8

Solution 4 : Problem analysis and execution:

We assess the square root to lie between 120 and 130, with squares 14400 and 16900. Also, the target value of 15876 is nearer to 16900, that is, beyond the midpoint value of 125 (square of which is 15625).

Among the choice values only 4 and 6 can produce a 6 at unit's position after squared.

So the number must be 126, and not 124 as 126 is beyond midpoint value of 125 while square of 124 is less than square of 125 itself.

Answer: c: 6.

Key concepts used: Square root midpoint assessment technique -- unit's digit power concept -- free reource use of choice values -- number system concepts -- mathematical reasoning.

Note: This should be by far the quickest way to arrive at the correct answer all based on basic concepts and sound mathematical reasoning. To test whether our method worked correctly we resort to algebraic technique of finding square,


Problem 5.

The value of $120+3 \text{ of }5\div{\left[7\times{2}\left\{10\div{5}\left(24-10\times{2}+\overline{7+3\times{10}\div{5}}\right)\right\}\right]}$ is,

  1. 120.03
  2. 116.04
  3. 125
  4. 118

Solution 5 : Problem analysis and execution:

The overlined expression within the first brackets is evaluated first as per BODMAS rule resulting in 13. The simplified expression after full evaluation of the expression within first brackets is,

$E=120+3 \text{ of }5\div{\left[7\times{2}\left\{10\div{5}(17)\right\}\right]}$.

Now we evaluate the value of the expression within the curly brackets to be 34 so that we can write the expression within the third brackets as a product of factors,

$E=120+3 \text{ of }5\div{\left[7\times{2}\times{34}\right]}$



Analyzing the choice values we decide the answer to be 120.03 and not any of the other three, 125, 116.04 or 118 by applying number estimation skill.

Answer: Option a: 120.03.

Key concepts used: BODMAS rule -- Efficient simplification -- Number estimation skill.

Problem 6.

Find the value of $\displaystyle\frac{1\displaystyle\frac{7}{9}\text{ of }\displaystyle\frac{27}{64}}{\displaystyle\frac{11}{12}\times{9\displaystyle\frac{9}{11}}}\div{\displaystyle\frac{4\displaystyle\frac{4}{7}\text{ of }\displaystyle\frac{21}{160}}{2\displaystyle\frac{5}{6}\div{2\displaystyle\frac{2}{15}}}}$.

  1. $\displaystyle\frac{421}{2443}$
  2. $\displaystyle\frac{425}{2434}$
  3. $\displaystyle\frac{425}{2344}$
  4. $\displaystyle\frac{425}{2304}$

Solution 6 : Problem analysis and execution:

The numerator of the first term is simplified mentally to $\displaystyle\frac{3}{4}$, the denominator simply to 9, the numerator of the second term to $\displaystyle\frac{3}{5}$ and the denominator of the second term to,


So the target expression is simplied to,




Answer: Option d : $\displaystyle\frac{425}{2304}$.

Key concepts used: Problem breakdown into evaluation of the numerators and the denominators of the two fraction terms -- efficient simplification -- mental maths.

Problem 7.

The value of $5\displaystyle\frac{1}{2}-\left[2\displaystyle\frac{1}{3}\div{\left\{\displaystyle\frac{3}{4}-\displaystyle\frac{1}{2}\left(\displaystyle\frac{2}{3}-\overline{\displaystyle\frac{1}{6}-\displaystyle\frac{1}{8}}\right)\right\}}\right]$ is,

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

Solution 7 : Problem analysis and execution:

By the BODMAS rule, we need to evaluate the expression inside the first brackets first, and within the first brackets the overlined subtraction first. Taking a shortcut, the expression within the first brackets is multiplied and divided by 24, the LCM of 3, 6 and 8,












Answer: Option a: $\displaystyle\frac{1}{6}$.

Key concepts used: BODMAS rule -- Simplififying fraction subtraction and addition by multiplying the expression within the first brackets by the LCM of the denominators of the fractions thus converting the fractions into simple integers, an efficient simplification technique -- simplifying the subtraction at the last step by subtracting the integer part and the fraction part separately, another efficient simplification technique.

Problem 8.

The smallest number that should be added to the number 8958 so that the result is a perfect square is,

  1. 67
  2. 69
  3. 79
  4. 77

Solution 8 : Problem analysis and execution:

To determine the desired perfect square just exceeding 8958 we will use the square root range estimation technique adapted for this problem.

As a first step, we fix easily obtainable upper and lower limits of the range within which the target will lie.

We fix a convenient easily obtainable upper limit of a perfect square exceeding (not just exceeding)  target number 8958 as, $(100)^2=10000$ and the lower limit as, $(90)^2=8100$.

The value 8958 lies a little below the midpoint of the range of 10000 and 8100. As we have to exceed 8958, we decide to choose first, $(95)^2=8100+2(90)5+25=9025$. It just exceeds $8958$, and by our number estimation skill, we presume no other perfect square between 9025 and 8958 can exist.

So we need to add, $9025-8958=67$ to 8958 to get our perfect square.

Still to satisfy our curiosity we can check quickly, $(94)^2=8100+2(90)4+16=8836$.

Answer: Option a: 67.

Key concepts used: Square root range estimation technique -- number system concepts -- basic algebraic concepts -- number estimation skill -- mathematical reasoning.

Problem 9.

The largest number of 5 digits which is a perfect square is,

  1. 99999
  2. 99764
  3. 99976
  4. 99856

Solution 9 : Problem analysis and execution:

To form the estimation range the two perfect square limits of which are on both sides of 100000, the lowest six digit number, we will start with lower limit, $(310)^2=96100$ and upper limit, $(320)^2=102400$.

With this easily obtained working range, we observe the upper limit of 102400 to be fairly closer to the target point of 100000 compared to the lower limit of 96100. Thus we will start our test with, $(317)^2$.

Applying our algebraic method of quick square estimation, we form an estimation of the square as,

$(310)^2+2(310)(7) +49=96100+4340+49\gt 100000$.

We need to test one step down, $(316)^2=(310)^2+2(310)6+36=96100+3600+120+36=99856$.

Answer: d: 99856.

Key concepts used: Square root range estimation technique -- number system concepts -- mathematical reasoning -- basic algebraic concepts -- efficient simplification -- algebraic square estimation -- enumeration technique.

Problem 10.

The simplified value of $\displaystyle\frac{3\sqrt{2}}{\sqrt{3}+\sqrt{6}}-\displaystyle\frac{4\sqrt{3}}{\sqrt{6}+\sqrt{2}}+\displaystyle\frac{\sqrt{6}}{\sqrt{3}+\sqrt{2}}$ is,

  1. $\sqrt{2}$
  2. $\sqrt{3}-\sqrt{2}$
  3. $0$
  4. $\displaystyle\frac{1}{\sqrt{2}}$

Solution 10 : Problem analysis and execution:

The simplest way is to rationalize each of the three denominators,





Answer: Option c:0.

Key concepts used: Surd rationalization -- efficient simplification, we have not tried to factorize the surd terms at all and kept all of them fully under square root.

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.

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