SSC CGL level Solution Set 89 on number system 11 | SureSolv

You are here

For guided learning and practice on Number system and HCF LCM, follow Comprehensive Suresolv Number system HCF LCM Guide with all articles

SSC CGL level Solution Set 89 on number system 11

89th SSC CGL level Solution Set, 11th on topic Number System

SSC-CGL-Solution-set-89-Number-system-11

In this 89th solution set of 10 practice problem exercise for SSC CGL exam and 11th on topic Number System, a few of the problems especially on fractions should pose a bit of challenge to the student.

Solutions are all concept and method based and mostly in mind. The problem solving approach should prove valuable to the students. Its paired question set can be used as a mini-mock test on number system, even for competitive tests other than SSC CGL.

We should mention that in an MCQ test, you need to deduce the answer in shortest possible time and select the right choice. You do not have to write the steps. Writing takes up valuable seconds that you can save by solving problems in mind and writing as little as possible. This is what we call Solving in mind.

All solutions are focused towards solving in mind and so should prove to be especially valuable to the student.

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.


89th solution set - 10 problems for SSC CGL exam: 11th on topic Number System - time 12 mins

Problem 1.

A fraction is greater than its reciprocal by $\displaystyle\frac{9}{20}$. What is the fraction?

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

Solution 1: Problem analysis and solving in mind by factors and multiples concept

We'll go ahead to solve the problem mentally.

Imagine the fraction to be $\displaystyle\frac{a}{b}$. So its reciprocal will be, $\displaystyle\frac{b}{a}$ and the denominator of the difference between the two will be $ab$.

What is $ab$? It is the product of the numerator and denominator of the fraction,

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

The denominator of the difference being 20, $ab$ must be a multiple of 20. This is use of factors multiples concept.

As 20 has two prime factors 4 and 5 ($20=4\times{5}$), $a$ and $b$ must take these two values. At this point we are not saying which of $a$ and $b$ will have these two values.

Note: If $ab$ is a larger multiple of 20 other than the first multiple 20 itself, the multiple factor, say $x$ of 20 must be the factor of both numerator and denominator and so will cancel out. So $a$ and $b$ can take only two values either 5 or 4.

We'll apply the Second logic now for deciding which of the two will be 5 and which one 4.

And that also is simple logic—as $a \gt b$ by assumption, $a$ must be 5 and $b$ must be 4.

The fraction is, $\displaystyle\frac{5}{4}$.

Verify that $a^2-b^2=5^2-4^2=25-16=9$. It satisfies perfectly the given condition,

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

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

Key concepts used: Factors and multiples concept -- Mathematical reasoning -- Result verification -- Solving in mind.

Result verification takes very little time and it makes you confident about your solution.

Problem 2.

A number is greater than 58 times its reciprocal by $\displaystyle\frac{3}{4}$. What is the number?

  1. $8$
  2. $-8$
  3. $-12$
  4. $12$

Solution 2: Problem analysis and solving in mind by mathematical reasoning and choice value test

We will solve the problem mentally but will imagine the required number to be $x$.

So the difference will be,

$x-\displaystyle\frac{58}{x}=\displaystyle\frac{x^2-58}{x}=\frac{3}{4}$.

It is easy to visualize this relation.

So the denominator $x$ will be a multiple of 4, the denominator of the difference.

At this point we'll not bother about the sign of the answer and test out two absolute values 8 and 12 of the choices.

With $x=12$, the numerator $144-58=86$ will be greater than denominator 12 and will violate the difference value $\displaystyle\frac{3}{4}$ which is less than 1.

With $x=8$, the numerator will be $64-58=6$ and the result will be,

$\displaystyle\frac{6}{8}=\frac{3}{4}$.

You have got the absolute value of the answer as 8. Now the sign is to be determined.

$x$ cannot be negative because numerator and the result $\displaystyle\frac{3}{4}$ both are positive.

It takes very little time to reach the solution if you use this approach of mathematical reasoning.

Otherwise taking the conventional approach, you can factorize the equation into its two factors,

$4x^2-3x-232=(x-8)(4x+29)$,

And reason that $x=-\frac{29}{4}$ is not a choice value.

We don't recommend this conventional approach.

Answer: Option a: 8.

Key concepts used: Mathematical reasoning -- Choice value test -- Solving in mind.

Problem 3.

The ten's digit of a two digit number is larger than the unit's digit by 7. If we subtract 63 from the number, the new number obtained is a number formed by interchange of the digits. The new number formed in this way don't have any factor common with the original number. Find the original number.

  1. 92
  2. 29
  3. 70
  4. 18

Solution 3: Problem analysis and solving in mind by mathematical reasoning and choice value test

As ten's digit is larger than the unit's digit by 7, both first and third choice values 92 and 70 may be the answer.

We'll test the second condition of subtraction of 63 and interchange of digits for the two,

$92-63=29$, and

$70-63=07$.

Again both satisfies the second condition.

When taking up the third condition, we find that the digit interchanged new number 29 has no common factor with its parent 92, but 7 is a factor of 70.

So 92 is the answer.

Answer: Option a: 92.

Key concepts used: Mathematical reasoning -- Choice value test -- Solving in mind.

Problem 4.

The reciprocal of the sum of reciprocals of $\displaystyle\frac{2}{9}$ and $\displaystyle\frac{7}{10}$ is,

  1. $\displaystyle\frac{83}{90}$
  2. $\displaystyle\frac{14}{83}$
  3. $\displaystyle\frac{83}{14}$
  4. $\displaystyle\frac{90}{83}$

Solution 4: Problem analysis and execution

First step is to evaluate the sum of reciprocals,

$\displaystyle\frac{9}{2}+\displaystyle\frac{10}{7}=\displaystyle\frac{83}{14}$.

Its reciprocal is, $\displaystyle\frac{14}{83}$.

Answer: Option b: $\displaystyle\frac{14}{83}$.

Key concepts used: Precise problem definition -- Solving in mind.

Apparently the problem is very easy, though its potential to confuse you lies in its use of 'reciprocal' multiple times. Careful understanding the problem statement will result in quick and correct answer.

Problem 5.

Which one is the largest among the fractions, $\displaystyle\frac{5}{113}$, $\displaystyle\frac{7}{120}$, $\displaystyle\frac{13}{125}$ and $\displaystyle\frac{17}{160}$

  1. $\displaystyle\frac{13}{145}$
  2. $\displaystyle\frac{7}{120}$
  3. $\displaystyle\frac{5}{113}$
  4. $\displaystyle\frac{17}{160}$

Solution 5: Problem analysis, key pattern and method discovery

Just a glance at the fractions tells you that there is practically no useful common factor betwen the denominators or the numerators for getting LCM of either of those easily and quickly.

The two fraction comparison method you know of are:

  1. Equalize the denominators to their LCM and select the fraction with LARGEST transformed numerator as the largest fraction. In this case this approach is not practicable.
  2. Equalize the numerators to their LCM. The fraction with the SMALLEST transformed denominator will be the largest fraction. In this case, even this approach is time consuming and calculation intensive.

Okay, then what will you do? How to compare the fractions quickly with least amount of calculation?

With the confidence that there must be a way to quick solution, you think again and take no time to discover the pattern that all four numerators are simple two digit numbers that appear in your school number table, and also the denominators are all close together three digit numbers less than 200.

This is the key pattern discovery and you decide to divide each denominator with its numerator. The resultant denominator of smallest value will belong to the largest fraction. This is the key method associated with the key pattern you have discovered.

Effectively you have decided to equalize the numerators—not to their LCM, but to just unit value 1.

Solution 5: Quick solution of fraction comparison problem by the new method of dividing the denominators by corresponding numerators

For the first fraction you get,

$\displaystyle\frac{5}{113}=\displaystyle\frac{1}{\displaystyle\frac{113}{5}}=\frac{1}{22.6}$.

This is easy and quick. You have multiplied 113 by 2 and divided by 10.

For the second fraction,

$\displaystyle\frac{7}{120}=\displaystyle\frac{1}{\displaystyle\frac{120}{7}}=17.1$, a straight and easy division.

For the third fraction,

$\displaystyle\frac{13}{145}=\displaystyle\frac{1}{\displaystyle\frac{145}{13}}=11.1$, again a quick division.

And for the fourth fraction,

$\displaystyle\frac{17}{160}=\displaystyle\frac{1}{\displaystyle\frac{160}{17}}=9.4$.

The fourth fraction $\displaystyle\frac{17}{160}$ with the smallest result of dividing the denominator by numerator turns out to be the largest one.

It doesn't take you more than 30 to 35 seconds as the divisions you find simple.

Answer: Option d: $\displaystyle\frac{17}{160}$.

Key concepts used: Strategic problem analysis -- Fraction range fitting -- Inclusion in a range problem -- Key pattern and method discovery -- Fraction comparison techniques -- Comparison of fractions by equalizing the numerators to unit value 1 on division of denominator by numerator -- Quick solution.

You needed to write at the most just the results of the four divisions.

Problem 6.

The sum of three consecutive natural numbers each divisible by 5 is 225. The middle number among them is,

  1. 75
  2. 80
  3. 85
  4. 70

Solution 6: Problem analysis and execution:

First divide the sum of 225 and each number by 5 so that the sum of three resultant consecutive numbers is,

$\displaystyle\frac{225}{5}=45$.

This is the sum of three consecutive natural numbers (original numbers divided by 5).

Average of the three must be the middle number 15 so that total of the three is, $3\times{15}=45$.

The original middle number is then,

$5\times{15}=75$.

Answer: Option a: 75.

Key concepts used: Sum of three consecutive natural numbers is three times the average middle number -- Defactorization by dividing the three multiples of 5 with 5 -- Solving in mind.

Problem 7.

A rational number lying between $\displaystyle\frac{1}{2}$ and $\displaystyle\frac{3}{5}$ is,

  1. $\displaystyle\frac{3}{5}$
  2. $\displaystyle\frac{2}{5}$
  3. $\displaystyle\frac{11}{20}$
  4. None of these

Solution 7: Problem analysis and fraction fitting within a range

You have to fit a rational fraction within the given range.

It means the target fraction must be less than the upper range limit $\displaystyle\frac{3}{5}$ and greater than lower range limit $\displaystyle\frac{1}{2}$.

The first choice value being equal to the upper range limit, it is an invalid choice.

The second choice $\displaystyle\frac{2}{5}$ is less than the lower range limit $\displaystyle\frac{1}{2}=\frac{2.5}{5}$.

$\displaystyle\frac{1}{2}$ is equivalent to $\displaystyle\frac{10}{20}$ and so the third choice value $\displaystyle\frac{11}{20}$ is greater than the lower range limit.

Test now whether it is less than the upper range limit $\displaystyle\frac{3}{5}$.

Converting the denominator of this upper limit to 20, the limit gets transformed to $\displaystyle\frac{12}{20}$ which is greater than $\displaystyle\frac{11}{20}$.

The third choice value satisfies both the required conditions and so lies within the given range.

Answer: Option c: $\displaystyle\frac{11}{20}$.

Key concepts used: Fraction range fitting -- Fraction comparison by denominator equalization -- Solving in mind.

Problem 8.

$\left(999\displaystyle\frac{999}{1000}\times{7}\right)$ is equal to,

  1. $7000\displaystyle\frac{7}{1000}$
  2. $6683\displaystyle\frac{7}{1000}$
  3. $6993\displaystyle\frac{7}{1000}$
  4. $6999\displaystyle\frac{993}{1000}$

Solution 8: Problem analysis and execution:

This is a mixed fraction multiplication problem.

To solve this problem with minimal calculation we apply the technique of separating out the mixed fraction into two parts,

$999\displaystyle\frac{999}{1000}=999+\displaystyle\frac{999}{1000}$.

Result of the product of 999 and 7 is 6993.

The numerator of fraction part multiplied by 7 is also 6993 which is 6000 plus 993.

So the final result is,

$6993+6+\displaystyle\frac{993}{1000}=6999\displaystyle\frac{993}{1000}$.

Answer: Option d: $6999\displaystyle\frac{993}{1000}$.

Key concepts used: Mixed fraction arithmetic -- Mixed fraction breakup technique -- Solving in mind.

Problem 9.

The number 323 has,

  1. three prime factors
  2. two prime factors
  3. five prime factors
  4. no prime factor

Solution 9: Problem solving execution

Basically this is problem of factorization.

First test is for divisibility by 3. Integer sum $3+2+3=8$ is not divisible by 3. So 323 doesn't have 3 as a factor.

Next test is for divisibility by 7. Directly dividing 323 by 7 you find no success.

You get same result by dividing next with 11, and then 13.

While testing with 17 for a factor you meet with first success,

$\displaystyle\frac{323}{17}=19$.

17 and 19 both being primes, option b is the answer.

Answer: Option b: two prime factors.

Key concepts used: Factorization -- Divisibility tests -- Prime numbers -- Solving in mind.

Problem 10.

A, B, C and D purchase a gift worth Rs.60. A pays $\frac{1}{2}$ of what others are paying. B pays $\frac{1}{3}$rd of what others are paying and C pays $\frac{1}{4}$th of what others are paying. What is the amount paid by D?

  1. Rs.13
  2. Rs.16
  3. Rs.15
  4. Rs.14

Solution 10: Quick solution by directly evaluating three variables in three steps

Assume $A$, $B$, $C$ and $D$ to be the amounts paid by A, B, C and D. 

So, $A+B+C+D=60$.

By first statement,

$A=\frac{1}{2}(B+C+D)=\frac{1}{2}(60-A)$,

Or, $\frac{3}{2}A=30$,

Or, $A=20$.

By second statement,

$B=\frac{1}{3}(A+C+D)=\frac{1}{3}(60-B)$,

Or, $\frac{4}{3}B=20$,

Or, $B=15$.

By third statement,

$C=\frac{1}{4}(A+B+D)=\frac{1}{4}(60-C)$,

Or, $\frac{5}{4}C=15$,

Or, $C=12$.

So, $D=60-(20+15+12)=60-47=13$.

Answer: Option a: Rs.13.

Key concepts used: Fraction arithmetic -- Direct approach -- Substitution -- Solving in mind.


Guided help on Number system, HCF LCM in Suresolv

To get the best results out of the extensive range of articles of tutorials, questions and solutions on Number system and HCF LCM in Suresolv, follow the guide,

Suresolv Number system, HCF LCM Reading and Practice Guide for SSC CHSL, SSC CGL, SSC CGL Tier II, Bank PO and Other Competitive exams.

The guide list of articles is up-to-date.


Share