SSC CGL level Solution Set 55, Number System 9

55th SSC CGL level Solution Set, 9th on topic Number System

SSC CGL solution set 55 number system 9

This is the 55th solution set of 10 practice problem exercise for SSC CGL exam and 9th on topic Number System. 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 corresponding test yet, you should refer to the companion, SSC CGL level Question Set 55 on Number System 9 and then after taking the test in prescribed time, return to go through the detailed concepts in the solution processes here.

55th solution set- 10 problems for SSC CGL exam: 9th on topic Number System - time 15 mins

Problem 1.

The smallest number that can be added to 803642 in order to obtain a multiple of 11 is,

  1. 1
  2. 7
  3. 9
  4. 4

Solution 1 : Problem analysis

If the divisor were a larger number for which no divisibility test is known, we would have to divide the given number by the divisor and find the shortfall to the next multiple by subtracting the remainder from the divisor. We will show this conventional method as an alternate solution.

In this case though a well known divisibility test for 11 operates on the sum of alternate digits. Taking this path, we will solve the problem much faster avoiding any division.

Solution 1 : Problem solving execution

The 11 divisibility test says,

If the difference between the sums of alternate digits of a number is either 0 or divisible by 11, the number itself will be divisible by 11.

In our problem, for the given number 803642, the sum of odd digits starting from unit's digit is 8 and sum of even digits is 15. The shortfall is 7.

By adding 7 to the unit's digit we can make this difference 0 and the number divisible by 11. As we are adding 7 to the unit's digit where 7 is less than 11, it is the smallest number that can be added to the number 803642 to make it divisible by 11.

Note: We could have added also $(7+11)=18$ (moving to next multiple of 11, so that the difference of sum of alternate digits is 11), or 70000 (adding the shortfall 7 to the 5th digit) as two of the many ways we could have made the result divisible by 11. But among all such possibilities, 7 is the smallest number that serves the purpose.

Answer: b: 7.

Alternate solution

Dividing 803642 by 11 mentally we get the quotient as, 73058 and remainder as 4. So adding the shortfall $(11-4)=7$ to 803642, we would reach the next 73059th multiple of 11, and 7 is thus the smallest such number.

This alternate solution is also quick, but not as quick as the first divisibility test method. In actual exam saving of seconds count.

Furthermore, the first solution being simpler, chances of error are less. One might make mistakes in a division operation, chances of error in a simple addition is much less.

This is application of Many ways technique of solving a problem in more than one way. This helps comparison of various methods for solving the same problem and improves problem solving skill.

Key concepts used: Divisibility tests -- Place value concept -- Basic number domain concepts -- Factors multiples concept -- Shortfall concept -- Remainder concept -- Many ways technique.

Problem 2.

A six digit number is formed by repeating a three digit number twice. For example, two such numbers are, 123123 and 728728. Any number of this form will always be divisible by,

  1. 7 only
  2. 13 only
  3. 1001
  4. 11 only

Solution 2 : Problem analysis and execution

We understand that any problem involving a pattern of digits as given needs breaking up of the number using Place value mechanism.

So if such a number as described in general is, $abcabc$, where the repeating 3 digit number is $abc$, using place value concept we can express the number as,

$abcabc=10^5a+10^4b+10^3c+10^2a+10b+c$

$=(10^2a+10b+c)(10^3+1)$

$=1001(100a+10b+c)$.

So the number is divisible by 1001, but not by 7, 11 or 13 alone as all of these three are factors of 1001.

Answer: Option c : 1001.

Key concepts used: Place value mechanism -- Break up of a number -- Basic algebra concepts -- Basic factors multiples concept.

Problem 3.

The product of two positive numbers is 11520 and the result of their division is $\displaystyle\frac{9}{5}$. The difference between the numbers is,

  1. 70
  2. 60
  3. 64
  4. 74

Solution 3 : Problem analysis and execution:

The fraction form of the division result urges us to use the Basic ratio concept and introduce the cancelled out HCF, say $x$ as a multiplicative factor to both numerator and denominator of the result of division to generate the actual values of the two quantities represented by the numerator and the denominator.

By this action the actual value of the larger number becomes $9x$ and the smaller $5x$.

The product of the two numbers is then,

$45x^2=11520$.

With a quick mental division of the number 11520 by 45 we get,

$x^2=256$,

Or, $x=16$,

So the difference between the two numbers is,

$9x-5x=4x=64$.

We didn't even calculate the individual values of the numbers. That's efficient simplification.

Answer: Option c: 64.

Key concepts used: Basic ratio concepts -- Problem analysis -- Efficient simplification.

Problem 4.

If $a$ and $b$ are two odd positive integers, by which of the following positive integers is $(a^4-b^4)$ is always divisible?

  1. 12
  2. 3
  3. 6
  4. 8

Solution 4 : Problem analysis and execution:

We decide this problem needs expansion of the given expression in $a$ and $b$ in terms of multiplicative factors and then apply odd even number concept on the factors.

Expanding the given expression we get,

$a^4-b^4=(a^2+b^2)(a^2-b^2)$

$=(a^2+b^2)(a+b)(a-b)$.

What can we certainly say on these three factors?

The first factor $(a^2+b^2)$ is sum of squares of two odd numbers each of which should be odd and so the sum should be even and must have a factor of 2.

The second and third factors are the sum and subtraction of two odd numbers, and so each is even having a factor of 2.

All three terms of the product expansion then have a 2 as a factor. Three 2s multiplied together then makes 8 as the number that will always divide the given expression.

Answer: d: 8.

Key concepts used: Basic algebra concepts -- basic factor multiples concept -- odd even number concept.

Problem 5.

The number 45 is written as a sum of four numbers so that when 2 is added to the first number, 2 is subtracted from the second number, the third number is multiplied by 2 and the fourth divided by 2, the results we get are the same. The four numbers are,

  1. 8, 12, 10, 15
  2. 1, 8, 15, 21
  3. 2, 12, 5, 26
  4. 8, 12, 5, 20

Solution 5 : Problem analysis and execution:

From the nature of the problem, we decide immediately to use the free resources of the choice value sets and test the given conditions suitably on the sets.

Looking at the four conditions, we decide the first and the second won't result in significant separation between the tested number pair. So instead, the addition and the multiplication operations are chosen as the test conditions.

This single two condition test invalidates the first three sets of numbers, leaving the fourth set as the only choice. This choice satisfies the two conditions. We don't use any other test.

Answer: Option d: 8, 12, 5, 20.

Key concepts used: Problem analysis -- Problem solving strategy -- Principle of free resource use -- Choice value test -- Logic analysis.

Problem 6.

The fractions $\displaystyle\frac{1}{3}$, $\displaystyle\frac{4}{7}$, and $\displaystyle\frac{2}{5}$ written in ascending order is,

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

Solution 6 : Problem analysis and execution:

There are two methods of comparison of fractions that use equalizing the base of numerators or the denominators to their LCM value by multiplying the numerator and the denominator pairs of the fractions by suitable numbers. But each of these two methods takes a little bit of time.

If the fractions conform to the pattern of equal difference between the denominator and numerator, a third rich fraction comparison concept can be applied to sequence the fractions in a few seconds.

We find the last two fractions $\displaystyle\frac{4}{7}$, and $\displaystyle\frac{2}{5}$ conform to the pattern of equal difference between the denominator and the numerator as 3, and also identify that if we multiply and divide the first fraction $\displaystyle\frac{1}{3}$ by 2, it will make the numerators of the first and the third fraction equal as 2.

This makes the third fraction $\displaystyle\frac{2}{5}$ larger than the transformed first $\displaystyle\frac{2}{6}$ as its denominator 5 is smaller than the resultant denominator 6 of the first. We have applied numerator base equalization technique.

Now we apply the rich concept of equal difference between the denominator and numerator on the second and third fractions to identify the second (with larger denominator) as the largest.

By the hybrid method, identifying two useful patterns we have successfully sequenced the three fractions in a few seconds as,

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

Answer: Option d : $\displaystyle\frac{1}{3} \lt \displaystyle\frac{2}{5} \lt \displaystyle\frac{4}{7}$.

Key concepts used: Key pattern identification -- Fraction comparison -- Base equalization technique -- Hybrid method -- Rich fraction comparison concept.

Mechanism behind the rich fraction comparison concept

Let us formally state the rich fraction comparison concept,

In a set of fractions, if the difference between the denominator and numerator for each fraction is of the same value, the fraction with largest denominator will be the largest, the fraction with the second largest denominator will be the second largest and so on.

The two fractions considered in our problem are,

$\displaystyle\frac{4}{7}$, and $\displaystyle\frac{2}{5}$ with difference between denominator and numerator for each as 3.

We subtract each of the fractions from 1 getting,

$\displaystyle\frac{3}{7}$, and $\displaystyle\frac{3}{5}$.

The second result is the larger (numerators equal, second denominator smaller), and as it is the result of subtraction from 1, the actual fraction $\displaystyle\frac{4}{7}$ is the larger.

In mathematical form,

$1-\displaystyle\frac{2}{5} \gt 1-\displaystyle\frac{4}{7}$.

Cancelling 1 and transposing,

$\displaystyle\frac{4}{7} \gt \displaystyle\frac{2}{5}$.

Problem 7.

The numerator of a fraction is 4 less than its denominator. If the numerator is decreased by 2 and the denominator is increased by 1 the denominator becomes eight times the numerator. The fraction is,

  1. $\displaystyle\frac{5}{9}$
  2. $\displaystyle\frac{3}{7}$
  3. $\displaystyle\frac{13}{17}$
  4. $\displaystyle\frac{4}{8}$

Solution 7 : Problem analysis and execution:

Proceeding conventionally we get two equations between the two unknowns, the numerator and the denominator, from the two given conditions. Solving the two equations we can find the numerator and the denominator and hence the fraction. But like most conventional solutions, this one is also time taking and inefficient.

Instead, we decide to use the free resource of the choice values and test each choice value with the second condition. in a few seconds we identify $\displaystyle\frac{3}{7}$ to satisfy both the given conditions and so our solution.

Answer: Option b: $\displaystyle\frac{3}{7}$.

Key concepts used: Problem solving strategy -- Solving linear equations -- Principle of free resource use -- Choice value test.

Problem 8.

The greatest number among $1.2\times{0.83}$, $1.02-\displaystyle\frac{0.6}{24}$, $0.7+\sqrt{0.16}$, and $\sqrt{1.44}$ is,

  1. $1.02-\displaystyle\frac{0.6}{24}$
  2. $1.2\times{0.83}$
  3. $0.7+\sqrt{0.16}$
  4. $\sqrt{1.44}$

Solution 8 : Problem analysis and execution:

Examining the four values we evaluate the simplest one,

$\sqrt{1.44}=1.2$.

Immediately comparing this value with the first value, $1.2\times{0.83}$, we could decide this second value to be smaller than $1.2$ as it equals $1.2$ multiplied by a factor less than 1. This is a basic number comparison concept.

Next we compare, $1.2$ with the second value $1.02-\displaystyle\frac{0.6}{24}$. This value is even less than $1.02$ and so $1.2$ is larger.

Lastly we compare $1.2$ with $0.7+\sqrt{0.16}=0.7+0.4=1.1$ giving $\sqrt{1.44}$ as the largest.

Answer: Option d:$\sqrt{1.44}$.

Key concepts used: Simplification technique, simplify immediately what is easiest to simplify -- Basic number comparison concept -- Efficient simplification, we have used least amount of calculation.

Problem 9.

$(49)^{15} - 1$ is exactly divisible by,

  1. 50
  2. 51
  3. 8
  4. 29

Solution 9 : Problem analysis and execution:

By Binomial theorem,

$(x+1)^n=x^n + {^n}C_{1}x^{n-1} + {^n}C_{2}x^{n-2} +......+ {^n}C_{n-1}x +1$.

So,

$(x+1)^n-1$ is always divisible by $x$.

In our case,

$(49)^{15}-1=(48+1)^{15}-1$ which is always divisible by 48.

Among the choices 8 is a factor of 48, and so is our answer.

Answer: Option c: 8.

Key concepts used: Binomial theorem -- Factor multiple concept -- Choice value test -- Basic algebra concept.

Alternate solution

$x^n-1$ is always divible by $x-1$ where $n$ is odd. In this case then $(49)^{15} -1$ is always divisible by $49-1=48$.

Problem 10.

The greatest fraction among $\displaystyle\frac{2}{3}$, $\displaystyle\frac{5}{6}$, $\displaystyle\frac{11}{15}$, and $\displaystyle\frac{7}{8}$ is,

  1. $\displaystyle\frac{5}{6}$
  2. $\displaystyle\frac{7}{8}$
  3. $\displaystyle\frac{11}{15}$
  4. $\displaystyle\frac{2}{3}$

Solution 10 : Problem analysis and execution:

We find that the three fractions, $\displaystyle\frac{2}{3}$, $\displaystyle\frac{5}{6}$, and $\displaystyle\frac{7}{8}$ each has equal difference between denominator and numerator as 1 and thus these three are suitable for applying the rich fraction comparison concept. By the rich concept, the fraction with the largest denominator, $\displaystyle\frac{7}{8}$ is the largest among the three.

Now for comparing $\displaystyle\frac{7}{8}$ with $\displaystyle\frac{11}{15}$ we equalize the base denominator to the LCM 120 so that the numerators become 105 and 88 respectively. So the first, $\displaystyle\frac{7}{8}$ is the larger among the two and hence it is the largest among all four given fractions.

Answer: Option b: $\displaystyle\frac{7}{8}$.

Key concepts used: Key pattern identification -- Problem solving strategy -- Hybrid fraction comparison approach -- Base equalization technique -- Denominator equalization technique -- Rich fraction comparison concept.


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