SSC CGL level Solution Set 14, Arithmetic Number System | SureSolv

SSC CGL level Solution Set 14, Arithmetic Number System

Fourteenth SSC CGL level Solution Set, topic Arithmetic Number System

SSC CGL Solution set14 number system

This is the fourteenth solution set of 10 practice problem exercise for SSC CGL exam on topic Arithmetic Number System. Students must complete the corresponding question set in prescribed time first and then only refer to the solution set.

It is emphasized here that answering in MCQ test is not at all the same as answering in a school test where you need to derive the solution in perfectly elaborated steps.

In MCQ test instead, you need basically to deduce the answer in shortest possible time and select the right choice. None will ask you about what steps you followed.

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 of the topics
  • 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 head rather than on paper.

Actual problem solving is done in the fourth layer. But how to do that?

You need to use your problem solving abilities to gain an edge in competition.

If you have not taken the corresponding test yet, take the test by referring to SSC CGL level Question Set 14 on Number system and come back to go through the solution.

Fourteenth solution set- 10 problems for SSC CGL exam: topic Arithmetic Number System - time 12 mins

Q1. Find the second largest prime number that will divide 5329 with 6 added to it.

  1. 5
  2. 11
  3. 93
  4. 97

Solution: Analysis of the problem tells us that the number 5329 with 6 added to it, that is, 5335 will have at least two prime factors and the desired factor is the second largest of all such prime factors of 5335.

Second stage reasoning tells us, unless we factorize the number 5335 we can't say which one is the second largest one.

Factorization of 5335

To quickly reduce a number by factorization we try to check divisibility by 9 and then 11. In this case 9 fails but 11 test is a success, we spot the pattern of divisibility by 11, that is, sum of alternate digits are equal.

Dividing by 11 yields, 485. Now only we divide by 5, the second factor, resulting in, 97 which is a prime number we know by experience.

But you can test also by checking divisibility first by 7, as it is already not divisible by 3. The test fails and as already we have tested for 11 square of which 121 is larger then 97, we know for sure that 97 is a prime number.

So, $5335 = 5\times{11}\times{97}$.

The second largest prime factor of 5335 is then 11.

Answer: Option b: 11.

Key concepts used: Quick factorization technique by starting from 9 then 11.

Q2. Find the missing digit in 4*379, if the five digit number is to be divisible by 9.

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

Solution: For the number to be divisible by 9 the integer sum or the sum of the digits of the number must be divisible by 9. Without the missing digit the sum is 23. Nearest multiple of 9 is 27 and the next is 36 which is larger than 9. So 27 is the target and the missing digit must be 4.

Answer: Option c : 4 .

Key concepts used: Divisibility rule of 9.

Q3. By which largest prime number should 5894 be divided to leave 5 as remainder?

  1. 13
  2. 151
  3. 153
  4. 17

Solution: The number to be factorized is 5894 less 5, that is, 5889. First factor is 3, result quotient 1963. Second factor we get 13, result quotient 151 which is a prime number as it can't be divided by the next prime divisor candidate 17 because $17^2=289$ is much larger than 151.

So, $5889 = 3\times{13}\times{151}$.

Answer: Option b: 151.

Key concepts used: Factorization and divisibility.

Q4. $\displaystyle\frac{1}{20} + \displaystyle\frac{1}{30} + \displaystyle\frac{1}{42} + \displaystyle\frac{1}{56} + \displaystyle\frac{1}{72} + $

$\hspace{30mm}\displaystyle\frac{1}{90} + \displaystyle\frac{1}{110} + \displaystyle\frac{1}{132}$ is,

  1. $\displaystyle\frac{1}{6}$
  2. $\displaystyle\frac{1}{8}$
  3. $\displaystyle\frac{1}{10}$
  4. $\displaystyle\frac{1}{7}$


Problem analysis

If you try to deduce the long fraction sum in conventional way, it will be quite complex, error-prone and time consuming.

So we look for patterns that will aid in quick addition of the fraction expression.

First clue

Looking at the terms we find our first clue. Each consecutive two terms have the denominators with a large HCF of 10, 14, 18 and 22. Looking for HCF between the denominators was natural as, we can factor out the large HCF and form the sum of two small fractions. It will be a simple process.

Let's see how.

$\displaystyle\frac{1}{20} + \displaystyle\frac{1}{30} + \displaystyle\frac{1}{42} + \displaystyle\frac{1}{56} + \displaystyle\frac{1}{72} +$

$\hspace{30mm}\displaystyle\frac{1}{90} + \displaystyle\frac{1}{110} + \displaystyle\frac{1}{132}$

$=\displaystyle\frac{1}{10}\left(\displaystyle\frac{1}{2} + \displaystyle\frac{1}{3}\right) + \displaystyle\frac{1}{14}\left(\displaystyle\frac{1}{3} + \displaystyle\frac{1}{4}\right) + $

$\hspace{15mm}\displaystyle\frac{1}{18}\left(\displaystyle\frac{1}{4} + \displaystyle\frac{1}{5}\right) + \displaystyle\frac{1}{22}\left(\displaystyle\frac{1}{5} + \displaystyle\frac{1}{6}\right)$

$=\displaystyle\frac{1}{10}\left(\displaystyle\frac{5}{6}\right) + \displaystyle\frac{1}{14}\left(\displaystyle\frac{7}{12}\right) + $

$\hspace{20mm}\displaystyle\frac{1}{18}\left(\displaystyle\frac{9}{20}\right) + \displaystyle\frac{1}{22}\left(\displaystyle\frac{11}{30}\right)$

$=\displaystyle\frac{1}{12} + \displaystyle\frac{1}{24} + \displaystyle\frac{1}{40} + \displaystyle\frac{1}{60}$

$=\displaystyle\frac{1}{12}\left(\displaystyle\frac{3}{2}\right) + \displaystyle\frac{1}{20}\left(\displaystyle\frac{5}{6}\right)$

$=\displaystyle\frac{1}{8} + \displaystyle\frac{1}{24} = \displaystyle\frac{4}{24}$


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

Key concepts used: Detecting patterns of large HCF in pairs of fraction denominators -- quick factorization of the denominators helped -- pairing the factor addition by factoring out the HCFs reduced the size and complexity quickly .

Q5. If the sum of squares of three consecutive positive numbers is 365, their sum is,

  1. 33
  2. 45
  3. 37
  4. 30


To be mathematically rigorous let's assume the first number as $n$ and the two others then are, $n+1$ and $n+ 2$. Squaring and summing up gives,

$n^2 + n^2 + 2n + 1 + n^2 + 4n + 4 $

$\hspace{25mm}3n^2 + 6n + 5 = 365$,

Or, $3n^2 + 6n - 360 = 0$,

Or, $n^2 + 2n -120 = 0$,

Or, $(n + 12)(n - 10) = 0$.

As $n$ is positive, $n = 10$, the numbers are, 10, 11 and 12 and their sum is 33.

Answer: Option a: 33.

Key concepts used: By problem analysis we understood at the very beginning that $n^2$ will be less than one third of the sum, that is less than about 121. So we could have started enumerating or testing with the hypothesis of $n = 10$ and would have arrived at the solution in a few seconds.

The treatment shown is mathematically rigorous and helps to clarify the concept of what really happens in the problem to a great extent.

Simplification technique

We generally use a simple strategy of simplifying the present state of affairs and then take the next step. This is, Simplification technique.

In any deduction, as soon as we notice a common factor that can be eliminated, we eliminate and reduce the expression and then only take the next step.

Factorizing the expression, $n^2 + 2n - 120 = 0$ is much easier to comprehend than factorizing the expression, $3n^2 + 6n - 360$. Though simplification technique seems to be a trivial and obvious step, if you declare it as a problem solving technique and use it as a habit, your speed and accuracy of deductions will improve a lot.

Q6. How many digits altogether are required to write the numbers from 1 to 60?

  1. 60
  2. 111
  3. 109
  4. 112


Leaving aside the first 9 single digit numbers, we find that the number set 10 to 19 form one set of two digit numbers, 20 to 29 the second set and 50 to 59 the fifth and last set, altogether 5 sets of 10 two digits numbers. The number of digits of these 5 ten sets is then $50\times{2}=100$.

We have left behind the first nine natural numbers and the last of the lot 60. Together these contribute another 11 digits.

So total number of digits is 111.

Answer: Option b : 111.

Key concepts used: Careful enumeration of the number of digits first identifying similar sets of numbers and keeping track of the exceptions.

One should be careful in dealing with this type of problems.

Q7. The value of $0.\overline{2} + 0.\overline{3} + 0.\overline{32}$ is,

  1. $0.\overline{82}$
  2. $0.\overline{87}$
  3. $0.\overline{77}$
  4. $0.\overline{86}$

Conventional solution:

We feel the need to use the method of conversion of a non-terminating recurring decimal to a fraction for converting each of the three decimals to fractions, add the fractions and convert the result back again to decimal form.

Let's see how.

Method of converting a non-terminating recurring decimal to a fraction

Let, $x=0.\overline{2}$,

Or, multiplying both sides by 10,

$10x = 2.\overline{2} = 2 + 0.\overline{2} = 2 + x$,

Or, $9x = 2$,

Or, $x = \displaystyle\frac{2}{9}$.

We know that, $0.\overline{3} = \displaystyle\frac{1}{3}$.

Let's convert the third decimal to fraction,

Let $z = 0.\overline{32}$,

Or, $100z = 32.\overline{32} = 32 + z$,

Or, $99z = 32$,

Or, $z = \displaystyle\frac{32}{99}$.

Adding the three fractions,

$S = \displaystyle\frac{2}{9} + \displaystyle\frac{1}{3} +\displaystyle\frac{32}{99}$

$=\displaystyle\frac{5}{9} +\displaystyle\frac{32}{99}$,

$=\displaystyle\frac{87}{99} = \frac{29}{33}$.

By dividing, we find,

$S = 0.878787... = 0.\overline{87}$.

Answer: Option b: $0.\overline{87}$.

Faster method by direct addition

By analyzing the first two decimals we can immediately form the sum as,

$0.2222... + 0.3333... = 0.5555...$.

As the two repeating digits of the third decimal are 3 and 2 adding them with 5 won't create any carry over and make things complicated. So the final sum will be,

$S = 0.5555... + 0.3232... = 0.8787... $

$= 0.\overline{87}$.

Naturally you should follow this method rather than the conventional method.

Key concepts used: First solution: Method of converting a non-terminating recurring decimal to fraction -- fraction conversion -- fraction addition -- converting a decimal to fraction. This is a time consuming and cumbersome solution.

Second solution: by observing the simplicity in terms of no carry-over in summing up any digit position, expansion of the decimals and direct summing up. This solution is much faster.

Q8. If 1, 3, 5, 7, 9......, 99 and 128 are multiplied together, the number of zeros on the right of the product will be,

  1. 7
  2. 6
  3. 8
  4. 9


A 0 at the right of a number represents one number of factor of 10. A factor of 10 is created by multiplying 5 with 2. We have 7 numbers of 2s in 128 and no other number in the product set has 2 as a factor as all are odd numbers.

So the limit of number of zeros or factors of 10 is established by this 7 number of 2s, there being enough number of nineteen 5s in the other factors.

Answer: Option a: 7.

Key concepts used: Mechanism of creation of a 0 on the right of a number -- finding the limiting factor that applies in the specific problem for creation of a factor of 10.

Q9. The product of two numbers is 120 and sum of their squares is 289. The difference between the numbers is,

  1. 6
  2. 7
  3. 8
  4. 9


Here we detect the possibility of applying the concept of algebraic relation,

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

As $a^2 + b^2$ and $ab$ are given we have,

$(a - b)^2 = 289 - 2\times{120} = 49$, and

$a - b = 7$.

Answer: Option b: 7.

Key concepts used: Use of algebraic relationship of $(a - b)^2$ -- we find often that application of algebraic concepts leads us to the solution of number problems quickly.

This is true for Trigonometric problems also, and that is why Algebra is so important.

Q10. A natural number is multiplied by 18 and a second natural number is multiplied by 21 and then their products are summed up. Which one can be the sum?

  1. 2009
  2. 2008
  3. 2007
  4. 2005


There are two unknown natural numbers that you can't find out. So we must resort to other means of analyzing the nature of the numbers that we can see, namely, 18 and 21. The immediate specialty that we notice between the two is, both are 3 divisible. That means the sum should also be 3 divisible.

Let's now examine the choice numbers and find only 2007 to be 3 divisible and others are not.

So only 2007 can be expressed as a sum of products of 18 and 21 with natural number factors of the two.

Answer: Option c: 2007.

Key concepts used: Identification of useful pattern of 3 divisibility in the given numbers -- applying the concept that if sum $S = n_1\times{18} + n_2\times{21}$, where $n_1$ and $n_2$ are positive integers, whatever common factors 18 and 21 will have will also be the factor of the sum $S$ -- use of 3 divisibility concept.

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