SSC CGL level Solution Set 3, Arithmetic Number System

Third SSC CGL level Solution Set, topic Arithmetic Number System

SSC CGL level Arithmetic Number System Solution Set 3

This is the third 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 most basic 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 happens in items 3 and 4 above. But how to do that?

You need to use your problem solving abilities only. There is no other recourse.

If you have not taken the corresponding test yet, take the test by referring to the SSC CGL level Question Set 3 on Number System and come back to go through the solution.

Third solution set- 10 problems for SSC CGL exam: topic Arithmetic Number System - time 20 mins

Q1. If $3^{4x}\times{9^{6x}} = 81^8$, $x=?$,

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

Solution: Let us first jot down the three basic concepts in indices that we would need for solving this problem. There are a few more such basic concepts.

  1. In an equation $a^x = a^y$, as the term bases are equal, the powers $x$ and $y$ must be equal.
  2. In a product of same base, powers are added up, $a^x\times{a^y} = a^{x+y}$.
  3. If a base term in power is raised to another power, effectively the two powers are multiplied together, $(a^x)^y = a^{xy}$.

Knowing that the best technique for solving indices sums is to equalize the base, we first determine that the target base to which all base terms need to be transformed is 3. This is because, all terms are in various powers of 3.

With this knowledge first we transform the LHS to get,

$3^{4x}\times{9^{6x}} = 3^{4x}\times{(3^2)^{6x}} = 3^{4x}\times{3^{12x}}$.

We have applied the third basic concept in indices here. Now we apply the second basic concept in Indices problems, $a^x\times{a^y} = a^{x+y}$ giving,

$3^{4x}\times{9^{6x}} = 3^{4x}\times{3^{12x}} = 3^{16x}$

So from the given equation,

$3^{16x} = 81^8 = 3^{32}$, again applying the third basic concept in indices problems, $(a^x)^y = a^{xy}$. We knew $81=3^4$.

As the bases are equal on both sides of the equation, by the first basic concept in indices, the powers must be equal. So,

$16x = 32$

Or, $x = 2$.

Answer: Option a: 2.

Key concepts used: Using base equalization technique complex LHS is transformed in terms of lowest base 3 -- same for RHS -- bases being equal, powers will be equal. Use of three basic concepts of indices sums, if $a^x=a^y$, $x$ and $y$ are equal, $a^x\times{a^y} = a^{x+y}$ and $(a^x)^y = a^{xy}$.

Q2. Which one of the following is a perfect square as well as a cube?

  1. 81
  2. 125
  3. 343
  4. 64

Solution: To be a perfect square and a cube, the power should be product of 2 and 3, that is, 6. Such a number must then be an integer raised to the power 6.

Knowing nature of powers of small numbers, we find, 125 and 343 unsuitable as these are, $125 = 5^3$ and $343 = 7^3$. 81 also goes out as it is, $81 = 3^4$. Answer then is $64=2^6$.

Answer: Option d : 64 .

Key concepts used: From given requirement to determination of exact power requirement -- enumeration or testing of choices for condition fulfillment.

Q3. If the digits of a two digit number are interchanged in position, the absolute value of the difference of the two numbers would be always divisible by,

  1. 5
  2. 7
  3. 9
  4. 11

Solution: Assuming the original number to be $N_{xy}$, new number is $N_{yx}$, where $x$ and $y$ are the number forming digts. Breaking up the two numbers using place value concept, we get,

$N_{xy} = 10x + y$, and $N_{yx} = 10y + x$.

Subtracting the second from the first (as we are working with absolute value, sign is irrelevant),

\begin{align} N_{xy} - N_{yx} & = (10x + y) - (10y + x) \\ & = 9x - 9y = 9(x - y) \end{align}

Thus the final result after subtraction will always be divisible by 9.

Answer: Option c: 9.

Key concepts used: Forming the new number by digit reversal -- Breaking up the numbers using place value concept -- simplification -- factor concept.

Q4. If two bells chime at intervals of 4 mins and 15 mins, after how long would they chime together first if they had chimed together in the beginning?

  1. 26 mins
  2. 52 mins
  3. 1 hour 18 minutes
  4. 1 hour

Solution: The two bells will chime together only when a multiple of 4 minutes becomes equal to a second multiple of 15 minutes. In other words, it will be the elapsed time after the start which is the lowest common multiple of 4 minutes and 15 minutes. In fact the two bells will continue to chime together at every such LCM elapsed time.

LCM of 4 and 15 is, 60 as 4 and 15 have no common factors.

It will take exactly 1 hour from the beginning for the two bells to chime together again.

Answer: Option d: 1 hour.

Key concepts used: Converting the chiming together problem to an LCM problem -- finding the LCM.

Q5. How many numbers from 20 to 50 have no number from 2 to 10 as factors?

  1. 0
  2. 14
  3. 7
  4. 4

Solution: The next larger prime number to 9 is 11 (10 is a multiple of 2 and 5 and we are primarily concerned with prime factors) and $11^2 = 121$.

By the factorization process termination principle, any number less than 121 that is not divisible by prime numbers starting from 2 to 9, must be a prime number. This is additionally a simple procedure to find prime numbers.

As the given target range is less than 121, we need to find basically the prime numbers from 20 to 50 to solve our problem.

These are, 23, 29, 31, 37, 41, 43, and 47, a total of 7 numbers.

Answer: Option c: 7.

Key concepts used: Divisibility and prime number concepts --- enumerating prime numbers.

Q6. $\displaystyle\frac{256\times{256} - 144\times{144}}{112}$ is,

  1. 420
  2. 400
  3. 360
  4. 360

Solution: Wherever we find an expression of the form $a^2 - b^2$, we would think of using the relation,

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

as the key resource for solving the problem. Applying this concept here we get,

\begin{align} & \displaystyle\frac{256\times{256} - 144\times{144}}{112}\\ & = \displaystyle\frac{(256 + 144)(256 - 144)}{112}\\ & = \displaystyle\frac{400\times{112}}{112} = 400. \end{align}

Answer: Option b : 400.

Key concepts used: Detecting and using the frequently used relationship $a^2 - b^2 = (a + b)(a - b)$.

Q7. If the digits in the unit's and ten's places of a three digit number are interchanged, the new number formed is found to be larger than the original by 63. All possible values that the unit's digit of the original number can take are,

  1. 7, 8, 9
  2. 2, 7, 9
  3. 0, 1, 2
  4. 1, 2, 8

Solution: Using place value concepts and assuming the original number as $N_{xyz}$, where $x$, $y$ and $z$ are the digits, the new number will be $N_{xzy}$. Thus,

\begin{align} N_{xzy} - N_{xyz} & = 100x + 10z + y \\ & \qquad - 100x - 10y - z \\ & = 9(z - y) \\ & = 63 \end{align}

Or, $z - y = 7$

Or, $z = 7 + y$

Or in other words, the value of $z$ starts from 7 with $y=0$ and can take the additional values 8 and 9 only.

Answer: Option a: 7, 8, 9.

Key concepts used: Forming the new number by digit interchange -- breaking up the number using place value mechanism concept -- simplification -- forming an equation with target $z$ on the RHS -- enumerating possible values of $z$ depending on possible values of $y$.

Q8. The sum of all the three digit numbers, each of which on division by 5 leaves a remainder 3 is,

  1. 180
  2. 1550
  3. 6995
  4. 99090

Solution: The numbers start with 103 and each three digit 100 numbers contain 20 such numbers. There are nine such hundreds, last number being 998. So total number of these numbers is, $9\times{20}=180$. So desired sum is,

$S = (103 + 108 + .... + 993 + 998)$, total number of terms being 180.

This is an Arithmetic Progression with first and last terms 103 and 998, and number of terms 180. The formula for the sum of such a progression is,

$S = \frac{n}{2}(f + l)$, where $n$ is the number of terms, $f$ is the first term and $l$ is the last term. In our case,

\begin{align} S & = (103 + 108 + .... + 993 + 998) \\ &= \frac{180}{2}(103 + 998) \\ & = 90\times{1101} \\ & = 99090 \end{align}.

Answer: Option d: 99090.

Key concepts used: Identifying the series as an Arithmetic Progression with first term 103 and last 998 -- finding the number of terms -- formula of an Arithmetic Progression.

Q9. Unit's digit in $264^{102} + 264^{103}$ is,

  1. 0
  2. 4
  3. 6
  4. 8

Solution: \begin{align} 264^{102} + 264^{103} & = 264^{102}(1 + 264) \\ & = 265\times{264^{102}} \end{align}.

As 264 is even, all its powers are even and multiplying the unit's digit 5 of 265 by this even unit's digit of $264^{102}$, the unit's digit of the result will be 0.

Answer: Option a: 0.

Key concepts used: Simplification of given expression to transform it to a product of two numbers -- unit's digit of any power of an even number is an even digit -- unit's digit of product of two numbers is unit's digit of product of the individual unit's digits of the numbers -- product of 5 and 2 results in unit's digit 0.

Q10. Unit's digit of $2137^{754}$ is,

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

Solution: Unit's digit of any power $n$ of a number is always the unit's digit of power $n$ of the unit's digit of the number. Simplifying, we get the desired unit's digit as, unit's digit of $7^{754}$.

7's powers starting from power 1 are, 7, 49, 343, 2401. That is, the unit's digits of 7's powers starting from 1 are 7, 9, 3 and 1. As you can predict the next unit's digit of $7^5$ will be again 7 and the next for $7^6$, again 9 and so on. In other words, the unit's digits of the powers of 7 follows a four number cycle, 7, 9, 3 and 1. In our expression power 754 when divided by 4 remainder is 2, and so the given expression will have unit's digit as the 2nd member of the cycle which is 9.

Answer: Option d: 9.

Key concepts used: Simplification of given expression by unit's digit basic concept -- finding the cycle of unit's digit of 7's powers -- finding the modulo of power 754 as 2 to get the desired unit's digit as second member of the 4 digit cycle.


Other resources on Number system and related topic

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