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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.

You may watch video solutions as below,

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 or LCM of 4 minutes and 15 minutes.

In fact the two bells will continue to chime together at every such elapsed time that is a multiple of LCM of 4 minutes and 15 minutes.

Think over.

The synchronization graphic shown below will help you to visualize the situation.

ssc-cgl-solution-3-number-system-1-q4-lcm-synchronization.png

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: 2, 3, 5 and 7 are the four prime factors in the set of numbers from 2 to 10.

The smallest non-prime number having none of these 4 factors is $11^2 = 121$. So in the range from 20 to 50, the numbers that do not have any factor in the set of numbers from 2 to 10 must be prime numbers. Think it over.

You can easily enumerate these prime numbers in the range from 20 to 50 as, 23, 29, 31, 37, 41, 43 and 47, a total of 7 numbers.

Answer: Option c: 7.

Key concepts used: Divisibility, factors 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. 380

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: First we'll find how many such three digit numbers are there.

To do it systematically, divide the whole three digit number range of 100 to 999 in 9 smaller ranges, 100 to 199, 200 to 299, 300 to 399 and on to 900 to 999.

Target numbers in the range of 100 to 199 are 103, 108, 113....up to 198, a total of 20 such numbers. Verify if you need.

In each of the nine such ranges ending at 999, behavior will be same and such numbers will count to 20.

So such numbers in total are $9\times{20} = 180$.

Sum of these 180 numbers is,

$S = (103 + 108 + .... + 993 + 998)$.

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}.

You may avoid the last calculation and get the answer by number estimation.

But what will you do if you don't remember what is an arithmetic progression!

Solution to Problem 8 without using concept of arithmetic progression

You can solve this problem quickly even if you don't know what is arithmetic progression let alone its formula.

Look at the series of 180 numbers (103+108+....+998).

Final result will be 180 times average of these 180 numbers.

You can conclude with certainty from the series that average of the sum must be much greater than 103.

Now mentally divide the choice values by 18 after stripping rightmost digit of the choice values.

First choice value 180 is too low.

For second choice value 1550, quotient is 8 and for the third choice value 6995, quotient is less than 40.

Both quotients being less than 103, fourth choice 99090 must be the answer.

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 for sum of an Arithmetic Progression -- Use of average concept and mathematical reasoning to solve without using arithmetic progression formula.

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: Let's state the first principle of Unit's digit of a number say, ABCDU raised to the power say, N,

$\text{Unit's digit of }(ABCDU)^N=\text{Unit's digit of }(U)^N$.

This is because when original number is multiplied with itself N times, its unit digit is multiplied ONLY with itself also N times without any carry over into it.

So,

$\text{Unit's digit of }2137^{754}= \text{ Unit's digit of }7^{754}$.

Let's now explain the second principle of unit's digit of a single digit number raised to the positive integer power of N.

When a single digit, not 0 or 1, is raised to any positive integer power, the unit's digit of the result takes one of the few fixed values specific to the single digit number.

We say—the unit's digit cycles through these few fixed values. To make the concept clear let's show you the result of raising 7 to the power 1, 2, 3, 4 and 5.

  • $7^1=7 \Rightarrow \text{Unit's digit }7$
  • $7^2=49 \Rightarrow \text{Unit's digit }9$
  • $7^3=343 \Rightarrow \text{Unit's digit }3$
  • $7^4=2401 \Rightarrow \text{Unit's digit }1$
  • $7^5=16807\Rightarrow \text{Unit's digit cycled back to 7}$

The unit's digit becomes respectively 7, 9, 3, 1 and then 7 again. The cycle of length 4 is completed at 1. The fixed set of values are 7, 9, 3 and 1.

Method: To use this concept in finding the unit's digit of $7^{754}$ you have to divide the power 754 by the cycle length of 4. Remainder will be 2.

Rule for unit's digit

  • for remainder 1: Unit's digit is 7, first digit in cycle.
  • for remainder 2: Unit's digit is 9, second digit in cycle.
  • for remainder 3: Unit's digit is 3, third digit in cycle.
  • for remainder 0: Unit's digit is 1, fourth digit in cycle.

So, unit's digit for $2137^{754}$ is 9 as the remainder is 2.

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 -- Principles on Unit's digit.


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