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SSC CGL level Solution Set 15, Number System

SSC CGL level Number System Solution Set 15

15th set of Solutions to 10 number system questions for SSC CGL

The 10 number system questions for SSC CGL in the 15th question set are now solved.

Concepts needed to solve the questions in brief are,

Division, unit digit of powers, ratio, factorization, HCF, divisibility, surds, and place value.

While solving each problem, attempt has always been to get the answer as quickly as possible. Speed and accuracy are at the heart of success in any MCQ test, isn't it?

This will give you a good exposure on how to solve number system questions in any competitive exam.

For best results,

  1. Do take the test like the actual test with timer on.
  2. After the test, score your performance from the answers, and,
  3. Use these solutions to be clear on your doubts, and know how to answer the questions quick.

And then use this knowledge to practice on more sets of number system questions.

To practice, you may use the tutorials, questions, and solutions on number system in the guide,

Suresolv Guide for Number System - SSC CGL, SSC CHSL and more.

Solutions to 10 selected number system questions - 15th set for SSC CGL - time to answer was 12 mins

Problem 1.

The product of two positive integers is 11520 and their ratio is 9 : 5. Find their difference.

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

Solution to Problem 1:

Ratio of two numbers being 9 : 5, introduce the cancelled out HCF $x$ as factor of both the ratio terms to express their actual values as, $9x$ and $5x$ respectively.

This is the HCF reintroduction technique indispensable for solving majority of ratio problems.

Product of the two numbers can now be expressed as,

$9x\times{5x} = 45x^2 = 11520$

Or, $x^2 = 256$,

Or, $x = 16$,

And the difference of the two values,

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

Answer: Option c: 64.

Key concepts used: HCF reintroduction technique -- Ratio concepts -- Solving in mind.

The question is easily solved in mind.

Problem 2.

A number when divided by 3 leaves a remainder of 1 and when the quotient is divided by 2 it leaves a remainder of 1 again. What will be the remainder when the number is divided by 6?

  1. 2
  2. 3
  3. 4
  4. 5

Solution to Problem 2:

Using Euler's division lemma, results of the first division expressed as,

$p = 3q + 1$, where $p$ is divided by quotient $q$ is the quotient with 1 as the remainder.

Similarly for the second division,

$q = 2q_1 + 1$, where $q_1$ is now the quotient when earlier quotient $q$ is divided by 2 with remainder again as 1.

Substitute this expression of $q$ in the first equation,

$p = 3(2q_1 + 1) + 1$,

Or, $p= 6q_1 + 4$.

This equation conforms to Euler's division lemma and,

It expresses the relation between divisor $p$, dividend 6, quotient $q_1$ and remainder 6.

In other words, the remainder in dividing the original number $p$ by 6 is 4.

Answer: Option c : 4.

Key concepts used: Basic relationship between the dividend, divisor, quotient and the remainder -- Euler's division lemma -- Solving in mind.

If you know how to use the basic relation of division in the form of Euler's division lemma, you would be able to get the answer wholly in mind.

Problem 3.

When a positive integer is divided by a second positive integer the quotient is 1. When the second number is divided by the remainder of the first division, the quotient is 2 and remainder 0. What is the ratio between the first and the second number?

  1. 2 : 3
  2. 3 : 5
  3. 2 : 5
  4. 3 : 2

Solution to Problem 3:

Work backwards from the second division to conclude that,

The second number is twice the remainder of first division.

It means, $n=2r$, where $n$ is the second number and $r$ is the remainder of first division.

Now express the operation of the first division by Euler's division lemma as,

$m=n+r$,

where $m$, $n$ and $r$ are the first number, second number and the remainder of first division respectively.

Substitute $r=\frac{1}{2}n$ in this equation of first division to get the relation between the two numbers as,

$m=n+\frac{1}{2}n=\frac{3}{2}n$,

Or, $2m=3n$,

Or, $m : n=3:2$.

Answer: Option d: 3 : 2.

Key concepts used: Most basic concepts of what happens in a division and the relationships between the various components involved in a division -- Euler's division lemma -- Ratio concepts -- Solving in mind.

What actually happens - the conceptual visualization

The second number added to the remainder forms the first number.

As the second number equals two numbers of remainders,

The first number is two numbers of remainders plus another remainder, that is 3 numbers of remainders.

Recall that the second number equals two numbers of remainders.

So the ratio of the first number to the second is, 3 : 2.

The remainder in this case is the HCF between the two numbers.

Problem 4.

50 persons are putting chocolates in 50 boxes. The first person puts 1 chocolate in each of the 50 boxes, the second person puts 2 chocolates in every second boxes, the third person puts 3 chocolates in every third boxes, and finally the 50th person puts 50 chocolates in the 50th box. At the end of the operation, how many chocolates did the 50th box have?

  1. 96
  2. 43
  3. 93
  4. 78

Solution to Problem 4:

Problem analysis

Make a quick examination of how many chocolates are put in which boxes by each person.

1st person puts 1 chocolate in every box. So 50th box gets 1 chocolate from the 1st person.

The 2nd person puts 2 chocolates in box numbers 2, 4, 6, 8,.....and 50. So 50th box gets 2 chocolates from the 2nd person.

The 3rd person puts 3 chocolates in box numbers 3, 6, 9.....and 48. Ahhh, this is the key pattern identified,

The 3rd person won't put any chocolate in the 50th box.

But why?

Simply because 3 cannot divide 50, it is not a factor of 50.

At this point (if you have not visualized the happenings and identified the key pattern earlier), you would clearly know what is happening.

The 50th box will get chocolates from ONLY THE PERSONS WHOSE SEQUENCE NUMBER can divide 50. These are ALL the factors of 50.

So the total number of chocolates in the 50th box at the end will be sum of ALL FACTORS of 50.

To find the factors easily, first form the breakup of 50 as a product of its prime factors as,

$50=2\times{5}\times{5}$.

All the factors of 50 will then be, 1, 2, 5, 10, 25 and 50. Don't forget 1 and 50.

And the required total,

$1 + 2 + 5 + 10 + 25 + 50 = 93$.

The factors are only a few.

Answer: c: 93.

Key concepts used: Event mapping -- Visualization -- Key pattern identification -- Converting initial problem definition to sum of factor form -- Finding all factors of a number -- Prime factors -- Solving in mind.

For a moment, you may find the problem confusing.

But be bold, shrug off the confusion, think what actually is happening.

Once you discover the key pattern, solution takes a few tens of seconds.

Problem 5.

$12345679\times{72}$ equals

  1. 999999998
  2. 88888888
  3. 898989898
  4. 888888888

Solution to Problem 5:

This is clearly a test of how much you know about divisibility rules.

As a start, you know 72 has factors, 8, 9. And hence 8 and 9 are both factors of the resultant product also.

We'll now test each choice options for divisibility of 9 (if integer sum of a number is divisible by 9, the number will also be divisible by 9).

First choice value test for 999999998:

As the integer sum (the sum of the digits of the number) is 80 which is not divisible by 9, the first choice number is not divisible by 9 and hence can't be the resultant product.

Second choice value test for 88888888:

Integer sum of 64, not divisible by 9. So this choice also is not the answer.

Third choice value test for 898989898:

The integer sum of 76 again has no factor of 9. This can't be the answer.

Fourth choice must be the answer, but we'll show nevertheless that it is divisible by 9,

The integer sum of the fourth choice 888888888 is 72, divisible by 9. But this doesn't prove that it is indeed the resultant product.

Let's be more certain by first dividing this choice value by 8, the other factor of 72, and get the result as,

$111111111$.

Now multiply the large given factor $12345679$ by 9, the second factor of 72,

$9\times{12345679} = 111111111$.

Though this is not necessary, you are feel happy that you are totally certain of the answer discounting the remote possibility that the question is erroneous.

Answer: Option d: 888888888.

Key concepts used: Using divisibility by 9 to eliminate three choices -- Selection by elimination -- Factors multiples concept -- Integer sum -- Solving in mind.

Problem 6.

The unit digit of the expression, $216^{5192} + 25^{4317} + 97^{7892}$ is,

  1. 1
  2. 2
  3. 3
  4. 5

Solution to Problem 6:

This is a very awkward looking problem to solve, if you have not solved such a problem earlier.

Use common sense, basic concepts and reasoning to make the first conclusion,

Unit digit of $(pqr)^n$ = Unit digit of $r^n$.

In other words, unit digit of a number raised to a high power of $n$ would be the unit digit of $n$th power of the unit digit of the number itself.

This is a basic concept in number system.

Apply this concept to get the first breakthrough result,

Unit digit of $216^{5192} =$ Unit digit of $6^{5192}$.

Moving ahead one more step, discover the second key pattern,

For all powers of 6, the unit digit remains fixed at 6.

So the unit digit contribution of the first term is 6.

Similarly, unit digit contribution from the second term of the given sum is 5.

For the contribution of the third term you have to evaluate the unit's digit of $7^{7892}$.

Though mathematically it is infeasible to evaluate the result of raising 7 to the power of 7892, use your reasoning,

The resultant unit digit can't have many values other than 1 to 9. We can easily see the unit digit for the first few powers of 7 then.

These are,

Unit digit of $7$ is 7.

Unit digit of $7^2=49$ is 9.

Unit digit of $7^3$ is unit digit of $7\times{9}=63$, that is 3.

Unit digit of $7^4$ is unit digit of $7\times{3}=21$, that is 1.

And the final breakthrough discovery,

Unit digit of $7^5$ is unit digit of $7\times{1}=7$.

That means, when the power of 7 is increased by 1 from 1 to 2, 3, 4, 5 and so on,

The unit digit of the result will be cycling sequentially through the 4 values, 7, 9, 3 and lastly 1.

This mechanism in fact is the key technique in finding unit digit of any integer power of any integer.

What will then be the unit digit of $7^{7892}$?

Obviously it will be one of the four digits 7, 9, 3 and 1, but which one?

Again visualize what actually happens to quickly form the key critical rule-set,

Divide 7892 by 4 (there are four sequentially possible values) and find the remainder.

If remainder is 0, unit digit of the result is the fourth value 1 in the cycle of four values, 7, 9, 3, 1.

If remainder is 1, unit digit of the result is the first value 7 in the cycle of four values, 7, 9, 3, 1.

If remainder is 2, unit digit of the result is the second value 9 in the cycle of for values, 7, 9, 3, 1.

And if remainder is 3, unit digit of the result is the third value 3 in the cycle of four values, 7, 9, 3, 1.

Divide 7892 by 4 to get remainder 0 (divide last two digits 92 by 4).

So the unit digit of the third term is 1.

Apply now the last of your basic conceptual reasoning,

Unit digit of sum of three numbers is the unit digit of sum of unit digits of the three numbers.

It follows, unit digit of the the given sum is, unit digit of,

$6+5+1=12 \Rightarrow 2$.

Answer: Option b : 2.

Key concepts used: Breaking up the problem into smaller parts -- Unit digit evaluation technique for high powers -- Unit digit for a sum of numbers -- Cycle of values of unit digit of high power.

This is not an easy problem to solve unless you are very clear about the unit digit evaluation concepts.

Problem 7.

What is the value relationship between the irrational numbers, $\sqrt{7} - \sqrt{5}$, $\sqrt{5} - \sqrt{3}$ and $3 - \sqrt{7}$?

  1. $3 - \sqrt{7} \lt \sqrt{5} - \sqrt{3} \lt \sqrt{7} - \sqrt{5}$
  2. $\sqrt{5} - \sqrt{3} \lt 3 - \sqrt{7} \lt \sqrt{7} - \sqrt{5}$
  3. $3 - \sqrt{7} \lt \sqrt{7} - \sqrt{5} \lt \sqrt{5} - \sqrt{3}$
  4. $3 - \sqrt{7} \gt \sqrt{7} - \sqrt{5} \gt \sqrt{5} - \sqrt{3}$

Solution to Problem 7:

In this subtraction form of surd expressions, you can't be mathematically certain about which one of the given surd expressions is larger or smaller than another.

Searching for some pattern in the three expressions, you detect one minor pattern, especially when you transform the third given expression and place it in the beginning as,

$\sqrt{9} - \sqrt{7}$, $\sqrt{7} - \sqrt{5}$, and $\sqrt{5} - \sqrt{3}$.

$\sqrt{7}$ is common between the first and second expressions and $\sqrt{5}$ is common between the second and third expressions.

But unfortunately these are in opposite signs.

At this point you realize, if you could have converted these surd expressions with the same terms but in addition form, evaluation of relative value would be easy.

And in fact, this conversion of a two term surd subtractive sum to additive sum is easily possible by Surd rationalization technique.

Here it works perfectly as you identify the second key pattern in the six surd terms of the three expressions,

Difference between the two numbers under square root for each expression is 2.

Rationalize the three given expressions by multiplying and dividing the first, the second and the third by $(\sqrt{9}+\sqrt{7})$, $(\sqrt{7}+\sqrt{5})$, and, $(\sqrt{5}+\sqrt{3})$ respectively.

The results you would get are,

$\displaystyle\frac{9-7}{\sqrt{9} + \sqrt{7}} = \displaystyle\frac{2}{\sqrt{9} + \sqrt{7}}$,

$\displaystyle\frac{7-5}{\sqrt{7} + \sqrt{5}} = \displaystyle\frac{2}{\sqrt{7} + \sqrt{5}}$, and,

$\displaystyle\frac{5-3}{\sqrt{5} + \sqrt{3}} = \displaystyle\frac{2}{\sqrt{5} + \sqrt{3}}$.

Which one will be the largest?

By basic inequality concept you know that between two fractions with same numerator, (the equal numerator is critical here), the fraction with smaller denominator will be the larger of the two.

Now it easy for you to decide on the desired inequality relation,

$(3 - \sqrt{7}) \lt (\sqrt{7} - \sqrt{5}) \lt (\sqrt{5} - \sqrt{3})$.

Answer: Option c: $3 - \sqrt{7} \lt \sqrt{7} - \sqrt{5} \lt \sqrt{5} - \sqrt{3}$.

Key concepts used: Detecting pattern of a common term between two pairs of given surd expressions -- Utilizing the pattern of common terms by rationalizing the expressions thus converting the subtraction expressions to summation expressions -- Establishing value relationship between inverted form of three expressions -- Inverting the expressions inverts the nature of value relationship which was the target -- Solving in mind.

You may wonder, how is it possible to solve this awkward problem in mind!

Knowing how to apply the rich concept of ranking subtractive two term surd expressions, it should take just a few tens of seconds to arrive at the answer.

In the three two-term surd expressions we would identify the patterns,

Two equal adjoining common terms between first-second and second-third terms, and,

Equal difference of the two numbers under square roots for each term.

These two patterns will decide $3-\sqrt{7}$ as the smallest and $\sqrt{5}-\sqrt{3}$ as the largest.

Problem 8.

The remainders of division of the numbers 6666666, 7777777 and 8888888 by 9 will be,

  1. 2, 4, 8
  2. 6, 4, 2
  3. 8, 3, 1
  4. 4, 6, 8

Problem 8.

The remainders of division of the numbers 6666666, 7777777 and 8888888 by 9 will be,

  1. 2, 4, 8
  2. 6, 4, 2
  3. 8, 3, 1
  4. 4, 6, 8

Solution to Problem 8:

Strategy decided: Break up each of the given numbers as a sum of place values of the digits, divide each of the terms in a sum by 9 to find the remainder, sum up the remainders and divide this sum one last time by 9 to get the desired remainder.

Let's show the steps.

Express the first number broken up as a sum of place values of the digits,

$6666666 = 6\times{10^6} + 6\times{10^5} + 6\times{10^4} + 6\times{10^3} + 6\times{10^2} + 6\times{10} + 6$.

So when 6666666 is divided by 9, effectively 9 divides each of the seven terms of the sum. Each division will result in a remainder. Let's see what the remainder are.

Excluding the unit digit, all the other terms have a factor of 10's power.

The special property of 10's powers when divided by 3 or 9 is, the remainder for division of any power of 10 by 3 or 9 will be 1.

In the expanded expression, this remainder 1 multiplied with the place digit will leave out only the place digit in the summation for evaluating final remainder.

In other words, remainder for division of 6666666 by 9 will be equal to the remainder for dividing the following sum by 9,

$6 + 6 + 6 + 6 + 6 + 6 + 6 = 42$.

This sum is the integer sum of the number and the remainder on division by 9 will be 6.

Similarly for the number 7777777, integer sum is 49 and on division by 9 remainder will be 4.

For the third number 8888888, integer sum is 56 and remainder on division by 9 will be 2.

Answer: Option b: 6, 4, 2.

Key concepts used: Expressing a number as a sum of place values -- Basic division and remainder concepts - Using division of integer sum technique for finding remainder instead of checking divisibility. This method of finding remainder will work directly for division by 3 and 9.

If you are clear about the divisibility concepts and place value break up technique, you would be able to solve this problem also in no time.

Problem 9.

If the number of employees of a company is a prime number less than 500, the ratio of the number of male employees to the number of rest of the employees may be,

  1. 101 : 88
  2. 87 : 100
  3. 85 : 98
  4. 97 : 84

Solution to Problem 9:

Let's assume number of male employees to be $m$ and the rest $n$.

The only way to proceed is to first convert each choice ratio in the form of desired ratio of male employees to total employees as,

$\displaystyle\frac{m}{m+n}$,

And test which total in the denominator is a prime number as stated in the question.

The choice ratios converted in this way will be,

For option a: $\displaystyle\frac{101}{189}$,

For option b: $\displaystyle\frac{87}{187}$,

For option c: $\displaystyle\frac{85}{183}$, and,

For option d: $\displaystyle\frac{97}{181}$.

The denominators of these four ratios represent the total number of employees that must be a prime number.

By divisibility test of 3, 189 and 183 are easily found to be multiples of 3. These are invalid choices.

The first choice of 189 is also a multiple of 7. So the fourth choice must be the answer.

For your confidence let's make a quick test to see whether the denominator 181 is indeed a prime number.

For the fourth choice, 181 is not divisible by 3, 5, 7 or 11. It is not also divisible by 13. The next larger prime divisor is 17 and as $17^2 = 289$ exceeds 181, it must be a prime number.

Answer: Option d: 97 : 84.

Key concepts used: Use of ratio concept to test the total for prime from the two parts of each ratio choice -- Ingenious use of available information and basic concepts.

Observe that the given condition, "less than 500" has not been used at all.

Problem 10.

A group of 630 children is arranged in rows such that each row contains 3 children more than the row behind it. Which of the following as number of rows is not possible in this case?

  1. 4
  2. 5
  3. 3
  4. 6

Solution to Problem 10:

Let number of children in the back row be $a$.

For rows 4 then the total number of children in the four rows is,

$a + (a + 3) + (a + 6) +( a + 9)= 4a + 18 = 630$,

Or, $4a = 612$, a number divisible by 4.

So 4 is a valid number of rows.

For 5 rows the total number of children is,

$a + (a + 3) + (a + 6) + (a + 9) + (a + 12) = 5a + 30 = 630$,

Or, $5a = 600$, a number divisible by 5.

So 5 is a valid number of rows.

For 3 rows the total number of children is,

$a + (a + 3) + (a + 6) = 3a + 9 = 630$,

Or, $3a = 621$, a number divisible by 3.

So 3 is a valid number of rows.

For 6 rows the total number of children is,

$a + (a + 3) + (a + 6) + (a + 9) + (a + 12) + (a + 15) = 6a + 45 = 630$,

Or, $6a = 585$, a number not a multiple of 6 and so $a$ will be a fraction, an invalid condition.

Answer: Option d: 4.

Key concepts used: Enumeration of the total number of children based on given condition. This way the problem can easily be solved without knowing what arithmetic progression is. This is problem solving.


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