## Fifteenth SSC CGL level Solution Set, topic Number System

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

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.

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

**Problem 1.**

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

- 60
- 70
- 64
- 74

** Solution to Problem 1:**

Ratio of two numbers being 9 : 5, we can assume the numbers to be $9x$ and $5x$ where $x$ is the HCF that has been canceled out while expressing the ratio in proper fraction form. This is **Rich ratio concept.**

Then the product of the numbers is,

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

Or, $x^2 = 256$,

Or, $x = 16$,

The difference is,

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

**Answer:** c: 64.

**Key concepts used:** Rich ratio concepts.

**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?

- 2
- 3
- 4
- 5

**Solution to Problem 2:**

For the first division,

$p = 3q + 1$, where $p$ is the number, $q$ is the quotient when divided by 3 and 1 is the remainder.

For the second division similarly,

$q = 2q_1 + 1$, where this time 2 divided the previous quotient $q$ producing the second quotient $q_1$ and remainder 1.

Let us substitute the value of $q$ in the first expression,

$p = 3(2q_1 + 1) + 1 = 6q_1 + 4$.

So when $p$ is divided by 6 the remainder will be 4.

**Answer:** Option c : 4 .

**Key concepts used:** Basic relationship between the dividend, divisor, quotient and the remainder.

**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?

- 2 : 3
- 3 : 5
- 2 : 5
- 3 : 2

**Solution to Problem 3:**

By the second statement, * working backwards* we conclude that the second number is twice the remainder. Now when we examine the first division we find that the first number is formed from 1 number of second number and the remainder. Restating, we can say, the first number consists of two numbers of remainders plus another remainder, that is 3 numbers of remainders.

As the second number equals two numbers of remainders, the ratio of the first number to the second is, 3 : 2.

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

**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 -- ratio concepts.

**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?

- 96
- 43
- 93
- 78

**Solution:**

**Problem analysis**

When we start thinking in terms of each person putting chocolates in the 50th box one by one we find that the first person puts 1, second person puts 2, the third person puts 0 and this goes on this way.

It is rather a long process. But if you tread on this path you may quickly realize that just finding out the factors of 50 in the range of 1 to 50 and adding them up will produce the result. In this way, the result is,

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

The factors are not many.

**Answer:** c: 93.

**Key concepts used:** Converting initial problem definition to sum of factor form -- factorization.

**Problem 5.**

$12345679\times{72}$ equals

- 999999998
- 88888888
- 898989898
- 888888888

**Solution:**

72 has factors, 8, 9.

As the integer sum, the sum of the digits of the number, is 80 for the first number it can't be the result as it doesn't have 9 as a factor.

Checking the second choice 88888888, as the integer sum of 64 doesn't have 9 as a factor, this also can't be the result by the same logic.

Similarly for the third number choice 898989898, the integer sum of 76 having no factor of 9, this also can't be the result by the same logic.

But by the integer sum of 72 of the fourth choice 888888888, we find this possibly the result.

Let's be more certain by dividing this choice by 8, one of the factors of 72, and getting $111111111$. Now we multiply the large given factor $12345679$ by 9, the second factor of 72, getting the same result of $9\times{12345679} = 111111111$. We can do these two calculations easily without using pen and paper.

**Answer:** Option d: 888888888.

**Key concepts used:** Avoiding the large calculation, it is found by divisibility checking of the choice values for the factors of 72 that we may quickly get the possible result by eliminating the first three choice values. Additionally, confirmation by factor splitting is found also to be easy for verification of the result.

#### Problem 6.

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

- 1
- 2
- 3
- 5

**Solution:**

As, $216 = 6^3$ and for all powers of 6 the unit's digit remains fixed at 6, the unit's digit contribution of the first term is 6.

Similarly as, $25 = 5^2$ and for all powers of 5 the unit's digit remains fixed at 5, the unit's digit contribution of the second term is 5.

When evaluating the unit's digit contribution of the third term we use the **rich concept of unit's digit contribution**,

For any number raised to a power, the unit's digit of the result will be determined by the unit's digit of the number raised to the same power.

For example, units digit of $17^2$ will be unit's digit of $7^2=49$ as 9.

Applying this rich concept on the third term, we have,

Unit's digit of $97^{7892} = $ unit's digit of $7^{7892}$.

Now our final job is to find out a pattern in the unit's digit for all powers of 7.

As actual calculation of large powers is not possible, we resort to enumeration technique and actually find the unit's digit for the first few powers of 7 starting from power 1,

Unit's digit for $7^1=7$ is 7,

Unit's digit for $7^2=49$ is 9,

Unit's digit for $7^3=343$ is 3,

Unit's digit for $7^4=2401$ is 1.

In fact calculating the numbers 343 and 2401 were not required. Multiplying the unit's digit of 49 by 7 would have given us 3 as the unit's digit of $7^3$ and similarly, multiplying 3 by 7 would have given us the unit's digit 1 for $7^4$.

As soon as the unit's digit reaches the value of 1 we know the next unit's digit in this case will be $7\times{1} = 7$, and the cycle 7, 9 , 3 and 1 will be repreated for powers 5, 6, 7 and 8.

This is the pattern we were searching for.

Now we can say, for every 4 sequential natural number powers of 7 starting from power 1, the unit's place values will follow a cycle of values 7, 9, 3 and 1. **This is the rich concept of unit's digit cycle of values** for various powers of the base number.

Now dividing the power $7892$ by 4 we get remainder 0 and realize that this power is the fourth in a cycle of 4, and so so will have the fourth unit's digit value of 1.

Finally then, the unit's digit of the target expression being sum of unit's digits of the terms of the expression, which is another **rich concept in the topic of unit's digit evaluation**, we get the desired result as,

unit's digit of target expression = unit's digit of $6 + 5 + 1 = 12$, which is 2.

**Answer:** Option b : 2.

**Key concepts used:** Breaking up the problem into smaller parts -- using rich concepts in the area of unit's digit evaluation for various powers of integers.

** Problem 7.**

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

- $3 - \sqrt{7} \lt \sqrt{5} - \sqrt{3} \lt \sqrt{7} - \sqrt{5}$
- $\sqrt{5} - \sqrt{3} \lt 3 - \sqrt{7} \lt \sqrt{7} - \sqrt{5}$
- $3 - \sqrt{7} \lt \sqrt{7} - \sqrt{5} \lt \sqrt{5} - \sqrt{3}$
- $3 - \sqrt{7} \gt \sqrt{7} - \sqrt{5} \gt \sqrt{5} - \sqrt{3}$

**Solution:**

In this subtraction form of surd expressoons, we 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, we detect one minor pattern, especially when we transform the third 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 we realize, if we could have converted these surd expressions with the same terms but in a summation of terms form, relative vaue evaluation would have been possible.

In surds this is possible by inverting the expressions and rationalizing. If we invert all three and compare that would be just equivalent to the inverse of comparing direct expressions.

By inverting and reationalizing the first expression, $a=\sqrt{9} - \sqrt{7}$, we get,

$\displaystyle\frac{1}{a} = \displaystyle\frac{\sqrt{9} + \sqrt{7}}{9 - 7} = \displaystyle\frac{\sqrt{9} + \sqrt{7}}{2}$ and similarly the other two,

$\displaystyle\frac{1}{b} = \displaystyle\frac{\sqrt{7} + \sqrt{5}}{7 - 5} = \displaystyle\frac{\sqrt{7} + \sqrt{5}}{2}$,

$\displaystyle\frac{1}{c} = \displaystyle\frac{\sqrt{5} + \sqrt{3}}{5 - 3} = \displaystyle\frac{\sqrt{5} + \sqrt{3}}{2}$.

Between first and second expression now, $\sqrt{7}$ is common and the second term of the first $\sqrt{9}$ is larger than the second term $\sqrt{5}$ of the second expression. So we can conclude,

$\displaystyle\frac{1}{a} \gt \displaystyle\frac{1}{b}$.

Similarly comparing second and third expression, we can conclude,

$\displaystyle\frac{1}{b} \gt \displaystyle\frac{1}{c}$, and joining the two,

$\displaystyle\frac{1}{a} \gt \displaystyle\frac{1}{b} \gt \displaystyle\frac{1}{c}$,

Or inverting,

$a \lt b \lt c$,

The desired relation then is,

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

**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 -- for utilizing the common terms, inverting and rationalizing the expressions thus converting the subtraction expressions to summmation expressions -- establishing value relationship between inverted form of three expressions -- inverting the expressions inverts the nature of value relationship which we targeted for.

** Problem 8.**

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

- 2, 4, 8
- 6, 4, 2
- 8, 3, 1
- 4, 6, 8

** Solution:**

In our treatment of * Number system and operations on numbers*, we had used the technique of expressing a number as a sum of face values while explaining the divisibility rule of 9. The same method will also give us the remainder when divided by 9.

Let us repeat the technique for the first number.

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

$\hspace{20mm} + 6\times{10^3} + 6\times{10^2} + 6\times{10} + 6$.

When we divide this number by 9, the division can be taken as term by term division of the expanded expression by 9. Excluding the unit's digit all the other terms have a factor of 10s power.

The special property of 10s 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 our 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 of division of 6666666 by 9 will be equivalent to remainder by dividing,

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

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

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

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

**Answer:** Option b: 6, 4, 2.

**Key concepts used:** Expressing a number as a sum of face 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.

**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,

- 101 : 88
- 87 : 100
- 85 : 98
- 97 : 84

**Solution:**

If the ratio of two parts of a whole is $a : b$, then the whole is $a + b$. Using this concept we get the total number of employees for each of the options as $\text{numerator} + \text{denominator}$.

With this new knowledge, we will now examine the sum of numerator and denominator for each of the choice values to check which one is a prime number.

For the first choice, $a + b = 189$ which is divisible by 3 as its integer sum 18 is divisible by 3.

For the second choice, $a + b = 187$ which is divisible by 11 as sum of its 1st and 3rd digits equals the middle digit 8. This choice is also not a prime number.

For the third choice, $a + b = 183$ again divisible by 3.

For the fourth choice, $a + b = 181$. This is not divisible by 3, 5, 7 or 11. It is not also divisible by 13, and as $17^2 = 289$ exceeds 181, this is then 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.

** 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?

- 4
- 5
- 3
- 6

**Solution:**

By the problem definition, the number of children in the rows form an Arithmetic progression of the form,

$a, a + 3, a + 6,....\text{[}a + (n - 1)\times{3}\text{]}$, where $a$ is the number of children in the back row and is the first term of the series, 3 is the the common difference and $n$ is the number of rows.

The sum of such an arithmetic progression is,

$S = an + \displaystyle\frac{3n(n - 1)}{2} = 630$.

Let us test the feasibility for value of $n=4$,

$630 = 4a + 18$, Or, $4a = 612$, a multiple of 4 and so, $a$ is an integer and $n=4$ is feasible.

For $n=5$,

$630 = 5a + 30$, Or, $5a = 600$, a multiple of 5 and so, $a$ is an integer and $n=5$ is feasible.

For $n=3$,

$630 = 3a + 9$, Or, $5a = 621$, a multiple of 3 and so $a$ is an integer and $n=3$ is feasible.

For $n=6$,

$630 = 6a + 45$, Or, $6a = 585$, not a multiple of 6, thus $a$ becomes a fraction. Thus 6 rows is not possible.

**Answer:** Option d: 6.

**Key concepts used:** Identification of the series as an arithmetic progression -- using the formula for sum of such a progression with common difference 3 testing each value of number of rows identifying 6 as the value that results in a fractional number of children in the back row.

We have used here the * free resource of the choice values* for reaching the solution.

#### Alternate solution

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 + $

$\hspace{10mm} 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.

#### Recommendation

Generally we are against using any formula blindly. Usually working from the basic and rich concepts produce faster solution. But in this case we would recommend, if you can identify the series of numbers of children in rows as an arithmetic progression and use this concept and associated simple formula for total of $n$ terms in an arithmetic progression, solution will be faster and less tedious.

### Other resources on Number system and related topic

You may refer to our other useful resources on number system and other related topic especially algebra.

#### Valuable guidelines with links to resources on SSC CGL

**7 steps for sure success in SSC CGL tier 1 and tier 2 Competitive tests**

#### Tutorials

**Numbers and Number system and basic mathematical operations**

**Factorization or finding out factors**

**Basic and Rich Algebra concepts for elegant solutions of SSC CGL problems - ****though on Algebra, it should be useful.**

#### Question sets and Solution sets on Number system

**SSC CGL level Question Set 15 on Number System**

**SSC CGL level Question set 14 on Number System**

**SSC CGL level Solution Set 14 on Number System**

**SSC CGL level Question Set 7 on Arithmetic Number Sysem**

**SSC CGL level Solution Set 7 on Arithmetic Number System**

**SSC CGL level Question Set 3 on Arithmetic Number Sysem**

**SSC CGL level Solution Set 3 on Arithmetic Number System**