## 30th SSC CGL level Solution Set, 6th on topic Number System

This is the 30th solution set of 10 practice problem exercise for SSC CGL exam and 6th on topic Number System. Students must complete the corresponding question set in prescribed time first and then only refer to the solution set for extracting maximum benefits from this resource.

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.

If you have not taken this test yet, you may refer to the * 30th SSC CGL level Question Set on Number System* to take the test before going ahead with its solution here.

### 30th solution set- 10 problems for SSC CGL exam: 6th on topic Number System - time 12 mins

**Problem 1.**

Thrice the square of a natural number decreased by four times the number is equal to 50 more than the number. The number is,

- 4
- 5
- 6
- 10

** Solution 1 : Problem analysis and execution:**

One option is to assume the unknown number as $x$, and form the equation, $3x^2 - 4x = x + 50$ and solve the quadratic equation for the value of $x$. That is a conventional deductive approach. To be successful in any high level competitive exam, number problems cannot be solved by this conventional deductive approach.

**Alternatively** let us intelligently test the option values for satisfying the conditions.

While testing, we will test only the unit's digit. **This is our special technique in action.**

Unit's digit for 10 is satisfying the operations on it but square of 10 is too high.

For 6, the unit's digit of the result is, $3\times{6^2} -4 = 4$, not 6.

For 4, the unit's digit of the result is, $3\times{4^2} - 6 = 2$, not 4.

But for 5, the unit's digit is, $3\times{5} - 0 = 5$. The actual calculation is, $3\times{25} - 20= 55 = 50 + 5$.

As the result must be equal to 50 plus the number, its unit's digit must be equal to the unit's digit of the number, that is, except 10, the number itself.

**Answer:** b: 5.

**Key concepts used:** Pattern recognition and new technique use -- **unit's digit behavior analysis.**

**Problem 2.**

The difference between two positive numbers is 3. If sum of their squares is 369, then the sum of the numbers is,

- 25
- 81
- 27
- 33

**Solution 2 : Problem analysis and execution:**

As far as our idea of squares of small two digit numbers goes, $15^2 + 12^2 = 369$.

**Method:** $16^2=256$ is too near to 369 because $13^2=169$ does not add up to 369. Next lower pair would be 15 and 12.

**Answer:** Option c : 27 .

**Key concepts used:** Sense of squares of small two digit numbers.

**Problem 3.**

A number consists of two digits such that the digit in the ten's place is less than the digit in the unit's place by 2. Three times the number added to $\displaystyle\frac{6}{7}$th of the reverse of the number is 108. The sum of the digits of the number is,

- 6
- 8
- 9
- 7

**Solution 3 : Problem analysis and execution:**

The conventional approach is to assume variables for unit's and ten's digits, form equations and solve.

**Alternatively we will use mathematical reasoning and sense of number system.**

As three times the number added to something is 108, * the number is less than one third of 108,* that is,

*.*

**less than 36**By the first condition, the possibilities then are,

35, 24 and 13.

35 is too near to 36 and is infeasible and three times 13 plus its full reverse 31 can't reach 108.

**24 satisfies all conditions.**

**Answer:** Option a: 6.

**Key concepts used:** From problem definition * limiting number of possible solutions* --

**testing each of the few solutions.****Problem 4.**

Of three numbers, the second is twice the first and thrice the third. If the average of the numbers is 44, the difference between the first number and the third number is,

- 18
- 12
- 6
- 24

**Solution 4 : Problem analysis and execution:**

If the first number is $A$, the second is $2A$ and the third is $\frac{2}{3}A$. So the sum of three numbers is,

$A + 2A + \frac{2}{3}A = \frac{11}{3}A = 3\times{44}$.

So, first number, $A=36$ and third number $=\frac{2}{3}A = 24$ and the difference between the two, 12.

**Answer:** b: 12.

**Key concepts used:** Converting all numbers in terms of a single number -- average.

**Problem 5.**

A two digit number is five times the sum of its digits. If 9 is added to the number, the digits interchange their positions. The sum of the digits of the number is,

- 11
- 9
- 6
- 7

**Solution 5 : Problem analysis and execution:**

*First conclusion* is, **the number is a multiple of 5.**

*Second conclusion* is, as * adding 9 reverses the number, the difference between the digits is 1 and unit's digit is non-zero.* For example, adding 9 to 12 will make it 21, but adding 9 to 13 can't reverse it.

This is two digit **number reversal property.**

Automatically 45 comes to mind because * it satisfies both the conditions* and also

**it is five times the sum of its digits.****Answer:** Option b: 9.

**Key concepts used:** Mathematical reasoning in number domain -- **number reversal property.**

#### Problem 6.

In a three digit number, the digit at the hundred's place is two times the digit at the unit's place and sum of the three digits is 18. If the digits are reversed the number is reduced by 396. The difference of hundred's and ten's digits of the number is,

- 5
- 2
- 1
- 3

**Solution 6 : Problem analysis and execution:**

From the first condition, the possible pairs of hundred's and unit's digits are,

(2, 1), (4, 2), (6, 3) and (8, 4).

As the sum of the three digits is 18, the first two pairs turn out to be infeasible and three digit possibilities are reduced to,

(6, 9, 3) and (8, 6, 4).

By the third condition, 693 is eliminated because of its odd unit's digit **(we haven't calculated, ***w*e have analyzed the behavior of the unit's digit).

**864 satisfies all conditions.**

**Answer:** Option b : 2.

**Key concepts used:** Mathematical reasoning in number domain -- formation of possible digit combinations by * limiting number of possible solutions* --

**unit's digit behavior analysis technique.**** Problem 7.**

If the sum of two numbers are multiplied by each number separately, the products so obtained are 247 and 114. The sum of the numbers is,

- 20
- 19
- 21
- 23

**Solution 7 : Problem analysis and execution:**

We will use factorization and detect the common factor between the two. That will be the sum of the two numbers.

$247 = 13\times{19}$, and

$114 = 6\times{19}$.

So the common factor is 19 and it is the desired sum of the two numbers which are 6 and 13.

**Answer:** Option b: 19.

** Key concepts used:** Mathematical reasoning in number domain -- factorization.

** Problem 8.**

A two digit number is three times the sum of its digits. If 45 is added to the number, its digits are interchanged. Sum of the digits of the number is,

- 11
- 7
- 5
- 9

**Solution 8 : Problem analysis and execution:**

By first statement, $3(x + y) = 10x + y$, Or, $7x = 2y$. * Only 27 fits the bill* and satisfies the other condition of digit reversal also.

The first statement itself is clinching. 27 is the only two digit number satisfying this condition.

**Answer:** Option d: 9.

**Key concepts used:** Using the condition to form an equation between the digits of the number identifying it uniquely at one go -- using **place value concept.**

**Problem 9.**

The numbers 2272 and 875 are divided by a three digit number giving same remainders. The sum of the digits of this three digit number is,

- 12
- 13
- 10
- 11

**Solution 9 : Problem analysis and execution:**

If we take the difference of the two given numbers, the remainders will be canceled out and the difference will also be divisible by the three digit number.

If $n_1=m_1q_1 + r_1$ and $n_2 = m_1q_2 + r_1$, then, $n_1 - n_2 = m_1(q_1 - q_2)$, that is, $m_1$ will be a factor of $n_1 - n_2$.

Thus, the difference of the two numbers is,

$2272 - 875 = 1397 = 11\times{127}$.

So 127 is the three digit factor of the two numbers giving same remainder.

**Answer:** Option c: 10.

**Key concepts used:** Basic concepts of a division, forming quotient, and remainder.

** Problem 10.**

A fraction having denominator 30 and lying between $\frac{5}{8}$ and $\frac{7}{11}$ is,

- $\frac{19}{30}$
- $\frac{18}{30}$
- $\frac{20}{30}$
- $\frac{21}{30}$

**Solution 10 : Problem analysis and execution:**

First by comparison we determine relative values of the two bounding fractions,

$\displaystyle\frac{7}{11} \gt \displaystyle\frac{5}{8}$, as $\displaystyle\frac{56}{88} \gt \displaystyle\frac{55}{88}$.

So by problem definition, the desired fraction needs to be larger than $\displaystyle\frac{5}{8}$ but smaller than $\displaystyle\frac{7}{11}$.

As a strategy we will test out the option values against the boundary fractions. While comparing fractions we will use both numerator equalization and denominator equalization methods.

Taking $\displaystyle\frac{21}{30}$ we equalize the numerator of $\displaystyle\frac{7}{11}$ transforming it to $\displaystyle\frac{21}{33}$ and being smaller than the option choice, this choice falls outside the higher limit of the range.

Similarly we equalize the numerator of $\displaystyle\frac{20}{30}$ with $\displaystyle\frac{5}{8}$ transforming it to $\displaystyle\frac{20}{32}$ to find the choice larger and so a probable. But comparing its minimized form $\displaystyle\frac{2}{3}$ with $\displaystyle\frac{7}{11}$ we find the former, $\displaystyle\frac{22}{33}$ larger than the latter, the upper limit of the range, $\displaystyle\frac{21}{33}$. Thus this choice value is also out of the range.

Testing $\displaystyle\frac{18}{30}$ or $\displaystyle\frac{3}{5}$ against $\displaystyle\frac{5}{8}$, we find $\displaystyle\frac{5}{8} = \displaystyle\frac{25}{40} \gt \displaystyle\frac{24}{40} = \displaystyle\frac{18}{30}$. So this fraction is also out of range.

The choice must then be the last fraction at option a.

Testing it against $\displaystyle\frac{5}{8}$,

$\displaystyle\frac{19}{30}=\frac{152}{240} \gt \displaystyle\frac{150}{240}=\frac{5}{8}$, and comparing it with $\displaystyle\frac{7}{11}$,

$\displaystyle\frac{7}{11}=\frac{210}{330} \gt \displaystyle\frac{209}{330}=\frac{19}{30}$.

**Answer:** Option a: $\displaystyle\frac{19}{30}$.

**Key concepts used:** Fraction comparison using minimum possible steps.

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

#### Alternate solution

Six decimal equivalents can be formed and then compared with each other.

Either way, this problem takes a bit of time and is recommended to be avoided in case of time shortage.

#### 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.**

#### How to solve number system problems quickly in a few steps

**Componendo dividendo applied on number system and ratio proportion problems**

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

**SSC CGL Tier II level Solution Set 29 on Number System 1**

**SSC CGL Tier II level Question Set 29 on Number System 1**

**SSC CGL level Solution Set 89 on Number System 11**

**SSC CGL level Question Set 89 on Number System 11**

**SSC CGL level Question Set 71 on Number System 10**

**SSC CGL level Solution Set 71 on Number System 10**

**SSC CGL level Solution Set 55, Number System 9**

**SSC CGL level Question Set 55, Number System 9**

**SSC CGL level Question Set 54 on Number System 8**

**SSC CGL level Solution Set 54 on Number System 8**

**SSC CGL level Question Set 46 on Number System 7**

**SSC CGL level Solution Set 46 on Number System 7**

**SSC CGL level Question Set 30 on Number System 6**

**SSC CGL level Question Set 28 on Arithmetic Number System**

**SSC CGL level Solution Set 28 on Arithmetic Number System**

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

**SSC CGL Level Solution Set 15 on Arithmetic Number System**

**SSC CGL level Question Set 14 on Arithmetic Number Sysem**

**SSC CGL level Solution Set 14 on Arithmetic 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 Solution Set 3 on Arithmetic Number System**

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

### Solved Question sets on Number System, HCF and LCM for SSC CHSL

**SSC CHSL level Solved Question set 3 on Number system 1**

**SSC CHSL level Solved Question set 4 on Number system 2**

**SSC CHSL level Solved Question set 5 on HCF and LCM 1**

**SSC CHSL level Solved Question set 6 on HCF and LCM 2**