## 2nd Solved set of SSC CHSL Number system Questions

2nd set of 10 SSC CHSL Number system questions with answers. To be used for timed practice. Solutions show how to solve the questions quickly in mind.

Should be *useful for prelims of other competitive exams also.*

It contains,

**10 chosen SSC CHSL Number system Questions**to be answered in 15 minutes,**Answers**to the questions, and,**Quick solutions**to the questions.

To absorb the **concepts and techniques for quick solutions** explained, you should solve many such problems *using the conceptual analytical approach.*

*Number system questions are an important component in any competitive exam all over the world, and solving number system questions quickly oftentimes depends heavily on conceptual analytical approach rather than conventional rote approaches.*

### 2nd set of 10 SSC CHSL Number System questions: Answering time 15 mins

**Q1. **A number consists of two digits and the digit in the ten's place exceeds that in the unit's place by 5. If 5 times the sum of digits be subtracted from the number, the digits of the number are reversed. Then the sum of digits of the number is,

- 13
- 7
- 9
- 11

**Q2.** Half of 1 percent written as a decimal is,

- 0.2
- 0.005
- 0.02
- 0.05

**Q3. **Ram left $\displaystyle\frac{1}{3}$rd of his property to his widow and $\displaystyle\frac{3}{5}$th of the remainder to his daughter. He gave the rest to his son who received Rs.6400. How much was the original property worth?

- Rs.1600
- Rs.16000
- Rs.32000
- Rs.24000

**Q4. **The product of digits of a two digit number is 24. If we add 45 to the number, the new number obtained is a number formed by interchanging the digits. What is the number?

- 38
- 54
- 83
- 45

**Q5.** The number $a$ divides 228 leaving a remainder 18. The biggest two digit value of $a$ is,

- 30
- 70
- 21
- 35

**Q6.** The sum and product of two numbers are 11 and 18 respectively. The sum of their reciprocals is,

- $\displaystyle\frac{11}{18}$
- $\displaystyle\frac{11}{2}$
- $\displaystyle\frac{18}{11}$
- $\displaystyle\frac{2}{11}$

**Q7.** Of the three numbers, the second is twice the first and is also thrice the third. If the average of these three numbers is 44, the largest number is,

- 24
- 36
- 108
- 72

**Q8.** Divide 37 into two parts so that 5 times one part and 11 times the other are together 227. The value of the two parts are,

- 15, 22
- 30, 7
- 20, 17
- 25, 12

**Q9.** 47 is added to the product of 71 and an unknown number. The new number is divisible by 7 giving the quotient 98. The unknown number is then a multiple of,

- 2
- 7
- 3
- 5

**Q10.** The value of $\lambda$ for which the expression $x^3+x^2-5x+\lambda$ is divisible by $(x-2)$ is,

- $2$
- $-3$
- $4$
- $-2$

### Answers to the questions

**Q1. Answer:** Option c: 9.

**Q2. Answer:** Option b : 0.005.

**Q3. Answer:** Option d: Rs.24000.

**Q4. Answer:** Option a: 38.

**Q5. Answer:** Option b: 70.

**Q6. Answer:** Option a : $\displaystyle\frac{11}{18}$.

**Q7. Answer:** Option d: 72.

**Q8. Answer:** Option b: 30, 7.

**Q9. Answer:** Option c: 3.

**Q10. Answer:** Option d: $-2$.

### Quick solutions to the 2nd set of 10 SSC CHSL Number system questions - answering time was 15 mins

**Q1. **A number consists of two digits and the digit in the ten's place exceeds that in the unit's place by 5. If 5 times the sum of digits be subtracted from the number, the digits of the number are reversed. Then the sum of digits of the number is,

- 13
- 7
- 9
- 11

** Solution 1: Problem analysis and solution by pattern identification and using the pattern**

Let $x$ and $y$ be the ten's and unit's digits of the number.

So by **place value mechanism**, the number is represented by,

$(10x+y)$.

By the first condition given,

$(x-y)=5$.

And by the second given condition,

$(10x+y)-5(x+y)=10y+x$.

When the digits of the number are reversed it is represented by,

$10y+x$.

At this point, with the objective of using the value, $(x-y)=5$, we examine the second equation.

Recall that, you don't have to find the values of $x$ and $y$, but only need to find the value of $(x+y)$.

So, can we express the second equation in terms of $(x-y)$ and $(x+y)$? If we could, we would get the solution immediately.

Just a bit of examination, and you find the way to do it. Take the terms on the RHS to LHS, regroup, and you will get,

$10(x-y)-(x-y)-5(x+y)=0$,

Or, $5(x+y)=9(x-y)=45$, as $(x-y)=5$.

So, $(x+y)=9$.

**Answer:** Option c: 9.

**Key concepts used: *** Place value mechanism* --

*--*

**Pattern identification***.*

**Problem solver's solution**This solution is quick and takes fewest number of steps. This is what we call a Problem solver's solution.

You may easily get the values of $x$ and $y$ individually from two linear equations, but that will take a few extra steps and a bit of extra time to solve.

**Q2.** Half of 1 percent written as a decimal is,

- 0.2
- 0.005
- 0.02
- 0.05

**Solution 2: Problem solving using basic percentage to decimal conversion concept**

When you encounter a value as a percentage, say, $x$%, the actual value of the term will be $x$ divided by 100 and expressed as a decimal. In other words,

$x\text{%}=\displaystyle\frac{x}{100}=0.01\times{x}$.

So, $1\text{%}=0.01$.

Dividing this by 2 you would get half percent as,

$\displaystyle\frac{1}{2}\text{%}=\displaystyle\frac{0.01}{2}=0.005$

**Answer:** Option b: 0.005.

**Key concepts used:** **Basic percentage concept -- Percentage to decimal conversion -- Decimal division -- Solving in mind****.**

**Q3. **Ram left $\displaystyle\frac{1}{3}$rd of his property to his widow and $\displaystyle\frac{3}{5}$th of the remainder to his daughter. He gave the rest to his son who received Rs.6400. How much was the original property worth?

- Rs.1600
- Rs.16000
- Rs.32000
- Rs.24000

**Solution 3: Problem analysis and solution by fraction arithmetic**

After giving $\displaystyle\frac{1}{3}$rd of his property to his widow the leftover portion is,

$1-\displaystyle\frac{1}{3}=\displaystyle\frac{2}{3}$.

The $\displaystyle\frac{3}{5}$th of this remaining portion given to his daughter is,

$\displaystyle\frac{2}{3}\times{\displaystyle\frac{3}{5}}=\frac{2}{5}$.

Leftover portion of the property is now,

$\displaystyle\frac{2}{3}-\displaystyle\frac{2}{5}=\displaystyle\frac{4}{15}$.

The final remaining portion given to his son is Rs.6400 and it equals the $\displaystyle\frac{4}{15}$th portion of the total property value,

$\displaystyle\frac{4}{15}\times{\text{ Total property value }}=6400$.

The original worth of the property is then,

$6400\times{\displaystyle\frac{15}{4}}=\text{Rs.}24000$

**Answer:** Option d: Rs.24000.

**Key concepts used: Fraction arithmetic -- Fraction of fractional remaining portion -- Multistage fractional portion sharing problem.**

**Q4. **The product of digits of a two digit number is 24. If we add 45 to the number, the new number obtained is a number formed by interchanging the digits. What is the number?

- 38
- 54
- 83
- 45

**Solution 4: Problem solving using Factor analysis and mathematical reasoning**

Assume the ten's and unit's digits of the number to be $x$ and $y$.

So, it can be expressed as $(10x+y)$ and when a new number is formed by interchanging the two digits, the new number as, $(10y+x)$.

By the first statement,

$x\times{y}=24$.

And, by the second statement,

$(10x+y)+45=(10y+x)$,

Or, $9(y-x)=45$.

Or, $y-x=5$.

As the first equation $x\times{y}=24$ is not linear, you don't have two linear equations to solve for $x$ and $y$ easily.

First we'll show you a **quick method of solution using factor analysis** of 24 to break it up into two parts that will satisfy the second linear equation.

The prime factors of 24 are,

$24=2\times{2}\times{2}\times{3}$.

* As the digit difference 5 is odd, one of the two digits must be odd.* Out of the four prime factors only 3 is odd, so it must be one of the digits, and the other 8. We have used inviolable

**mathematical reasoning.**

The combination of digits 3 and 8 satisfy all the conditions.

As $(y-x)$ is a positive number, $y \gt x$ giving $y=8$ and $x=3$.

The number is 38.

**Verify:** Add 45 to it to get, $38+45=83$, the digits reversed.

**Answer:** Option a: 38.

**Key concepts used: Digit reversal problem -- Factor analysis -- Basic number system concept of odd number property -- Mathematical reasoning -- Solving in mind -- Factoring a quadratic equation.**

#### Alternate conventional deductive solution by factoring a quadratic equation

This is not an efficient solution, but it uses standard maths taught.

You have,

$x\times{y}=24$, and,

$y-x=5$,

Or, $y=x+5$.

Substitute $y$ in first equation,

$x(x+5)=24$,

Or, $x^2+5x-24=0$.

Factorize the quadratic equation by splitting the numeric constant term $(-24)$ into two parts, one negative and the other positive so that, the sum of the two parts is $(+5)$.

Looking closely you can easily identify $(+8)$ and $(-3)$ as the two parts sum of which is $(+5)$.

The quadratic equation is factorized as,

$x^2+5x-24=0$,

Or, $(x+8)(x-3)=0$.

As $x$ can't be negative it must be 3, and so $y=8$.

This type of *conventional solution in competitive tests* is * never recommended* when time to solve is crucial for high score.

**Q5.** The number $a$ divides 228 leaving a remainder 18. The biggest two digit value of $a$ is,

- 30
- 70
- 21
- 35

**Solution 5: Problem analysis and solving by Euclid's division lemma concept and factor analysis**

As remainder is 18 and dividend is 228, with $a$ as the divisor, the division remainder relation is,

$228=qa+18$, where $q$ is the quotient.

Simplifying,

$qa=210$.

The other condition on the value of $a$ is—it is a two digit number, which means,

$10 \leq a \leq 99$.

To get the largest possible value of $a$, 210 need to be factorized into two parts, one—the largest two digit number possible, and the other—product of the rest of the factors.

We would get a clear idea of the solution from the prime factor product of 210 which is,

$210=2\times{3}\times{5}\times{7}$.

To regroup the four factors into two parts—one smallest possible so the other is the largest possible number less than 100, first try 2 and resultant 105—not a solution.

So, taking up the next larger factor 3, you get the solution this time as,

$210=3\times{70}$.

Largest possible value of $a$ is 70.

**Answer:** Option b: 70.

*Key concepts used:* **Euclid's division relation -- Factor analysis -- Number system concepts -- Mathematical reasoning -- Solving in mind****.**

**Q6.** The sum and product of two numbers are 11 and 18 respectively. The sum of their reciprocals is,

- $\displaystyle\frac{11}{18}$
- $\displaystyle\frac{11}{2}$
- $\displaystyle\frac{18}{11}$
- $\displaystyle\frac{2}{11}$

**Solution 6: Problem analysis and solution by factor analysis**

The prime factors of 18 are,

$18=2\times{3}\times{3}=a\times{b}$. where $a$ and $b$ are the two numbers.

Also given,

$a+b=11$.

Select the obvious choice values of $a$ and $b$ by observation as,

$a=2$, and $b=9$.

These values satisfy both the given conditions.

Sum of their reciprocals is,

$\displaystyle\frac{1}{a}+\displaystyle\frac{1}{b}=\displaystyle\frac{1}{2}+\displaystyle\frac{1}{9}=\displaystyle\frac{11}{18}$.

**Answer:** Option a : $\displaystyle\frac{11}{18}$.

**Key concepts used:** *Factor analysis -- Fraction arithmetic -- Solving in mind.*

**Q7.** Of the three numbers, the second is twice the first and is also thrice the third. If the average of these three numbers is 44, the largest number is,

- 24
- 36
- 108
- 72

**Solution 7: Problem analysis and Solving by Problem definition and total from average concept**

Assume the three numbers as $x$, $y$ and $z$.

By the first statement,

$y=2x=3z$.

So,

$x=\displaystyle\frac{1}{2}y$, and

$z=\displaystyle\frac{1}{3}y$.

Total of the three numbers is then,

$x+y+z=y\left(\displaystyle\frac{1}{2}+1+\displaystyle\frac{1}{3}\right)=\displaystyle\frac{11}{6}y$.

As average of 3 numbers is 44, the total is three times the average, which is,

$3\times{44}=132=\displaystyle\frac{11}{6}y$.

So, $y=12\times{6}=72$.

And $y$ is the largest of the three by problem definition.

**Answer:** Option d: 72.

** Key concepts used:** ** Fraction arithmetic **--

*--*

**Solution by elimination of two variables to form an equation in one variable**

**Total as number of items multiplied by average, Total from average concept -- Solving in mind.****Q8.** Divide 37 into two parts so that 5 times one part and 11 times the other are together 227. The value of the two parts are,

- 15, 22
- 30, 7
- 20, 17
- 25, 12

** Solution 8: Problem analysis and Solution by solving two linear equations**

Let the value of two parts be $x$ and $y$.

By the first statement, as the sum of two parts is 37,

$x+y=37$.

Similarly by the second statement,

$5x+11y=227$.

Multiply the first linear equation by 5 and subtract the result from the second equation. The result is,

$6y=227-185=42$,

So, $y=7$, and,

$x=37-y=30$.

**Answer:** Option b: 30, 7.

**Key concepts used:** **Formation and solution of two linear equations by elimination.**

**Q9.** 47 is added to the product of 71 and an unknown number. The new number is divisible by 7 giving the quotient 98. The unknown number is then a multiple of,

- 2
- 7
- 3
- 5

**Solution 9: Problem analysis and Solving by division and quotient concept**

Let the unknown number be $x$.

By the first statement the new number formed is,

$71x+47$.

This number is divisible by 7 with quotient 98. Divisibility means remainder is zero.

So,

$71x+47=7\times{98}=686$,

Or, $71x=686-47=639$.

So, $x=9$.

This new number is a multiple of 3.

**Answer:** Option c: 3.

**Key concepts used:** * Divisibility concept -- Multiple concept* --

**Solving in mind.****Q10.** The value of $\lambda$ for which the expression $x^3+x^2-5x+\lambda$ is divisible by $(x-2)$ is,

- $2$
- $-3$
- $4$
- $-2$

**Solution 10: Problem analysis and Solving by continued factor extraction technique**

In case of division of one algebraic expression by another it is easiest to adopt the technique of continued factor extraction.

As $(x-2)$ has to be a factor of the second expression, we would take out this factor of $(x-2)$ starting from the term with the largest power of $x$ and compensating for the resultant second term created by multiplication of $(x-2)$.

At each step the largest power of $x$ will be taken care of so that the power of $x$ reduces by 1 in the remaining portion of the expression.

Let us show you by applying the technique for best explanation.

The given expression is,

$x^3+x^2-5x+\lambda$

$=x^2(x-2)+(2x^2)+(x^2-5x+\lambda$)

$=x^2(x-2)+3x^2-5x+\lambda$

At this **first step**, the **term with highest power** $x^3$ is **taken care of** by forming a factor of $(x-2)$ out of it.

This leaves only the **remaining portion of the expression with one less term and one less power** of $x$ to be factorized now.

The term $2x^2$ is added to compensate for $-2x^2$ needed to form the first factor.

The expression to factorize in the **second step** is then,

$x^2(x-2)+3x^2-5x+\lambda$

$=x^2(x-2)+3x(x-2)+6x-5x+\lambda$

$=x^2(x-2)+3x(x-2)+x+\lambda$.

So, $\lambda=-2$, for $(x-2)$ to be a factor of the larger expression.

Quotient will be,

$x^2+3x+1$.

**Answer: **Option d: $-2$.

**Key concepts used: ****Division of algebraic expressions -- Continued factor extraction method.**

### Guided help on Number system and HCF LCM in Suresolv

To get the best results out of the extensive range of articles of **tutorials**, **questions** and **solutions** on **Number system and HCF LCM **in Suresolv, *follow the guide,*

**The guide list of articles is up-to-date.**

### SSC CHSL level Question and Solution sets

#### Work and Time, Pipes and Cisterns

**SSC CHSL level Solved Question set 1 on Work time 1**

**SSC CHSL level Solved Question set 2 on Work time 2**

#### Number System, HCF and LCM

**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**

#### Surds and Indices

**SSC CHSL level Solved Question set 7 on Surds and Indices 1**

**SSC CHSL level Solved Question set 8 on Surds and Indices 2**

**SSC CHSL level Solved Question set 17 on Surds and indices 3**

#### Mixture or Alligation

**SSC CHSL level Solved Question set 9 on Mixture or Alligation 1**

**SSC CHSL level Solved Question set 10 on Mixture or Alligation 2**

#### Algebra

**SSC CHSL level Solved Question set 11 on Algebra 1**

**SSC CHSL level Solved Question set 12 on Algebra 2**

**SSC CHSL level Solved Question set 13 on Algebra 3**

**SSC CHSL level Solved Question set 14 on Algebra 4**

**5th set of Solved algebra questions for SSC CHSL 18**

**6th set of solved algebra questions for SSC CHSL 19**

#### Trigonometry

**SSC CHSL level Solved Question set 15 on Trigonometry 1**

**SSC CHSL level Solved Question set 16 on Trigonometry 2**