## 5th SSC CPO Solved Question Set, 1st on Number system

This is the fifth set of 10 questions for SSC CPO CISF Delhi Police SI and ASI and 1st on Number system. All problems are solved by specially quick methods based on basic concepts and discovered patterns. The solved question set contains,

**Question set**on**Number system for SSC CPO**to be answered in 15 minutes (10 chosen questions)**Answers**to the questions, and**Detailed conceptual solutions**to the questions.

For maximum gains, the test should be taken first, and then the solutions are to be read.

**IMPORTANT:** To absorb the concepts, techniques and reasoning explained in the solutions fully and apply those in solving problems on Number system quickly, one must solve many problems in a systematic manner using the conceptual analytical approach.

Learning by doing is the best learning. There is no other alternative towards achieving excellence.

### 5th Question set - 10 problems for SSC CPO exam: 1st on topic Number system - answering time 15 mins

**Q1.** How many positive factors of 40 are there?

- $8$
- $6$
- $4$
- $3$

**Q2.** What is the the maximum value of $F$ in the equation, $5E9+2F8+3G7=1114$, where $E$, $F$, $G$ each stands for any digit?

- $8$
- $5$
- $9$
- $7$

**Q3.** A boy added all natural numbers from 1 to 10. However, he added one number twice due to which the sum became 58. What was the number he added twice?

- $8$
- $3$
- $7$
- $4$

**Q4.** If $(7^{19}+2)$ is divided by 6, the remainder is,

- $5$
- $3$
- $1$
- $2$

**Q5.** The sum of first sixty numbers from one to sixty is divisible by,

- $59$
- $13$
- $60$
- $61$

**Q6.** If $[n]$ denotes greatest integer $\lt n$ and $(n)$ denotes smallest integer $\gt n$, where $n$ is any real number, then, $\left(1\frac{1}{5}\right)\times{\left[1\frac{1}{5}\right]}-\left(1\frac{1}{5}\right) \div \left[1\frac{1}{5}\right]+(1.5)$ is,

- $1.5$
- $3.5$
- $2$
- $2.5$

**Q7.** The rational number between $\displaystyle\frac{1}{2}$ and $\displaystyle\frac{3}{5}$ is,

- $\displaystyle\frac{4}{7}$
- $\displaystyle\frac{2}{5}$
- $\displaystyle\frac{1}{3}$
- $\displaystyle\frac{2}{3}$

**Q8.** If two numbers $x$ and $y$ are separately divided by a number $d$, remainders obtained are 4375 and 2986 respectively. If the sum of the numbers, that is, $(x+y)$ is divided by the same number $d$, remainder obtained is 2361. The value of the number is,

- $4000$
- $5000$
- $7361$
- $2542$

**Q9.** The value of $x$ in the equation, $0.\dot{3} +0.\dot6+0.\dot7+0.\dot8=x$ is,

- $53$
- $2.35$
- $2\displaystyle\frac{3}{10}$
- $2\displaystyle\frac{2}{3}$

**Q10.** On multiplying a number by 7, all the digits in the product appear as 3's. The smallest such number is,

- $47649$
- $48619$
- $47619$
- $47719$

### Answers to the questions

**Q1. Answer:** Option a: $8$.

**Q2. Answer:** Option c: $9$.

**Q3. Answer:** Option b: $3$.

**Q4. Answer:** Option b: $3$.

**Q5. Answer:** Option d: $61$.

**Q6. Answer:** Option c: $2$.

**Q7. Answer**: Option a: $\displaystyle\frac{4}{7}$.

**Q8. Answer:** Option b: $5000$.

**Q9. Answer**: Option d: $2\displaystyle\frac{2}{3}$.

**Q10. Answer:** Option c: $47619$.

### 5th solution set - 10 problems for SSC CPO exam: 1st on topic Number system - answering time 15 mins

**Q1.** How many positive factors of 40 are there?

- $8$
- $6$
- $4$
- $3$

#### Solution 1: Immediate solution by prime factorization and counting combination of factors, Non-prime factor identification

Prime factors of 40 are given by,

$40=2\times{2}\times{2}\times{5}$.

Prime factors are 4 in number.

Non-prime factors are formed by combining 2 to 3 (not 4) of the prime factors. These are,

$2\times{2}=4$,

$2\times{5}=10$, both 4 and 10 are from $(4\times{10}=40)$,

$2\times{2}\times{2}=8$, 5 and 8 are from $(5\times{8}=40)$, and,

$4\times{5}=20$, 2 and 20 are from $(2\times{20}=40)$.

So, total number of factors (non-unique) of 40 is, 8.

**Answer:** Option a: $8$.

**Key concepts used:** **Prime factorization of an integer and combining prime factors -- ***Non-prime factor identification -- ***Solving in mind.**

**Q2.** What is the the maximum value of $F$ in the equation, $5E9+2F8+3G7=1114$, where $E$, $F$, $G$ each stands for any digit?

- $8$
- $5$
- $9$
- $7$

#### Solution 2: Quick solution by concept of addition and carry-over

Addition of the three numbers is,

$\text{ 5E9}$

$\text{ 2F8}$

$\text{ 3G7}$

-- -- -- --

$\text{1114}$, with carry-over 2 at the unit's digit position.

In forming the ten's place digit of the sum 1, the carry-over of $2$ will be added to $E+F+G$ to form $11$ with a second carry-over of $1$ to make the sum of the hundred's place digits plus carry-over 1, $5+2+3+1=11$,

$E+F+G+2=11$.

So, $F$ will attain maximum value of $11-2=9$ only when the other two ten's place digits $E$ and $G$ are both 0.

**Answer:** Option c: $9$.

**Key concepts used:** **Basic addition and carry-over concepts -- Mathematical reasoning -- Solving in mind.**

**Q3.** A boy added all natural numbers from 1 to 10. However, he added one number twice due to which the sum became 58. What was the number he added twice?

- $8$
- $3$
- $7$
- $4$

#### Solution 3: Immediate solution by sum of first $n$ natural numbers concept

The sum of first $n$ natural numbers is,

$\displaystyle\frac{n(n+1)}{2}=55$ for $n=10$.

As mistaken sum is 58, the twice added digit must be,

$58-55=3$.

**Answer:** Option b: $3$.

**Key concepts used:** **Sum of first $n$ natural numbers -- Solving in mind.**

**Q4.** If $(7^{19}+2)$ is divided by 6, the remainder is,

- $5$
- $3$
- $1$
- $2$

#### Solution 4: Immediate solution by key pattern identification and Binomial theorem concept of expansion of $(x+1)^n$

The first breakthrough is by identifying,

$7^{19}+2=(6+1)^{19}+2$.

Binomial theorem formula gives the expansion of $(x+1)^n$ as,

$(x+)^n=x^n+^nC_1x^{n-1}+^nC_2x^{n-2}+.......+^nC_{n-1}x^1+1$.

You don't have to remember the formula, remembering only the basic concept is enough. The **core concept of Binomial theorem** required in this case is,

Expansion of $(x+1)^n$ consists of $n+1$ terms, first term power of $x$ is $n$ and last term power of $x$ is $0$.

In other words, in the expansion, all the terms except the last must have $x$ in it, and so the sum of these terms, that is, sum of all terms except the last term of 1, will be divisible by $x$.

In our case then the remainder of dividing $7^{19}=(6+1)^{19}$ will be 1.

Final remainder will then be, $1+2=3$.

**Answer:** Option b: $3$.

**Key concepts used:** **Key pattern identification of forming $7^{19}=(6+1)^{19}$ -- Binomial theorem concept --***-- Core concept of Binomial theorem --*** Solving in mind.**

**Q5.** The sum of first sixty numbers from one to sixty is divisible by,

- $59$
- $13$
- $60$
- $61$

#### Solution 5: Solve quickly by sum of first $n$ natural numbers concept and choice value test

Sum of first 60 natural numbers is,

$\displaystyle\frac{60(60+1)}{2}=30\times{61}$.

61 at option d will be the factor dividing the sum fully.

**Answer:** Option d: $61$.

**Key concepts used:** **Sum of first $n$ natural numbers concept -- ***Choice value test -- ***Solving in mind.**

**Q6.** If $[n]$ denotes greatest integer $\lt n$ and $(n)$ denotes smallest integer $\gt n$, where $n$ is any real number, then, $\left(1\frac{1}{5}\right)\times{\left[1\frac{1}{5}\right]}-\left(1\frac{1}{5}\right) \div \left[1\frac{1}{5}\right]+(1.5)$ is,

- $1.5$
- $3.5$
- $2$
- $2.5$

#### Solution 6: Solving in mind by decoding the specially defined operators

By the two given specially defined operators,

$\left(1\frac{1}{5}\right)=2$, and,

${\left[1\frac{1}{5}\right]}=1$.

And the target expression,

$\left(1\frac{1}{5}\right)\times{\left[1\frac{1}{5}\right]}-\left(1\frac{1}{5}\right) \div \left[1\frac{1}{5}\right]+(1.5)$

$=2\times{1}-\displaystyle\frac{2}{1}+2=2$.

**Answer:** Option c: $2$.

**Key concepts used:** **Decoding specially defined operators carefully -- ***Coded operation -- ***Solving in mind.**

**Q7.** The rational number between $\displaystyle\frac{1}{2}$ and $\displaystyle\frac{3}{5}$ is,

- $\displaystyle\frac{4}{7}$
- $\displaystyle\frac{2}{5}$
- $\displaystyle\frac{1}{3}$
- $\displaystyle\frac{2}{3}$

#### Solution 7: Quick solution by fraction comparison of choice values with range limits, Fraction range fitting

This is problem of placing a fraction in a range defined by a smaller and a larger fraction.

The choice value that will be your answer must be greater than the lower limit of the range, that is, $\gt \displaystyle\frac{1}{2}$.

And must be smaller than the upper limit of the range, that is, $\lt \displaystyle\frac{3}{5}$.

You have to compare each choice value fraction with the two range limit fractions to find the solution that satisfies both the conditions.

Basically then, you have to compare 8 pairs of fractions, not an easy task in case of awkward fraction numerator and denominator values.

But in this simple case, the first fraction directly satisfies both the conditions and turns out to be the quick solution,

$\displaystyle\frac{1}{2} \lt \displaystyle\frac{4}{7} \lt \displaystyle\frac{3}{5}$.

$\displaystyle\frac{4}{7}-\displaystyle\frac{1}{2}=\displaystyle\frac{8-7}{14}=\frac{1}{14}$, and

$\displaystyle\frac{3}{5}-\displaystyle\frac{4}{7}=\displaystyle\frac{21-20}{35}=\frac{1}{35}$.

Both these direct subtractions can be done mentally with ease.

**Note:** Fraction comparison and fraction range problems are treated in details in our knowledge article,

**How to solve difficult fraction comparison problems in a few steps.**

**Answer:** Option a: $\displaystyle\frac{4}{7}$.

**Key concepts used:** **Fraction range placement problem -- Fraction comparison -- ***Fraction range fitting -- ***Solving in mind.**

**Q8.** If two numbers $x$ and $y$ are separately divided by a number $d$, remainders obtained are 4375 and 2986 respectively. If the sum of the numbers, that is, $(x+y)$ is divided by the same number $d$, remainder obtained is 2361. The value of the number is,

- $4000$
- $5000$
- $7361$
- $2542$

#### Solution 8: Immediate solution by basic division and remainder concept and Euclid's division lemma

By basic division and remainder concept or Euclid's division lemma,

$x=p\times{d}+4375$, where $p$ is the assumed quotient, and

$y=q\times{d}+2986$, where $q$ is the assumed quotient.

Adding,

$(x+y)=(p+q)d+7361$.

So the remainder of dividing $(x+y)$ by $d$, 2361, will be the remainder of dividing $7361$ by $d$ as $d$ is already a factor of the first term $(p+q)d$.

By Euclid's division lemma again then,

$7361=r\times{d}+2361$, where assumed quotient $r$ must be 1.

This is because, by basic division and remainder concepts, none of the two remainders $4375$ and $2986$ (sum of which is $7361$) of the first two divisions can be greater than $d$,

So, $d=7361-2362=5000$.

Calculations being simple, if your basic concepts are ok, the problem can easily be solved in mind.

**Answer:** Option b: $5000$.

**Key concepts used:** **Basic division and remainder concepts -- Euclid's division lemma -- Solving in mind.**

**Q9.** The value of $x$ in the equation, $0.\dot{3} +0.\dot6+0.\dot7+0.\dot8=x$ is,

- $53$
- $2.35$
- $2\displaystyle\frac{3}{10}$
- $2\displaystyle\frac{2}{3}$

#### Solution 9: Quick solution by converting repeating non-terminating decimal to fraction

Assume the repeating non-terminating decimal $0.\dot{3}$ as $z$,

$z= 0.\dot{3}=0.333333...$.

Multiply the equation by 10,

$10z=3+z$, Or, $9z=3$,

Or, $z=\displaystyle\frac{3}{9}=\frac{1}{3}$.

Similarly,

$0.\dot6=0.666666....=\displaystyle\frac{6}{9}=\frac{2}{3}$,

$0.\dot7=0.777777....=\displaystyle\frac{7}{9}$, and,

$0.\dot8=0.888888....=\displaystyle\frac{8}{9}$.

Summing up,

$0.\dot{3} +0.\dot6+0.\dot7+0.\dot8=x$,

Or, $x=\displaystyle\frac{1}{3}+\displaystyle\frac{2}{3}+\displaystyle\frac{7}{9}+\displaystyle\frac{8}{9}$

$=1+\displaystyle\frac{15}{9}=2\displaystyle\frac{2}{3}$.

**Note:** Quick formula is, $0.\dot{z}=\displaystyle\frac{z}{9}$, where $z$ can take values 1 to 8.

**Answer:** Option d: $2\displaystyle\frac{2}{3}$.

**Key concepts used:** **Conversion of repeating non-terminating decimal to rational fraction -- Solving in mind.**

**Q10.** On multiplying a number by 7, all the digits in the product appear as 3's. The smallest such number is,

- $47649$
- $48619$
- $47619$
- $47719$

#### Solution 10: Quick solution by key pattern identification and directly dividing the product $333333$ by $7$

On first thought, the problem may seem to be a little difficult. But if you think for a moment, you would discover the key pattern,

When a five digit number, as in the choices (with most significant digit of all four choices 4), is multiplied by single digit number 7, and the product has the digits all 3's, the product must be the six digit number $333333$.

So, just divide $333333$ by $7$ and get your answer.

Dividing product $333333$ by $7$,

$333333 \div 7=47619$

**Answer:** Option c: $47619$.

**Key concepts used:** **Key pattern identification from choice value analysis to form the product number -- Direct division of the product by 7 -- Solving in mind.**

**Note:** If the problem were, "On multiplying a number by 7, all the digits in the product appear as 3's. Find the smallest such number.", then also would have got the same answer by dividing $333333...$ by $7$. In this case, you couldn't have identified the product to be $333333$ by choice value analysis. Nevertheless, the product would certainly have the first six digits as $333333$ followed by more numbers of 3's.

### End note

Observe that, each of the problems could be quickly and cleanly solved in minimum number of steps using special key patterns and methods in each case.

This is the hallmark of quick problem solving:

**Concept**based**pattern and method**formation, and,**Identification of the key pattern**and**use of the method**associated with it. Every special pattern has its own method, and not many such patterns are there.

Important is the **concept based pattern identification and use of quick problem solving method.**

### SSC CPO level Question and Solution sets

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**SSC CPO level Solved Question set 1 on Algebra 1**

**SSC CPO level Solved Question set 2 on Algebra 2**

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**SSC CPO level Solved Question set 3 on Trigonometry 1**

**SSC CPO level Solved Question set 4 on Trigonometry 2**

#### Number system

**SSC CPO level Solved Question set 5 on Number system 1**

#### Surds and Indices

**SSC CPO level Solved Question set 6 on Surds and indices 1**