## 6th SSC CHSL Solved Question Set, 2nd on HCF and LCM

This is the 6th solved question set of 10 practice problem exercise for SSC CHSL exam and the 2nd on topic HCF and LCM. It contains,

**Question set on HCF and LCM**for SSC CHSL 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 referred to. But more importantly, to absorb the concepts, techniques and reasoning explained in the solutions, 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.

### 6th Question set - 10 problems for SSC CHSL exam: 2nd on topic HCF and LCM - answering time 15 mins

**Q1. **If $P=2^3.3^{10}.5$ and $Q=2^5.3.7$, then HCF of $P$ and $Q$ is,

- $2.3.5.7$
- $3.2^3$.
- $2^5.3^{10}.5.7$
- $2^2.3^7$

**Q2.** If $x:y$ be the ratio of two whole numbers and $z$ be their HCF, then LCM of the numbers is,

- $xyz$
- $yz$
- $\displaystyle\frac{xz}{y}$
- $\displaystyle\frac{xy}{z}$

**Q3. **The greatest number that divides 122 and 243 leaving remainders 2 and 3 respectively, is,

- 120
- 30
- 24
- 12

**Q4. **Find the least multiple of 23, which when divided by 18, 21 and 24 leaves the remainder 7, 10 and 13 respectively.

- 3013
- 3036
- 3002
- 3024

**Q5.** What least value should be added to 1812 to make it divisible by 7, 11 and 14?

- 12
- 72
- 36
- 154

**Q6.** Three numbers are in the ratio 2: 3 : 4 and their HCF is 12. The LCM of the numbers is,

- 96
- 144
- 72
- 192

**Q7.** The greatest 4 digit number which is exactly divisible by each one of the numbers 12, 18, 21 and 28 is,

- 9828
- 9928
- 9288
- 9882

**Q8.** The HCF and LCM of two numbers are 44 and 264 respectively. If the first number is divided by 2, the quotient is 44. The other number is,

- 147
- 132
- 264
- 528

**Q9.** The number between 4000 and 5000 that is divisible by each of 12, 18, 21 and 32 is,

- 4023
- 4032
- 4203
- 4302

**Q10.** A number $x$ is divisible by 7. When the number is divided by 8, 12 and 16 it leaves a remainder 3 in each case. The least value of $x$ is,

- 147
- 148
- 149
- 150

### Answers to the questions

**Q1. Answer:** Option b: $3.2^3$.

**Q2. Answer:** Option a : $xyz$.

**Q3. Answer:** Option a: 120.

**Q4. Answer:** Option d: 3024.

**Q5. Answer:** Option c: 36.

**Q6. Answer:** Option b : 144.

**Q7. Answer:** Option a: 9828.

**Q8. Answer:** Option b: 132.

**Q9. Answer:** Option b: 4032.

**Q10. Answer:** Option a: 147.

### 6th solution set - 10 problems for SSC CHSL exam: 2nd on topic LCM and HCF - answering time 15 mins

**Q1. **If $P=2^3.3^{10}.5$ and $Q=2^5.3.7$, then HCF of $P$ and $Q$ is,

- $2.3.5.7$
- $3.2^3$.
- $2^5.3^{10}.5.7$
- $2^2.3^7$

** Solution 1: Problem analysis and solution by Identifying largest common factors in powers of each prime factor**

First Identify largest factor in 2's power as $2^3$ common to both $P$ and $Q$.

Next, identify that 5 and 7 are not appearing in both the numbers.

At the last, identify largest factor in 3's power as $3$ common between $P$ and $Q$.

So, HCF of $P$ and $Q$ is given by,

$3.2^3$.

**Answer:** Option b: $3.2^3$.

**Key concepts used: ****Identifying largest common factor in powers of each prime factor*** for finding HCF* --

*.*

**Solving in mind****Q2.** If $x:y$ be the ratio of two whole numbers and $z$ be their HCF, then LCM of the numbers is,

- $xyz$
- $yz$
- $\displaystyle\frac{xz}{y}$
- $\displaystyle\frac{xy}{z}$

**Solution 2: Problem solving using basic ratio concept and HCF LCM product concept**

In a ratio, you can always reintroduce the cancelled out HCF by multiplying both the ratio variable values by the same variable representing the cancelled out HCF and get the actual values of the ratio variables.

The cancelled out HCF in this case is, $z$. Apply HCF reintroduction to get the two numbers as,

$xz$, and,

$yz$, so that their ratio still remains $x:y$.

Now we'll use the relation of,

Product of two numbers is equal to the product of their HCF and LCM.

By this relation, you get,

$xyz^2=zp$, where $p$ is the LCM of the two numbers.

Finally then,

$p=xyz$.

**Answer:** Option a: $xyz$.

**Key concepts used:** **Basic ratio concept -- HCF reintroduction technique -- HCF LCM product -- Solving in mind****.**

**Q3. **The greatest number that divides 122 and 243 leaving remainders 2 and 3 respectively, is,

- 120
- 30
- 24
- 12

**Solution 3: Problem analysis and solution by pattern identification, remainder concept and HCF concept**

Identify the pattern that if you subtract 2 and 3 from 122 and 243 respectively, the results become 120 and 240, HCF of these two is 120.

So, when 120 divides 122 and 243, remainders left are 2 and 3 respectively, and 120 being the HCF of 120 and 240, it is the greatest such number.

**Answer:** Option a: 120.

**Key concepts used: Key pattern identification -- Division and remainder concepts -- HCF concept -- Solving in mind.**

**Q4. **Find the least multiple of 23, which when divided by 18, 21 and 24 leaves the remainder 7, 10 and 13 respectively.

- 3013
- 3036
- 3002
- 3024

**Solution 4: Problem solving using redefining the problem and Choice value test**

Analyze the problem for a moment and redefine it to a simpler form of the problem,

Find the least number which is divisible by 23 but leaves a remainder of 7, 10 and 13 respectively when divided by 18, 21 and 24.

Now just divide each of the choice values by 23 mentally and see **if there is only one multiple of 23.**

With this objective, you can quickly identify **3013 at option a as the only multiple of 23.**

As in a correctly formed Multiple choice question, there cannot be more than one correct answer, divisibility of only 3013 by 23 confirms it as the answer.

**Answer:** Option a: 3013.

**Verification:**

For satisfaction, you may test whether 3013 satisfies the remainder conditions.

Dividing 3013 by 18 you get,

$3013=18\times{167}+7$.

Dividing 3013 by 21 you get,

$3013=21\times{143}+10$, and

Dividing 3013 by 24 you get,

$3013=24\times{125}+13$.

All conditions satisfied.

**Answer:** Option a: 3013.

Let us see how this problem can be solved by deductive approach. It will take a bit more time, but you would have the satisfaction of solving the problem conceptually.

#### Solution to Problem 4: Solving by pattern identification and Remainder from Shortfall in multiple of LCM concept

The deductive solution to this problem hinges on the key pattern discovered,

$18-7=11$,

$21-10=11$, and

$24-13=11$.

If your starting point is a multiple of **LCM of 18, 21 and 24**, and then if you **subtract 11 from it to form a single number point**, when you divide this number thus formed by 18, 21 and 24, you will **have excess from the previous multiple of 18, 21 and 24 respectively by 7, 10 and 13, and will reach the starting multiple in each case by adding 11**.

So 7, 10 and 13 will be the remainders as required (still it may not be divisible by 23, but that is the second part of this interesting problem).

Algebraically, if this number is represented by $(x-11)$, $x$ being the starting multiple, you will have,

$(x-11)=(x-18)+7$, remainder 7, previous multiple $(x-18)$

$(x-11)=(x-21)+10$, remainder 10, and

$(x-11)=(x-24)+13$, remainder 13,

where $(x-18)$, $(x-21)$ and $(x-24)$ are the previous multiples respectively.

**Note:** *Carefully go through the concept to understand it fully.*

The concept graphic is shown below.

Let us now form the LCM.

LCM of 18, 21 and 24 will be derived by the numbers in the form of product of factors in powers of prime numbers,

$18=2\times{3^2}$,

$21=3\times{7}$, and

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

LCM is,

$2^3\times{3^2}\times{7}=504$.

If you form the trial target as,

$504\times{1}-11=493=18\times{28}-11$, the result of dividing this number by 18 will be,

$493=18\times{27}+(18-11)=18\times{27}+7$, remainder 7 condition satisfied.

By taking the number backwards from the LCM by 11, you have ensured that this number is in excess of 7 from the previous multiple of 18 so that remainder becomes 7 as desired.

The same mechanism will apply when dividing 493 by 21 and 24 resulting in remainders 10 and 13 respectively.

But the question remains—is the number 493 divisible by 23? It is not. We have to test higher multiples of 504, subtract 11 and test for divisibility by 23.

You may test each multiple of LCM by multiplying it with 2, 3, 4.., subtracting 11 from the result and dividing by 23. This takes a bit more time than taking 504 directly to the zone of 3000 plus by multiplying it with 6 and subtracting 11 from it,

$504\times{6}-11=3024-11=3013$.

Now divide this number by 23. It is divisible by 23. But still it is not proved that this is the least such number.

To be confirmed that **it is indeed the least such number mathematically**, first divide LCM by 23 to find the shortfall 2 for each multiple of 504,

$504=23\times{22}-2$.

Each successively increasing multiple of 504 will add 2 to the accumulating shortfall. Subtraction of 11 to be done after forming the multiple of 504 reduces the shortfall to be accumulated by the multiples to,

$23-11=12$.

So the total shortfall required to make the number divisible by 23 will be accumulated with the sixth multiple of 504,

$6\times{504}-11=23\times{132}-(12+11)=23\times{131}$.

**Shortfall concept** is a powerful one and you can use it for solving varieties of awkward problems.

**Answer:** Option a: 3013.

**Key concepts used: Pattern identification -- Least Shortfall in multiple of LCM -- Divisibility -- Shortfall accumulation with multiples.**

**Q5.** What least value should be added to 1812 to make it divisible by 7, 11 and 14?

- 12
- 72
- 36
- 154

**Solution 5: Problem analysis and solving by Shortfall compensation concept**

For the smallest number to be divisible by 7, 11 and 14, it must be divisible by the LCM of 7, 11 and 14.

The LCM of 7, 11 and 14 is,

$11\times{14}=154$.

You have to find the multiple of 154 that exceeds 1812 and the previous multiple is less than 1812.

An estimate of the number can be obtained by dividing 180 by 15 getting 12 as quotient and then multiplying 154 by 12 to get,

$154\times{12}=1800+48=1848$.

The multiple of 154 previous to 1848 is 1694 and it is less than 1812.

This means, 1848 is the least multiple of 154 that exceeds 1812.

And being a multiple of their LCM, 1848 will be divisible by 7, 11 and 14.

The number to be added to 1812 to make it 1848 is,

$1848-1812=36$.

36 compensates the shortfall from the next multiple.

**Answer:** Option c: 36.

*Key concepts used:* **LCM concept -- Least multiple exceeding a number concept -- Shortfall compensation concept -- Solving in mind****.**

**Q6.** Three numbers are in the ratio 2: 3 : 4 and their HCF is 12. The LCM of the numbers is,

- 96
- 144
- 72
- 192

**Solution 6: Problem analysis and solution by HCF reintroduction technique, Ratio concept and LCM concept**

Get the actual values of the three numbers by multiplying each by the cancelled out HCF 12. The numbers are then,

$2\times{12}=24$,

$3\times{12}=36$, and,

$4\times{12}=48$.

You could do this because while forming the minimal form ratio with no common factor, $2:3:4$, the highest factor 12 common to the three numbers was eliminated.

To find the LCM of the three numbers, express 24, 36 and 48 in the form of product of powers of prime numbers,

$24=2^3\times{3}$,

$48=2^4\times{3}$, and

$36=2^2\times{3^2}$.

LCM is then,

$9\times{16}=144$.

**Answer:** Option b : 144.

**Key concepts used:** *Ratio concept -- HCF reintroduction technique -- LCM by expressing numbers in terms of powers of prime factors -- Solving in mind.*

**Q7.** The greatest 4 digit number which is exactly divisible by each one of the numbers 12, 18, 21 and 28 is,

- 9828
- 9928
- 9288
- 9882

**Solution 7: Problem analysis and Solving by LCM and greatest multiple of 4 digits by taking one step back from least multiple of 5 digits**

First get the LCM by expressing the numbers in terms of powers of primes factors,

$12=2^2\times{3}$,

$18=2\times{3^2}$.

$21=3\times{7}$, and

$28=2^2\times{7}$.

The LCM is,

$2^2\times{3^2}\times{7}=252$.

Multiply 252 by 40 to get,

$40\times{252}=10080$ which is the least multiple of 5 digits divisible by the four given numbers as 80 in excess of 10000 is less than 252.

So subtracting one number of 252 from 10080 you will get the greatest multiple of 4 digits divisible by the four numbers, 12, 18, 21 and 28 as,

$10080-252=9828$.

**Answer:** Option a: 9828.

** Key concepts used:** ** LCM **--

*-- Greatest 4 digit multiple of LCM by taking one step back and subtracting the LCM from the Least 5 digit multiple*

**Least 5 digit multiple of LCM**

**-- Solving in mind.****Q8.** The HCF and LCM of two numbers are 44 and 264 respectively. If the first number is divided by 2, the quotient is 44. The other number is,

- 147
- 132
- 264
- 528

** Solution 8: Problem analysis and Solution by HCF LCM product**

If the first number is divided by 2 the quotient is 44. So the first number is,

$2\times{44}=88$.

As product of the HCF and LCM is equal to the product of the two numbers, the second number is,

$\displaystyle\frac{\text{HCF}\times{\text{LCM}}}{\text{First number}}$

$\displaystyle\frac{44\times{264}}{44}=264$.

**Answer:** Option c: 264.

**Key concepts used:** **HCF LCM product -- Solving in mind.**

**Q9.** The number between 4000 and 5000 that is divisible by each of 12, 18, 21 and 32 is,

- 4023
- 4032
- 4203
- 4302

**Solution 9: Problem analysis and Solving by Least Multiple of LCM in a range concept**

For a number to be divisible by 12, 18, 21 and 32, it must be a multiple of the LCM of these four numbers. Let us first find the LCM.

Get the LCM by expressing the fours numbers in the form of product of powers in prime factors,

$12=2^2\times{3}$,

$18=2\times{3^2}$,

$21=3\times{7}$, and

$32=2^5$.

So the LCM is,

$2^5\times{3^2}\times{7}=288\times{7}=2016$.

Have a look at the choice values and just double this LCM to get,

$2\times{2016}=4032$ as the desired number.

**Answer:** Option b: 4032.

**Key concepts used:** * LCM by expressing in powers of prime factors -- Multiple of LCM in a range *--

**Solving in mind.****Q10.** A number $x$ is divisible by 7. When the number is divided by 8, 12 and 16 it leaves a remainder 3 in each case. The least value of $x$ is,

- 147
- 148
- 149
- 150

**Solution 10: Problem analysis and Solving by fixed remainder for multiple of LCM concept**

Fixed remainder for multiple of LCM concept in this case can be stated as,

A number that leaves a fixed remainder of 3 for division by 8, 12 and 16 must be a multiple of the LCM of the three numbers plus 3.

Get the LCM of 8, 12 and 16 by expressing the numbers in terms of powers of prime factors,

$8=2^3$,

$12=2^2\times{3}$, and

$16=2^4$.

Otherwise also it is easy identify 48 as the LCM of 8, 12 and 16 mentally.

Multiply 48 by 3 to get 144 in the range of the four choice values. Add 3 to 144 to form 147.

And test whether 147 is divisible by 7. It is so with quotient 21.

Desired number is 147.

**Answer: **Option a: 147.

**Key concepts used:** **Fixed remainder for multiple of LCM concept -- LCM -- Multiple of LCM in a range -- Solving in mind.**

### 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**