# QUESTIONS ON LCM & HCF (PART-II)

#### QUERY 11

**If the sum of two numbers is 55 and the HCF and LCM of these numbers are 5 and 120 respectively, then the sum of the reciprocals of the numbers is equal to?**

A) ^{55}⁄_{601}

B) ^{601}⁄_{55}

C) ^{11}⁄_{120}

D) ^{120}⁄_{11}

**MAHA GUPTA
**Let the numbers be a and b.

Then, a + b = 55

As the product of the numbers is equal to the product of their HCF and LCM, so ab = 5 × 120 = 600

Now, sum of the reciprocal of the numbers = ^{1}⁄_{a }+ ^{1}⁄_{b }= ^{a+b}⁄_{ab}

Now substituting values of a+b and ab in the above

^{1}⁄_{a }+ ^{1}⁄_{b }= ^{55}⁄_{600 = 11⁄120 }(option ‘C’)

#### QUERY 12

**A worker was engaged for a certain number of days. But he remained absent for some days and was paid Rs 1387. Had he worked all the days, he would have earned Rs 1752. How many days did he work for?**

A) 19

B) 21

C) 24

D) 29

**MAHA GUPTA**

Amount got by the worker for the days he worked = Rs 1387

Amount for unpaid days = 1752 – 1387 = Rs 365

It’s obvious that the daily wages was a fixed certain amount, means it must divide 1387 and 365 completely, and also it must be the greatest divisor of both. So we need to find the HCF of 1387 and 365.

Now, HCF of 1387 and 365 = 73

Therefore the number of days he worked for = ^{1387}⁄_{73} = 19 (option ‘A’)

#### QUERY 13

**In finding HCF of two numbers by the division method the quotients are 1, 2 and 3 and the last divisor is 15, what are the numbers?**

A) 105, 120

B) 90, 120

C) 105, 150

D) 120, 150

**MAHA GUPTA**

We start such questions from rear end.

The last quotient is 3 and the last divisor is 15, means the last remainder = 15 × 3 = 45

Obviously 45 will also be the second last divisor with 2 as quotient and 15 being the remainder

Hence the second last remainder = 45 × 2 + 15 = 105

Now, obviously 105 will also be the first divisor with 1 as quotient and 45 being the remainder

Hence the other number = 105 × 1 + 45 = 150

We see the required number are 105 and 150 (option ‘C’)

#### QUERY 14

**Four prime numbers are written in ascending order. The product of the first three numbers is 385 and that of the last three numbers is 1001. What is the largest number? Also find the sum of all the numbers.**

A) 7, 24

B) 11, 36

C) 13, 36

D) 17, 45

**MAHA GUPTA**

Let the numbers be a, b, c, d

So, according to the question

a × b × c = 385

b × c × d = 1001

We see product of b and c i.e. b × c is common to both, and as all numbers are co-prime, so b × c is the LCF of both 385 and 1001

Therefore, b × c = HCF of 385 and 1001 = 77

As all the numbers are co-prime, therefore b = 7 and c = 11

Now, b × c × d = 1001

=> 77d = 1001

=> d = 1001/77 = 13 (option ‘C’)

Also a × b × c = 385

=> 77a = 385

=> a = 385/77 = 5

Therefore, sum of all the numbers = a + b + c + d = 5 + 7 + 11 + 13 = 36 (option ‘C’)

ANOTHER METHOD

Let the numbers be a, b, c, d

So, according to the question

a × b × c = 385

b × c × d = 1001

We see product of b and c i.e. b × c is common to both, and as all numbers are co-prime, so b × c is the LCF of both 385 and 1001

Therefore, b × c = HCF of 385 and 1001 = 77

As all the numbers are co-prime, therefore b = 7 and c = 11

Now according to the question

_{a*b*c⁄b*c*d }= ^{385}⁄_{1001}

Dividing both sides by b*c i.e. 77

=> ^{a}⁄_{d}_{ }= ^{5}⁄_{13}

=> a = 5 and d = 13 (option ‘C’)

Therefore, sum of all the numbers = a + b + c + d = 5 + 7 + 11 + 13 = 36 (option ‘C’)

#### QUERY 15

**HCF and LCM of 77, 99 and x are 11 and 3465 respectively, the least value of x is?**

A) 11

B) 35

C) 45

D) 55

**MAHA GUPTA**

As the HCF of the numbers is 11, obviously 11 must be a factor of each of these numbers.

So, let x = 11a

Now, LCM of 77, 99, 11a = 7*9*11*a

But LCM is already given 3465

Hence, 7*9*11*a = 3465

=> a = 5

Therefore least value of x = 11a = 11*5 = 55 (option ‘D’)

#### QUERY 16

**The least multiple of 13, which on dividing by 4, 5, 6, 7 and 8 leaves the remainder 2 in each case is?**

A) 2520

B) 842

C) 2522

D) 840

**MAHA GUPTA**

According to the question, obviously, the required number must be 2 more than the multiplication of the LCM of 4, 5, 6, 7 and 8 with an unknown number

Let that unknown number = x

Now LCM of 4, 5, 6, 7, 8 = 840

Hence, the required number = 840*x + 2

But this number is multiple of 13, means divisible by 13

Now, (840*x + 2) ÷ 13 = 64x*13 + (8x + 2); where 8x + 2 is the remainder of this division

But their must not remain a remainder as the number is multiple of 13

Hence, 8x + 2 is divisible by 13

But it’s divisible by 13 when x = 3

Therefore, the required number = 840*x + 2 = 840*3 + 2 = 2522 (option ‘C’)

TRICK

Begin with the given answer options

We see 2520 – 2 = 12518, and 840 – 2 = 838 are not divisible by 5 (So option A and option D are rejected.)

842 is not divisible by 13 (So option B is also rejected.)

Hence 2522 (option ‘C’) is correct.

#### QUERY 17

**Two 3-digit numbers have their HCF 29 and LCM 4147. The sum of numbers is?**

A) 966

B) 696

C) 669

D) 666

**MAHA GUPTA**

As the HCF of the numbers is 29, let the numbers be 29x and 29y.

As the product of the numbers is equal to the product of their HCF and LCM

29x × 29y = 29 × 4147

=> x × y = 143

=> x = 11 and y = 13

Therefore, the sum of the numbers = 29x + 29y = 29*11 + 29*13

= 29(11 + 13)

= 696 (option ‘B’)

#### QUERY 18

**The sum of two numbers is 528 and their HCF is 33. Find the number of pairs of such numbers.**

A) 1

B) 2

C) 3

D) 4

**MAHA GUPTA**

As the HCF of the numbers is 33, let the numbers be 33x and 33y

As the sum of the number is given 528

33x + 33y = 528

=> x + y = 16

Therefore, possible values of x and y = (1, 15), (2, 14), (3, 13), (4, 12), (5, 11), (6, 10), (7, 9)

But (2, 14), (4, 12), (6, 10) are not valid as they are not co-prime numbers.

Hence, 4 pairs are possible (option ‘D’)

#### QUERY 19

**The HCF and LCM of two numbers are 13 and 455 respectively. If one of the numbers lies between 75 and 125, then that number is?**

A) 78

B) 91

C) 104

D) 117

**MAHA GUPTA**

As the HCF of the numbers is 21, let the numbers are 13x and 13y

As the product of the numbers is equal to the product of their HCF and LCM

13x × 13y = 13 × 455

=> x × y = 35

=> x = 5 and y = 7

Therefore the numbers = 13x and 13y

= 13*5 and 13*7

= 65, 91

But the required number is between 75 and 125, therefore the answer 91 (option ‘B’)

SHORT

LCM = 455 = 5 × 7 × 13

But 13 is the HCF of the numbers

Therefore, numbers are 13*5 = 65 and 13*7 = 91

But the required number is between 75 and 125, therefore the answer 91 (option ‘B’)

#### QUERY 20

**The least multiple of 23, which on dividing by 18, 21, 24 leaves the remainders 7, 10, 13 respectively is?**

A) 3013

B) 3024

C) 3002

D) 3036

**MAHA GUPTA
**LCM of 18, 21 and 24 = 504

Remainder in each case = 18 – 7 = 21 – 10 = 24 – 13 = 11 less than the respective quotient

When difference between the divisor and remainder is the same, the number is equal to the LCM of the numbers minus that difference.

Therefore, the required number = 504x – 11

But this number is multiple of 23, means divisible by 23

Now, (504x – 11) ÷ 23 = 21x*23 + (21x – 11); where 21x – 11 is the remainder of this division

But their must not remain a remainder as the number is multiple of 23

Hence, 21x – 11 is divisible by 23

But it’s divisible by 13 when x = 6

Therefore, the required number = 504*x – 11 = 504*6 – 11 = 3013 (option ‘A’)

TRICK

LCM of 18, 21, 24 = 504

All the given answer options are around 3000 which is approximately 6 times 504.

Therefore the required number = 504*6 – 11 = 3013 (option ‘A’)