Testbook presents a wide range of efficacious HCF MCQs Quiz with answers and brief explanations. In addition to the HCF objective questions, some tips and tricks are also given to assist you in constructive preparation. HCF (Highest Common Factor) is a vital part of mathematics and recruitment examinations. Entrance exams such as GRE, MAT, Bank and Railway Exams have HCF as a major section in their tests. Do solve these HCF Questions to master this section.

Option 3 : 1

**GIVEN: **

The product of two numbers is 1521 and the HCF of these numbers is 13.

**CONCEPT:**

HCF: The **highest common factor** (**HCF**) is found by finding all common factors of two numbers and selecting the largest one.

**CALCULATION:**

Suppose the numbers are 13a and 13b as the HCF of these numbers is 13.

We can write:

13a × 13b = 1521

⇒ ab = 9

**∴ Only possible pair is 13, 117**

**Mistake Points**According to question,

ab = 9

For a = 1 and b = 9

The numbers will be 13 and 117 and their HCF will be 13

Here we will not consider a = 3 and b = 3.

The numbers will be 39 and 39.

Here HCF would be 39 which does not satisfy the given condition.

HCF of two number is 4 and the sum of those two numbers is 36. Find how many such pair of number is possible.

Option 3 : 3

**Given:**

HCF of two number is 4 and the sum of those two numbers is 36.

**Concept Used:**

Concept of HCF

HCF is the least common factor among two or more numbers.

**Calculation:**

HCF of two number is 4

Let, those numbers be 4x and 4y where x and y are prime to each other

Accordingly,

4x + 4y = 36

⇒ 4(x + y) = 36

⇒ (x + y) = 9

Now,

9 = 8 + 1

9 = 7 + 2

9 = 6 + 3

9 = 5 + 4

In all these cases only (8,1); (7,2); and (5,4) are prime to each other. So, such three pair is possible

**∴ Such three pair of number is possible.**

Option 4 : 4

**Given- **

Given numbers = 56, 216, 28

**Concept Used-**

The greatest number which divides a given set of numbers is the HCF of that given set of numbers.

**Calculation- **

56 = 7 × 2^{3}

216 = 2^{3} × 3^{3}

28 = 2^{2} × 7

**∴ HCF of 56, 216, 28 = 2 ^{2} = 4.**

Option 1 : (7^{19} + 1)

**Given:**

Two numbers are (757 + 1) and (7133 + 1)

**Concept Used:**

HCF (a^{m} + 1) and (a^{n} + 1) = a^{hcf(m, n)} + 1

**Calculation:**

H.C.F {(757 + 1), (7133 + 1)}

According to he formula,

⇒ H.C.F = 7^{hcf (57, 133)} + 1

**∴ H.C.F is (7 ^{19} + 1).**

Option 4 : 16

**Formula:**

The HCF of two or more than two numbers is the greatest number that divides each of them exactly.

**Calculation:**

12 = 2 × 2 × 3

15 = 3 × 5

21 = 3 × 7

HCF of 12, 15 and 21 = 3

Required number of rows = 12 / 3 + 15 / 3 + 21 / 3

⇒ 4 + 5 + 7

⇒ 16

Option 3 : 504

**Given:**

Ratio of numbers = 7 ∶ 11

HCF = 28

**Calculation:**

Let the numbers be 7x and 11x

HCF of 7x and 11x is x

HCF = x = 28

The numbers will be 7 × 28 and 11 × 28

⇒ The numbers will be 196 and 308

Sum of numbers = 196 + 308

⇒ Sum of numbers = 504

**∴ Sum of numbers is 504**

Option 1 : 6

**Given:**

LCM = 40 times HCF

The product of the numbers is 1440

**Formula used:**

LCM × HCF = Product of numbers

**Calculation:**

Let HCF of the number be 'H'

⇒ LCM = 40H

Now,

LCM × HCF = Product of numbers

⇒ 40H × H = 1440

⇒ H^{2} = 1440/40 = 36

⇒ H = 6

∴ HCF = 6

Option 3 : 72

**Given: **

Three numbers are 108, x, and 144,

HCF of these numbers is 12.

**Concept:**

H.C.F. of two or more factors is the greatest factor that divides each of them exactly.

**Calculation: **

Factor of 108 = 2 × 2 × 3 × 3 × 3

Factor of 144 = 2 × 2 × 2 × 2 × 3 × 3

By taking option (c),

Factor of 72 = 2 × 2 × 2 × 3 × 3

If we take the value of x is 72, then HCF of the given numbers becomes 36

**∴ The value of x will not be 72.**

Option 2 : 2^{3} × 3^{1} × 5^{1}

**Given:**

Numbers are 720 & 600.

**Concept used:**

First, break the numbers into prime factors.

Pick the least common power of each prime number from both the numbers.

**Explanation:**

720 = 2^{4} × 3^{2} × 5^{1}

600 = 2^{3} × 3^{1} × 5^{2}

Least power of '2' is 3.

Least power of '3' is 1.

Least power of '5' is 1.

⇒ HCF = 23 × 31 × 5^{1}

**∴ The largest common factor of 720 & 600 is 23 × 31 × 5 ^{1}**

** Confusion Points** You have to find the largest common factor and LCM stands for Lowest common multiple. HCF stands for Highest/largest common factor.

Option 2 : 38

In a flower garden, there are three types of flower plants, namely 38 Guldawdi, 114 Chameli and 76 Surajmukhi.

**Concept Used**:

to keep minimum number of rows the possible number of flower plants in a row will be HCF of the different variety of the plant

**Calculation**:

38 = 2 × 19

114 = 2 × 3 × 19

76 = 2^{2} × 19

⇒ HCF (38, 114, 76) = 2 × 19 = 38

∴ Required possible number of flower plants in a row = 38

Option 4 : \(3 \over 150\)

**Given:**

We have to find the HCF of \({9 \over 10}, {3 \over 25}, {6 \over 15}, {12 \over 5}\)

**Concept Used:**

HCF of fraction = (HCF of numerator)/(LCM of denominator)

**Calculation:**

HCF of \({9 \over 10}, {3 \over 25}, {6 \over 15}, {12 \over 5}\)

⇒ \({HCF\ (9,\ 3,\ 6,\ 12) \over LCM\ (10,\ 25,\ 15,\ 5)}\)

HCF of numerators is 3

LCM of denominators is 150

Required HCF is 3/150

**∴ HCF of \({9 \over 10}, {3 \over 25}, {6 \over 15}, {12 \over 5}\) is (3/150).**

__Additional Information__

We can make (3/150) fraction into a small form i.e (1/50), But we have to see options also.

In the given option (1/50) is not present. If (1/50) will be present in place of (3/150), then we can mark (1/50) as the answer. Which is totally correct.

Sometimes we have to check options also to mark the correct answers. In this case, we are marking (3/150) according to the options provided.

Option 2 : 12

**Given:**

When 29, 43 and 71 divided by the number, it leaves the remainder 5, 7 and 11 respectively.

**Concept Used:**

Concept of HCF

**Calculation:**

29 - 5 = 24

43 - 7 = 36

71 - 11 = 60

Now,

24 = 2 × 2 × 2 × 3

36 = 2 × 2 × 3 × 3

60 = 2 × 2 × 3 × 5

HCF of 24, 36 and 60 is 2 × 2 × 3 = 12

**∴ The required greatest number is 12.**

Option 3 : 20 cm

**Solution:**

**Given:**

Lengths are 12 m, 20 cm, and 4 m 20 cm.

**Formula:**

1 m = 100 cm

**Calculation:**

12 m = 1200 cm

4 m 20 cm = 420 cm

H.C.F of 1200, 420, 20

20 = 20 × 1

1200 = 20 × 6 × 10

420 = 20 × 21

H.C.F = 20

**∴** **T****he largest possible length of the scale is 20 cm**

Option 2 : 2xy

**Given:**

The terms are 4x2y, 16xy3, 2x3y3

**Calculation: **

Factors of 4x2y = **2** × 2 × **x** × x × **y**

Factors of 16xy3 = **2** × 2 × 2 × 2 × **x** × **y** × y × y

Factors of 2x3y^{3} = **2** × **x** × x × x × **y** × y × y

Common factors = 2 × x × y = 2xy

**∴ HCF is 2xy**

Three chocolate bars weighing 600 grams, 420 grams and 780 grams were cut into small pieces of uniform weight. If 2 such bars are given to one person, then what will be the minimum number of such persons between whom all the bars can be distributed?

Option 3 : 15

**Solution:**

**Given:** Weight of three chocolate bars = 600 gm, 420 gm, 780 gm

**Concept used:** For minimum number of persons we distribute chocolate bars weighing according to Highest Common Factor.

**Calculation:** H.C.F. of 600, 420 and 780 = 60

Total number of pieces = (600 + 420 + 780) / 60

⇒ 1800 / 60

⇒ 30

Number of persons = 30 / 2

∴ 15

Option 2 : 48

**Concept:**

The required number will be Highest common factor (HCF) of (52 – 4) and (104 – 8)

**Calculation:**

HCF of 48 and 96

48 = 2^{4} × 3

96 = 2^{5} × 3

∴ HCF = 2^{4} × 3 = 48

Option 4 : 2

**Given:**

The product of two number is 1083 and their HCF is 19

**Concept:**

The HCF of two or more numbers is the greatest factor that divides each of them exactly.

**Calculation:**

Let the number be 19a and 19b

The product of the two number = 1083

⇒ 19a × 19b = 1083

⇒ a × b = 1083/361

⇒ a × b = 3

Possible pairs are (1, 3), (3, 1)

Option 1 : 5/18

**Given:**

The given fractions are 25/9 and 5/18.

**Formula used:**

HCF of fraction = (HCF of numerators)/(LCM of denominators)

**Calculation:**

HCF of numerator = HCF of (25, 5) = 5

And, LCM of denominator = LCM of (9, 18) = 18

∴ HCF of (25/9 and 5/18) = HCF of (25, 10)/LCM of (9, 18) = 5/18Option 2 : 37

**Given****:**

Number of mangoes = 111

Number of bananas = 148

**Concept:**

The HCF of two or more than two numbers is the greatest number that divides each of them exactly.

**Calculation:**

HCF of 111 and 148

111 = 3 × 37

148 = 2 × 2 × 37

HCF of 111 and 148 = 37

**∴ Number of children is 37**

Option 1 : 20

**Given:**

HCF of two numbers = 1/45^{th} of LCM of two numbers

The sum of HCF and LCM = 184

One number of the two numbers = 36

**Formula:**

Product of two numbers = Product of HCF and LCM of two numbers

**Calculation:**

Let LCM of two numbers be 45x,

HCF of two numbers = 45x × [1/45] = x

According to the question

45x + x = 184

⇒ 46x = 184

⇒ x = 184/46

⇒ x = 4

HCF of two numbers = 4

LCM of two numbers = 45 × 4 = 180

Let other number be y

According to the question

36 × y = 4 × 180

⇒ y = (4 × 180)/36

⇒ y = 20