{"id":6907,"date":"2021-12-28T08:30:55","date_gmt":"2021-12-28T08:30:55","guid":{"rendered":"https:\/\/stgwebsite.mindspark.in\/studymaterial\/?page_id=6907"},"modified":"2022-01-03T06:38:02","modified_gmt":"2022-01-03T06:38:02","slug":"lcm-and-hcf-questions-with-solutions","status":"publish","type":"page","link":"https:\/\/stgwebsite.mindspark.in\/studymaterial\/math-concepts\/lcm-and-hcf-questions-with-solutions\/","title":{"rendered":"LCM and HCF Questions with solutions"},"content":{"rendered":"

[et_pb_section fb_built=”1″ admin_label=”Section” module_class=”mainsec” _builder_version=”4.10.4″ _module_preset=”default” background_color=”#e0f2fd” z_index=”1″ custom_padding=”5px||5px||true|false” locked=”off” collapsed=”off” global_colors_info=”{}”][et_pb_row column_structure=”3_5,2_5″ custom_padding_last_edited=”on|phone” _builder_version=”4.10.8″ _module_preset=”default” background_color=”#FFFFFF” width=”100%” max_width=”1310px” custom_padding=”|51px|40px|51px|false|true” custom_padding_tablet=”” custom_padding_phone=”|40px|30px|40px|false|true” border_radii=”on|10px|10px|10px|10px” global_colors_info=”{}”][et_pb_column type=”3_5″ admin_label=”Column L” _builder_version=”4.9.10″ _module_preset=”default” global_colors_info=”{}”][et_pb_text admin_label=”Acute Angles
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LCM and HCF Questions with solutions<\/h1>\n

[\/et_pb_text][et_pb_text admin_label=”Text” _builder_version=”4.13.1″ _module_preset=”default” text_font_size=”16px” header_2_font=”|600|||||||” header_2_text_color=”#a01414″ header_3_font=”|600|||||||” custom_padding=”15px|15px|54px|4px|false|false” global_colors_info=”{}”]<\/p>\n

LCM and HCF Questions – Solved Examples<\/b><\/h2>\n

<\/b><\/p>\n

<\/h3>\n

1. Find the HCF and LCM of 30 and 45 using the prime factorisation method.<\/strong><\/p>\n

Solution:\u00a0<\/strong><\/p>\n

Prime factorisation of 30<\/span><\/p>\n

\"\"<\/span><\/p>\n

<\/span><\/p>\n

30 = 2 \u00d7 3 \u00d7 5<\/span><\/p>\n

<\/span><\/p>\n

Prime factorisation of 45<\/span><\/p>\n

\"\"<\/p>\n

<\/span><\/p>\n

45 = 3 \u00d7 3 \u00d7 5<\/span><\/p>\n

<\/span><\/p>\n

<\/b><\/p>\n

HCF of 30 and 45<\/b><\/p>\n

3 and 5 are the common prime factors of 30 and 45.<\/span><\/p>\n

HCF = 3 \u00d7 5 <\/span>= 15<\/span><\/p>\n

LCM of 30 and 45<\/b><\/p>\n

Common prime factors \u2013 (3, 5)<\/span><\/p>\n

Uncommon prime factors \u2013 (2, 3)<\/span><\/p>\n

LCM of 30 and 45 = Product of common factors \u00d7 Product of uncommon factors<\/span><\/p>\n

\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 = <\/span>(3 \u00d7 <\/span>5)<\/span>(2 \u00d7 <\/span>3)<\/span><\/p>\n

\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 = 15 \u00d7 6<\/span><\/p>\n

\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 = 90<\/span><\/p>\n

<\/b><\/p>\n

2. Find the HCF and LCM of 50 and 80 by division method<\/strong><\/p>\n

Solution:<\/strong><\/p>\n

HCF of 50 and 80<\/b><\/p>\n

Step 1:\u00a0<\/strong><\/p>\n

Divide the greater number 80 by the smaller number 50. <\/span>
<\/span>80 = 50 \u00d7 1 + 30<\/span><\/p>\n

Step 2:\u00a0<\/strong><\/p>\n

Now 50 is the dividend and 30 is the divisor.<\/span>
<\/span>50 = 30 \u00d7 1 + 20<\/span><\/p>\n

Step 3:<\/strong><\/p>\n

Now 30 is the dividend and 20 is the divisor.<\/span>
<\/span>30 = 20 \u00d7 1 + 10<\/span><\/p>\n

Step 4:<\/strong><\/p>\n

Now 20 is the dividend and 10 is the divisor.<\/span><\/p>\n

20 = 10 \u00d7 2 + 0<\/span><\/p>\n

Hence, HCF of 50 and 80 is 10<\/span><\/p>\n

\"\"<\/p>\n

<\/b><\/p>\n

LCM of 50 and 80<\/b><\/p>\n

Step 1: <\/b><\/p>\n

Divide the numbers 50 and 80 by the smallest possible prime factor of at least one of these numbers.<\/p>\n

Here 2 is the divisor and the quotient becomes 25 and 40.<\/p>\n

Step 2: <\/b><\/p>\n

Divide the numbers by 2 again.<\/p>\n

Since 2 is not a factor 25, keep the number as it is in the next step.<\/span><\/p>\n

Step 3: <\/b><\/p>\n

Now 25 and 20 are the dividends.<\/p>\n

Divide these numbers by 2.<\/span>
<\/span><\/p>\n

2 is not a factor of 25, so it is kept as it is in the next step.<\/span><\/p>\n

Step 4:<\/b><\/p>\n

Now 25 and 10 are the dividends.<\/span>
<\/span><\/p>\n

Now 5 is the divisor as it is the smallest prime number which is a factor of these numbers.<\/span><\/p>\n

Step 5:<\/b><\/p>\n

Repeat these steps till we get 1 and 1 as the quotients.<\/span><\/p>\n

Step 6:<\/b><\/p>\n

Multiply all the divisors on the left side of the table to find the LCM of 50 and 80.<\/span><\/p>\n

<\/span><\/p>\n

<\/span><\/p>\n

\"\"<\/p>\n

<\/span><\/p>\n

LCM of 50 and 80 = <\/span>2 \u00d7 <\/span>2 \u00d7 <\/span>2 \u00d7 <\/span>2 \u00d7 <\/span>5 \u00d7 <\/span>5 <\/span>= 400<\/span><\/p>\n

Hence, 400 is the LCM of 50 and 80.<\/span><\/p>\n

<\/b><\/p>\n

3. Find the HCF of 55, 75 and 90.<\/strong><\/p>\n

Solution:<\/strong><\/p>\n

Prime factorisation of 55, 75 and 90.<\/span><\/p>\n

\"\"\u00a0 \u00a0 \u00a0\"\"\u00a0 \u00a0 \"\"\u00a0<\/span><\/p>\n

<\/span><\/p>\n

55 = <\/span>5 \u00d7 <\/span>11<\/span><\/p>\n

75 = <\/span>3 \u00d7 <\/span>5 \u00d7 <\/span>5<\/span><\/p>\n

90 = <\/span>2 \u00d7 <\/span>3 \u00d7 <\/span>3 \u00d7 <\/span>5<\/span><\/p>\n

The common factor of 55, 75 and 90 is 5 only.<\/span><\/p>\n

Hence 5 is the HCF of 55,75 and 90<\/span><\/p>\n

<\/b><\/p>\n

4. Find the HCF of 9 and 25.<\/strong><\/p>\n

Solution:<\/strong><\/p>\n

Prime factorisation of 9 and 25<\/span><\/p>\n

9 = 3 \u00d7 3<\/span><\/p>\n

25 = 5 \u00d7 5<\/span><\/p>\n

There is no common prime factor of 9 and 25.<\/span><\/p>\n

Hence 1 is the HCF of 9 and 25.<\/span><\/p>\n

<\/strong><\/span><\/h3>\n

5. What is the LCM of 15,45 and 80?<\/strong><\/p>\n

Solution:<\/strong><\/p>\n

Using the division method, we found the following table:<\/span><\/p>\n

<\/span><\/p>\n

\"\"<\/p>\n

<\/span><\/p>\n

Referring to the above table,<\/span><\/p>\n

LCM of 15, 45 and 80 = Product of the quotients<\/span><\/p>\n

\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 = <\/span>2 \u00d7 <\/span>2 \u00d7 <\/span>2 \u00d7 <\/span>2 \u00d7 <\/span>3 \u00d7 <\/span>3 \u00d7 <\/span>5<\/span><\/p>\n

\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 = 720<\/span><\/p>\n

Hence 720 is the LCM of 15,45 and 80.<\/span><\/p>\n

<\/span><\/p>\n

6. 3 friends Ashok, Mohit and Vickey are building a tower with the help of blocks having height 5 cm, 8 cm and 10 cm respectively. All the three towers built by them have the same height. What is the height of the tower and how many blocks are used by Ashok, Mohit and Vickey?<\/strong><\/p>\n

Solution:\u00a0<\/strong><\/p>\n

Height of Ashok\u2019s block = 5 cm\u00a0<\/span><\/p>\n

Height of Mohit\u2019s block = 8 cm<\/span><\/p>\n

Height of Vickey\u2019s block = 10 cm<\/span><\/p>\n

Since all three of them make a tower having the same height, we need to find the LCM of 5, 8 and 10 to know the height of the tower.<\/span><\/p>\n

\"\" <\/span><\/p>\n

Height of the tower = LCM of 5, 8 and 10\u00a0<\/span><\/p>\n

\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 = <\/span>2 \u00d7 <\/span>2 \u00d7 <\/span>2 \u00d7 <\/span>5<\/span><\/p>\n

\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 = 40 cm<\/span><\/p>\n

<\/span><\/p>\n

Number of blocks Ashok needs =\\frac{40 \\mathrm{~cm}}{5 \\mathrm{~cm}}=8<\/span><\/span><\/p>\n


Number of blocks Mohit needs =\\frac{40 \\mathrm{~cm}}{8 \\mathrm{~cm}}=5<\/span><\/span><\/p>\n


Number of blocks Vickey needs =\\frac{40 \\mathrm{~cm}}{10 \\mathrm{~cm}}=4<\/span><\/span><\/p>\n

<\/span><\/p>\n

Hence, Ashok, Mohit and Vickey build towers of height 40 cm with the help of 8,5 and 4 blocks respectively.<\/span><\/p>\n

<\/span><\/p>\n

7. Find the greatest 4-digit number exactly when by 10,15 and 18 leaves no remainder.<\/b><\/p>\n

Solution:<\/strong><\/p>\n

To find the greatest 4-digit number exactly divisible by 10, 15 and 18, we need to find the LCM of these three numbers first.<\/span><\/p>\n

<\/span><\/p>\n

\"\"<\/span><\/p>\n

<\/span><\/p>\n

LCM of 10, 15 and 18 = <\/span> Product of the quotients<\/span><\/p>\n

\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0= <\/span>2 \u00d7 <\/span>3 \u00d7 <\/span>3 \u00d7 <\/span>5<\/span><\/p>\n

\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0= 90<\/span><\/p>\n

The greatest 4-digit number which can be divided exactly by 10, 15 and 18 is the greatest 4-digit multiple of 90.<\/span><\/p>\n

Greatest 4-digit number = 9999<\/span><\/p>\n

\"\"<\/p>\n

When we divide 9999 by 90, we get 9 as the remainder<\/span><\/p>\n

So, the greatest 4-digit multiple of 90 = 9999 – 9 = 9990<\/span><\/p>\n

Hence, the greatest 4-digit number exactly when by 10, 15 and 18 leaves no remainder is 9990.<\/span><\/p>\n

<\/b><\/p>\n

8. Four tankers contain 350 litres,400 litres,430 litres and 480 litres of oil respectively. What is the maximum capacity of the container which can measure the oil in these four tankers in exact numbers?<\/b><\/p>\n

Solution:<\/strong><\/p>\n

The maximum capacity of the container which can measure the oil in tankers an exact number of times is the HCF of 350 litres, 400 litres, 430 litres and 480 litres.<\/span><\/p>\n

<\/span><\/p>\n

\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \"\"\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \"\"<\/span><\/p>\n

\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \"\"\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \"\" \u00a0<\/span><\/p>\n

The common factors of 350, 400, 430 and 480 are 2 and 5 only.<\/span><\/p>\n

HCF = 2 \u00d7 5 <\/span>= 10<\/span><\/p>\n

Hence the maximum capacity of the container is 10 litres.<\/span><\/p>\n

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Explore Other Topics<\/h1>\n

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\n
Geometry<\/a><\/div>\n
Trigonometry<\/a><\/div>\n
Operations<\/a><\/div>\n
Numbers<\/a><\/div>\n<\/div>\n

[\/et_pb_text][et_pb_text admin_label=”Related Concepts
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Related Concepts<\/h1>\n

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\n
Prime Factorisation Method<\/a><\/div>\n
LCM and HCF<\/a><\/div>\n
HCF<\/a><\/a><\/div>\n<\/div>\n

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Frequently Asked Questions\u00a0<\/span><\/span><\/h1>\n
    <\/ol>\n

    Q1. What is LCM?
    <\/strong><\/h3>\n

    Ans: <\/strong>The smallest number of all the common multiples of two or more numbers is known as the LCM of these numbers.<\/strong><\/span><\/span><\/span><\/p>\n

    Q2. What is HCF?
    <\/strong><\/h3>\n

    Ans: <\/strong>The greatest number of all the common factors of two or more numbers is known as the HCF of these numbers.<\/span><\/strong><\/p>\n

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