Maharashtra Board Class 7 Chapter 3 Set 14 HCF and LCM Solutions

Question 1. Choose the right option.
(i) The HCF of 120 and 150 is -
(A) 30
(B) 45
(C) 20
(D) 120
Answer: (A) 30
Hint:
\( 120 = 2 \times 2 \times 2 \times 3 \times 5 \)
\( 150 = 2 \times 3 \times 5 \times 5 \)
\( \therefore \) HCF of 120 and 150 = \( 2 \times 3 \times 5 = 30 \)
(ii) The HCF of this pair of numbers is not 1.
(A) 13,17
(B) 29,20
(C) 40, 20
(D) 14, 15
Answer: (C) 40, 20
Hint:
\( 40 = 2 \times 2 \times 2 \times 5 \)
\( 20 = 2 \times 2 \times 5 \)
\( \therefore \) HCF of 40 and 20 = \( 2 \times 5 = 10 \)
In simple words: The HCF (Highest Common Factor) is the largest number that divides two or more numbers without leaving a remainder. For multiple-choice questions, verify the options by finding the prime factors or using the division method.

🎯 Exam Tip: For HCF questions, remember to find the common prime factors and multiply them. For pairs of numbers where HCF is not 1, look for numbers that share common divisors other than 1.

Question 2. Find the HCF and LCM.
(i) 14,28
(ii) 32,16
(iii) 17,102,170
(iv) 23,69
(v) 21,49,84
Answer:
(i) \( 14 = 2 \times 7 \)
\( 28 = 2 \times 14 = 2 \times 2 \times 7 \)
\( \therefore \) HCF of 14 and 28 = \( 2 \times 7 = 14 \)
LCM of 14 and 28 = \( 2 \times 2 \times 7 = 28 \)
(ii) \( 32 = 2 \times 16 = 2 \times 2 \times 8 = 2 \times 2 \times 2 \times 4 = 2 \times 2 \times 2 \times 2 \times 2 \)
\( 16 = 2 \times 8 = 2 \times 2 \times 4 = 2 \times 2 \times 2 \times 2 \)
\( \therefore \) HCF of 32 and 16 = \( 2 \times 2 \times 2 \times 2 = 16 \)
\( \therefore \) LCM of 32 and 16 = \( 2 \times 2 \times 2 \times 2 \times 2 = 32 \)
(iii) \( 17 = 17 \times 1 \)
\( 102 = 2 \times 51 = 2 \times 3 \times 17 \)
\( 170 = 2 \times 85 = 2 \times 5 \times 17 \)
\( \therefore \) HCF of 17, 102 and 170 = \( 17 \)
\( \therefore \) LCM of 17, 102 and 170 = \( 17 \times 2 \times 3 \times 5 = 510 \)
(iv) \( 23 = 23 \times 1 \)
\( 69 = 3 \times 23 \)
\( \therefore \) HCF of 23 and 69 = \( 23 \)
\( \therefore \) LCM of 23 and 69 = \( 23 \times 3 = 69 \)
(v) \( 21 = 3 \times 7 \)
\( 49 = 7 \times 7 \)
\( 84 = 2 \times 42 = 2 \times 2 \times 21 = 2 \times 2 \times 3 \times 7 \)
\( \therefore \) HCF of 21, 49 and 84 = \( 7 \)
\( \therefore \) LCM of 21, 49 and 84 = \( 7 \times 3 \times 7 \times 2 \times 2 = 588 \)
In simple words: To find HCF and LCM, first find the prime factorization of each number. HCF is the product of common prime factors with the lowest power, and LCM is the product of all prime factors with the highest power.

🎯 Exam Tip: Always list the prime factors systematically. For HCF, identify only common factors. For LCM, include all factors, using the highest power for any repeated factor across the numbers.

Question 3. Find the LCM.
(i) 36, 42
(ii) 15, 25, 30
(iii) 18, 42, 48
(iv) 4, 12, 20
(v) 24, 40, 80, 120
Answer:
(i) 36, 42

\( \therefore \) LCM of 36 and 42 = \( 2 \times 3 \times 2 \times 3 \times 7 = 252 \)
(ii) 15, 25, 30

\( \therefore \) LCM of 15, 25 and 30 = \( 5 \times 3 \times 5 \times 2 = 150 \)
(iii) 18, 42, 48

\( \therefore \) LCM of 18,42 and 48 = \( 2 \times 3 \times 2 \times 2 \times 3 \times 7 \times 2 = 1008 \)
(iv) 4, 12, 20

\( \therefore \) LCM of 4, 12 and 20 = \( 2 \times 2 \times 3 \times 5 = 60 \)
(v) 24, 40, 80, 120

\( \therefore \) LCM of 24, 40, 80 and 120 = \( 2 \times 2 \times 2 \times 5 \times 3 \times 2 = 240 \)
In simple words: To find the LCM using the division method, write the numbers in a row and divide them by common prime factors. Continue dividing until no two numbers share a common prime factor, then multiply all divisors and remaining numbers.

🎯 Exam Tip: When calculating LCM for multiple numbers, use the ladder method. Be careful to divide by all prime factors, even if they only divide one of the numbers, carrying others down. Ensure all factors are included in the final product.

Question 4. Find the smallest number which when divided by 8,9,10,15,20 gives a remainder of 5 every time.
Answer: Here, the smallest number for division is LCM of 8, 9, 10,15 and 20.
\( 8 = 2 \times 2 \times 2 \)
\( 9 = 3 \times 3 \)
\( 10 = 2 \times 5 \)
\( 15 = 3 \times 5 \)
\( 20 = 2 \times 2 \times 5 \)
LCM of given numbers = \( 2 \times 2 \times 2 \times 3 \times 3 \times 5 = 360 \)
\( \therefore \) Required, smallest number = LCM + Remainder = \( 360 + 5 = 365 \)
\( \therefore \) The required smallest number is 365.
In simple words: To find a number that leaves a specific remainder when divided by several numbers, first find the LCM of those divisors, then add the remainder to the LCM.

🎯 Exam Tip: When a problem asks for the "smallest number" that leaves a certain remainder, it's a strong indicator to calculate the LCM of the given divisors and then adjust it with the remainder.

Question 5. Reduce the fractions \( \frac{348}{319}, \frac{221}{247}, \frac{437}{551} \) to the lowest terms.
Answer:
(i)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र यूक्लिडियन एल्गोरिथम का उपयोग करके 348 और 319 का महत्तम समापवर्तक (HCF) ज्ञात करने की प्रक्रिया को दर्शाता है। 348 को 319 से भाग देने पर शेषफल 29 प्राप्त होता है, फिर 319 को 29 से भाग देने पर शेषफल 0 प्राप्त होता है, जिससे HCF 29 है।
HCF of 348 and 319 = 29
\( \frac{348}{319} = \frac{348 \div 29}{319 \div 29} = \frac{12}{11} \)
(ii)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र यूक्लिडियन एल्गोरिथम का उपयोग करके 247 और 221 का महत्तम समापवर्तक (HCF) ज्ञात करने की प्रक्रिया को दर्शाता है। 247 को 221 से भाग देने पर शेषफल 26 प्राप्त होता है, फिर 221 को 26 से भाग देने पर शेषफल 13 प्राप्त होता है, और अंततः 26 को 13 से भाग देने पर शेषफल 0 प्राप्त होता है, जिससे HCF 13 है।
HCF of 221 and 247 = 13
\( \frac{221}{247} = \frac{221 \div 13}{247 \div 13} = \frac{17}{19} \)
(iii)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र यूक्लिडियन एल्गोरिथम का उपयोग करके 551 और 437 का महत्तम समापवर्तक (HCF) ज्ञात करने की प्रक्रिया को दर्शाता है। 551 को 437 से भाग देने पर शेषफल 114 प्राप्त होता है, फिर 437 को 114 से भाग देने पर शेषफल 95 प्राप्त होता है, 114 को 95 से भाग देने पर शेषफल 19 प्राप्त होता है, और अंततः 95 को 19 से भाग देने पर शेषफल 0 प्राप्त होता है, जिससे HCF 19 है।
HCF of 437 and 551 = 19
\( \frac{437}{551} = \frac{437 \div 19}{551 \div 19} = \frac{23}{29} \)
In simple words: To reduce a fraction to its lowest terms, find the HCF of the numerator and the denominator, and then divide both by their HCF. The Euclidean algorithm (long division method) is often used to find the HCF of larger numbers.

🎯 Exam Tip: Always show the steps for finding the HCF using the division method, especially for larger numbers. Dividing both numerator and denominator by their HCF ensures the fraction is in its simplest form.

Question 6. The LCM and HCF of two numbers are 432 and 72 respectively. If one of the numbers is 216, what is the other?
Answer: Here, LCM = 432, HCF = 72, First number = 216
First number \( \times \) Second number = LCM \( \times \) HCF
\( \therefore \) 216 \( \times \) Second number = \( 432 \times 72 \)
\( \therefore \) Second number = \( \frac{432 \times 72}{216} = 432 \times \frac{72}{216} = 432 \times \frac{1}{3} = 144 \)
\( \therefore \) The other number is 144.
In simple words: The product of two numbers is equal to the product of their HCF and LCM. Using this fundamental formula, you can find an unknown number if the other number, HCF, and LCM are given.

🎯 Exam Tip: Remember the fundamental relationship: Product of two numbers = HCF \( \times \) LCM. This formula is crucial for solving problems involving finding an unknown number given the other factors.

Question 7. The product of two two-digit numbers is 765 and their HCF is 3. What is their LCM?
Answer: Here, HCF = 3, Product of the given numbers = 765
Now, HCF \( \times \) LCM = Product of the given numbers
\( \therefore \) 3 \( \times \) LCM = 765
\( \therefore \) LCM = \( \frac{765}{3} = 255 \)
\( \therefore \) The LCM of the two two-digit numbers is 255.
In simple words: Similar to Question 6, apply the formula that states the product of two numbers equals the product of their HCF and LCM to easily calculate the LCM when the product and HCF are known.

🎯 Exam Tip: This problem directly tests your understanding of the HCF-LCM product rule. Clearly state the formula and substitute the given values to avoid errors.

Question 8. A trader has three bundles of string 392 m, 308 m and 490 m long. What is the greatest length of string that the bundles can be cut up into without any left over string?
Answer: The required greatest length of the string is the highest common factor (HCF) of 392, 308 and 490.
\( \therefore \) \( 392 = 2 \times 2 \times 2 \times 7 \times 7 \)
\( 308 = 2 \times 2 \times 7 \times 11 \)
\( 490 = 2 \times 7 \times 7 \times 5 \)
\( \therefore \) HCF of 392, 308 and 490 = \( 2 \times 7 = 14 \)
\( \therefore \) The required greatest length of the string is 14 m.

In simple words: When a problem asks for the "greatest length" or "maximum capacity" for an equal division without remainder, it indicates that you need to find the HCF of the given quantities.

🎯 Exam Tip: Look for keywords like "greatest," "largest," or "maximum" in word problems, as they often imply finding the HCF. Factorization or the division method can be used to determine the HCF of multiple numbers.

Question 9. Which two consecutive even numbers have an LCM of 180?
Answer: LCM of two consecutive even numbers = 180
But, HCF of two consecutive even numbers = 2
Now, product of the given number = HCF \( \times \) LCM
\( = 2 \times 180 = 360 \)
To find the two consecutive even numbers, we have to factorize 360.
\( 360 = 2 \times 2 \times 2 \times 3 \times 3 \times 5 \)
\( 360 = (2 \times 3 \times 3) \times (2 \times 2 \times 5) \)
\( = 18 \times 20 \)
\( \therefore \) The two consecutive even numbers whose LCM is 180 are 18 and 20.
In simple words: Consecutive even numbers always have an HCF of 2. Use the HCF-LCM product formula to find the product of the numbers, then factorize that product to identify the two consecutive even numbers.

🎯 Exam Tip: Remember that the HCF of any two consecutive even numbers is always 2. This fact, combined with the product formula, simplifies the process of finding the numbers.