NCERT Solutions Class 7 Mathematics Ganita Prakash 2 Chapter 03 Finding Common Ground

Question 1. List all the factors of the following numbers: (a) 90 (b) 105 (c) 132 (d) 360 (this number has 24 factors) (e) 840 (this number has 32 factors)
Answer:
(a) Start by breaking down 90 using prime factorization: \( 90 = 2 \times 3 \times 3 \times 5 \). The complete list of factors is 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, and 90. There are 12 factors in total.
(b) Break down 105: \( 105 = 3 \times 5 \times 7 \). The complete list of factors is 1, 3, 5, 7, 15, 21, 35, and 105. There are 8 factors in total.
(c) Break down 132: \( 132 = 2 \times 2 \times 3 \times 11 \). The complete list of factors is 1, 2, 3, 4, 6, 11, 12, 22, 33, 44, 66, 132. There are 12 factors in total.
(d) Break down 360: \( 360 = 2 \times 2 \times 2 \times 3 \times 3 \times 5 \). The complete list of factors is 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 18, 20, 24, 30, 36, 40, 45, 60, 72, 90, 120, 180, 360. There are 24 factors in total.
(e) Break down 840: \( 840 = 2 \times 2 \times 2 \times 3 \times 5 \times 7 \). The complete list of factors is 1, 2, 3, 4, 5, 6, 7, 8, 10, 12, 14, 15, 20, 21, 24, 28, 30, 35, 40, 42, 56, 60, 70, 84, 105, 120, 140, 168, 210, 280, 420, 840. There are 32 factors in total.
In simple words: To find all factors of a number, break it into prime factors first. Then write down all the different ways you can combine these prime factors - each combination gives you one factor.

Exam Tip: Always verify the total count of factors by using the formula: if \( n = p_1^{a_1} \times p_2^{a_2} \times ... \), then the number of factors equals \( (a_1 + 1)(a_2 + 1)... \). This saves time checking your list.

Question 1. Find the common factors and the HCF of the following numbers: (a) 50, 60 (b) 140, 275 (c) 77, 725 (d) 370, 592 (e) 81, 243 How do we directly find the HCF without listing all the factors?
Answer:
(a) Work out the prime factorization: \( 50 = 2 \times 5 \times 5 \) and \( 60 = 2 \times 2 \times 3 \times 5 \). The shared prime factors are 2 and 5, so HCF(50, 60) = \( 2 \times 5 = 10 \).
(b) Work out the prime factorization: \( 140 = 2 \times 2 \times 5 \times 7 \) and \( 275 = 5 \times 5 \times 11 \). The only shared prime factor is 5, so HCF(140, 275) = 5.
(c) Work out the prime factorization: \( 77 = 7 \times 11 \) and \( 725 = 5 \times 5 \times 29 \). These numbers share no prime factors, so HCF(77, 725) = 1 (they are co-prime).
(d) Work out the prime factorization: \( 370 = 2 \times 5 \times 37 \) and \( 592 = 2 \times 2 \times 2 \times 2 \times 37 \). The shared prime factors are 2 and 37, so HCF(370, 592) = \( 2 \times 37 = 74 \).
(e) Work out the prime factorization: \( 81 = 3 \times 3 \times 3 \times 3 \) and \( 243 = 3 \times 3 \times 3 \times 3 \times 3 \). The shared prime factors are four 3's, so HCF(81, 243) = \( 3 \times 3 \times 3 \times 3 = 81 \).
In simple words: To find the HCF directly, write both numbers as products of primes. Then multiply together only the prime factors that appear in both numbers.

Exam Tip: Write the prime factorizations side by side and circle or box the common prime factors - this visual method helps avoid errors and is faster than listing all factors.

Question 1. Find the HCF of the following numbers: (a) 24, 180 (b) 42, 75, 24 (c) 240, 378 (d) 400, 2500 (e) 300, 800
Answer:
(a) Prime factorizations: \( 24 = 2 \times 2 \times 2 \times 3 \) and \( 180 = 2 \times 2 \times 3 \times 3 \times 5 \). The shared prime factors are 2, 2, and 3, so HCF(24, 180) = \( 2 \times 2 \times 3 = 12 \).
(b) Prime factorizations: \( 42 = 2 \times 7 \times 3 \), \( 75 = 5 \times 5 \times 3 \), and \( 24 = 2 \times 2 \times 3 \times 2 \). The only shared prime factor among all three is 3, so HCF(42, 75, 24) = 3.
(c) Prime factorizations: \( 240 = 2 \times 2 \times 2 \times 2 \times 3 \times 5 \) and \( 378 = 2 \times 3 \times 3 \times 3 \times 7 \). The shared prime factors are 2 and 3, so HCF(240, 378) = \( 2 \times 3 = 6 \).
(d) Prime factorizations: \( 400 = 2 \times 2 \times 2 \times 2 \times 5 \times 5 \) and \( 2500 = 2 \times 2 \times 5 \times 5 \times 5 \times 5 \). The shared prime factors are 2, 2, 5, and 5, so HCF(400, 2500) = \( 2 \times 2 \times 5 \times 5 = 100 \).
(e) Prime factorizations: \( 300 = 2 \times 2 \times 3 \times 5 \times 5 \) and \( 800 = 2 \times 2 \times 2 \times 2 \times 5 \times 5 \). The shared prime factors are 2, 2, 5, and 5, so HCF(300, 800) = \( 2 \times 2 \times 5 \times 5 = 100 \).
In simple words: Write out the prime factorization for each number. The HCF is the product of all the prime factors that show up in every single number.

Exam Tip: When working with more than two numbers, keep checking that each prime factor you select appears in ALL the numbers, not just in some of them.

Question 2. Consider the numbers 72 and 144. Suppose they are factorised into composite numbers as: 72 = 6 × 12 and 144 = 8 × 18. Seeing this, can one say that these two numbers have no common factor other than 1? Why not?
Answer: No, you cannot claim that 72 and 144 have no common factor other than 1 just by looking at these composite factorizations. The reason is that composite numbers themselves contain prime factors. Here, 72 = 6 × 12 and 144 = 8 × 18, but 6, 12, 8, and 18 are all composite - they are not prime factorizations. When you break them down into primes: \( 72 = 2 \times 2 \times 2 \times 3 \times 3 \) and \( 144 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 \), you can see they share many prime factors. The HCF(72, 144) = \( 2 \times 2 \times 2 \times 3 \times 3 = 72 \). Since the HCF equals 72, which is much larger than 1, these numbers have many common factors besides just 1.
In simple words: You must use prime factorization to find the real HCF. Breaking a number into composite factors does not show you all the shared factors - you have to go all the way down to prime numbers.

Exam Tip: Always insist on prime factorization when finding HCF. Answers based on composite factorization alone are incomplete and can lead to wrong conclusions about common factors.

Question 1. Find the LCM of the following numbers: (a) 30, 72 (b) 36, 54 (c) 105, 195, 65 (d) 222, 370
Answer:
(a) Prime factorizations: \( 30 = 2 \times 3 \times 5 \) and \( 72 = 2 \times 2 \times 2 \times 3 \times 3 \). For the LCM, take the highest power of each prime: three 2's, two 3's, and one 5. So LCM(30, 72) = \( 2 \times 2 \times 2 \times 3 \times 3 \times 5 = 8 \times 9 \times 5 = 360 \).
(b) Prime factorizations: \( 36 = 2 \times 2 \times 3 \times 3 \) and \( 54 = 2 \times 3 \times 3 \times 3 \). For the LCM, take the highest power of each prime: two 2's and three 3's. So LCM(36, 54) = \( 2 \times 2 \times 3 \times 3 \times 3 = 4 \times 27 = 108 \).
(c) Prime factorizations: \( 105 = 3 \times 5 \times 7 \), \( 195 = 3 \times 5 \times 13 \), and \( 65 = 5 \times 13 \). For the LCM, take the highest power of each prime across all three numbers: one 3, one 5, one 7, and one 13. So LCM(105, 195, 65) = \( 3 \times 5 \times 7 \times 13 = 1365 \).
(d) Prime factorizations: \( 222 = 2 \times 3 \times 37 \) and \( 370 = 2 \times 5 \times 37 \). For the LCM, take the highest power of each prime: one 2, one 3, one 5, and one 37. So LCM(222, 370) = \( 2 \times 3 \times 5 \times 37 = 1110 \).
In simple words: To find the LCM, write out prime factorizations. Then take each prime number and use the highest number of times it appears in any one number - multiply all of these highest powers together.

Exam Tip: Line up the prime factorizations vertically and match each prime across all numbers. Circle the highest power of each prime, then multiply those circled powers - this method prevents missing a prime or using a lower power than you should.

Question 1. Make a general statement about the HCF for the following pairs of numbers. You could consider examples before coming up with general statements. Look for possible explanations of why they hold. (a) Two consecutive even numbers (b) Two consecutive odd numbers (c) Two even numbers (d) Two consecutive numbers (e) Two co-prime numbers
Answer:
(a) Two Consecutive Even Numbers
Examples: (2, 4) gives HCF = 2; (6, 8) gives HCF = 2; (10, 12) gives HCF = 2.
General Statement: The HCF of any two consecutive even numbers is always 2.
Reason: All even numbers are divisible by 2. Consecutive even numbers differ by exactly 2, so they never share any other prime factors besides 2. Therefore, 2 is their only common factor greater than 1.

(b) Two Consecutive Odd Numbers
Examples: (3, 5) gives HCF = 1; (7, 9) gives HCF = 1; (11, 13) gives HCF = 1.
General Statement: The HCF of any two consecutive odd numbers is always 1.
Reason: Consecutive odd numbers differ by 2 and cannot share any common even factor. Since they are odd and separated by 2, they also do not share odd prime factors. This makes them co-prime.

(c) Two Even Numbers
Examples: (4, 10) gives HCF = 2; (8, 12) gives HCF = 4; (14, 20) gives HCF = 2.
General Statement: The HCF of two even numbers is always even.
Reason: Since every even number contains 2 as a prime factor, the HCF must include 2. If the two numbers share additional common prime factors, the HCF becomes a larger even number that is a multiple of 2.

(d) Two Consecutive Numbers
Examples: (7, 8) gives HCF = 1; (14, 15) gives HCF = 1; (20, 21) gives HCF = 1.
General Statement: The HCF of any two consecutive numbers is always 1.
Reason: When two numbers differ by exactly 1, they cannot share any common prime factors. If they shared a factor greater than 1, that same factor would have to divide their difference, which is 1 - but no number greater than 1 divides 1.

(e) Two Co-prime Numbers
Examples: (4, 9) gives HCF = 1; (5, 8) gives HCF = 1; (7, 10) gives HCF = 1.
General Statement: The HCF of two co-prime numbers is always 1.
Reason: Co-prime numbers are defined by having no common prime factors other than 1. Therefore, their HCF must be 1 by definition.
In simple words: An HCF of 1 means the two numbers share no prime factors. An even HCF means both numbers must be even. Different types of number pairs follow different HCF rules.

Exam Tip: Learn these patterns - they often appear in proofs and multiple-choice questions. Being able to recall these general statements saves time on exam day.

Question 2. The LCM of 3 and 24 is 24 (it is one of the two given numbers). (a) Find more such number pairs where the LCM is one of the two numbers. (b) Make a general statement about such numbers. Describe such number pairs using algebra.
Answer:
(a) Here are more pairs where the LCM equals one of the two numbers:
(i) LCM of 2 and 4 is 4. Prime factorization: \( 2 = 2 \) and \( 4 = 2 \times 2 \). LCM(2, 4) = \( 2 \times 2 = 4 \).
(ii) LCM of 5 and 10 is 10. Prime factorization: \( 5 = 5 \) and \( 10 = 2 \times 5 \). LCM(5, 10) = \( 2 \times 5 = 10 \).
(iii) LCM of 6 and 12 is 12. Prime factorization: \( 6 = 2 \times 3 \) and \( 12 = 2 \times 2 \times 3 \). LCM(6, 12) = \( 2 \times 2 \times 3 = 12 \).
(iv) LCM of 7 and 49 is 49. Prime factorization: \( 7 = 7 \) and \( 49 = 7 \times 7 \). LCM(7, 49) = 49.
(v) LCM of 10 and 100 is 100. Prime factorization: \( 10 = 2 \times 5 \) and \( 100 = 2 \times 2 \times 5 \times 5 \). LCM(10, 100) = \( 2 \times 2 \times 5 \times 5 = 100 \).

(b) General Statement: For any two positive integers a and b where a is smaller than b, the LCM equals b if and only if a divides b evenly (with no remainder). In other words, the smaller number must be a factor of the larger number. In the original example, the LCM of 3 and 24 is 24 because 3 divides 24 evenly (24 ÷ 3 = 8).

Algebraic Description: Let two positive integers be a and b, where a < b. The LCM(a, b) = b if and only if b is a multiple of a. Using algebra, this can be written as: LCM(a, b) = b
\[ \iff b = k \times a \]
where k is a positive integer.
In simple words: If a smaller number divides into a larger number with no leftover, then the LCM of both numbers is just the larger number.

Exam Tip: Check divisibility first before calculating LCM - if one number divides the other, the LCM is simply the larger number, saving you calculation time.

Question 3. Make a general statement about the LCM for the following pairs of numbers. You could consider examples before coming up with these general statements. Look for possible explanations of why they hold. (a) Two multiples of 3 (b) Two consecutive even numbers (c) Two consecutive numbers (d) Two co-prime numbers
Answer:
(a) Two Multiples of 3
Examples: LCM(6, 9) = 18; LCM(9, 12) = 36; LCM(12, 18) = 36.
Observation: The LCM of two multiples of 3 is also a multiple of 3.
General Statement: The LCM of two multiples of 3 is always a multiple of 3.
Reason: Since both numbers are divisible by 3, any number they both divide into (their common multiple) must also be divisible by 3. Therefore, 3 appears as a prime factor in the LCM.

(b) Two Consecutive Even Numbers
Examples: LCM(2, 4) = 4; LCM(6, 8) = 24; LCM(10, 12) = 60.
Observation: The LCM of two consecutive even numbers is half their product.
General Statement: For consecutive even numbers 2n and 2n + 2, the LCM equals:
\[ \text{LCM}(2n, 2n + 2) = \frac{2n \times (2n + 2)}{2} = n(2n + 2) = 2n^2 + 2n \]
Reason: Consecutive even numbers always share exactly one factor of 2 (and no higher powers of 2 in all cases), but their other prime factors differ. This overlap of one factor of 2 means the LCM is smaller than their full product.

(c) Two Consecutive Numbers
Examples: LCM(7, 8) = 56; LCM(9, 10) = 90; LCM(10, 11) = 110.
Observation: The LCM of two consecutive numbers equals their product.
General Statement: The LCM of any two consecutive numbers equals their product.
Reason: Consecutive numbers share no common prime factors other than 1 (they are always co-prime). Since they have no overlapping prime factors, the smallest number that both divide into is simply their product.

(d) Two Co-prime Numbers
Examples: LCM(4, 9) = 36; LCM(5, 8) = 40; LCM(7, 10) = 70.
Observation: The LCM of two co-prime numbers equals their product.
General Statement: The LCM of two co-prime numbers equals their product.
Reason: By definition, co-prime numbers share no common prime factors except 1. Without any overlapping prime factors, the smallest common multiple must be their product. Note: Co-prime numbers are any two natural numbers with no common factor other than 1.
In simple words: If two numbers share a prime factor, their LCM is less than their product. If they share no prime factors, their LCM equals their product.

Exam Tip: These four patterns cover most LCM questions - memorize them and verify each with one example to speed up your answers.

Question 1. In the two rows below, colours repeat as shown. When will the black stars meet next?
Answer: Find the positions where black stars appear in each row. In the first row, black stars are at positions 4 and 10, with a spacing of 6 positions between them. So black stars in row one are at: 4, 10, 16, 22, 28, and so on. In the second row, black stars are at positions 4 and 8, with a spacing of 4 positions between them. So black stars in row two are at: 4, 8, 12, 16, 20, and so on. The first position after 4 where both rows have a black star is position 16. Therefore, the black stars meet again at the 16th position.
In simple words: List all the positions of black stars in each row. Find the first position (after the starting point) that appears in both lists - that is when they meet next.

Exam Tip: This is an LCM problem. The gap between stars in each row gives you the numbers - find their LCM, then add it to the first meeting position.

Question 2. (a) Is \( 5 \times 7 \times 11 \times 11 \) a multiple of \( 5 \times 7 \times 7 \times 11 \times 2 \)? (b) Is \( 5 \times 7 \times 11 \times 11 \) a factor of \( 5 \times 7 \times 7 \times 11 \times 2 \)?
Answer:
(a) For a number to be a multiple of another, all prime factors of the second number must appear in the first number with at least the same frequency. Let \( a = 5 \times 7 \times 11 \times 11 \) and \( b = 5 \times 7 \times 7 \times 11 \times 2 \). Number a does not contain the prime factor 2, which appears in b. Also, a contains only one 7, while b contains two 7's. Therefore, a is not a multiple of b.

(b) For a number to be a factor of another, the second number must contain all prime factors of the first number with at least the same frequency. Let \( a = 5 \times 7 \times 11 \times 11 \) and \( b = 5 \times 7 \times 7 \times 11 \times 2 \). Number b contains only one 11, while a contains two 11's. Since b does not have enough 11's, b cannot contain all factors of a. Therefore, a is not a factor of b.
In simple words: For a multiple, the second number must be "inside" the first number (with all its prime pieces). For a factor, the first number must be "inside" the second number (with all its prime pieces).

Exam Tip: Count each prime factor carefully - mismatches in the count of any single prime disqualify the relationship immediately.

Question 3. Find the HCF and LCM of the following (state your answers in the form of prime factorisations): (a) \( 3 \times 3 \times 5 \times 7 \times 7 \) and \( 12 \times 7 \times 11 \) (b) 45 and 36
Answer:
(a) Rewrite \( 12 \times 7 \times 11 \) as \( 2 \times 2 \times 3 \times 7 \times 11 \). Now comparing \( 3 \times 3 \times 5 \times 7 \times 7 \) and \( 2 \times 2 \times 3 \times 7 \times 11 \): The shared prime factors are 3 and 7 (each appearing once). HCF = \( 3 \times 7 = 21 \). For the LCM, take the highest power of each prime: two 3's, two 2's, one 5, two 7's, and one 11. LCM = \( 2 \times 2 \times 3 \times 3 \times 5 \times 7 \times 7 \times 11 \).

(b) Work out prime factorizations: \( 45 = 3 \times 3 \times 5 \) and \( 36 = 2 \times 2 \times 3 \times 3 \). The shared prime factors are two 3's. HCF(45, 36) = \( 3 \times 3 = 9 \). For the LCM, take the highest power of each prime: two 2's, two 3's, and one 5. LCM(45, 36) = \( 2 \times 2 \times 3 \times 3 \times 5 = 180 \).
In simple words: For HCF, multiply the prime factors that both numbers share. For LCM, multiply each prime factor the maximum number of times it appears in either number.

Exam Tip: Always write composite numbers in prime factorization form first - this prevents calculation errors and makes comparing factors straightforward.

Question 4. Find two numbers whose HCF is 1, and LCM is 66.
Answer: Use the relationship: for any two numbers a and b, \( a \times b = \text{HCF}(a, b) \times \text{LCM}(a, b) \). Given HCF = 1 and LCM = 66, we get \( a \times b = 1 \times 66 = 66 \). You need to find pairs of numbers whose product is 66. The factor pairs of 66 are: (1, 66), (2, 33), (3, 22), and (6, 11). All of these pairs satisfy the conditions - they have HCF equal to 1 (since one divides the LCM and the other times they equal 66) and LCM equal to 66. Any one of these pairs is a valid answer: (1, 66), (2, 33), (3, 22), or (6, 11).
In simple words: Use the formula linking HCF and LCM to find the product. Then list all pairs of numbers with that product.

Exam Tip: Remember the formula \( a \times b = \text{HCF}(a, b) \times \text{LCM}(a, b) \) - it connects the four quantities and solves many problems where two are known.

Question 5. A cowherd took all his cows to graze in the fields. The cows can go to a crossing with 3 gates. An equal number of cows passed through each gate. Later, at another crossing with 5 gates again an equal number of cows passed through each gate. The same happened at the third crossing with 7 gates. If the cowherd had fewer than 200 cows, how many cows did he have? (Based on the folklore mathematics from Karnataka).
Answer: Since an equal number of cows passed through each gate at every crossing, the total number of cows must be divisible by 3, by 5, and by 7. This means the number of cows must be a common multiple of these three numbers. Calculate LCM(3, 5, 7): since these are all prime numbers with no common factors, LCM(3, 5, 7) = \( 3 \times 5 \times 7 = 105 \). So the number of cows must be a multiple of 105. The multiples of 105 are: 105, 210, 315, and so on. The problem states the cowherd had fewer than 200 cows. Among the multiples of 105, only 105 is less than 200. Therefore, the cowherd had 105 cows.
In simple words: If the cows divide evenly at each crossing, the total must divide evenly by the number of gates at each crossing. Find the LCM of all gate numbers, then find which multiple of this LCM fits the given constraint.

Exam Tip: Translate "equal number through each gate" into mathematical language: the total is divisible by the gate count - this signals an LCM problem.

Question 6. The length, width, and height of a box are 12 cm, 18 cm, and 36 cm, respectively. Which of the following-sized cubes can be packed in this box without leaving gaps? (a) 9 cm (b) 6 cm (c) 4 cm (d) 3 cm (e) 2 cm
Answer: For cubes to pack without gaps, the side length must divide evenly into all three dimensions of the box. Find HCF(12, 18, 36). Prime factorizations: \( 12 = 2 \times 2 \times 3 \), \( 18 = 2 \times 3 \times 3 \), and \( 36 = 2 \times 2 \times 3 \times 3 \). The common prime factors are one 2 and one 3, so HCF = \( 2 \times 3 = 6 \) cm. The largest cube that fits without gaps has a side of 6 cm. However, any factor of the HCF will also work: (a) 9 cm - No, 9 does not divide 12 evenly. (b) 6 cm - Yes, 6 divides 12, 18, and 36 exactly. (c) 4 cm - No, 4 does not divide 18 evenly. (d) 3 cm - Yes, 3 is a factor of the HCF and divides all three dimensions. (e) 2 cm - Yes, 2 is a factor of the HCF and divides all three dimensions. The sizes that work are (b) 6 cm, (d) 3 cm, and (e) 2 cm. The largest is (b) 6 cm.
In simple words: The cube's side length must divide all three box dimensions evenly. Find the HCF of the dimensions - that gives the largest cube. All factors of the HCF also work.

Exam Tip: Test each option by dividing it into all three dimensions - if any dimension leaves a remainder, that cube size fails. This direct method is faster than calculating HCF first.

Question 7. Among the numbers below, which is the largest number that perfectly divides both 306 and 36? (a) 36 (b) 612 (c) 18 (d) 3 (e) 2 (f) 360
Answer: The largest number that divides both 306 and 36 is their HCF. Find the prime factorizations: \( 306 = 2 \times 3 \times 3 \times 17 \) and \( 36 = 2 \times 2 \times 3 \times 3 \). The shared prime factors are one 2 and two 3's. HCF(306, 36) = \( 2 \times 3 \times 3 = 18 \). Therefore, option (c) 18 is the correct answer.
In simple words: The "largest number that divides both" means you want their HCF - find the prime factors they share.

Exam Tip: When a question asks for the "largest divisor" or "largest number that divides," always calculate the HCF - this is what the question is really asking for.

Question 8. Find the smallest number that is divisible by 3, 4, 5, and 7, but leaves a remainder of 10 when divided by 11.
Answer: First, find the smallest number divisible by 3, 4, 5, and 7: LCM(3, 4, 5, 7) = \( 3 \times 4 \times 5 \times 7 = 420 \). Any number meeting the first condition can be written as \( 420k \) where k is a whole number. Now apply the second condition: the number must leave a remainder of 10 when divided by 11. Test \( N = 420 \times 1 = 420 \): dividing, \( 420 = 11 \times 38 + 2 \), so the remainder is 2, not 10. For a remainder of 10 instead of 2, you need to add \( 10 - 2 = 8 \) to the remainder, which means multiplying k by 5: \( 2k = 10 \) gives \( k = 5 \). Therefore, \( N = 420 \times 5 = 2100 \).
In simple words: Find the LCM for the first part. Then use that LCM as a "base" and find its multiples - adjust k until the remainder condition is satisfied.

Exam Tip: When a problem has both "divisible by" and "remainder" conditions, start with LCM for the divisibility part, then adjust using the remainder condition.

Question 9. Children are playing 'Fire in the Mountain.' When the number 6 was called out, no one got out. When the number 9 was called out, no one got out. But when the number 10 was called out, some people got out. How many children could have been playing initially? (a) 72 (b) 90 (c) 45 (d) 3 (e) 36 (f) None of these
Answer: Interpret the rules: when a number k is called, children form groups of k, and "no one got out" means the total divides evenly by k (no remainder). So the total number N must be divisible by 6 and by 9, but not divisible by 10. For N to be divisible by both 6 and 9, it must be divisible by LCM(6, 9) = 18. So N is a multiple of 18 that is NOT a multiple of 10. Check each option: (a) 72 = 18 × 4 - divisible by 18 and not by 10 → possible. (b) 90 = 18 × 5 - divisible by 18 but IS divisible by 10 → not possible. (c) 45 - not divisible by 18 → not possible. (d) 3 - not divisible by 18 → not possible. (e) 36 = 18 × 2 - divisible by 18 and not by 10 → possible. (f) None of these - not applicable. The numbers that work are 36 and 72, so either (a) or (e) is correct, or (f) if the answer must be unique.
In simple words: "No one got out" means the total divides evenly by that number. Find the LCM of the "no one got out" numbers. Then check which options are multiples of that LCM but NOT multiples of the "some got out" number.

Exam Tip: This game-based problem hides an LCM/divisibility question. Translate the rules first, then work systematically through the options.

Question 10. Tick the correct statement(s). The LCM of two different prime numbers (m, n) can be: (a) Less than both numbers (b) In between the two numbers (c) Greater than both numbers (d) Less than m × n (e) Greater than m × n
Answer: For two different prime numbers m and n, the LCM is their product \( m \times n \), since prime numbers share no common factors except 1. Evaluate each statement: (a) Less than both numbers - False, because \( m \times n > m \) and \( m \times n > n \) for any primes m and n. (b) In between the two numbers - False, because \( m \times n \) is always larger than either individual prime. (c) Greater than both numbers - True, the LCM equals \( m \times n \), which is greater than both m and n individually. (d) Less than \( m \times n \) - False, because the LCM IS equal to \( m \times n \). (e) Greater than \( m \times n \) - False, because the LCM equals \( m \times n \), not greater. Therefore, only option (c) is correct.
In simple words: When two numbers are prime and different, their LCM always equals their product - nothing less, nothing more.

Exam Tip: Memorize that LCM of different primes = their product. This fact answers many multiple-choice questions quickly.

Question 11. A dog is chasing a rabbit that has a head start of 150 feet. It jumps 9 feet every time the rabbit jumps 7 feet. In how many leaps does the dog catch up with the rabbit?
Answer: In each round of jumping, the dog closes the gap. Each time the rabbit jumps 7 feet, the dog jumps 9 feet, gaining \( 9 - 7 = 2 \) feet. The dog needs to close the initial 150-foot head start. The number of leaps required: \( \frac{150}{2} = 75 \) leaps.
In simple words: The dog gains 2 feet with every jump. Divide the head start by the gain per jump to find how many jumps are needed.

Exam Tip: Focus on the gap closed per cycle (dog leap minus rabbit leap) - this is the key to solving chase problems quickly.

Question 12. What is the smallest number that is a multiple of 1, 2, 3, 4, 5, 6, 8, 9, 10? Do you remember the answer from Grade 6, Chapter 5?
Answer: To find the smallest number divisible by all these numbers, calculate LCM(1, 2, 3, 4, 5, 6, 8, 9, 10). Write prime factorizations: \( 1 = 1 \), \( 2 = 2 \), \( 3 = 3 \), \( 4 = 2 \times 2 \), \( 5 = 5 \), \( 6 = 2 \times 3 \), \( 8 = 2 \times 2 \times 2 \), \( 9 = 3 \times 3 \), \( 10 = 2 \times 5 \). The highest powers are three 2's, two 3's, and one 5. LCM = \( 2 \times 2 \times 2 \times 3 \times 3 \times 5 = 8 \times 9 \times 5 = 360 \). Yes, this matches the answer from Grade 6, Chapter 5, Prime Time.
In simple words: Find the prime factorization for each number. Take the highest power of each prime. Multiply them all together.

Exam Tip: When finding LCM of many numbers, organize by listing primes vertically and circling the highest power in each column - this prevents skipping any prime.

Question 13. Here is a problem posed by the ancient Indian Mathematician Mahaviracharya (850 C.E.). Add together \( \frac{8}{15}, \frac{1}{20}, \frac{7}{36}, \frac{11}{63} \) and \( \frac{1}{21} \). What do you get? How can we find this sum efficiently?
Answer: To add fractions, find a common denominator. The denominators are 15, 20, 36, 63, and 21. Find their prime factorizations: \( 15 = 3 \times 5 \), \( 20 = 2 \times 2 \times 5 \), \( 36 = 2 \times 2 \times 3 \times 3 \), \( 63 = 3 \times 3 \times 7 \), \( 21 = 3 \times 7 \). The LCM of these denominators is \( 2 \times 2 \times 3 \times 3 \times 5 \times 7 = 1260 \). Convert each fraction: \( \frac{8}{15} = \frac{8 \times 84}{1260} \), \( \frac{1}{20} = \frac{1 \times 63}{1260} \), \( \frac{7}{36} = \frac{7 \times 35}{1260} \), \( \frac{11}{63} = \frac{11 \times 20}{1260} \), \( \frac{1}{21} = \frac{1 \times 60}{1260} \). The sum is: \( \frac{8 \times 84 + 1 \times 63 + 7 \times 35 + 11 \times 20 + 1 \times 60}{1260} = \frac{672 + 63 + 245 + 220 + 60}{1260} = \frac{1260}{1260} = 1 \).
In simple words: Find the LCM of all the denominators - this is your common denominator. Convert each fraction, add the numerators, and simplify.

Exam Tip: The fact that the answer is exactly 1 is remarkable - this ancient problem was designed to show a beautiful result. Always check if fractions sum to a whole number.