Finding Common GroundClass 7 Mathematics NCERT Solutions

To Find: All factors of the given numbers.

Solution: We find the prime factorisation first and then list all possible combinations of these prime factors (including 1).

(a) 90 Prime factorisation: $90 = 2 \times 3 \times 3 \times 5 = 2^1 \times 3^2 \times 5^1$. Factors are formed by taking combinations of these primes: 1 2 3, $3 \times 3 = 9$ 5 $2 \times 3 = 6$ $2 \times 5 = 10$ $3 \times 5 = 15$ $2 \times 3 \times 3 = 18$ $3 \times 3 \times 5 = 45$ $2 \times 3 \times 5 = 30$ $2 \times 3 \times 3 \times 5 = 90$ Factors of 90: 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90.

(b) 105 Prime factorisation: $105 = 3 \times 5 \times 7$. Factors are: 1 3, 5, 7 $3 \times 5 = 15$ $3 \times 7 = 21$ $5 \times 7 = 35$ $3 \times 5 \times 7 = 105$ Factors of 105: 1, 3, 5, 7, 15, 21, 35, 105.

(c) 132 Prime factorisation: $132 = 2 \times 66 = 2 \times 2 \times 33 = 2^2 \times 3 \times 11$. Factors are: 1 2, $2 \times 2 = 4$ 3 11 $2 \times 3 = 6$ $2 \times 11 = 22$ $2 \times 2 \times 3 = 12$ $2 \times 2 \times 11 = 44$ $3 \times 11 = 33$ $2 \times 3 \times 11 = 66$ $2 \times 2 \times 3 \times 11 = 132$ Factors of 132: 1, 2, 3, 4, 6, 11, 12, 22, 33, 44, 66, 132.

(d) 360 Prime factorisation: $360 = 36 \times 10 = (6 \times 6) \times (2 \times 5) = (2 \times 3 \times 2 \times 3) \times 2 \times 5 = 2^3 \times 3^2 \times 5^1$. Factors: 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 18, 20, 24, 30, 36

To Find: The size of the tiles Sameeksha should buy.

Solution: For the square tiles to cover the floor exactly without being cut, the side length of the tile must be a factor of both the length and the breadth of the room.

Factors of the breadth (12 ft) are: 1, 2, 3, 4, 6, 12. Factors of the length (16 ft) are: 1, 2, 4, 8, 16.

The common factors of 12 and 16 are 1, 2, and 4. So, the possible side lengths of the square tiles are 1 ft, 2 ft, or 4 ft.

To use as few tiles as possible, the area of each tile must be as large as possible. This means the side length of the tile must be the largest possible.

The largest common factor of 12 and 16 is their Highest Common Factor (HCF).

HCF(12, 16) = 4.

Therefore, the side length of the largest square tile that can be used is 4 ft.

Final Answer: Sameeksha should buy square tiles of size 4 ft by 4 ft.

To Find: The number of tiles required to cover the floor.

Solution: Area of the room = Length $\times$ Breadth = $16 \text{ ft} \times 12 \text{ ft} = 192 \text{ sq ft}$. Area of one tile = Side $\times$ Side = $4 \text{ ft} \times 4 \text{ ft} = 16 \text{ sq ft}$.

Number of tiles = $\frac{\text{Area of the room}}{\text{Area of one tile}}$ Number of tiles = $\frac{192}{16} = 12$.

Alternatively, Number of tiles along the length = $\frac{\text{Length of room}}{\text{Side of tile}} = \frac{16}{4} = 4$. Number of tiles along the breadth = $\frac{\text{Breadth of room}}{\text{Side of tile}} = \frac{12}{4} = 3$. Total number of tiles = $4 \times 3 = 12$.

Final Answer: She should purchase 12 tiles of size 4 ft by 4 ft.

Solution: If the length of the tile could be a fractional number, the problem remains the same: to find the largest number that can divide both 12 and 16 exactly. This is still the definition of the Greatest Common Divisor (GCD) or Highest Common Factor (HCF).

The HCF of two integers is always an integer. For 12 and 16, the HCF is 4. There is no larger number, fractional or whole, that divides both 12 and 16 without a remainder.

For example, if we tried a tile of size 4.5 ft, it would not fit exactly along either the 12 ft side ($12 \div 4.5 = 2.66...$) or the 16 ft side ($16 \div 4.5 = 3.55...$).

Therefore, even if fractional lengths were allowed, the largest possible square tile would still have a side length that is the HCF of the room's dimensions.

Final Answer: No, the answer would not change. The largest possible tile size would still be 4 ft.

To Find: The weight of each bag.

Solution: To use as few bags as possible, the weight of each bag must be as large as possible. The weight of each bag must be a common factor of both 84 kg and 108 kg, so that the rice from both farms can be packed into these bags without any remainder.

We need to find the Highest Common Factor (HCF) of 84 and 108.

First, find the prime factorization of each number: $84 = 2 \times 42 = 2 \times 2 \times 21 = 2^2 \times 3 \times 7$ $108 = 2 \times 54 = 2 \times 2 \times 27 = 2^2 \times 3 \times 3 \times 3 = 2^2 \times 3^3$

To find the HCF, we take the lowest power of each common prime factor. The common prime factors are 2 and 3. The lowest power of 2 is $2^2$. The lowest power of 3 is $3^1$.

HCF(84, 108) = $2^2 \times 3^1 = 4 \times 3 = 12$.

So, the largest possible weight for each bag is 12 kg.

Final Answer: The weight of each bag should be 12 kg.

Given: Pairs of numbers where treasures are kept.

To Find: The longest jump size to land on both numbers in each pair. This is equivalent to finding the Highest Common Factor (HCF) of each pair.

Solution: (a) 14 and 30 Factors of 14: 1, 2, 7, 14 Factors of 30: 1, 2, 3, 5, 6, 10, 15, 30 Common factors: 1, 2 HCF(14, 30) = 2. The longest jump size is 2.

(b) 7 and 11 7 and 11 are both prime numbers. Their only common factor is 1. HCF(7, 11) = 1. The longest jump size is 1.

(c) 30 and 50 Factors of 30: 1, 2, 3, 5, 6, 10, 15, 30 Factors of 50: 1, 2, 5, 10, 25, 50 Common factors: 1, 2, 5, 10 HCF(30, 50) = 10. The longest jump size is 10.

(d) 28 and 42 Factors of 28: 1, 2, 4, 7, 14, 28 Factors of 42: 1, 2, 3, 6, 7, 14, 21, 42 Common factors: 1, 2, 7, 14 HCF(28, 42) = 14. The longest jump size is 14.

Final Answer: (a) 2 (b) 1 (c) 10 (d) 14

To Find: Whether the longest jump size is the same as the HCF and explain why.

Solution: Yes, the longest jump size is the same as the HCF of the two numbers.

Explanation: For Jumpy to land on a number (say 'a') starting from 0 with a jump size of 'j', the number 'a' must be a multiple of the jump size 'j'. This is the same as saying that the jump size 'j' must be a factor of 'a'.

For Jumpy to land on both numbers ('a' and 'b'), the jump size 'j' must be a factor of both 'a' and 'b'. This means 'j' must be a common factor of 'a' and 'b'.

The question asks for the longest possible jump size. This corresponds to the highest or greatest common factor of the two numbers.

Therefore, the longest jump size is the Highest Common Factor (HCF) of the two numbers.

To Find: The prime factorisation of 105 and 30.

Solution: We will use the division method for prime factorisation.

For 105: We start by dividing by the smallest prime factor, which is 3. $105 \div 3 = 35$ Now, we find the smallest prime factor of 35, which is 5. $35 \div 5 = 7$ 7 is a prime number, so we stop. The prime factors are 3, 5, and 7. So, the prime factorisation of 105 is $3 \times 5 \times 7$.

For 30: We start by dividing by the smallest prime factor, which is 2. $30 \div 2 = 15$ Now, we find the smallest prime factor of 15, which is 3. $15 \div 3 = 5$ 5 is a prime number, so we stop. The prime factors are 2, 3, and 5. So, the prime factorisation of 30 is $2 \times 3 \times 5$.

Final Answer: The prime factorisation of 105 is $3 \times 5 \times 7$. The prime factorisation of 30 is $2 \times 3 \times 5$.

To Find: The prime factorisation of 1200.

Solution: Using the division method: $1200 \div 2 = 600$ $600 \div 2 = 300$ $300 \div 2 = 150$ $150 \div 2 = 75$ $75 \div 3 = 25$ $25 \div 5 = 5$ $5 \div 5 = 1$

The prime factors are 2, 2, 2, 2, 3, 5, 5.

Final Answer: The prime factorisation of 1200 is $2 \times 2 \times 2 \times 2 \times 3 \times 5 \times 5$, or $2^4 \times 3 \times 5^2$.

Given: The number 840 and a potential factor 28. Prime factorisation of 840 is $2 \times 2 \times 2 \times 3 \times 5 \times 7$.

To Find: Whether 28 is a factor of 840.

Solution: A number is a factor if its prime factors are a 'subpart' of the prime factorisation of the larger number. The prime factorisation of 28 is $2 \times 2 \times 7$. The prime factorisation of 840 is $2 \times 2 \times 2 \times 3 \times 5 \times 7$.

We can see that the prime factors of 28 ($2, 2, 7$) are present in the prime factorisation of 840. We can write $840 = (2 \times 2 \times 7) \times (2 \times 3 \times 5) = 28 \times 30$. Since 840 can be expressed as 28 multiplied by an integer (30), 28 is a factor of 840.

Final Answer: Yes, 28 is a factor of 840.

Given: $840 = (2 \times 2 \times 7) \times (2 \times 3 \times 5)$

To Find: The number that 28 ($2 \times 2 \times 7$) must be multiplied by to get 840.

Solution: From the rearranged prime factorisation, we can see that the remaining factors are $2 \times 3 \times 5$. $2 \times 3 \times 5 = 30$. So, $28 \times 30 = 840$.

Final Answer: It should be multiplied by 30.

To Find: Whether 14 is a factor of 840 and why.

Solution: Yes, 14 is a factor of 840.

Reason: The prime factorisation of 14 is $2 \times 7$. The prime factorisation of 840 is $2 \times 2 \times 2 \times 3 \times 5 \times 7$. The prime factors of 14 ($2, 7$) are present in the prime factorisation of 840. We can write $840 = (2 \times 7) \times (2 \times 2 \times 3 \times 5) = 14 \times 60$. Since 840 can be expressed as 14 multiplied by an integer (60), 14 is a factor of 840.

To Find: Whether $2 \times 2 \times 2 = 8$ is a factor of 840 and why.

Solution: Yes, 8 is a factor of 840.

Reason: The prime factorisation of 8 is $2 \times 2 \times 2$. The prime factorisation of 840 is $2 \times 2 \times 2 \times 3 \times 5 \times 7$. The prime factors of 8 ($2, 2, 2$) are present in the prime factorisation of 840. We can write $840 = (2 \times 2 \times 2) \times (3 \times 5 \times 7) = 8 \times 105$. Since 840 can be expressed as 8 multiplied by an integer (105), 8 is a factor of 840.

To Find: Whether $3 \times 3 \times 3 = 27$ is a factor of 840 and why.

Solution: No, 27 is not a factor of 840.

Reason: The prime factorisation of 27 is $3 \times 3 \times 3$. The prime factorisation of 840 is $2 \times 2 \times 2 \times 3 \times 5 \times 7$. The prime factorisation of 840 contains the prime factor 3 only once. The prime factorisation of 27 requires the prime factor 3 to appear three times. Since the prime factors of 27 are not a subpart of the prime factors of 840, 27 is not a factor of 840.