Answer :
Let's solve each question step by step.
### Question 1:
There are 392 chocolates in a box and 728 chocolates in another box. What is the highest number of chocolates that can be picked out at a time from both boxes so that both will become empty at the same time?
To solve this, we need to determine the Greatest Common Divisor (GCD) of the two numbers, 392 and 728. The GCD is the largest number that divides both 392 and 728 exactly without leaving a remainder.
To find the GCD:
1. Prime Factorization of 392:
- [tex]\( 392 = 2^3 \times 7^2 \)[/tex]
2. Prime Factorization of 728:
- [tex]\( 728 = 2^3 \times 7 \times 13 \)[/tex]
3. Identify the common factors with the smallest power:
- Both 392 and 728 share the common prime factors [tex]\(2\)[/tex] and [tex]\(7\)[/tex].
4. Take the lowest power of the common prime factors:
- GCD = [tex]\( 2^3 \times 7 = 8 \times 7 = 56 \)[/tex]
Thus, the highest number of chocolates that can be picked from both boxes so that both are emptied at the same time is 56.
### Question 2:
A rectangular ground is 18 metres long and 12 metres wide. What is the length of the biggest square marble needed to pave it with square marbles of the same size?
To find the largest square marble tiles that can be used, we must determine the GCD of the length and width of the rectangular ground.
1. Length of the ground: 18 meters
2. Width of the ground: 12 meters
To find the GCD:
1. Prime Factorization of 18:
- [tex]\( 18 = 2 \times 3^2 \)[/tex]
2. Prime Factorization of 12:
- [tex]\( 12 = 2^2 \times 3 \)[/tex]
3. Identify the common factors with the smallest power:
- Both 18 and 12 share the common prime factors [tex]\(2\)[/tex] and [tex]\(3\)[/tex].
4. Take the lowest power of the common prime factors:
- GCD = [tex]\( 2^1 \times 3^1 = 2 \times 3 = 6 \)[/tex]
Thus, the length of the biggest square marble needed to pave the ground is 6 meters.
### Question 3:
A rectangular court-yard is 12.5 metres long and 7.5 metres wide. What is the length of the biggest square slate of the same size needed to pave it?
To solve this, let's find the GCD of the lengths, considering they need to be handled as fractions.
1. Length of the court-yard: 12.5 meters = [tex]\( \frac{125}{10} \)[/tex] = [tex]\( \frac{25 \times 5}{10} \)[/tex] = [tex]\( 5 \times \frac{25}{10} \)[/tex] = [tex]\( 5 \times 2.5 \)[/tex]
2. Width of the court-yard: 7.5 meters = [tex]\( \frac{75}{10} \)[/tex] = [tex]\( \frac{15 \times 5}{10} \)[/tex] = [tex]\( 5 \times \frac{15}{10} \)[/tex] = [tex]\( 5 \times 1.5 \)[/tex]
To find the GCD of 12.5 and 7.5:
- Both values can be simplified by dealing with their integer equivalents:
- [tex]\( 125 \)[/tex] and [tex]\( 75 \)[/tex]
- GCD of [tex]\( 125 \)[/tex] and [tex]\( 75 \)[/tex] is [tex]\( 25 \)[/tex]
Thus, the GCD is [tex]\( 25 \)[/tex] when brought back into fractional terms:
- [tex]\(5 \times 0.5 \)[/tex] = 2.5 meters
Therefore, the length of the biggest square slate needed to pave the court-yard is 2.5 meters.
So, the answers for the questions are:
1. 56 chocolates
2. 6 meters
3. 2.5 meters
### Question 1:
There are 392 chocolates in a box and 728 chocolates in another box. What is the highest number of chocolates that can be picked out at a time from both boxes so that both will become empty at the same time?
To solve this, we need to determine the Greatest Common Divisor (GCD) of the two numbers, 392 and 728. The GCD is the largest number that divides both 392 and 728 exactly without leaving a remainder.
To find the GCD:
1. Prime Factorization of 392:
- [tex]\( 392 = 2^3 \times 7^2 \)[/tex]
2. Prime Factorization of 728:
- [tex]\( 728 = 2^3 \times 7 \times 13 \)[/tex]
3. Identify the common factors with the smallest power:
- Both 392 and 728 share the common prime factors [tex]\(2\)[/tex] and [tex]\(7\)[/tex].
4. Take the lowest power of the common prime factors:
- GCD = [tex]\( 2^3 \times 7 = 8 \times 7 = 56 \)[/tex]
Thus, the highest number of chocolates that can be picked from both boxes so that both are emptied at the same time is 56.
### Question 2:
A rectangular ground is 18 metres long and 12 metres wide. What is the length of the biggest square marble needed to pave it with square marbles of the same size?
To find the largest square marble tiles that can be used, we must determine the GCD of the length and width of the rectangular ground.
1. Length of the ground: 18 meters
2. Width of the ground: 12 meters
To find the GCD:
1. Prime Factorization of 18:
- [tex]\( 18 = 2 \times 3^2 \)[/tex]
2. Prime Factorization of 12:
- [tex]\( 12 = 2^2 \times 3 \)[/tex]
3. Identify the common factors with the smallest power:
- Both 18 and 12 share the common prime factors [tex]\(2\)[/tex] and [tex]\(3\)[/tex].
4. Take the lowest power of the common prime factors:
- GCD = [tex]\( 2^1 \times 3^1 = 2 \times 3 = 6 \)[/tex]
Thus, the length of the biggest square marble needed to pave the ground is 6 meters.
### Question 3:
A rectangular court-yard is 12.5 metres long and 7.5 metres wide. What is the length of the biggest square slate of the same size needed to pave it?
To solve this, let's find the GCD of the lengths, considering they need to be handled as fractions.
1. Length of the court-yard: 12.5 meters = [tex]\( \frac{125}{10} \)[/tex] = [tex]\( \frac{25 \times 5}{10} \)[/tex] = [tex]\( 5 \times \frac{25}{10} \)[/tex] = [tex]\( 5 \times 2.5 \)[/tex]
2. Width of the court-yard: 7.5 meters = [tex]\( \frac{75}{10} \)[/tex] = [tex]\( \frac{15 \times 5}{10} \)[/tex] = [tex]\( 5 \times \frac{15}{10} \)[/tex] = [tex]\( 5 \times 1.5 \)[/tex]
To find the GCD of 12.5 and 7.5:
- Both values can be simplified by dealing with their integer equivalents:
- [tex]\( 125 \)[/tex] and [tex]\( 75 \)[/tex]
- GCD of [tex]\( 125 \)[/tex] and [tex]\( 75 \)[/tex] is [tex]\( 25 \)[/tex]
Thus, the GCD is [tex]\( 25 \)[/tex] when brought back into fractional terms:
- [tex]\(5 \times 0.5 \)[/tex] = 2.5 meters
Therefore, the length of the biggest square slate needed to pave the court-yard is 2.5 meters.
So, the answers for the questions are:
1. 56 chocolates
2. 6 meters
3. 2.5 meters
