Answer :
Sure, let's go through each of the questions step by step:
### Question 1: Simplify the expression
Since the specific expression to simplify is missing, we can't provide the steps or the solution here. Please include the expression next time for a full solution.
### Question 2: Factorize \( 4 p q r^2 + 6 p^2 q r^2 - 2 p q^2 r^2 \)
To factorize \( 4 p q r^2 + 6 p^2 q r^2 - 2 p q^2 r^2 \):
1. Identify the common factors in each term:
- All terms have \( 2 p q r^2 \).
2. Factor out the common factor \( 2 p q r^2 \):
[tex]\[ 4 p q r^2 + 6 p^2 q r^2 - 2 p q^2 r^2 = 2 p q r^2(2 + 3 p - q) \][/tex]
So, the factorized form is:
[tex]\[ 2 p q r^2 (3 p - q + 2) \][/tex]
### Question 3: Simplify the common factors in \( acx - 1 - x + 6a \)
To simplify \( acx - 1 - x + 6a \):
1. Group the terms:
[tex]\[ acx - x + 6a - 1 \][/tex]
2. Factor out the common factor \( x \) from the first two terms:
[tex]\[ x(ac - 1) + 6a - 1 \][/tex]
So, the simplified form is:
[tex]\[ acx + 6a - x - 1 \][/tex]
### Question 4: Calculate the number of days required for two tractors working 10 hours a day
Given:
- Three tractors working 8 hours each can plough the field in 5 days:
1. Total work done by three tractors in one day:
[tex]\[ \text{Work in one day} = 3 \times 8 = 24 \text{ tractor-hours} \][/tex]
2. Total work required (in tractor-hours):
[tex]\[ \text{Total work} = 5 \times 24 = 120 \text{ tractor-hours} \][/tex]
3. Work done by two tractors working 10 hours a day:
[tex]\[ \text{Daily work by two tractors} = 2 \times 10 = 20 \text{ tractor-hours} \][/tex]
4. Number of days required by two tractors:
[tex]\[ \text{Days required} = \frac{120}{20} = 6 \text{ days} \][/tex]
So, the number of days required is \( 6 \) days.
### Question 5: Find the square roots using tables
a) Square root of 9521695:
[tex]\[ \sqrt{9521695} = 3085.7243882109756 \][/tex]
b) Square root of 0.0456:
[tex]\[ \sqrt{0.0456} = 0.21354156504062624 \][/tex]
### Question 6: Find the squares using tables
a) Square of 893.689:
[tex]\[ (893.689)^2 = 798680.028721 \][/tex]
b) Square of 40.43:
[tex]\[ (40.43)^2 = 1634.5849 \][/tex]
### Question 7: Evans' average water consumption per day
1. Total water consumed per week:
[tex]\[ \text{Water per week} = 86.8 \text{ L} \][/tex]
2. Number of days in a week: \(7 \)
3. Average consumption per day:
[tex]\[ \text{Average consumption} = \frac{86.8}{7} = 12.4 \text{ L/day} \][/tex]
Evans' average water consumption per day is \( 12.4 \text{ L/day} \).
### Question 8: Evaluate the BODMAS expression
Evaluate:
[tex]\[ 4 \frac{1}{3} - \left(5 \frac{1}{5} + \frac{4}{5} \times \frac{25}{3}\right) \times \left(8 \frac{26}{5} + \frac{4}{5} \times \frac{25}{2}\right) \][/tex]
Breaking it down:
1. Convert to improper fractions:
[tex]\[ 4 \frac{1}{3} = \frac{13}{3} \][/tex]
[tex]\[ 5 \frac{1}{5} = \frac{26}{5} \][/tex]
[tex]\[ 8 \frac{26}{5} = \frac{66}{5} \][/tex]
2. Evaluate the innermost expression:
[tex]\[ \frac{4}{5} \times \frac{25}{3} = \frac{100}{15} = \frac{20}{3} \][/tex]
[tex]\[ \frac{4}{5} \times \frac{25}{2} = \frac{100}{10} = 10 \][/tex]
3. Substitute back:
[tex]\[ \frac{13}{3} - \left(\frac{26}{5} + \frac{20}{3}\right) \times \left(\frac{66}{5} + 10\right) \][/tex]
4. Simplify the fractions inside the parentheses:
[tex]\[ \left(\frac{26}{5} + \frac{20}{3}\right) = \left(\frac{78}{15} + \frac{100}{15}\right) = \frac{178}{15} \][/tex]
[tex]\[ \left(\frac{66}{5} + 10\right) = \left(\frac{66}{5} + \frac{50}{5}\right) = \frac{116}{5} \][/tex]
5. Multiply the two main terms:
[tex]\[ \frac{178}{15} \times \frac{116}{5} = \frac{20648}{75} \][/tex]
6. Final subtraction:
[tex]\[ \frac{13}{3} - \frac{20648}{75} \][/tex]
After calculating, the final result is:
[tex]\[ -174.77333333333334 \][/tex]
This completes all the steps and solutions.
### Question 1: Simplify the expression
Since the specific expression to simplify is missing, we can't provide the steps or the solution here. Please include the expression next time for a full solution.
### Question 2: Factorize \( 4 p q r^2 + 6 p^2 q r^2 - 2 p q^2 r^2 \)
To factorize \( 4 p q r^2 + 6 p^2 q r^2 - 2 p q^2 r^2 \):
1. Identify the common factors in each term:
- All terms have \( 2 p q r^2 \).
2. Factor out the common factor \( 2 p q r^2 \):
[tex]\[ 4 p q r^2 + 6 p^2 q r^2 - 2 p q^2 r^2 = 2 p q r^2(2 + 3 p - q) \][/tex]
So, the factorized form is:
[tex]\[ 2 p q r^2 (3 p - q + 2) \][/tex]
### Question 3: Simplify the common factors in \( acx - 1 - x + 6a \)
To simplify \( acx - 1 - x + 6a \):
1. Group the terms:
[tex]\[ acx - x + 6a - 1 \][/tex]
2. Factor out the common factor \( x \) from the first two terms:
[tex]\[ x(ac - 1) + 6a - 1 \][/tex]
So, the simplified form is:
[tex]\[ acx + 6a - x - 1 \][/tex]
### Question 4: Calculate the number of days required for two tractors working 10 hours a day
Given:
- Three tractors working 8 hours each can plough the field in 5 days:
1. Total work done by three tractors in one day:
[tex]\[ \text{Work in one day} = 3 \times 8 = 24 \text{ tractor-hours} \][/tex]
2. Total work required (in tractor-hours):
[tex]\[ \text{Total work} = 5 \times 24 = 120 \text{ tractor-hours} \][/tex]
3. Work done by two tractors working 10 hours a day:
[tex]\[ \text{Daily work by two tractors} = 2 \times 10 = 20 \text{ tractor-hours} \][/tex]
4. Number of days required by two tractors:
[tex]\[ \text{Days required} = \frac{120}{20} = 6 \text{ days} \][/tex]
So, the number of days required is \( 6 \) days.
### Question 5: Find the square roots using tables
a) Square root of 9521695:
[tex]\[ \sqrt{9521695} = 3085.7243882109756 \][/tex]
b) Square root of 0.0456:
[tex]\[ \sqrt{0.0456} = 0.21354156504062624 \][/tex]
### Question 6: Find the squares using tables
a) Square of 893.689:
[tex]\[ (893.689)^2 = 798680.028721 \][/tex]
b) Square of 40.43:
[tex]\[ (40.43)^2 = 1634.5849 \][/tex]
### Question 7: Evans' average water consumption per day
1. Total water consumed per week:
[tex]\[ \text{Water per week} = 86.8 \text{ L} \][/tex]
2. Number of days in a week: \(7 \)
3. Average consumption per day:
[tex]\[ \text{Average consumption} = \frac{86.8}{7} = 12.4 \text{ L/day} \][/tex]
Evans' average water consumption per day is \( 12.4 \text{ L/day} \).
### Question 8: Evaluate the BODMAS expression
Evaluate:
[tex]\[ 4 \frac{1}{3} - \left(5 \frac{1}{5} + \frac{4}{5} \times \frac{25}{3}\right) \times \left(8 \frac{26}{5} + \frac{4}{5} \times \frac{25}{2}\right) \][/tex]
Breaking it down:
1. Convert to improper fractions:
[tex]\[ 4 \frac{1}{3} = \frac{13}{3} \][/tex]
[tex]\[ 5 \frac{1}{5} = \frac{26}{5} \][/tex]
[tex]\[ 8 \frac{26}{5} = \frac{66}{5} \][/tex]
2. Evaluate the innermost expression:
[tex]\[ \frac{4}{5} \times \frac{25}{3} = \frac{100}{15} = \frac{20}{3} \][/tex]
[tex]\[ \frac{4}{5} \times \frac{25}{2} = \frac{100}{10} = 10 \][/tex]
3. Substitute back:
[tex]\[ \frac{13}{3} - \left(\frac{26}{5} + \frac{20}{3}\right) \times \left(\frac{66}{5} + 10\right) \][/tex]
4. Simplify the fractions inside the parentheses:
[tex]\[ \left(\frac{26}{5} + \frac{20}{3}\right) = \left(\frac{78}{15} + \frac{100}{15}\right) = \frac{178}{15} \][/tex]
[tex]\[ \left(\frac{66}{5} + 10\right) = \left(\frac{66}{5} + \frac{50}{5}\right) = \frac{116}{5} \][/tex]
5. Multiply the two main terms:
[tex]\[ \frac{178}{15} \times \frac{116}{5} = \frac{20648}{75} \][/tex]
6. Final subtraction:
[tex]\[ \frac{13}{3} - \frac{20648}{75} \][/tex]
After calculating, the final result is:
[tex]\[ -174.77333333333334 \][/tex]
This completes all the steps and solutions.
