CBSE Class 7 Mathematics Exponents And Powers Worksheet Set 06

Question Q1. Find the area of a square park whose perimeter is 320 m.
Answer:
The perimeter of a square is given by the formula:
\( \text{Perimeter} = 4 \times \text{side} \)
Given that the perimeter is \( 320\text{ m} \):
\( 4 \times \text{side} = 320\text{ m} \)
\( \implies \text{side} = \frac{320}{4} = 80\text{ m} \)

Now, calculate the area of the square park:
\( \text{Area} = \text{side} \times \text{side} \)
\( \text{Area} = 80 \times 80 = 6400\text{ m}^2 \)

Therefore, the area of the square park is \( 6400\text{ m}^2 \).
In simple words: Divide the perimeter by 4 to find the length of one side, then multiply that side length by itself to get the area.

Exam Tip: Always make sure to write the unit of area as square units, such as \( \text{m}^2 \), to avoid losing marks.

Question Q2. A wall 4.84 m long and 3.1 m high is to be covered with rectangular tiles of size 22cm by 10cm .Find the cost of the tiles at the rate of Rs.1.50 per tile .
Answer:
First, convert the dimensions of the wall from meters to centimeters:
- \( \text{Length of the wall} = 4.84\text{ m} = 4.84 \times 100 = 484\text{ cm} \)
- \( \text{Height of the wall} = 3.1\text{ m} = 3.1 \times 100 = 310\text{ cm} \)

Now, calculate the total area of the wall:
\( \text{Area of the wall} = \text{Length} \times \text{Height} \)
\( \text{Area of the wall} = 484 \times 310 = 150040\text{ cm}^2 \)

Next, calculate the area of one rectangular tile:
\( \text{Area of one tile} = 22\text{ cm} \times 10\text{ cm} = 220\text{ cm}^2 \)

Find the total number of tiles needed to cover the wall:
\( \text{Number of tiles} = \frac{\text{Area of the wall}}{\text{Area of one tile}} \)
\( \text{Number of tiles} = \frac{150040}{220} = 682\text{ tiles} \)

Now, find the total cost of these tiles at the rate of Rs 1.50 per tile:
\( \text{Total cost} = 682 \times 1.50 = \text{Rs } 1023 \)

Therefore, the total cost of the tiles is Rs 1023.
In simple words: Convert the wall size to centimeters, find the areas of both the wall and a tile, divide them to see how many tiles you need, and then multiply by the cost of one tile.

Exam Tip: Never divide areas with different units. Always convert meters to centimeters before starting your calculations.

Question Q3. A table cover 4m x 2m , is spread on a meeting table. If 25 cm of the table cover is hanging all around the table, find the cost of painting the table top at Rs.2.25 per square meter .
Answer:
The dimensions of the table cover are \( 4\text{ m} \) by \( 2\text{ m} \).
The cover hangs down by \( 25\text{ cm} \) on all sides. Convert this to meters:
\( 25\text{ cm} = 0.25\text{ m} \)

Since the cover hangs from both ends of the length and width, we subtract \( 0.25\text{ m} \) twice from each dimension of the cover to find the size of the tabletop:
- \( \text{Length of the tabletop} = 4\text{ m} - (2 \times 0.25\text{ m}) = 4 - 0.5 = 3.5\text{ m} \)
- \( \text{Width of the tabletop} = 2\text{ m} - (2 \times 0.25\text{ m}) = 2 - 0.5 = 1.5\text{ m} \)

Now, calculate the area of the tabletop:
\( \text{Area of the tabletop} = 3.5 \times 1.5 = 5.25\text{ m}^2 \)

Calculate the total cost of painting the tabletop:
\( \text{Total cost} = 5.25 \times 2.25 = \text{Rs } 11.8125 \approx \text{Rs } 11.81 \)

Therefore, the cost of painting the tabletop is approximately Rs 11.81.
In simple words: Subtract the hanging parts from both sides to get the real table size, find its area, and multiply that area by the price per square meter.

Exam Tip: Remember to subtract the hanging width twice (once for each side) from both the length and width of the cover.

Question Q4. A rectangular park is 45 m long and 30 m wide. A path 2.5 m wide is constructed outside the park. Find the area of the park also find the cost of cementing the path at the rate of Rs.150 per meter square.
Answer:
First, find the area of the rectangular park:
\( \text{Area of the park} = 45 \times 30 = 1350\text{ m}^2 \)

Since a path of width \( 2.5\text{ m} \) is built outside the park, we add this width to both sides of the length and width to find the outer dimensions:
- \( \text{Outer length} = 45 + (2 \times 2.5) = 50\text{ m} \)
- \( \text{Outer width} = 30 + (2 \times 2.5) = 35\text{ m} \)

Now, find the total area including the path:
\( \text{Total outer area} = 50 \times 35 = 1750\text{ m}^2 \)

Calculate the area of the path alone by subtracting the inner park area from the outer total area:
\( \text{Area of the path} = 1750 - 1350 = 400\text{ m}^2 \)

Find the total cost of cementing this path at Rs 150 per square meter:
\( \text{Total cost} = 400 \times 150 = \text{Rs } 60000 \)

Therefore, the area of the park is \( 1350\text{ m}^2 \) and the cost of cementing the path is Rs 60,000.
In simple words: Add the path width to both sides of the park to find the outer size. Subtract the inside park area from this outer area to get the path's area, then multiply by the cost.

Exam Tip: Draw a quick sketch of the park with its path to clearly visualize the outer and inner boundary lines.

Question Q6. A grassy plot is 80 m x 60 m. Two cross paths each 4m wide are constructed at right angles through the center of the plot such that each path is parallel to one of the sides of the plot. Find the total area of paths. Also find the cost of gravelling them at Rs. 5 per square m.
Answer:
The dimensions of the plot are \( 80\text{ m} \) by \( 60\text{ m} \), and the width of the paths is \( 4\text{ m} \).
- The area of the first path parallel to the length:
\( \text{Area}_1 = 80 \times 4 = 320\text{ m}^2 \)
- The area of the second path parallel to the width:
\( \text{Area}_2 = 60 \times 4 = 240\text{ m}^2 \)

Since these two paths cross at the center, they overlap. This overlap forms a small square with sides equal to the path's width:
\( \text{Area of overlapping square} = 4 \times 4 = 16\text{ m}^2 \)

Find the total area of the paths by adding the two paths and subtracting the overlap:
\( \text{Total area of paths} = 320 + 240 - 16 = 544\text{ m}^2 \)

Calculate the total cost of gravelling at Rs 5 per square meter:
\( \text{Total cost} = 544 \times 5 = \text{Rs } 2720 \)

Therefore, the total area of the paths is \( 544\text{ m}^2 \) and the cost of gravelling is Rs 2720.
In simple words: Find the area of each path as a simple rectangle. Subtract the tiny square in the middle where they cross so you don't count it twice, then multiply by the rate.

Exam Tip: Never forget to subtract the area of the overlapping central square when calculating the total area of two cross paths.

Question Q7. A plot of land is in the form of a right triangular region. The legs are of lengths 12m and 5m. Find the area of the plot.
Answer:
In a right-angled triangle, the two legs perpendicular to each other act as the base and the height.
Using the area formula:
\( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \)
Substitute the given lengths of \( 12\text{ m} \) and \( 5\text{ m} \):
\( \text{Area} = \frac{1}{2} \times 12 \times 5 = 6 \times 5 = 30\text{ m}^2 \)

Therefore, the area of the right triangular plot of land is \( 30\text{ m}^2 \).
In simple words: Multiply the two shorter sides of the right triangle and divide the result by 2 to get the area.

Exam Tip: The longest side (hypotenuse) is not used when calculating the area of a right-angled triangle.

Question Q8. The area of the triangular field is 225 m\(^2\) . Find its base if its height is 9m.
Answer:
The area of a triangle is given by the formula:
\( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \)
We are given the area as \( 225\text{ m}^2 \) and the height as \( 9\text{ m} \). Substitute these values:
\( 225 = \frac{1}{2} \times \text{base} \times 9 \)
\( \implies 225 \times 2 = \text{base} \times 9 \)
\( \implies 450 = \text{base} \times 9 \)
\( \implies \text{base} = \frac{450}{9} = 50\text{ m} \)

Therefore, the base of the triangular field is \( 50\text{ m} \).
In simple words: Multiply the area by 2 and then divide by the height to find the base.

Exam Tip: Rearrange the formula to \( \text{base} = \frac{2 \times \text{Area}}{\text{height}} \) as your first step to keep your work clean.

Question Q9. The area of a triangle is equal to that of a square, whose each side measures 70 m. Find the base of the triangle when the height is 98m.
Answer:
First, calculate the area of the square:
\( \text{Area of the square} = \text{side} \times \text{side} = 70 \times 70 = 4900\text{ m}^2 \)

Since the area of the triangle is equal to the area of the square:
\( \text{Area of the triangle} = 4900\text{ m}^2 \)

Now, use the triangle area formula with the given height of \( 98\text{ m} \) to find the base:
\( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \)
\( 4900 = \frac{1}{2} \times \text{base} \times 98 \)
\( \implies 4900 = 49 \times \text{base} \)
\( \implies \text{base} = \frac{4900}{49} = 100\text{ m} \)

Therefore, the base of the triangle is \( 100\text{ m} \).
In simple words: Find the area of the square by multiplying its side length by itself. Use this same area value for the triangle, then solve for the base using the height.

Exam Tip: Look for opportunities to simplify your math before multiplying large numbers; here, \( 98 \) divided by \( 2 \) simplifies to \( 49 \) very cleanly.

Question Q10 The base of a parallelogram is thrice its height. If the area is 876 cm\(^2\),find the base and height of the parallelogram.
Answer:
Let the height of the parallelogram be \( h \).
The base of the parallelogram is three times its height, which we write as \( 3h \).

The formula for the area of a parallelogram is:
\( \text{Area} = \text{base} \times \text{height} \)
Substitute the given values:
\( (3h) \times h = 876 \)
\( \implies 3h^2 = 876 \)
\( \implies h^2 = \frac{876}{3} = 292 \)
\( \implies h = \sqrt{292} \approx 17.09\text{ cm} \)

Now, calculate the base:
\( \text{Base} = 3h = 3 \times 17.09 \approx 51.26\text{ cm} \)

Therefore, the height is approximately \( 17.09\text{ cm} \) and the base is approximately \( 51.26\text{ cm} \).
In simple words: Use a variable for height and three times that for base. Multiply them, divide the area by 3, then find the square root to get the height.

Exam Tip: Be comfortable working with square roots for non-perfect square numbers in geometry worksheets.

Question Q11.If the area of a rhombus be 24 cm\(^2\) and one of the diagonals be 4cm, find the perimeter of the rhombus.
Answer:
The area of a rhombus is given by the formula:
\( \text{Area} = \frac{1}{2} \times d_1 \times d_2 \)
Given that the area is \( 24\text{ cm}^2 \) and one diagonal \( d_1 = 4\text{ cm} \):
\( 24 = \frac{1}{2} \times 4 \times d_2 \)
\( \implies 24 = 2 \times d_2 \)
\( \implies d_2 = 12\text{ cm} \)

The diagonals of a rhombus bisect each other at right angles, forming four small right-angled triangles inside. The legs of these right-angled triangles are half the lengths of the diagonals:
- \( \text{Leg}_1 = \frac{4}{2} = 2\text{ cm} \)
- \( \text{Leg}_2 = \frac{12}{2} = 6\text{ cm} \)

Now, use Pythagoras' theorem to calculate the side of the rhombus:
\( \text{side} = \sqrt{\text{Leg}_1^2 + \text{Leg}_2^2} = \sqrt{2^2 + 6^2} = \sqrt{4 + 36} = \sqrt{40} \approx 6.32\text{ cm} \)

Since all four sides of a rhombus are equal, the perimeter is:
\( \text{Perimeter} = 4 \times \text{side} \approx 4 \times 6.32 = 25.28\text{ cm} \)

Therefore, the perimeter of the rhombus is approximately \( 25.28\text{ cm} \).
In simple words: Find the second diagonal first. Use half of both diagonals with Pythagoras' theorem to find one side, then multiply by 4 to get the perimeter.

Exam Tip: Use the property that diagonals bisect each other at right angles to find side lengths in any rhombus problem.

Question Q12. Find the area of a rhombus having each side equal to 13cm and one of whose diagonal is 24cm.
Answer:
The sides of the rhombus are \( 13\text{ cm} \) long and one diagonal \( d_1 = 24\text{ cm} \).
Because the diagonals bisect each other at right angles, half of the first diagonal is:
\( \text{Half-diagonal}_1 = \frac{24}{2} = 12\text{ cm} \)

Using Pythagoras' theorem with the side length of \( 13\text{ cm} \) as the hypotenuse, we find the half-length of the other diagonal:
\( (\text{Half-diagonal}_2)^2 + 12^2 = 13^2 \)
\( \implies (\text{Half-diagonal}_2)^2 + 144 = 169 \)
\( \implies (\text{Half-diagonal}_2)^2 = 25 \)
\( \implies \text{Half-diagonal}_2 = 5\text{ cm} \)

So, the full length of the second diagonal is:
\( d_2 = 2 \times 5 = 10\text{ cm} \)

Now, calculate the area of the rhombus:
\( \text{Area} = \frac{1}{2} \times d_1 \times d_2 = \frac{1}{2} \times 24 \times 10 = 120\text{ cm}^2 \)

Therefore, the area of the rhombus is \( 120\text{ cm}^2 \).
In simple words: Use the side length and half of the given diagonal to find the other half-diagonal using Pythagoras' theorem. Double it to get the full second diagonal, then calculate the area.

Exam Tip: Memorize the common Pythagorean triple \( (5, 12, 13) \) to save valuable calculation time during exams.

Question Q13. The longer side of a parallelogram is 54cm and the corresponding altitude is 16cm. If the altitude corresponding to the shorter side is 24 cm, find the length of the shorter side.
Answer:
The area of a parallelogram can be calculated using any base and its corresponding height (altitude):
\( \text{Area} = \text{Longer base} \times \text{Corresponding altitude} \)
\( \text{Area} = 54 \times 16 = 864\text{ cm}^2 \)

Now, use the shorter side as the base with its given corresponding altitude of \( 24\text{ cm} \):
\( \text{Area} = \text{Shorter side} \times 24 \)
\( 864 = \text{Shorter side} \times 24 \)
\( \implies \text{Shorter side} = \frac{864}{24} = 36\text{ cm} \)

Therefore, the length of the shorter side is \( 36\text{ cm} \).
In simple words: Work out the area of the shape first using the longer side. Then divide this area by the shorter side's height to get the shorter side length.

Exam Tip: No matter which base and corresponding height you use, the calculated area of the parallelogram remains exactly the same.

Question Q14. A piece of wire in the form of a rectangle 8.9 cm long and 5.4 cm broad is reshaped and bent into the form of a circle. Find the radius of the circle.
Answer:
Since the same wire is reshaped, the perimeter of the rectangle is equal to the circumference of the circle.
First, find the perimeter of the rectangle:
\( \text{Perimeter} = 2 \times (\text{Length} + \text{Width}) \)
\( \text{Perimeter} = 2 \times (8.9 + 5.4) = 2 \times 14.3 = 28.6\text{ cm} \)

Set this equal to the circumference of the circle:
\( 2\pi r = 28.6 \)
\( \implies 2 \times \frac{22}{7} \times r = 28.6 \)
\( \implies \frac{44}{7} \times r = 28.6 \)
\( \implies r = \frac{28.6 \times 7}{44} = 4.55\text{ cm} \)

Therefore, the radius of the circle is \( 4.55\text{ cm} \).
In simple words: Calculate the total length of the wire using the rectangle's perimeter. Use this same length as the circle's outline to find the radius.

Exam Tip: Use \( \pi = \frac{22}{7} \) for problems that involve values like \( 28.6 \) to get clean, easy-to-simplify fractions.

Question Q15. The circumference of a circle exceeds its diameter by 30cm. Find the radius of the circle.
Answer:
Let the radius of the circle be \( r \).
The circumference of the circle is \( 2\pi r \) and its diameter is \( 2r \).
We are given that:
\( \text{Circumference} = \text{Diameter} + 30 \)
\( \implies 2\pi r = 2r + 30 \)

Divide the entire equation by 2:
\( \pi r = r + 15 \)
\( \implies \pi r - r = 15 \)
\( \implies r(\pi - 1) = 15 \)

Substitute \( \pi = \frac{22}{7} \):
\( r\left(\frac{22}{7} - 1\right) = 15 \)
\( \implies r \times \frac{15}{7} = 15 \)
\( \implies r = \frac{15 \times 7}{15} = 7\text{ cm} \)

Therefore, the radius of the circle is \( 7\text{ cm} \).
In simple words: Set up an equation with circumference and diameter. Simplify it using \( \pi = \frac{22}{7} \) to find that the radius is exactly 7 cm.

Exam Tip: Factoring out \( r \) from \( \pi r - r \) is a crucial algebraic step to solve this equation quickly.

Question Q16. In the given figure above , the radius of quarter circular plot taken is 2m and radius of the flower bed is 2m. Find the area of the remaining field.
Answer:
Based on the standard rectangular plot of \( 8\text{ m} \) by \( 6\text{ m} \) with four quarter circular plots at the corners and a central circular flower bed:
1. Calculate the area of the entire rectangular plot:
\( \text{Area of the rectangle} = 8 \times 6 = 48\text{ m}^2 \)
2. Calculate the combined area of the four quarter circular corner plots (which make 1 full circle):
\( \text{Area of 4 quadrants} = 4 \times \left(\frac{1}{4} \times \pi \times 2^2\right) = 4\pi \approx 12.57\text{ m}^2 \)
3. Calculate the area of the central circular flower bed of radius \( 2\text{ m} \):
\( \text{Area of flower bed} = \pi \times 2^2 = 4\pi \approx 12.57\text{ m}^2 \)
4. Subtract both areas from the total rectangular plot area:
\( \text{Remaining Area} = 48 - (12.57 + 12.57) = 48 - 25.14 = 22.86\text{ m}^2 \) (using \( \pi = 3.14 \))

Therefore, the area of the remaining field is \( 22.86\text{ m}^2 \).
In simple words: Calculate the area of the big rectangle first. Subtract the area of the four corners and the middle circle to find the space left over.

Exam Tip: Recognize that four corner quarter-circles combine to form exactly one full circle of the same radius.

Question Q17. The circumference of a circle is 44cm . Find the area of the circle.
Answer:
The formula for the circumference is:
\( \text{Circumference} = 2\pi r \)
Given that the circumference is \( 44\text{ cm} \):
\( 2 \times \frac{22}{7} \times r = 44 \)
\( \implies \frac{44}{7} \times r = 44 \)
\( \implies r = 7\text{ cm} \)

Now, use this radius to calculate the area of the circle:
\( \text{Area} = \pi r^2 = \frac{22}{7} \times 7 \times 7 = 154\text{ cm}^2 \)

Therefore, the area of the circle is \( 154\text{ cm}^2 \).
In simple words: Use the circumference to work out that the radius of the circle is 7 cm. Then, use that radius to find the area.

Exam Tip: Be ready for this classic question structure where you must solve for radius \( r \) first before calculating the area.

Question Q18. An ox in a kolhu ( an oil processing apparatus ) is tethered to a rope 3m long. How much distance does it cover in 14 rounds ?
Answer:
The length of the rope is the radius of the circular path, so \( r = 3\text{ m} \).
In one single round, the distance covered is equal to the circumference of the circle:
\( \text{Distance in 1 round} = 2\pi r = 2 \times \frac{22}{7} \times 3 = \frac{132}{7}\text{ m} \)

To find the total distance covered in 14 rounds:
\( \text{Total distance} = 14 \times \frac{132}{7} = 2 \times 132 = 264\text{ m} \)

Therefore, the total distance covered by the ox is \( 264\text{ m} \).
In simple words: The distance in one round is the circle's outer edge. Multiply this by 14 rounds to find the total distance.

Exam Tip: Keep your calculations as fractions until the very end to simplify easily with numbers like 14.

Question Q19. The diameter of a circular park is 84 metres. A 3.5 m wide road runs on the outside around it. Find the cost of constructing the road at Rs. 20 per m\(^2\)
Answer:
The diameter of the circular park is \( 84\text{ m} \), so its inner radius is:
\( r = \frac{84}{2} = 42\text{ m} \)

Since a road of \( 3.5\text{ m} \) width runs on the outside, the outer radius of the park including the road is:
\( R = 42 + 3.5 = 45.5\text{ m} \)

The area of the road is the difference between the outer and inner circular areas:
\( \text{Area of the road} = \pi R^2 - \pi r^2 = \pi(R^2 - r^2) \)
\( \text{Area of the road} = \frac{22}{7} \times (45.5^2 - 42^2) \)
\( \text{Area of the road} = \frac{22}{7} \times (45.5 - 42)(45.5 + 42) = \frac{22}{7} \times 3.5 \times 87.5 = 962.5\text{ m}^2 \)

Now, find the cost of constructing the road at Rs 20 per square meter:
\( \text{Total cost} = 962.5 \times 20 = \text{Rs } 19250 \)

Therefore, the cost of constructing the road is Rs 19,250.
In simple words: Find the inner and outer radius, subtract the small circle's area from the large circle's area to find the road's area, then multiply by the cost.

Exam Tip: Use the algebraic identity \( A^2 - B^2 = (A - B)(A + B) \) to quickly subtract large squares of decimal values.

Question Q20. The area of a circle is 100 times the area of another circle. What is the ratio of their circumferences ?
Answer:
Let the radius of the larger circle be \( R \) and the smaller circle be \( r \).
We are given that:
\( \text{Area of larger circle} = 100 \times \text{Area of smaller circle} \)
\( \implies \pi R^2 = 100 \pi r^2 \)
\( \implies R^2 = 100 r^2 \)
Taking the square root on both sides:
\( R = 10r \)

Now, find the ratio of their circumferences:
\( \frac{\text{Circumference of larger circle}}{\text{Circumference of smaller circle}} = \frac{2\pi R}{2\pi r} = \frac{R}{r} = \frac{10r}{r} = 10 \)

Therefore, the ratio of their circumferences is \( 10:1 \).
In simple words: Since the area is 100 times larger, the radius is 10 times larger. The circumference is also 10 times larger, making the ratio 10 to 1.

Exam Tip: The ratio of circumferences of two circles is always equal to the ratio of their radii.

Question Q21. Express the following as a rational number:
(a) \( (-3)^{-3} \)
(b) \( \left(\frac{-3}{5}\right)^{-2} \)
(c) \( \left(\frac{-2}{3}\right)^3 \)
(d) \( \left(\frac{-2}{3}\right)^{-2} \)
(e) \( \left(\frac{1}{2}\right)^{-4} \)
Answer:
We simplify each exponential term into its rational form:

(a) \( (-3)^{-3} = \frac{1}{(-3)^3} = \frac{1}{-27} = -\frac{1}{27} \)

(b) \( \left(\frac{-3}{5}\right)^{-2} = \left(\frac{5}{-3}\right)^2 = \frac{5^2}{(-3)^2} = \frac{25}{9} \)

(c) \( \left(\frac{-2}{3}\right)^3 = \frac{(-2)^3}{3^3} = -\frac{8}{27} \)

(d) \( \left(\frac{-2}{3}\right)^{-2} = \left(\frac{3}{-2}\right)^2 = \frac{3^2}{(-2)^2} = \frac{9}{4} \)

(e) \( \left(\frac{1}{2}\right)^{-4} = \left(\frac{2}{1}\right)^4 = 2^4 = 16 = \frac{16}{1} \)
In simple words: A negative power flips the fraction upside down. Evaluate the top and bottom powers to find the final simple fraction.

Exam Tip: Remember that any negative number raised to an even power becomes positive, while raised to an odd power remains negative.

Question Q22. Simplify the following and express the result as a rational number:
(a) \( (3/2)^{-3} \times (3/2)^{-2} \)
(b) \( (-5/3)^6 \times (-5/3)^{-4} \)
(c) \( \{ 6^{-1} + \left(\frac{3}{2}\right)^{-1} \}^{-1} \)
(d) \( (5^{-1} \div 3^{-1})^2 \times \left(\frac{27}{125}\right)^{-1} \)
(e) \( (5^{-1} - 7^{-1})^{-1} \div \left(\frac{7}{10}\right) \)
(f) \( [ (2/3)^{-1} - (1/2)^{-1} ]^{-1} \)
(g) \( (-2/5)^3 \div (-3/10)^4 \)
(h) \( \{ (1/2)^2 - (1/4)^3 \} \times 2^3 \)
(i) \( \{ (3^2 - 2^2) \div (1/5)^2 \} \)
(j) \( \{ (5^2)^3 \times 5^4 \} \div 5^7 \)
(k) \( \frac{3 \times 7^6 \times 11^8}{21^6 \times 11^5} \)
(l) \( \frac{2^3 \times 3^4 \times 4}{3 \times 32} \)
(m) \( \frac{12^4 \times 9^3 \times 4}{6^3 \times 8^2 \times 27} \)
(n) \( \frac{3^5 \times 10^5 \times 25}{5^7 \times 6^5} \)
Answer:
(a) Use exponent rules: \( a^m \times a^n = a^{m+n} \).
\( \left(\frac{3}{2}\right)^{-3 + (-2)} = \left(\frac{3}{2}\right)^{-5} = \left(\frac{2}{3}\right)^5 = \frac{32}{243} \).

(b) \( \left(-\frac{5}{3}\right)^{6 + (-4)} = \left(-\frac{5}{3}\right)^2 = \frac{25}{9} \).

(c) Simplify inside the bracket first:
\( \left\{ \frac{1}{6} + \frac{2}{3} \right\}^{-1} = \left\{ \frac{1 + 4}{6} \right\}^{-1} = \left\{ \frac{5}{6} \right\}^{-1} = \frac{6}{5} \).

(d) \( \left( \frac{1}{5} \div \frac{1}{3} \right)^2 \times \frac{125}{27} = \left( \frac{3}{5} \right)^2 \times \frac{125}{27} = \frac{9}{25} \times \frac{125}{27} = \frac{5}{3} \).

(e) \( \left( \frac{1}{5} - \frac{1}{7} \right)^{-1} \div \frac{7}{10} = \left( \frac{7 - 5}{35} \right)^{-1} \div \frac{7}{10} = \left( \frac{2}{35} \right)^{-1} \div \frac{7}{10} = \frac{35}{2} \times \frac{10}{7} = 25 \).

(f) \( \left[ \frac{3}{2} - 2 \right]^{-1} = \left[ -\frac{1}{2} \right]^{-1} = -2 \).

(g) \( -\frac{8}{125} \div \frac{81}{10000} = -\frac{8}{125} \times \frac{10000}{81} = -\frac{8 \times 80}{81} = -\frac{640}{81} \).

(h) \( \left\{ \frac{1}{4} - \frac{1}{64} \right\} \times 8 = \left\{ \frac{16 - 1}{64} \right\} \times 8 = \frac{15}{64} \times 8 = \frac{15}{8} \).

(i) \( \{ (9 - 4) \div \frac{1}{25} \} = 5 \div \frac{1}{25} = 5 \times 25 = 125 \).

(j) \( \{ 5^6 \times 5^4 \} \div 5^7 = 5^{10} \div 5^7 = 5^3 = 125 \).

(k) \( \frac{3 \times 7^6 \times 11^8}{(3 \times 7)^6 \times 11^5} = \frac{3 \times 7^6 \times 11^8}{3^6 \times 7^6 \times 11^5} = \frac{11^3}{3^5} = \frac{1331}{243} \).

(l) \( \frac{8 \times 81 \times 4}{3 \times 32} = \frac{32 \times 81}{3 \times 32} = \frac{81}{3} = 27 \).

(m) Write in prime bases: \( \frac{(2^2 \times 3)^4 \times (3^2)^3 \times 2^2}{(2 \times 3)^3 \times (2^3)^2 \times 3^3} = \frac{2^8 \times 3^4 \times 3^6 \times 2^2}{2^3 \times 3^3 \times 2^6 \times 3^3} = \frac{2^{10} \times 3^{10}}{2^9 \times 3^6} = 2 \times 3^4 = 162 \).

(n) Write in prime bases: \( \frac{3^5 \times (2 \times 5)^5 \times 5^2}{5^7 \times (2 \times 3)^5} = \frac{3^5 \times 2^5 \times 5^5 \times 5^2}{5^7 \times 2^5 \times 3^5} = \frac{2^5 \times 3^5 \times 5^7}{2^5 \times 3^5 \times 5^7} = 1 \).
In simple words: Solve what is inside brackets first. Use rules of indices like adding powers when multiplying and subtracting powers when dividing.

Exam Tip: Converting composite base numbers like 6, 8, 9, 12, 21 into their prime factors \( (2, 3, 7) \) is the easiest way to simplify complex index expressions.