Math worksheet with 25 questions on exponents and powers.
A worksheet titled "QUESTIONS" containing 25 math problems related to exponents, bases, exponents, and powers, including finding values, expressing numbers in exponential form, and simplifying expressions.
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Step-by-step solution for: CBSE Class 7 Mental Maths Exponents And Powers Worksheet
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Show Answer Key & Explanations
Step-by-step solution for: CBSE Class 7 Mental Maths Exponents And Powers Worksheet
Let's solve the problems step by step. Here is the detailed solution for each question:
---
- Base: The base is the number being raised to a power, which is \(7\).
- Exponent: The exponent is the power to which the base is raised, which is \(6\).
Answer: Base = \(7\), Exponent = \(6\).
---
- Base: The base is \(-41\).
- Exponent: The exponent is \(91\).
Answer: Base = \(-41\), Exponent = \(91\).
---
- The expression \((-P) \times (-P) \times (-P) \times (-P)\) can be written as \((-P)^4\).
Answer: \((-P)^4\).
---
- The expression \(8324 \times 8324 \times 8324 \times 8324\) can be written as \(8324^4\).
Answer: \(8324^4\).
---
- The square of \(21\) is \(21^2 = 21 \times 21 = 441\).
Answer: \(441\).
---
- The cube of \(11\) is \(11^3 = 11 \times 11 \times 11 = 1331\).
Answer: \(1331\).
---
- The square of \(-16\) is \((-16)^2 = (-16) \times (-16) = 256\).
Answer: \(256\).
---
- The cube of \(-50\) is \((-50)^3 = (-50) \times (-50) \times (-50) = -125000\).
Answer: \(-125000\).
---
- The exponent \(501\) is odd. When \(-1\) is raised to an odd power, the result is \(-1\).
Answer: \(-1\).
---
- First, express \(27\) as a power of \(3\): \(27 = 3^3\).
- So, \(27^2 = (3^3)^2 = 3^{3 \times 2} = 3^6\).
- Now, calculate \(3^3 \times 3^6\):
\[
3^3 \times 3^6 = 3^{3+6} = 3^9.
\]
Answer: \(9\).
---
- Express \(25\) and \(125\) as powers of \(5\):
\[
25 = 5^2 \quad \text{and} \quad 125 = 5^3.
\]
- So, \(25^5 = (5^2)^5 = 5^{2 \times 5} = 5^{10}\) and \(125^4 = (5^3)^4 = 5^{3 \times 4} = 5^{12}\).
- Now, calculate \(25^5 \times 125^4\):
\[
25^5 \times 125^4 = 5^{10} \times 5^{12} = 5^{10+12} = 5^{22}.
\]
Answer: \(22\).
---
- Note that \(10^7\) and \(10^4\) are not like terms, so they cannot be combined into a single power of \(10\). The expression remains as it is:
\[
10^7 + 10^4.
\]
- However, if the question intends to find a single power, it is not possible without additional context. Assuming the question is asking for the dominant term, the answer would be \(10^7\) (since \(10^7\) is much larger than \(10^4\)).
Answer: Not combinable into a single power without approximation.
---
- Use the property of exponents \(a^m \times a^n = a^{m+n}\):
\[
7^3 \times 7^7 = 7^{3+7} = 7^{10}.
\]
Answer: \(7^{10}\).
---
- Use the property of exponents \(a^m \times a^n = a^{m+n}\):
\[
4^{14} \times 4^4 = 4^{14+4} = 4^{18}.
\]
Answer: \(4^{18}\).
---
- Use the property of exponents \(a^m \div a^n = a^{m-n}\):
\[
4^{48} \div 4^6 = 4^{48-6} = 4^{42}.
\]
Answer: \(4^{42}\).
---
- Since the bases are the same, equate the exponents:
\[
2^x = x.
\]
- The only solution to this equation is \(x = 2\) (by inspection or solving).
Answer: \(2\).
---
- Express \(64\) as a power of \(4\):
\[
64 = 4^3.
\]
- So, \((64)^3 = (4^3)^3 = 4^{3 \times 3} = 4^9\).
- Equate the exponents:
\[
4^9 = 4^x \implies x = 9.
\]
Answer: \(9\).
---
- Simplify \(2^6 + 2^{-3}\):
\[
2^6 = 64 \quad \text{and} \quad 2^{-3} = \frac{1}{8},
\]
\[
2^6 + 2^{-3} = 64 + \frac{1}{8} = \frac{512}{8} + \frac{1}{8} = \frac{513}{8}.
\]
- Now, multiply by \(2^{14}\):
\[
\left( \frac{513}{8} \right) \times 2^{14} = \frac{513 \times 2^{14}}{8}.
\]
- Express \(8\) as a power of \(2\):
\[
8 = 2^3 \implies \frac{513 \times 2^{14}}{8} = \frac{513 \times 2^{14}}{2^3} = 513 \times 2^{14-3} = 513 \times 2^{11}.
\]
- Since \(513\) is not a power of \(2\), the expression cannot be simplified further into a single power of \(2\). However, if we assume the question intends the dominant term, the answer would be \(2^{11}\).
Answer: Not combinable into a single power without approximation.
---
- Calculate \((-16)^6\) and \((-16)^3\):
\[
(-16)^6 = 16^6 \quad \text{(even power, result is positive)},
\]
\[
(-16)^3 = -16^3 \quad \text{(odd power, result is negative)}.
\]
- Add these values:
\[
(-16)^6 + (-16)^3 = 16^6 - 16^3.
\]
- Multiply by \((-16)^{-3}\):
\[
\left[16^6 - 16^3\right] \times (-16)^{-3} = \left[16^6 - 16^3\right] \times \frac{1}{(-16)^3}.
\]
- Simplify:
\[
\left[16^6 - 16^3\right] \times \frac{1}{-16^3} = \frac{16^6}{-16^3} - \frac{16^3}{-16^3} = -16^{6-3} + 1 = -16^3 + 1.
\]
- Calculate \(-16^3\):
\[
-16^3 = -(4^2)^3 = -4^6 = -4096.
\]
- So:
\[
-16^3 + 1 = -4096 + 1 = -4095.
\]
Answer: \(-4095\).
---
- Express \(256\) and \(81\) as powers of smaller numbers:
\[
256 = 2^8 \quad \text{and} \quad 81 = 3^4.
\]
- So:
\[
\frac{256}{81} = \frac{2^8}{3^4}.
\]
Answer: \(\frac{2^8}{3^4}\).
---
- Express \(-27\) and \(125\) as powers of smaller numbers:
\[
-27 = -3^3 \quad \text{and} \quad 125 = 5^3.
\]
- So:
\[
\frac{-27}{125} = \frac{-3^3}{5^3} = -\left(\frac{3}{5}\right)^3.
\]
Answer: \(-\left(\frac{3}{5}\right)^3\).
---
- Use the property of exponents \(a^m \div a^n = a^{m-n}\):
\[
\left(\frac{-7}{5}\right)^{13} \div \left(\frac{-7}{5}\right)^{15} = \left(\frac{-7}{5}\right)^{13-15} = \left(\frac{-7}{5}\right)^{-2}.
\]
- Simplify:
\[
\left(\frac{-7}{5}\right)^{-2} = \left(\frac{5}{-7}\right)^2 = \left(\frac{5}{7}\right)^2 = \frac{25}{49}.
\]
Answer: \(\frac{25}{49}\).
---
- Any non-zero number raised to the power of \(0\) is \(1\):
\[
3^0 = 1, \quad 5^0 = 1, \quad 19^0 = 1.
\]
- So:
\[
3^0 + 5^0 + 19^0 = 1 + 1 + 1 = 3.
\]
Answer: \(3\).
---
- Calculate each term:
\[
7^0 = 1, \quad 3^0 = 1, \quad 8^0 = 1, \quad 5^0 = 1.
\]
- So:
\[
(7^0 + 3^0) \times (8^0 - 5^0) = (1 + 1) \times (1 - 1) = 2 \times 0 = 0.
\]
Answer: \(0\).
---
- Any non-zero number raised to the power of \(0\) is \(1\):
\[
4^0 = 1, \quad 6^0 = 1, \quad 100^0 = 1.
\]
- So:
\[
4^0 \times 6^0 \times 100^0 = 1 \times 1 \times 1 = 1.
\]
Answer: \(1\).
---
\[
\boxed{1}
\]
---
Question 1: Find the base and exponent of \(7^6\).
- Base: The base is the number being raised to a power, which is \(7\).
- Exponent: The exponent is the power to which the base is raised, which is \(6\).
Answer: Base = \(7\), Exponent = \(6\).
---
Question 2: Find the base and exponent of \((-41)^{91}\).
- Base: The base is \(-41\).
- Exponent: The exponent is \(91\).
Answer: Base = \(-41\), Exponent = \(91\).
---
Question 3: Give exponential form for \((-P) \times (-P) \times (-P) \times (-P)\).
- The expression \((-P) \times (-P) \times (-P) \times (-P)\) can be written as \((-P)^4\).
Answer: \((-P)^4\).
---
Question 4: Give exponential form for \(8324 \times 8324 \times 8324 \times 8324\).
- The expression \(8324 \times 8324 \times 8324 \times 8324\) can be written as \(8324^4\).
Answer: \(8324^4\).
---
Question 5: What is the square of 21?
- The square of \(21\) is \(21^2 = 21 \times 21 = 441\).
Answer: \(441\).
---
Question 6: What is the cube of 11?
- The cube of \(11\) is \(11^3 = 11 \times 11 \times 11 = 1331\).
Answer: \(1331\).
---
Question 7: What is the square of \(-16\)?
- The square of \(-16\) is \((-16)^2 = (-16) \times (-16) = 256\).
Answer: \(256\).
---
Question 8: What is the cube of \(-50\)?
- The cube of \(-50\) is \((-50)^3 = (-50) \times (-50) \times (-50) = -125000\).
Answer: \(-125000\).
---
Question 9: Find the value of \((-1)^{501}\).
- The exponent \(501\) is odd. When \(-1\) is raised to an odd power, the result is \(-1\).
Answer: \(-1\).
---
Question 10: \(3^3 \times 27^2 = 3^\square\).
- First, express \(27\) as a power of \(3\): \(27 = 3^3\).
- So, \(27^2 = (3^3)^2 = 3^{3 \times 2} = 3^6\).
- Now, calculate \(3^3 \times 3^6\):
\[
3^3 \times 3^6 = 3^{3+6} = 3^9.
\]
Answer: \(9\).
---
Question 11: \(25^5 \times 125^4 = 5^\square\).
- Express \(25\) and \(125\) as powers of \(5\):
\[
25 = 5^2 \quad \text{and} \quad 125 = 5^3.
\]
- So, \(25^5 = (5^2)^5 = 5^{2 \times 5} = 5^{10}\) and \(125^4 = (5^3)^4 = 5^{3 \times 4} = 5^{12}\).
- Now, calculate \(25^5 \times 125^4\):
\[
25^5 \times 125^4 = 5^{10} \times 5^{12} = 5^{10+12} = 5^{22}.
\]
Answer: \(22\).
---
Question 12: \(10^7 + 10^4 = 10^\square\).
- Note that \(10^7\) and \(10^4\) are not like terms, so they cannot be combined into a single power of \(10\). The expression remains as it is:
\[
10^7 + 10^4.
\]
- However, if the question intends to find a single power, it is not possible without additional context. Assuming the question is asking for the dominant term, the answer would be \(10^7\) (since \(10^7\) is much larger than \(10^4\)).
Answer: Not combinable into a single power without approximation.
---
Question 13: \(7^3 \times 7^7 = \_\_\_\_\_\_.\)
- Use the property of exponents \(a^m \times a^n = a^{m+n}\):
\[
7^3 \times 7^7 = 7^{3+7} = 7^{10}.
\]
Answer: \(7^{10}\).
---
Question 14: \(4^{14} \times 4^4 = \_\_\_\_\_\_.\)
- Use the property of exponents \(a^m \times a^n = a^{m+n}\):
\[
4^{14} \times 4^4 = 4^{14+4} = 4^{18}.
\]
Answer: \(4^{18}\).
---
Question 15: \(4^{48} \div 4^6 = \_\_\_\_\_\_.\)
- Use the property of exponents \(a^m \div a^n = a^{m-n}\):
\[
4^{48} \div 4^6 = 4^{48-6} = 4^{42}.
\]
Answer: \(4^{42}\).
---
Question 16: \(5^{(2^x)} = 5^x\). Find \(x\).
- Since the bases are the same, equate the exponents:
\[
2^x = x.
\]
- The only solution to this equation is \(x = 2\) (by inspection or solving).
Answer: \(2\).
---
Question 17: \((64)^3 = (4)^x\). Find \(x\).
- Express \(64\) as a power of \(4\):
\[
64 = 4^3.
\]
- So, \((64)^3 = (4^3)^3 = 4^{3 \times 3} = 4^9\).
- Equate the exponents:
\[
4^9 = 4^x \implies x = 9.
\]
Answer: \(9\).
---
Question 18: Find the value of \(x\) if \((2^6 + 2^{-3}) \times 2^{14} = 2^x\).
- Simplify \(2^6 + 2^{-3}\):
\[
2^6 = 64 \quad \text{and} \quad 2^{-3} = \frac{1}{8},
\]
\[
2^6 + 2^{-3} = 64 + \frac{1}{8} = \frac{512}{8} + \frac{1}{8} = \frac{513}{8}.
\]
- Now, multiply by \(2^{14}\):
\[
\left( \frac{513}{8} \right) \times 2^{14} = \frac{513 \times 2^{14}}{8}.
\]
- Express \(8\) as a power of \(2\):
\[
8 = 2^3 \implies \frac{513 \times 2^{14}}{8} = \frac{513 \times 2^{14}}{2^3} = 513 \times 2^{14-3} = 513 \times 2^{11}.
\]
- Since \(513\) is not a power of \(2\), the expression cannot be simplified further into a single power of \(2\). However, if we assume the question intends the dominant term, the answer would be \(2^{11}\).
Answer: Not combinable into a single power without approximation.
---
Question 19: Find the value of \(\left[(-16)^6 + (-16)^3\right] \times (-16)^{-3}\).
- Calculate \((-16)^6\) and \((-16)^3\):
\[
(-16)^6 = 16^6 \quad \text{(even power, result is positive)},
\]
\[
(-16)^3 = -16^3 \quad \text{(odd power, result is negative)}.
\]
- Add these values:
\[
(-16)^6 + (-16)^3 = 16^6 - 16^3.
\]
- Multiply by \((-16)^{-3}\):
\[
\left[16^6 - 16^3\right] \times (-16)^{-3} = \left[16^6 - 16^3\right] \times \frac{1}{(-16)^3}.
\]
- Simplify:
\[
\left[16^6 - 16^3\right] \times \frac{1}{-16^3} = \frac{16^6}{-16^3} - \frac{16^3}{-16^3} = -16^{6-3} + 1 = -16^3 + 1.
\]
- Calculate \(-16^3\):
\[
-16^3 = -(4^2)^3 = -4^6 = -4096.
\]
- So:
\[
-16^3 + 1 = -4096 + 1 = -4095.
\]
Answer: \(-4095\).
---
Question 20: Express \(\frac{256}{81}\) in the exponential form.
- Express \(256\) and \(81\) as powers of smaller numbers:
\[
256 = 2^8 \quad \text{and} \quad 81 = 3^4.
\]
- So:
\[
\frac{256}{81} = \frac{2^8}{3^4}.
\]
Answer: \(\frac{2^8}{3^4}\).
---
Question 21: Express \(\frac{-27}{125}\) in the exponential form.
- Express \(-27\) and \(125\) as powers of smaller numbers:
\[
-27 = -3^3 \quad \text{and} \quad 125 = 5^3.
\]
- So:
\[
\frac{-27}{125} = \frac{-3^3}{5^3} = -\left(\frac{3}{5}\right)^3.
\]
Answer: \(-\left(\frac{3}{5}\right)^3\).
---
Question 22: Find the value of \(\left(\frac{-7}{5}\right)^{13} \div \left(\frac{-7}{5}\right)^{15}\).
- Use the property of exponents \(a^m \div a^n = a^{m-n}\):
\[
\left(\frac{-7}{5}\right)^{13} \div \left(\frac{-7}{5}\right)^{15} = \left(\frac{-7}{5}\right)^{13-15} = \left(\frac{-7}{5}\right)^{-2}.
\]
- Simplify:
\[
\left(\frac{-7}{5}\right)^{-2} = \left(\frac{5}{-7}\right)^2 = \left(\frac{5}{7}\right)^2 = \frac{25}{49}.
\]
Answer: \(\frac{25}{49}\).
---
Question 23: Find the value of \(3^0 + 5^0 + 19^0\).
- Any non-zero number raised to the power of \(0\) is \(1\):
\[
3^0 = 1, \quad 5^0 = 1, \quad 19^0 = 1.
\]
- So:
\[
3^0 + 5^0 + 19^0 = 1 + 1 + 1 = 3.
\]
Answer: \(3\).
---
Question 24: Find the value of \((7^0 + 3^0) \times (8^0 - 5^0)\).
- Calculate each term:
\[
7^0 = 1, \quad 3^0 = 1, \quad 8^0 = 1, \quad 5^0 = 1.
\]
- So:
\[
(7^0 + 3^0) \times (8^0 - 5^0) = (1 + 1) \times (1 - 1) = 2 \times 0 = 0.
\]
Answer: \(0\).
---
Question 25: Find the value of \(4^0 \times 6^0 \times 100^0\).
- Any non-zero number raised to the power of \(0\) is \(1\):
\[
4^0 = 1, \quad 6^0 = 1, \quad 100^0 = 1.
\]
- So:
\[
4^0 \times 6^0 \times 100^0 = 1 \times 1 \times 1 = 1.
\]
Answer: \(1\).
---
Final Answer:
\[
\boxed{1}
\]
Parent Tip: Review the logic above to help your child master the concept of 7th grade math exponents worksheet.
