Test Yourself
Q1: Multiple Choice Type
i. The multiplicative inverse of \(\left(8^0+5^0\right)\left(8^0-5^0\right)\) is
We know: \(a^0 = 1\) for any non-zero a.
So,
\[
(8^0 + 5^0)(8^0 – 5^0) = (1 + 1)(1 – 1) = 2 \times 0 = 0
\]
Multiplicative inverse of 0 is undefined.
Answer: d. Undefined
ii. The value of \(\left[\left(\frac{1}{4}\right)^{-2}+\left(\frac{1}{3}\right)^{-2}\right]\div\left(\frac{1}{5}\right)^{-2}\) is:
\[
= \left[4^2 + 3^2\right] \div 5^2 = (16 + 9) \div 25 = 25 \div 25 = 1
\]Answer: b. 1
iii. If \(3^4\times9^3=9^n\), then the value of n is:
Note: \(9 = 3^2\), so
\[
3^4 \times (3^2)^3 = 3^4 \times 3^6 = 3^{10}
\]
Also,
\[
9^n = (3^2)^n = 3^{2n}
\Rightarrow 3^{10} = 3^{2n} \Rightarrow 2n = 10 \Rightarrow n = 5
\]Answer: a. 5
iv. If \(\left(\frac{5}{6}\right)^5\times\left(\frac{6}{5}\right)^{-4}=\left(\frac{5}{6}\right)^{3x}\), then value of x is:
Note:
\(\left(\frac{6}{5}\right)^{-4} = \left(\frac{5}{6}\right)^4\)
So,
\[
\left(\frac{5}{6}\right)^5 \times \left(\frac{5}{6}\right)^4 = \left(\frac{5}{6}\right)^9
\Rightarrow \left(\frac{5}{6}\right)^{3x} = \left(\frac{5}{6}\right)^9
\Rightarrow 3x = 9 \Rightarrow x = 3
\]Answer: d. 3
v. \(\left(\frac{2}{5}\right)^{-8}\div\left(\frac{2}{5}\right)^5\) is equal to:
Use law: \(a^m \div a^n = a^{m-n}\)
\[
= \left(\frac{2}{5}\right)^{-8 – 5} = \left(\frac{2}{5}\right)^{-13}
\]Answer: b. \(\left(\frac{2}{5}\right)^{-13}\)
vi. Statement 1: \(\left(x^0+y^0\right){(x+y)}^0=1,x,y\neq0\).
Statement2: \(\left(1+1\right)\left(1-1\right)=2\times0=0\).
Which of the following options is correct?
Checking Statement 1:
Since \(x \neq 0\) and \(y \neq 0\), we use the identity:
\[
x^0 = 1 \quad \text{and} \quad y^0 = 1 \quad \Rightarrow \quad x^0 + y^0 = 1 + 1 = 2
\]
Also,
\[
(x + y)^0 = 1 \quad \text{(as } x + y \neq 0 \text{ because } x, y \neq 0)
\]
So:
\[
(x^0 + y^0)(x + y)^0 = 2 \times 1 = 2 \neq 1
\]
Hence, **Statement 1 is false**.
Checking Statement 2:
\[
(1 + 1)(1 – 1) = 2 \times 0 = 0
\]
This is mathematically correct.
So, **Statement 2 is true**.
Answer: d. Statement 1 is false, and statement 2 is true.
vii. Assertion (A): \({(-100)}^3=-10,00,000\).
Reason (R): \({(-p)}^q=p^q\); if q is even.
Assertion (A): Let’s evaluate \((-100)^3\).
We calculate: \((-100) \times (-100) \times (-100) = -1,000,000\). Hence, Assertion (A) is correct as it equals \(-10,00,000\).
Reason (R): The general rule states:
For even exponents, \((-p)^q = p^q\). However, in the case of odd exponents, like 3, the negative sign is preserved:
\((-p)^3 = -p^3\). In this case, the reasoning is true but doesn’t directly explain Assertion (A), because it applies to even exponents.
Thus, both Assertion (A) and Reason (R) are correct, but Reason (R) is not the explanation for Assertion (A).
Answer: b. Both A and R are correct, and R is not the correct explanation for A.
viii. Assertion: \(\left(7^0+2^0\right)\left(7^0-2^0\right)=0\).
Reason (R): Any number raised to the power zero (0) is always equal to 1.
Step 1: Understanding the Assertion
The assertion is: \(\left(7^0 + 2^0\right)\left(7^0 – 2^0\right) = 0\).
We simplify this expression:
\(7^0 = 1\) and \(2^0 = 1\), so:
\(\left(1 + 1\right)\left(1 – 1\right) = (2) \times (0) = 0\).
Thus, the assertion is true: \(\left(7^0 + 2^0\right)\left(7^0 – 2^0\right) = 0\).
Step 2: Understanding the Reason (R)
The reason given is: “Any number raised to the power zero (0) is always equal to 1.”
This is a fundamental property in mathematics: \(a^0 = 1\) for any non-zero \(a\), so:
\(7^0 = 1\) and \(2^0 = 1\).
Therefore, the reason is also true.
Step 3: Conclusion
Both the assertion and the reason are true. Moreover, the reason explains the assertion.
Answer: a. Both A and R are correct, and R is the correct explanation for A.
ix. Assertion (A):\(\left(\frac{1}{5}\right)^{-5}\times\left(\frac{1}{2}\right)^{-5}={(10)}^{-5}\).
Reason (R): \(p^{-q}=\frac{1}{p^q}\ and\ \frac{1}{p^{-q}}=p^q,\ p\neq0\).
Step 1: Simplifying the Assertion
The assertion is: \(\left(\frac{1}{5}\right)^{-5} \times \left(\frac{1}{2}\right)^{-5} = \left(10\right)^{-5}\).
Using the negative exponent rule, we can simplify:
\(\left(\frac{1}{5}\right)^{-5} = 5^5\) and \(\left(\frac{1}{2}\right)^{-5} = 2^5\), so:
\(5^5 \times 2^5 = (5 \times 2)^5 = 10^5\), but the right-hand side is \(\left(10\right)^{-5} = \frac{1}{10^5}\).
Clearly, \(10^5 \neq \frac{1}{10^5}\), so the assertion is false.
Step 2: Understanding the Reason (R)
The reason is: \(p^{-q} = \frac{1}{p^q}\) and \(\frac{1}{p^{-q}} = p^q\), which are both true exponent rules.
The first part is correct: \(p^{-q} = \frac{1}{p^q}\), and the second part is also correct: \(\frac{1}{p^{-q}} = p^q\).
Step 3: Conclusion
Since the assertion is false and the reason is true.
Answer: d. A is false, but R is true.
x. Assertion (A): \({(p-q)}^{-1}\left(p^{-1}-q^{-1}\right)=-{(pq)}^{-1}.\)
Reason (R): \(a^{-1} and a^1\) are multiplication reciprocal to each other.
Let’s evaluate Assertion (A):
\[
\left(p – q\right)^{-1} = \frac{1}{p – q} \quad \text{and} \quad p^{-1} – q^{-1} = \frac{1}{p} – \frac{1}{q} = \frac{q – p}{pq}
\]
Multiplying the two expressions:
\[
\frac{1}{p – q} \times \frac{q – p}{pq} = -\frac{1}{pq}
\]
So, Assertion (A) is true.
Now, evaluate Reason (R):
Yes, \(a^{-1}\) and \(a^1\) are multiplicative reciprocals, so Reason (R) is true.
Since both Assertion (A) and Reason (R) are correct and steps in the verification of Assertion (A) utilize the concept of reciprocals and multiplicative inverses. The expression is simplified by using the relationship between a number and its inverse.
Answer: a. Both A and R are correct, and R is the correct explanation for A.
Q2: Evaluate:
i. \(\left(-\frac{3}{5}\right)^3\)
To evaluate \(\left(-\frac{3}{5}\right)^3\), we cube the numerator and denominator separately:
\(\left(-\frac{3}{5}\right)^3 = \frac{\left(-3\right)^3}{\left(5\right)^3} = \frac{-27}{125}\)
So, the value of \(\left(-\frac{3}{5}\right)^3\) is:
\(\frac{-27}{125}\)
ii. \(\left(\frac{2}{7}\right)^{-2}\)
When you have a negative exponent, it means we take the reciprocal of the base and change the sign of the exponent. So:
\(\left(\frac{2}{7}\right)^{-2} = \left(\frac{7}{2}\right)^2\)
Now, square both the numerator and denominator:
\(\left(\frac{7}{2}\right)^2 = \frac{7^2}{2^2} = \frac{49}{4}\)
So, the value of \(\left(\frac{2}{7}\right)^{-2}\) is:
\(\frac{49}{4}\)
Q3: Evaluate:
i. \(\left(-\frac{2}{5}\right)^4\times\left(-\frac{5}{2}\right)^2\)
To evaluate \(\left(-\frac{2}{5}\right)^4 \times \left(-\frac{5}{2}\right)^2\), we will evaluate each term separately:
First, calculate \(\left(-\frac{2}{5}\right)^4\):
\(\left(-\frac{2}{5}\right)^4 = \frac{(-2)^4}{5^4} = \frac{16}{625}\)
Now, calculate \(\left(-\frac{5}{2}\right)^2\):
\(\left(-\frac{5}{2}\right)^2 = \frac{(-5)^2}{2^2} = \frac{25}{4}\)
Now, multiply the two results:
\(\frac{16}{625} \times \frac{25}{4} = \frac{16 \times 25}{625 \times 4} = \frac{400}{2500} = \frac{4}{25}\)
So, the value of \(\left(-\frac{2}{5}\right)^4 \times \left(-\frac{5}{2}\right)^2\) is:
\(\frac{4}{25}\)
ii. \(\left(-\frac{3}{7}\right)^{-5}\times\left(\frac{7}{3}\right)^2\)
To evaluate \(\left(-\frac{3}{7}\right)^{-5} \times \left(\frac{7}{3}\right)^2\), let’s start by evaluating each term:
First, calculate \(\left(-\frac{3}{7}\right)^{-5}\):
A negative exponent means taking the reciprocal:
\(\left(-\frac{3}{7}\right)^{-5} = \left(-\frac{7}{3}\right)^5\)
Now, multiply the two results:
\(\left(-\frac{7}{3}\right)^{5} \times \left(\frac{7}{3}\right)^2 = \left(-\frac{7}{3}\right)^7\)
So, the value of \(\left(-\frac{3}{7}\right)^{-5} \times \left(\frac{7}{3}\right)^2\) is:
\(\left(-\frac{7}{3}\right)^7\)
Q4: Evaluate:
i. \(\left\{\left(-\frac{1}{2}\right)^{-2}\right\}^{-3}\)
To evaluate \(\left\{\left(-\frac{1}{2}\right)^{-2}\right\}^{-3}\), let’s break it down step by step:
First, evaluate \(\left(-\frac{1}{2}\right)^{-2}\):
\(\left(-\frac{1}{2}\right)^{-2} = \left(\frac{1}{-\frac{1}{2}}\right)^2 = \left(-2\right)^2 = 4\)
Now, we need to raise this result to the power of \(-3\):
\(4^{-3} = \frac{1}{4^3} = \frac{1}{64}\)
So, the value of \(\left\{\left(-\frac{1}{2}\right)^{-2}\right\}^{-3}\) is:
\(\frac{1}{64}\)
ii. \(\left[\left\{\left(-\frac{1}{5}\right)^2\right\}^{-2}\right]^{-1}\)
To evaluate \(\left[\left\{\left(-\frac{1}{5}\right)^2\right\}^{-2}\right]^{-1}\), let’s proceed step by step:
First, evaluate \(\left(-\frac{1}{5}\right)^2\):
\(\left(-\frac{1}{5}\right)^2 = \frac{1}{25}\)
Now, raise this result to the power of \(-2\):
\(\left(\frac{1}{25}\right)^{-2} = 25^2 = 625\)
Finally, take the reciprocal of the result:
\(625^{-1} = \frac{1}{625}\)
So, the value of \(\left[\left\{\left(-\frac{1}{5}\right)^2\right\}^{-2}\right]^{-1}\) is:
\(\frac{1}{625}\)
Q5: Evaluate:
i. \(\left(7^{-1}-8^{-1}\right)-\left(3^{-1}-4^{-1}\right)^{-1}\)
To evaluate \(\left(7^{-1}-8^{-1}\right)-\left(3^{-1}-4^{-1}\right)^{-1}\), we will evaluate each term step by step:
First, calculate \(7^{-1}\) and \(8^{-1}\):
\(7^{-1} = \frac{1}{7}\) and \(8^{-1} = \frac{1}{8}\)
So, \(7^{-1} – 8^{-1} = \frac{1}{7} – \frac{1}{8} = \frac{8 – 7}{56} = \frac{1}{56}\)
Now, calculate \(3^{-1}\) and \(4^{-1}\):
\(3^{-1} = \frac{1}{3}\) and \(4^{-1} = \frac{1}{4}\)
So, \(3^{-1} – 4^{-1} = \frac{1}{3} – \frac{1}{4} = \frac{4 – 3}{12} = \frac{1}{12}\)
Now, we need to take the reciprocal of \(\frac{1}{12}\):
\(\left(\frac{1}{12}\right)^{-1} = 12\)
Finally, subtract the results:
\(\frac{1}{56} – 12 = \frac{1}{56} – \frac{672}{56} = \frac{1 – 672}{56} = \frac{-671}{56}\)
So, the value of \(\left(7^{-1}-8^{-1}\right)-\left(3^{-1}-4^{-1}\right)^{-1}\) is:
\(\frac{-671}{56}\)
ii. \(5^{-7}\div5^{-10}\times5^{-5}\)
To evaluate \(5^{-7}\div5^{-10}\times5^{-5}\), let’s first use the laws of exponents:
We know that \(a^m \div a^n = a^{m – n}\) and \(a^m \times a^n = a^{m + n}\).
First, apply the division rule:
\(5^{-7} \div 5^{-10} = 5^{-7 – (-10)} = 5^{-7 + 10} = 5^{3}\)
Now, multiply by \(5^{-5}\):
\(5^{3} \times 5^{-5} = 5^{3 + (-5)} = 5^{-2}\)
So, the value of \(5^{-7}\div5^{-10}\times5^{-5}\) is:
\(5^{-2}\)
Q6: By what number should \({(-5)}^{-1}\) be divided to give the quotient \({(-25)}^{-1}\).
To find the number by which \(\left(-5\right)^{-1}\) should be divided to give \(\left(-25\right)^{-1}\), let’s break down the equation:
We are given:
\[
\frac{\left(-5\right)^{-1}}{x} = \left(-25\right)^{-1}
\]To isolate \(x\), multiply both sides by \(x\):
\[
\left(-5\right)^{-1} = x \times \left(-25\right)^{-1}
\]Now, simplify both sides:
\[
\frac{1}{-5}=x\times\frac{1}{-25}
\]So, the equation becomes:
\[
\frac{1}{-5} = x \times \frac{1}{-25}
\]To solve for \(x\), multiply both sides by \(-25\):
\[
\frac{1}{-5} \times (-25) = x
\]
\[
\frac{1 \times (-25)}{-5} = x
\]
\[
\frac{-25}{-5} = x
\]
\[
x = 5
\]So, \(\left(-5\right)^{-1}\) should be divided by 5 to give the quotient \(\left(-25\right)^{-1}\).
Answer: 5
Q7: Find n so that \(8^{11}\div8^5=8^{-3}\times8^{2n-1}\).
We are given the equation:
\[
8^{11} \div 8^5 = 8^{-3} \times 8^{2n-1}
\]
Let’s simplify both sides of the equation.
Step 1: Simplify the left-hand side
We know that when dividing powers of the same base, subtract the exponents. So:
\[
8^{11} \div 8^5 = 8^{11-5} = 8^6
\]
So, the left-hand side simplifies to \(8^6\).
Step 2: Simplify the right-hand side
Now, using the property of exponents, when multiplying powers of the same base, we add the exponents:
\[
8^{-3} \times 8^{2n-1} = 8^{-3 + (2n-1)} = 8^{2n – 4}
\]
So, the right-hand side simplifies to \(8^{2n – 4}\).
Step 3: Set the exponents equal
Since the bases are the same on both sides of the equation, we can set the exponents equal to each other:
\[
6 = 2n – 4
\]
Step 4: Solve for \(n\)
Add 4 to both sides:
\[
6 + 4 = 2n
\]
\[
10 = 2n
\]
Now, divide by 2:
\[
n = \frac{10}{2} = 5
\]
Therefore, the value of \(n\) is:
5
Q8: Find n so that \(9^{n+2}=240+9^n\).
We are given the equation:
\[
9^{n+2} = 240 + 9^n
\]Step 1: Express \(9^{n+2}\) in terms of \(9^n\)
Using the properties of exponents, we know that:
\[
9^{n+2} = 9^n \times 9^2
\]
Since \(9^2 = 81\), the equation becomes:
\[
9^n \times 81 = 240 + 9^n
\]
Step 2: Move all terms involving \(9^n\) to one side
To simplify the equation, subtract \(9^n\) from both sides:
\[
9^n \times 81 – 9^n = 240
\]
Factor out \(9^n\) from the left-hand side:
\[
9^n (81 – 1) = 240
\]
\[
9^n \times 80 = 240
\]Step 3: Solve for \(9^n\)
Divide both sides by 80:
\[
9^n = \frac{240}{80} = 3
\]Step 4: Solve for \(n\)
We know that \(9^n = 3\) can be written as:
\[
(3^2)^n = 3
\]
Simplifying this:
\[
3^{2n} = 3^1
\]
Since the bases are the same, we can equate the exponents:
\[
2n = 1
\]
Now, solve for \(n\):
\[
n = \frac{1}{2}
\]Therefore, the value of \(n\) is:
\(\frac{1}{2}\)
Q9: Find x, if:
i. \(3^{2x-1} = \left(27\right)^{x-3}\)
Step 1: Express 27 as a power of 3
Since \(27 = 3^3\), the equation becomes:
\[
3^{2x-1} = \left(3^3\right)^{x-3}
\]
Using the rule \((a^m)^n = a^{mn}\), we can simplify the right-hand side:
\[
3^{2x-1} = 3^{3(x-3)}
\]Step 2: Equate the exponents
Since the bases are the same, we can set the exponents equal to each other:
\[
2x – 1 = 3(x – 3)
\]
Simplifying the right-hand side:
\[
2x – 1 = 3x – 9
\]Step 3: Solve for \(x\)
Rearranging the equation:
\[
2x – 3x = -9 + 1
\]
Simplifying:
\[
-x = -8
\]
Therefore:
\[
x = 8
\]Answer: \(x = 8\)
ii. \(\frac{{25}^x \times 5^5 \times \left(125\right)^3}{5 \times \left(625\right)^4} = 125\)
Step 1: Express all numbers as powers of 5
– \(25 = 5^2\), so \(25^x = (5^2)^x = 5^{2x}\)<
br>
– \(125 = 5^3\), so \(125^3 = (5^3)^3 = 5^9\)
– \(625 = 5^4\), so \(625^4 = (5^4)^4 = 5^{16}\)
Substituting these into the equation:
\[
\frac{5^{2x} \times 5^5 \times 5^9}{5 \times 5^{16}} = 5^3
\]Step 2: Simplify the left-hand side
Use the rule \(a^m \times a^n = a^{m+n}\) to combine the powers of 5:
\[
\frac{5^{2x+5+9}}{5^{1+16}} = 5^3
\]
This simplifies to:
\[
\frac{5^{2x+14}}{5^{17}} = 5^3
\]Step 3: Simplify the fraction
Using the rule \(\frac{a^m}{a^n} = a^{m-n}\), we get:
\[
5^{2x + 14 – 17} = 5^3
\]
Simplifying the exponent:
\[
5^{2x – 3} = 5^3
\]Step 4: Equate the exponents
Since the bases are the same, we can set the exponents equal:
\[
2x – 3 = 3
\]
Solving for \(x\):
\[
2x = 6
\]
\[
x = 3
\]Answer: \(x = 3\)
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