CBSE Class 7 Mathematics Exponents And Powers Worksheet Set 05

Section-A

Multiple Choice questions

Question Q1. Which of the following is not in standard form?
(a) \( 6.2 \times 10^5 \)
(b) \( 72.6 \times 10^6 \)
(c) \( 0.26 \times 10^7 \)
(d) \( 6.0 \times 10^5 \)
Answer: (b) \( 72.6 \times 10^6 \) and (c) \( 0.26 \times 10^7 \)
In simple words: Standard form means having only one non-zero number before the decimal point. Both option (b) and option (c) do not follow this rule.

Exam Tip: Always make sure the decimal part of standard form is greater than or equal to 1, but strictly less than 10.

Question Q2. \( 8^2 \) is same as:
(a) \( 2^8 \)
(b) \( (2^6)^2 \)
(c) \( (2^3)^2 \)
(d) \( 8 \times 2 \)
Answer: (c) \( (2^3)^2 \)
In simple words: Write 8 as \( 2^3 \). Thus, \( 8^2 \) is the same as \( (2^3)^2 \).

Exam Tip: Remember to write composite numbers in their prime base forms to compare powers easily.

Question Q3. \( 3^5 \div 3^5 \) is equal to :
(a) 0
(b) 1
(c) 3
(d) 4
Answer: (b) 1
In simple words: Dividing any non-zero number by itself always gives 1.

Exam Tip: When dividing identical bases, subtract the powers: \( 3^{5-5} = 3^0 = 1 \).

Question Q4. If an angle is \( 105^\circ \), what is its supplement?
(a) \( 65^\circ \)
(b) \( 75^\circ \)
(c) \( 45^\circ \)
(d) \( 95^\circ \)
Answer: (b) \( 75^\circ \)
In simple words: Two angles are supplementary if they add up to \( 180^\circ \). Subtract \( 105^\circ \) from \( 180^\circ \) to get \( 75^\circ \).

Exam Tip: Always check if the question asks for complementary (adds to \( 90^\circ \)) or supplementary (adds to \( 180^\circ \)) angles.

Question Q5. In the given figure, a and b are
(a) Alternate angles
(b) Corresponding angles
(c) Vertically opposite angles
(d) Adjacent angles
Answer: (b) Corresponding angles
In simple words: Both angles sit in the same relative position at each intersection, making them corresponding angles.

Exam Tip: Corresponding angles are equal when the two lines cut by the transversal are parallel.

Question Q6. If an angle is \( 27^\circ \), what is its complement?
(a) \( 34^\circ \)
(b) \( 63^\circ \)
(c) \( 75^\circ \)
(d) \( 46^\circ \)
Answer: (b) \( 63^\circ \)
In simple words: Complementary angles always add up to \( 90^\circ \). Subtract \( 27^\circ \) from \( 90^\circ \) to find the other angle, which is \( 63^\circ \).

Exam Tip: Keep complementary and supplementary definitions clear in your mind to avoid mixing up the math.

Section-B

Question Q7. Write the following in exponential form:
(a) \( \frac{81}{625} \)
(b) \( \frac{1}{9} \times \frac{1}{9} \times \frac{1}{9} \)
Answer:
(a) Change both numbers to prime power bases:
\( 81 = 3^4 \) and \( 625 = 5^4 \).

\( \implies \frac{81}{625} = \frac{3^4}{5^4} = \left(\frac{3}{5}\right)^4 \)

(b) Since \( \frac{1}{9} \) is multiplied three times:

\( \implies \left(\frac{1}{9}\right)^3 \)
In simple words: Write each group of numbers as a base raised to a power.

Exam Tip: When a fraction is raised to a power, make sure to use parentheses to show the power applies to both numerator and denominator.

Question Q8. Simplify and express in exponential form:
\( \left(\frac{-1}{2}\right)^2 \times \left(\frac{-3}{4}\right)^2 \)
Answer: Using the law of exponents where \( a^m \times b^m = (a \times b)^m \):
\( \left(\frac{-1}{2}\right)^2 \times \left(\frac{-3}{4}\right)^2 = \left[ \left(-\frac{1}{2}\right) \times \left(-\frac{3}{4}\right) \right]^2 \)

\( \implies \left(\frac{3}{8}\right)^2 \)
In simple words: Since the powers match, multiply the inner fractions first to get \( \left(\frac{3}{8}\right)^2 \).

Exam Tip: Multiplying two negative numbers inside parentheses always results in a positive value.

Question Q9. If an angle is 4 times its complement, find the angles.
Answer: Let the complementary angle be \( x \).
Then, the other angle must be \( 4x \).
Since they are complementary, they sum to \( 90^\circ \):
\( x + 4x = 90^\circ \)

\( \implies 5x = 90^\circ \)

\( \implies x = 18^\circ \)
The complementary angle is \( 18^\circ \). The other angle is:
\( 4x = 4 \times 18^\circ = 72^\circ \).
So, the angles are \( 72^\circ \) and \( 18^\circ \).
In simple words: One angle is \( 72^\circ \), which is exactly four times the other angle of \( 18^\circ \).

Exam Tip: Set up a simple equation with \( x \) to solve word problems systematically.

Question Q10. In the given figure, m || n. Find the unknown angles.
Answer: Given that lines \( m \) and \( n \) are parallel, and line \( p \) is the transversal:
1. The angle \( a \) and the angle \( 72^\circ \) lie on a straight line, forming a linear pair:
\( a + 72^\circ = 180^\circ \)

\( \implies a = 180^\circ - 72^\circ = 108^\circ \)
2. The angle \( b \) and the angle \( 72^\circ \) are vertically opposite angles:

\( \implies b = 72^\circ \)
3. The angle \( c \) and the angle \( 72^\circ \) are corresponding angles because \( m \parallel n \):

\( \implies c = 72^\circ \)
Thus, the angles are \( a = 108^\circ \), \( b = 72^\circ \), and \( c = 72^\circ \).
In simple words: Use straight line and parallel line rules to get \( a = 108^\circ \), \( b = 72^\circ \), and \( c = 72^\circ \).

Exam Tip: Clearly write down the geometric reason (like "linear pair" or "corresponding angles") next to each step of your calculation.

Section-C

Question Q11. Simplify: \( \frac{2^3 \times 3^4 \times 4}{3 \times 32} \)
Answer: First, write 4 and 32 as base-2 powers:
\( 4 = 2^2 \) and \( 32 = 2^5 \).
Now substitute these into the expression:
\( \frac{2^3 \times 3^4 \times 2^2}{3^1 \times 2^5} = \frac{2^{3+2} \times 3^4}{2^5 \times 3^1} \)

\( \implies \frac{2^5 \times 3^4}{2^5 \times 3^1} \)
Cancel the common terms:
\( \implies 3^{4-1} = 3^3 = 27 \)
In simple words: Break numbers down into powers of 2. Cancel matching terms on the top and bottom to get 27.

Exam Tip: Expressing numbers as prime base exponents helps simplify rational algebraic expressions quickly.

Question Q12. Write \( 0.756 \times 10^8 \) in its usual form.
Answer: To convert from scientific notation to usual decimal form, move the decimal point 8 places to the right:
\( 0.756 \times 10^8 = 75,600,000 \)
In simple words: Shifting the decimal point 8 places to the right gives 75,600,000.

Exam Tip: Count the number of places carefully when shifting the decimal. Add zeros as placeholders if you run out of digits.

Question Q13. The difference in the measure of two complementary angles is 12°. Find the measure of these angles.
Answer: Let the two complementary angles be \( x \) and \( y \).
Since they are complementary, their sum is \( 90^\circ \):
\( x + y = 90^\circ \)
The difference between the two angles is \( 12^\circ \):
\( x - y = 12^\circ \)
Adding these two equations:
\( 2x = 102^\circ \)

\( \implies x = 51^\circ \)
Now, find the other angle:
\( y = 90^\circ - 51^\circ = 39^\circ \)
So, the angles are \( 51^\circ \) and \( 39^\circ \).
In simple words: The two angles add up to \( 90^\circ \) with a gap of \( 12^\circ \) between them. They are \( 51^\circ \) and \( 39^\circ \).

Exam Tip: Add the two equations together to quickly eliminate one of the variables and solve for the other.

Question Q14. Two angles of a linear pair are in the ratio 1: 3, find the angles.
Answer: Let the two angles be \( 1x \) and \( 3x \).
Since they form a linear pair, they add up to \( 180^\circ \):
\( x + 3x = 180^\circ \)

\( \implies 4x = 180^\circ \)

\( \implies x = \frac{180^\circ}{4} = 45^\circ \)
Now, calculate each angle:
- First angle = \( x = 45^\circ \)
- Second angle = \( 3x = 3 \times 45^\circ = 135^\circ \)
Thus, the angles are \( 45^\circ \) and \( 135^\circ \).
In simple words: The two angles lie on a straight line and add up to \( 180^\circ \). Their values are \( 45^\circ \) and \( 135^\circ \).

Exam Tip: Linear pair angles are always supplementary, which means they always sum up to \( 180^\circ \).

Section-D

Question Q15. Find the value of m, if \( 5^m \times 5^4 = 5^9 \)
Answer: Combine the exponents on the left side of the equation:
\( 5^{m+4} = 5^9 \)
Since the bases are equal, we can compare the exponents:
\( m + 4 = 9 \)

\( \implies m = 9 - 4 = 5 \)
So, the value of \( m \) is 5.
In simple words: Add the powers when multiplying matching bases. This shows \( m + 4 = 9 \), so \( m \) must be 5.

Exam Tip: Be sure not to multiply the bases together when applying exponent rules; the base remains unchanged.

Question Q16. Find the values of x, y and z in the given figure.
Answer: From the geometric relationships shown:
1. The horizontal and vertical axes are perpendicular, forming a right angle (\( 90^\circ \)) in the top-right quadrant:
\( x + 40^\circ = 90^\circ \)

\( \implies x = 90^\circ - 40^\circ = 50^\circ \)
2. The intersecting diagonal and vertical lines form vertically opposite angles:

\( \implies y = 40^\circ \)
3. Similarly, the diagonal and horizontal lines form vertically opposite angles:

\( \implies z = x = 50^\circ \)
Thus, the values are \( x = 50^\circ \), \( y = 40^\circ \), and \( z = 50^\circ \).
In simple words: The right angle helps find \( x = 50^\circ \). Using vertically opposite angles, we find \( y = 40^\circ \) and \( z = 50^\circ \).

Exam Tip: Identify straight lines in intersection figures to use vertically opposite angle relationships directly.

Question Q17. Simplify: \( (3^{11} \times 3^7) \div (3^8 \times 3^{10}) \)
Answer: Simplify the terms inside each parenthesis first:
\( 3^{11} \times 3^7 = 3^{11+7} = 3^{18} \)
\( 3^8 \times 3^{10} = 3^{8+10} = 3^{18} \)
Now perform the division:
\( 3^{18} \div 3^{18} = 3^{18-18} = 3^0 = 1 \)
In simple words: Both parts inside the parentheses simplify to \( 3^{18} \). Any non-zero number divided by itself is 1.

Exam Tip: Remember that any base raised to the power of 0 always equals 1.

Question Q18. If AC || DE, ∠D = 140° and ∠CBE = 43°, find x and y.
Answer: Using parallel lines and triangle angle properties:
1. Since \( AC \parallel DE \), and line \( BE \) acts as a transversal, the angle \( x \) and \( \angle CBE \) are alternate interior angles:
\( \implies x = \angle CBE = 43^\circ \)
2. At vertex \( D \), the interior angle \( \angle BDE \) and the given exterior angle \( 140^\circ \) form a linear pair on the line \( DE \):
\( \angle BDE + 140^\circ = 180^\circ \)

\( \implies \angle BDE = 180^\circ - 140^\circ = 40^\circ \)
3. In triangle \( \triangle BDE \), the sum of all three interior angles is \( 180^\circ \):
\( \angle BDE + \angle DBE + \angle DEB = 180^\circ \)

\( \implies 40^\circ + y + x = 180^\circ \)
Substitute the value \( x = 43^\circ \):
\( 40^\circ + y + 43^\circ = 180^\circ \)
\( 83^\circ + y = 180^\circ \)

\( \implies y = 180^\circ - 83^\circ = 97^\circ \)
Thus, the values are \( x = 43^\circ \) and \( y = 97^\circ \).
In simple words: The alternate interior angle gives \( x = 43^\circ \). Then, find the third angle of the triangle to get \( y = 97^\circ \).

Exam Tip: Try to use multiple parallel line properties to double-check your angles and confirm your final values.