CBSE Class 7 Mathematics Exponents And Powers Worksheet Set 02

EXPONATNTS (STANDARD FORMS)

1. Express each of the following numbers in standard form:

(i) 538
Solution:- A given number is said to be in standard form if it can be expressed as k × 10ⁿ, where k is a real number such that 1 ≤ k < 10and n is a positive integer.
Then, 538 = 5.38 × 10²

(ii) 6428000
Solution:- A given number is said to be in standard form if it can be expressed as k × 10ⁿ, where k is a real number such that 1 ≤ k < 10and n is a positive integer.
Then, 6428000 = 6.428 × 10⁶

(iii) 82934000000
Solution:- A given number is said to be in standard form if it can be expressed as k × 10ⁿ, where k is a real number such that 1 ≤ k < 10and n is a positive integer.
Then, 82934000000 = 8.2934 × 10¹⁰

(iv) 940000000000
Solution:- A given number is said to be in standard form if it can be expressed as k × 10ⁿ, where k is a real number such that 1 ≤ k < 10and n is a positive integer.
Then, 940000000000 = 9.4 × 10¹¹

2. Express each of the following numbers in standard form:

(i) Diameter of Earth = 12756000 m
Solution:- A given number is said to be in standard form if it can be expressed as k × 10ⁿ, where k is a real number such that 1 ≤ k < 10and n is a positive integer.
Then,
Diameter of Earth = 12756000 m = (1.2156 × 10⁷) m (in standard form)

(ii) Distance between Earth and Moon = 384000000 m
Solution:- A given number is said to be in standard form if it can be expressed as k × 10ⁿ, where k is a real number such that 1 ≤ k < 10and n is a positive integer.
Then,
Distance between Earth and Moon = 384000000 m = (3.84 × 108) m (in standard form)

(iii) Population of India in March 2001= 1027000000
Solution:- A given number is said to be in standard form if it can be expressed as k × 10ⁿ, where k is a real number such that 1 ≤ k < 10and n is a positive integer.
Then,
Population of India in March 2001= 1027000000 = (1.027 × 109) (in standard form)

(iv) Number of stars in a galaxy = 100000000000
Solution:- A given number is said to be in standard form if it can be expressed as k × 10ⁿ, where k is a real number such that 1 ≤ k < 10and n is a positive integer.
Then,
Number of stars in a galaxy = 100000000000 = (1 × 10¹¹) (in standard form)

(v) The present age of universe = 12000000000 years
Solution:- A given number is said to be in standard form if it can be expressed as k × 10ⁿ, where k is a real number such that 1 ≤ k < 10and n is a positive integer.
Then,
The present age of universe = 12000000000 = (1.2 × 10¹⁰) years (in standard form)

More Question

1. Find the value of the following:
(a) (-1/4)³
(b) (-2/7)²
(c) 34 X (-1)173

2 Simplify using laws of exponents:
(a) 53x X 25
(b) 25 X 34 X 5
(c) [( 52)3 X 54 ] ÷ 55 5x 3 X 16
(d) 20 X 5 + (-3)0 X 90 X 40

3. Express the following in standard form:
(a) 70,30,000
(b) 750000000000
(c) 8765.43

4. Convert into power notation:
(a) -1/343
(b) -1/81
(c) 512/1331

5. The age of the Earth is 4,600,000,000 years. Write the number in standard form.

Question 1. Find the value of the following:
(a) \( (-1/4)^3 \)
(b) \( (-2/7)^2 \)
(c) \( 3^4 \times (-1)^{173} \)
Answer:
Let us solve each part step-by-step:
(a) Multiply the fraction by itself three times:
\( \left(-\frac{1}{4}\right)^3 = \left(-\frac{1}{4}\right) \times \left(-\frac{1}{4}\right) \times \left(-\frac{1}{4}\right) = -\frac{1}{64} \)

(b) Multiply the fraction by itself two times:
\( \left(-\frac{2}{7}\right)^2 = \left(-\frac{2}{7}\right) \times \left(-\frac{2}{7}\right) = \frac{4}{49} \)

(c) Calculate the values of both terms and multiply:
\( 3^4 = 3 \times 3 \times 3 \times 3 = 81 \)
Since 173 is an odd number, any negative number raised to an odd power stays negative:
\( (-1)^{173} = -1 \)
\( 81 \times (-1) = -81 \)
In simple words: Multiply the numbers by themselves as many times as the exponent power shows. Remember that negative numbers become positive when multiplied an even number of times.

Exam Tip: Be careful with negative bases. An even power always gives a positive answer, while an odd power always gives a negative answer.

Question 2. Simplify using laws of exponents:
(a) \( \frac{5^{3x} \times 25}{5^x} \)
(b) \( \frac{2^5 \times 3^4 \times 5}{3 \times 16} \)
(c) \( [(5^2)^3 \times 5^4] \div 5^5 \)
(d) \( 2^0 \times 5 + (-3)^0 \times 9^0 \times 4^0 \)
Answer:
Let us use the laws of exponents to solve each part:
(a) Express 25 as a power of 5:
\( 25 = 5^2 \)
\( \frac{5^{3x} \times 5^2}{5^x} = \frac{5^{3x + 2}}{5^x} \)
Subtract the bottom power from the top power:
\( 5^{3x + 2 - x} = 5^{2x + 2} \)

(b) Express 16 as a power of 2:
\( 16 = 2^4 \)
\( \frac{2^5 \times 3^4 \times 5}{3^1 \times 2^4} \)
Subtract matching base powers:
\( 2^{5 - 4} \times 3^{4 - 1} \times 5 = 2^1 \times 3^3 \times 5 = 2 \times 27 \times 5 = 270 \)

(c) Solve the bracket first using power rules:
\( (5^2)^3 = 5^{2 \times 3} = 5^6 \)
\( [5^6 \times 5^4] \div 5^5 = 5^{6 + 4} \div 5^5 = 5^{10} \div 5^5 \)
Subtract powers when dividing:
\( 5^{10 - 5} = 5^5 \) (which equals 3125)

(d) Any non-zero number raised to the power of 0 is equal to 1:
\( 2^0 \times 5 + (-3)^0 \times 9^0 \times 4^0 = 1 \times 5 + 1 \times 1 \times 1 \)
\( = 5 + 1 = 6 \)
In simple words: Add powers when multiplying, and subtract them when dividing with the same base. Also, any number with a power of 0 is always 1.

Exam Tip: Always convert larger composite numbers (like 16 or 25) into powers of prime numbers first to make simplification easier.

Question 3. Expres he following in standard form:
(a) 70,30,000
(b) 750000000000
(c) 8765.43
Answer:
Standard form means writing a number as a value between 1 and 10 multiplied by a power of 10:
(a) Move the decimal point 6 places to the left:
\( 70,30,000 = 7.03 \times 10^6 \)

(b) Move the decimal point 11 places to the left:
\( 750000000000 = 7.5 \times 10^{11} \)

(c) Move the decimal point 3 places to the left:
\( 8765.43 = 8.76543 \times 10^3 \)
In simple words: Shift the decimal point so there is only one number before it, then count the jumps to find the power of 10.

Exam Tip: Be sure to count the jumps carefully from right to left to get the correct power of 10.

Question 4. Convert into power notation:
(a) \( -1/343 \)
(b) \( -1/81 \)
(c) \( 512/1331 \)
Answer:
We write the numerator and denominator as powers of the same exponent:
(a) We know that \( 343 = 7^3 \):
\( -\frac{1}{343} = -\frac{1}{7^3} = \left(-\frac{1}{7}\right)^3 \)

(b) We know that \( 81 = 3^4 \):
\( -\frac{1}{81} = -\frac{1}{3^4} = -\left(\frac{1}{3}\right)^4 \)

(c) We know that \( 512 = 8^3 \) and \( 1331 = 11^3 \):
\( \frac{512}{1331} = \frac{8^3}{11^3} = \left(\frac{8}{11}\right)^3 \)
In simple words: Find which numbers multiplied by themselves make the top and bottom values, then write them with powers.

Exam Tip: Memorize perfect cubes up to 10 (like \( 7^3 = 343 \)) to quickly solve power notation problems on tests.

Question 5. The age of the Earth is 4,600,000,000 years. Write the number in standard form.
Answer:
To write the age of the Earth in standard form, we shift the decimal point 9 places to the left:
\( 4,600,000,000 = 4.6 \times 10^9 \text{ years} \)
In simple words: Shifting the decimal point 9 places to the left gives us the standard value of \( 4.6 \times 10^9 \).

Exam Tip: Never forget to write the unit ("years") after your calculated standard form number to secure full marks.