Question 1

Find out the units digit in the value of 2224 ÷ 432? [Hint: 4 = 22]

Accepted Answer

To find the units digit of 2224 ÷ 432, first note that 4 = 22. This implies 4 is 2 squared. The question likely means to find the units digit of 2 raised to the power 224 divided by 4 raised to the power 32, or possibly 2^224 divided by 4^32. Since 4 = 2^2, 4^32 = (2^2)^32 = 2^(64). So the expression is 2^224 ÷ 2^64 = 2^(224-64) = 2^160. The units digit of powers of 2 cycle every 4 powers: 2^1=2, 2^2=4, 2^3=8, 2^4=6, then repeats. 160 mod 4 = 0, so units digit is 6. — Step 1: Express 4 as 2^2.
Step 2: Rewrite 4^32 as (2^2)^32 = 2^(64).
Step 3: Divide 2^224 by 2^64 to get 2^(224-64) = 2^160.
Step 4: Units digit of powers of 2 cycle every 4: 2,4,8,6.
Step 5: Since 160 mod 4 = 0, units digit is 6.

Question 2

There are 5 bottles in a container. Every day, a new container is brought in. How many bottles would be there after 40 days?

Accepted Answer

Each container has 5 bottles. Every day, one new container is added. After 40 days, number of containers = 40. Total bottles = 40 × 5 = 200 bottles. — Step 1: Number of containers after 40 days = 40.
Step 2: Each container has 5 bottles.
Step 3: Total bottles = 40 × 5 = 200.

Question 3

Write the given number as the product of two or more powers in three different ways. The powers can be any integers.
(i) 643
(ii) 1928
(iii) 32–5

Accepted Answer

For each number, express as product of powers in three different ways:

(i) 643
- 64 = 8^2 = 4^3 = 2^6
- So possible products:
  1) 8^2
  2) 4^3
  3) 2^6

(ii) 1928
- 1928 = 19 × 28
- 19 is prime, 28 = 2^2 × 7
- Possible products:
  1) 19 × 28
  2) (19 × 2)^2 × 7^{-1} (not integer power, so avoid)
  3) 19^1 × (2^2 × 7)^1

(iii) 32^{-5}
- 32 = 2^5
- So 32^{-5} = (2^5)^{-5} = 2^{-25}
- Possible products:
  1) 2^{-25}
  2) (2^{-5})^{5}
  3) (2^{-1})^{25} — Step 1: Factorize each number into prime powers.
Step 2: Express as product of powers in different ways using exponent rules.
Step 3: For negative powers, use reciprocal powers.

Question 4

Examine each statement below and find out if it is ‘Always True’, ‘Only Sometimes True’, or ‘Never True’. Explain your reasoning.
(i) Cube numbers are also square numbers.
(ii) Fourth powers are also square numbers.
(iii) The fifth power of a number is divisible by the cube of that number.
(iv) The product of two cube numbers is a cube number.
(v) q46 is both a 4th power and a 6th power (q is a prime number).

Accepted Answer

(i) Cube numbers are also square numbers.
- Only sometimes true. For example, 64 = 4^3 = 8^2 is both cube and square, but 8 = 2^3 is not a square.

(ii) Fourth powers are also square numbers.
- Always true. Because (n^2)^2 = n^4, so fourth powers are squares of squares.

(iii) The fifth power of a number is divisible by the cube of that number.
- Always true. Since n^5 = n^3 × n^2, n^5 is divisible by n^3.

(iv) The product of two cube numbers is a cube number.
- Always true. Since a^3 × b^3 = (ab)^3.

(v) q^46 is both a 4th power and a 6th power (q is a prime number).
- Always true. Because q^{46} = (q^{23})^2, but 46 is divisible by 2 but not by 4 or 6. Actually, 46 is not divisible by 4 or 6, so q^{46} is not a perfect 4th or 6th power. So this is Never True. — Step 1: Analyze each statement using properties of exponents and perfect powers.
Step 2: Provide examples or counterexamples.
Step 3: Conclude truth value accordingly.

Question 5

Simplify and write these in the exponential form.
(i) 10– 2 × 10– 5
(ii) 57 ÷ 54
(iii) 9– 7 ÷ 94
(iv) (13– 2) – 3
(v) m5n12(mn)9

Accepted Answer

(i) 10^{-2} × 10^{-5} = 10^{-2 + (-5)} = 10^{-7}

(ii) 5^7 ÷ 5^4 = 5^{7-4} = 5^3

(iii) 9^{-7} ÷ 9^4 = 9^{-7 - 4} = 9^{-11}

(iv) (13^{-2})^{-3} = 13^{-2 × (-3)} = 13^{6}

(v) m^5 n^{12} (mn)^9 = m^5 n^{12} m^9 n^9 = m^{5+9} n^{12+9} = m^{14} n^{21} — Use exponent laws:
- a^m × a^n = a^{m+n}
- a^m ÷ a^n = a^{m-n}
- (a^m)^n = a^{m×n}
- (ab)^n = a^n b^n

Question 6

If 122 = 144 what is
(i) (1.2)2
(ii) (0.12)2
(iii) (0.012)2
(iv) 1202

Accepted Answer

(i) (1.2)^2 = (12/10)^2 = 144/100 = 1.44

(ii) (0.12)^2 = (12/100)^2 = 144/10000 = 0.0144

(iii) (0.012)^2 = (12/1000)^2 = 144/1,000,000 = 0.000144

(iv) 120^2 = 14400 — Use the fact that 12^2 = 144 and adjust decimal places accordingly.
For (iv), 120^2 = (12 × 10)^2 = 12^2 × 10^2 = 144 × 100 = 14400.

Question 7

Circle the numbers that are the same —
24 × 36
64 × 32
610
182 × 62
624

Accepted Answer

Calculate each:
- 24 × 36 = 2^4 × 3^6
- 64 × 32 = 2^6 × 2^5 = 2^{11}
- 6^{10} = (2 × 3)^{10} = 2^{10} × 3^{10}
- 18^2 × 6^2 = (2 × 3^2)^2 × (2 × 3)^2 = 2^2 × 3^4 × 2^2 × 3^2 = 2^{4} × 3^{6}
- 6^{24} = (2 × 3)^{24} = 2^{24} × 3^{24}

Comparing:
24 × 36 = 2^4 × 3^6
182 × 62 = 2^4 × 3^6
So 24 × 36 and 18^2 × 6^2 are the same.
Others differ. — Step 1: Express each number in prime factorization.
Step 2: Compare the powers of prime factors.
Step 3: Identify which are equal.

Question 8

Identify the greater number in each of the following —
(i) 43 or 34
(ii) 28 or 82
(iii) 1002 or 2100

Accepted Answer

(i) 4^3 = 64, 3^4 = 81, so 3^4 is greater.

(ii) 2^8 = 256, 8^2 = 64, so 2^8 is greater.

(iii) 10^{02} = 10^2 = 100, 2^{100} is a very large number, so 2^{100} is greater. — Calculate or estimate each power and compare values.

Question 9

A dairy plans to produce 8.5 billion packets of milk in a year. They want a unique ID (identifier) code for each packet. If they choose to use the digits 0–9, how many digits should the code consist of?

Accepted Answer

Number of packets = 8.5 billion = 8.5 × 10^9.
Number of unique codes with d digits using digits 0-9 = 10^d.
We want 10^d ≥ 8.5 × 10^9.
Taking log base 10:
d ≥ log(8.5 × 10^9) = log(8.5) + 9 ≈ 0.93 + 9 = 9.93.
So minimum digits = 10. — Step 1: Calculate total packets.
Step 2: Number of codes with d digits = 10^d.
Step 3: Find smallest d such that 10^d ≥ number of packets.
Step 4: Use logarithms to solve for d.

Question 10

64 is a square number (82) and a cube number (43). Are there other numbers that are both squares and cubes? Is there a way to describe such numbers in general?

Accepted Answer

Yes, numbers that are both perfect squares and perfect cubes are perfect sixth powers.
Because if a number is both a square and a cube, it can be written as n^{2} = m^{3} = k.
The number must be of the form a^{6} because:
- Square: (a^3)^2 = a^6
- Cube: (a^2)^3 = a^6
So such numbers are perfect sixth powers. — Step 1: Let the number be x.
Step 2: If x is a perfect square, x = a^2.
Step 3: If x is a perfect cube, x = b^3.
Step 4: Then a^2 = b^3 = x.
Step 5: So x must be a perfect sixth power, x = c^6.

Question 11

A digital locker has an alphanumeric (it can have both digits and letters) passcode of length 5. Some example codes are G89P0, 38098, BRJKW, and 003AZ. How many such codes are possible?

Accepted Answer

Alphanumeric characters include 26 letters + 10 digits = 36 characters.
Length of code = 5.
Number of possible codes = 36^5 = 60,466,176. — Step 1: Count total possible characters = 26 + 10 = 36.
Step 2: For each of 5 positions, 36 choices.
Step 3: Total codes = 36 × 36 × 36 × 36 × 36 = 36^5.

Question 12

The worldwide population of sheep (2024) is about 10^9, and that of goats is also about the same. What is the total population of sheep and goats?
(i) 2 × 10^9
(ii) 10^11
(iii) 10^10
(iv) 10^18
(v) 2 × 10^9
(vi) 10^9 + 10^9

Accepted Answer

The population of sheep = 10^9
The population of goats = 10^9
Total population = 10^9 + 10^9 = 2 × 10^9
So correct options: (i), (v), and (vi) represent the same value. — Step 1: Add populations: 10^9 + 10^9 = 2 × 10^9.
Step 2: Recognize that options (i), (v), and (vi) are equivalent.
Step 3: Other options are incorrect.

Question 13

Calculate and write the answer in scientific notation:
(i) If each person in the world had 30 pieces of clothing, find the total number of pieces of clothing.
(ii) There are about 100 million bee colonies in the world. Find the number of honeybees if each colony has about 50,000 bees.
(iii) The human body has about 38 trillion bacterial cells. Find the bacterial population residing in all humans in the world.
(iv) Total time spent eating in a lifetime in seconds.

Accepted Answer

(i) World population ≈ 8 × 10^9
Pieces of clothing = 30 × 8 × 10^9 = 2.4 × 10^{11}

(ii) Bee colonies = 100 million = 1 × 10^8
Bees per colony = 5 × 10^4
Total bees = 1 × 10^8 × 5 × 10^4 = 5 × 10^{12}

(iii) Bacterial cells per human = 38 trillion = 3.8 × 10^{13}
World population = 8 × 10^9
Total bacteria = 3.8 × 10^{13} × 8 × 10^9 = 3.04 × 10^{23}

(iv) Average lifetime = 70 years ≈ 70 × 365 × 24 × 3600 seconds = 2.2 × 10^9 seconds
Assuming 1/3 of time spent eating: (1/3) × 2.2 × 10^9 ≈ 7.3 × 10^8 seconds — Use scientific notation and multiplication rules:
- Multiply coefficients and add exponents.
- Convert units to seconds for time calculation.

Question 14

What was the date 1 arab/1 billion seconds ago? Try This

Accepted Answer

1 arab = 1 billion = 10^9 seconds.
Convert seconds to days:
1 day = 24 × 3600 = 86400 seconds
Days = 10^9 ÷ 86400 ≈ 11574 days
Convert days to years:
Years = 11574 ÷ 365 ≈ 31.7 years
If today is 2024, then 1 billion seconds ago was approximately in 1992 or early 1993. — Step 1: Convert seconds to days by dividing by 86400.
Step 2: Convert days to years by dividing by 365.
Step 3: Subtract years from current year to find approximate date.

Power Play

Power Play — Study Notes

Introduction

Laws of Exponents

Negative Exponents and Zero Exponent

Practice Questions — Power Play

All 7 Chapters in Ganita Prakash Part-I