Class 8 Maths Ganita Prakash Part 1 Chapter 2: Power Play MCQs with Answers

Each fold multiplies the thickness by 2, so 10 folds give a factor of 2¹⁰.
Thickness after 10 folds = 0.001 × 2¹⁰ cm
= 0.001 × 1024 cm
= 1.024 cm.

Let the thickness before the 3 additional folds be t.
After 1 fold, thickness = 2t.
After 2 folds, thickness = 4t.
After 3 folds, thickness = 8t.
Therefore, the thickness becomes 8 times its earlier value.

5⁴ = 5 × 5 × 5 × 5.
The repeated factor is 5, so the base is 5.
There are 4 factors of 5, so the exponent is 4.
The value of the expression is 625.

A pair of factors (−1) × (−1) has product 1.
In (−1)⁵, two pairs leave one factor −1, so (−1)⁵ = −1.
In (−1)⁵⁶, all 56 factors form 28 pairs, so (−1)⁵⁶ = 1.

648 = 8 × 81.
8 = 2 × 2 × 2 = 2³.
81 = 3 × 3 × 3 × 3 = 3⁴.
Therefore, 648 = 2³ × 3⁴. Both 2 and 3 are prime numbers.

Number of baskets = 3 × 3 = 3².
Number of keys = 3³; number of rooms = 3⁴.
Number of tables = 3⁵; number of necklaces = 3⁶.
Number of diamonds = 3⁷ = 2187.
Each new level multiplies the previous count by 3.

p⁴ contains 4 factors of p, and p⁶ contains 6 factors of p.
Their product contains 4 + 6 = 10 factors of p.
p⁴ × p⁶ = p⁴⁺⁶ = p¹⁰.
The exponents add because the bases are the same.

(4³)² = 4³ × 4³.
There are two groups, each containing 3 factors of 4.
Therefore, the total number of factors of 4 is 3 × 2 = 6.
(4³)² = 4⁶ = 4096.

The numerator and denominator have the same exponent.
10⁴ ÷ 5⁴ = (10 ÷ 5)⁴
= 2⁴
= 2 × 2 × 2 × 2
= 16.
Each of the four factors of 10 is divided by a factor of 5.

Covered area on day 30 = 2 × covered area on day 29.
Covered area on day 29 = full area ÷ 2.
Therefore, the pond was half covered one day before it became fully covered, which was day 29.

After the first 4 days, number of lotuses = 1 × 2⁴ = 16.
During the next 4 days, this number is multiplied by 3⁴ = 81.
Final number = 2⁴ × 3⁴ = (2 × 3)⁴
= 6⁴ = 1296.

For each dress, there are 2 choices of hat and 3 choices of shoes.
Outfits with one dress = 2 × 3 = 6.
There are 7 dresses.
Total number of outfits = 7 × 6 = 42.

There are 10 choices for each position, including the first position.
The choice in one position does not restrict any other position because repetitions are allowed.
Number of passwords = 10 × 10 × 10 × 10 × 10
= 10⁵ = 1,00,000.

The numerator and denominator have the same non-zero base.
Dividing cancels 25 of the 100 factors of 2 in the numerator.
2¹⁰⁰ ÷ 2²⁵ = 2¹⁰⁰⁻²⁵
= 2⁷⁵.
Therefore, 75 factors of 2 remain.

Since x ≠ 0, the quotient x³ ÷ x³ is defined and equals 1.
By the exponent rule, x³ ÷ x³ = x³⁻³ = x⁰.
Therefore, x⁰ = 1 for every non-zero x.

A negative exponent gives the reciprocal of the corresponding positive power.
(−5)⁻³ = 1 ÷ (−5)³.
(−5)³ = (−5) × (−5) × (−5) = −125.
Therefore, (−5)⁻³ = 1 ÷ (−125) = −1/125.

The required ratio is 4² ÷ 4⁻².
4² ÷ 4⁻² = 4²⁻⁽⁻²⁾ = 4⁴ = 256.
Also, 4² = 16 and 4⁻² = 1/16.
Thus, 16 ÷ (1/16) = 256.

The bases are equal, so the exponents add.
2⁻⁴ × 2⁷ = 2⁻⁴⁺⁷ = 2³.
This also follows from 2⁻⁴ = 1/16 and 2⁷ = 128.
Their product is 128/16 = 8 = 2³.

In 561.903, the digit 3 is in the thousandths place.
Its value is 3/1000 = 0.003.
Since 1/1000 = 10⁻³, the required term is 3 × 10⁻³.
The digit 9 is in the tenths place, and 0 is in the hundredths place.

Scientific notation has the form a × 10ⁿ, where 1 ≤ a < 10. 5900 = 5.9 × 1000 = 5.9 × 10³. The coefficient 5.9 meets this condition. The other expressions equal 5900, but their coefficients do not meet the condition.

All four coefficients are at least 1 and less than 10.
1.496 × 10¹¹ = 0.1496 × 10¹².
Now all four numbers have the common factor 10¹².
Since 0.1496 is the smallest coefficient, 1.496 × 10¹¹ is the smallest number.

Each day adds 5 bottles to the total. The number of bottles does not multiply by 5 each day.
Total bottles received = number of days × bottles per day
= 40 × 5
= 200.
This is additive, or linear, growth.

1 km = 1,00,000 cm = 10⁵ cm.
Total distance = 3,84,400 × 10⁵ cm = 3.844 × 10¹⁰ cm.
Number of upward steps = total distance ÷ distance per step
= (3.844 × 10¹⁰) ÷ 20
= 1.922 × 10⁹.

Average ants per human = estimated number of ants ÷ estimated number of humans.
= (2 × 10¹⁶) ÷ (8 × 10⁹)
= (2/8) × 10¹⁶⁻⁹
= 0.25 × 10⁷
= 2.5 × 10⁶.
This is an approximate ratio based on the stated estimates.

1 day = 24 × 60 × 60 seconds = 86,400 seconds.
One million seconds = 10⁶ seconds.
Number of days = 10⁶ ÷ 86,400
= 11.574074…
Therefore, one million seconds is approximately 11.57 days, to the nearest hundredth.

One billion = 10⁹.
A thousand billion = 1000 × 10⁹
= 10³ × 10⁹
= 10³⁺⁹
= 10¹².
Therefore, one trillion is 10¹², or 1,000,000,000,000.

4³² = (2²)³² = 2⁶⁴.
2²²⁴ ÷ 4³² = 2²²⁴⁻⁶⁴ = 2¹⁶⁰.
The units digits of positive powers of 2 repeat as 2, 4, 8, 6.
Since 160 is divisible by 4, the units digit is 6.

The power of a product gives (mn)⁹ = m⁹n⁹.
Therefore, m⁵n¹²(mn)⁹ = m⁵n¹²m⁹n⁹
= m⁵⁺⁹n¹²⁺⁹
= m¹⁴n²¹.
The exponents are added separately for the factors with base m and those with base n.

A code with n positions has 10ⁿ possible forms.
Required number of codes = 8.5 × 10⁹.
With 9 positions, 10⁹ codes are possible, which is too few.
With 10 positions, 10¹⁰ codes are possible, which is enough.
Therefore, the minimum is 10 positions.

n⁶ = (n³)², so it is the square of the integer n³.
Also, n⁶ = (n²)³, so it is the cube of the integer n².
Therefore, n⁶ is always both a perfect square and a perfect cube.

The provided tables show the progressive thickness. After 26 folds, the thickness is approximately 670 m. Further down, it states, “After 30 folds, the thickness of the paper is about 10.7 km, the typical height at which planes fly”.

The chapter highlights the incredible growth: “It might be hard to digest the fact that after just 46 folds, the thickness is more than 7,00,000 km”.
This emphasizes the power of multiplicative growth discussed in the section.

“This is the power of multiplicative growth, also called exponential growth”.
This term is introduced to explain the rapid increase in thickness with each fold, where the quantity doubles.

The chapter defines exponential notation: “5⁴ is the exponential form of 625. Here, 4 is the exponent/power, and 5 is the base”.
This clarifies the role of each component in the exponential expression.

The section on exponential notation explains this directly: “n × n × n × n × n × n × n = n⁷ (read as ‘n raised to the power 7’ or ‘the 7th power of n’)”.
This is a core definition presented.

The chapter states that initial thickness ‘v’ becomes v × 2¹ after one fold, v × 2² after two folds, and so on.
Therefore, after 10 folds, the thickness will be v × 2¹⁰ or 2¹⁰v.

“(−4) × (−4) × (−4) = (−4)³ = −64”.
This illustrates the multiplication of a negative base an odd number of times, resulting in a negative product.

The riddle describes a hierarchy of three daughters, three baskets each, three keys per basket and three rooms per key.
The diagram and calculation confirm that the number of rooms is 3 × 3 × 3 × 3 = 3⁴.

After illustrating with examples like 3⁴ × 3³ = 3⁷ and p⁴ × p⁶ = p¹⁰, the chapter states the general rule: “nᵃ × nᵇ = n⁽ᵃ⁺ᵇ⁾ where a and b are counting numbers”.

“In general, (nᵃ)ᵇ = (nᵇ)ᵃ = n⁽ᵃˣᵇ⁾ = nᵃᵇ, where a and b are counting numbers”.
This rule applies to expressions with a base raised to one exponent, then that entire quantity raised to another.

The riddle states, “Since the number of lotuses doubles every day, the pond should be half covered on the 29th day”.
This is a classic example of working backward in an exponential growth scenario.

After an example like 2⁴ × 3⁴ = 6⁴, the chapter generalizes: “In general form, mᵃ × nᵃ = (mn)ᵃ, where a is a counting number”.
This demonstrates the combining of bases.

Its provides an example 10⁴/5⁴ and then states, “In general, we can show that mᵃ/nᵃ = (m/n)ᵃ”.
This is presented as a general rule for division of powers with the same exponent.

According to the chapter, “There are 10 options for the first digit (0 to 9). For each of these, there are 10 options for the second digit…Therefore the total number of combinations for a 2-digit lock is 10 × 10 = 100”.

Through examples of halving a line, the rule is derived: “2⁴ ÷ 2³ = 2⁽⁴⁻³⁾ = 2¹”.
It is then generalized as: “nᵃ ÷ nᵇ = n⁽ᵃ⁻ᵇ⁾, where n ≠ 0 and a and b are counting numbers and a > b”.

The chapter defines this by using the division rule:
“2⁰ = 2⁽⁴⁻⁴⁾ = 2⁴ ÷ 2⁴ = 1”.
It then states, “In general, 1 = x⁰ where x ≠ 0”.

It is demonstrates with examples like 2⁻¹ = 1/2 and 10⁻³ = 1/10³.
It generalizes this as: “n⁻ᵃ = 1/nᵃ and nᵃ = 1/n⁻ᵃ, where n ≠ 0”.

The chapter defines scientific notation: “In scientific notation or scientific form (also called standard form), we write numbers as x × 10ʸ, where x ≥ 1 and x < 10 is the coefficient and y, the exponent, is any integer".

In the section “A Pinch of History”, a table shows the modern naming conventions.
It states, “A hundred lakhs is a crore” and provides the exponential form as 10⁷.

The chapter introduces specific large number names: “The number 10¹⁰⁰ is also called a googol”.
It also mentions googolplex as 10ᵍᵒᵒᵍᵒˡ, indicating its immense scale.