Class 8 Maths Ganita Prakash Part 1 Chapter 1: A Square and a Cube MCQs with Answers

A locker changes state once for each factor of its number.
It remains open when the number of factors is odd. Only perfect squares have an odd number of factors.
The squares from 1 to 100 are 1², 2², …, 10².
Therefore, 10 lockers remain open.

The factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, and 36.
There are 9 factors, so locker 36 changes state 9 times.
The factor 6 is its own partner because 6 × 6 = 36, giving an odd factor count.

A positive integer has exactly two factors when it is prime. Those factors are 1 and the number itself.
The first five primes are 2, 3, 5, 7, and 11.
Therefore, these are the first five locker numbers that change state exactly twice.

Area of a square = Side × Side.
Area = (3/5) × (3/5)
= (3 × 3)/(5 × 5)
= 9/25 square unit.
Both the numerator and denominator are multiplied by themselves when the fraction is squared.

A perfect square can end only in 0, 1, 4, 5, 6, or 9.
The units digit of 2032 is 2, so it cannot be a perfect square.
The other numbers are squares: 1089 = 33², 576 = 24², and 1156 = 34².

The numbers 16 and 36 end in 6 and are perfect squares: 16 = 4² and 36 = 6².
However, 25 < 26 < 36, so 26 is not a perfect square. Therefore, a units digit of 6 does not determine whether the number is a square.

The units digit of a square depends on the units digit of the original number.
108 ends in 8, and 8² = 64, so 108² ends in 4.
The other numbers end in 4 or 6, whose squares both end in 6.

Let the integer be 1000m, where m does not end in zero.
Its square is (1000m)² = 10,00,000m².
Since m² does not end in zero, the six zeros supplied by 1000² are the only trailing zeros.
Therefore, the square ends in exactly six zeros.

The difference between consecutive squares is the next odd number.
The 36th odd number = 2 × 36 − 1 = 71.
36² = 35² + 71
= 1225 + 71 = 1296.
Therefore, the square of 36 is 1296.

16² = 256 and 17² = 289.
The required whole numbers run from 257 to 288.
Their number = 289 − 256 − 1 = 32.
The subtraction of 1 excludes the two endpoints from the count between the squares.

The pattern adds two consecutive triangular numbers.
After 6 and 10, the next pair is 10 and 15.
10 + 15 = 25 = 5².
Therefore, the next square is obtained by combining the triangular arrangements containing 10 dots and 15 dots.

324 = 2 × 2 × 3 × 3 × 3 × 3
= (2 × 3 × 3) × (2 × 3 × 3)
= 18².
Therefore, √324 = 18. The two identical groups of prime factors each have product 18.

A perfect square has prime factors that can all be arranged in identical pairs.
156 = (2 × 2) × 3 × 13.
The two factors 2 form a pair, but 3 and 13 are left unpaired.
Therefore, 156 is not a perfect square.

40² = 1600 and 50² = 2500, so √1936 lies between 40 and 50.
A square ending in 6 can have a positive integer root ending in 4 or 6.
44² = (40 + 4)² = 1600 + 320 + 16 = 1936.
Therefore, √1936 = 44.

15² = 225 and 16² = 256.
Since 225 < 250 < 256, the positive square root satisfies 15 < √250 < 16. Therefore, √250 lies between 15 and 16. It is not equal to 16 because 16² is greater than 250.

The cloth’s side length is √125 cm.
11² = 121 and 12² = 144.
Since 121 < 125 < 144, the cloth's side lies between 11 cm and 12 cm. Therefore, an 11 cm square fits, but a 12 cm square does not.

Divisibility by 4, 9, and 10 requires the factors 2², 3², and 5.
In a perfect square, every prime factor must occur in a pair.
The smallest suitable product is 2² × 3² × 5² = 900.
Also, 900 = 30² and is divisible by all three numbers.

9408 = 2⁶ × 3 × 7².
The factors 2 and 7 already occur in pairs, while the factor 3 is unpaired.
Multiplying by 3 gives 2⁶ × 3² × 7² = (2³ × 3 × 7)² = 168².
Therefore, the smallest multiplier is 3.

Each layer contains 4 × 4 = 16 unit cubes.
There are 4 such layers along the height.
Total unit cubes = 16 × 4 = 64.
Thus, 4³ = 4 × 4 × 4 = 64 unit cubes fill the larger cube.

If an integer ends in k zeros, it can be written as m × 10ᵏ, where m does not end in zero.
Its cube is m³ × 10³ᵏ.
Therefore, its number of trailing zeros is 3k, which is a multiple of 3.
Exactly two trailing zeros are impossible.

(−6)³ = (−6) × (−6) × (−6).
The product of the first two negative factors is 36.
Then 36 × (−6) = −216.
Therefore, (−6)³ = −216. A product of three negative factors is negative.

4/6 simplifies to 2/3.
(4/6)³ = (2/3) × (2/3) × (2/3)
= 8/27.
The numerator and denominator each appear three times in the product. Since 8 and 27 have no common factor greater than 1, this is the simplest form.

1³ + 12³ = 1 + 1728 = 1729.
9³ + 10³ = 729 + 1000 = 1729.
Thus, the two different pairs of positive cubes give the same sum.
This is the property of 1729 highlighted by the Hardy–Ramanujan number.

The ten numbers can be paired from the two ends.
91 + 109 = 93 + 107 = 95 + 105 = 97 + 103 = 99 + 101 = 200.
There are 5 pairs, so the sum is 5 × 200 = 1000 = 10³.

375 = 3 × 3 × 3 × 5 × 5 × 5
= (3 × 5) × (3 × 5) × (3 × 5)
= 15³.
Therefore, ∛3375 = 15. Each of the three identical groups of prime factors has product 15.

For a perfect cube, all prime factors can be arranged in triplets of identical factors.
500 = (2 × 2) × (5 × 5 × 5).
The factors 5 form a triplet, but only two factors 2 are present.
Therefore, 500 is not a perfect cube.

1323 = 3 × 3 × 3 × 7 × 7.
The factors 3 form a triplet, while the factors 7 need one more 7.
1323 × 7 = (3 × 7)³ = 21³ = 9261.
Therefore, the smallest multiplier is 7.

10648 = 2³ × 11³
= (2 × 11)³
= 22³.
Also, 22 × 22 × 22 = 484 × 22 = 10648.
Therefore, ∛10648 = 22, since multiplying 22 by itself three times gives the original number.

First differences: 7, 19, 37, 61, 91.
Second differences: 12, 18, 24, 30.
Third differences: 6, 6, 6.
Therefore, the third level of differences is constant at 6. Each level is obtained by subtracting each entry from the next entry in the previous level.

With 10 in the blank, the adjacent sums are:
3 + 6 = 9 = 3²
6 + 10 = 16 = 4²
10 + 15 = 25 = 5²
15 + 1 = 16 = 4².
Therefore, 10 makes every adjacent pair add to a perfect square.

In the locker puzzle, a locker remains open if it is toggled an odd number of times. The number of times a locker is toggled corresponds to the number of its factors.
Only square numbers have an odd number of factors because they have one factor that is multiplied by itself to equal the number, unlike other numbers where factors come in pairs.

The will states, “If all 100 of you answer it at the same time, you will share the ratnas equally. However, if you are the first one to solve the problem, you will get to keep the entire inheritance to yourself.
This clearly indicates that being the first to solve it grants the entire inheritance.

The minister explains that if a locker is toggled an odd number of times, it will be open.
Conversely, if it is toggled an even number of times, it will be closed. This directly dictates the final state of each locker.

“The number of times a locker is toggled is the same as the number of factors of the locker number”.
This is illustrated with locker 6, where its factors (1, 2, 3 and 6) correspond to the people who toggle it. 

After determining which lockers remain open, Khoisnam collects word clues and reads, “The passcode consists of the first five locker numbers that were touched exactly twice”.
The chapter then clarifies that lockers toggled twice are prime numbers.

A perfect square is a number that can be expressed as the product of a natural number with itself.
From the given examples, 16 is 4 multiplied by 4 (4²), making it a perfect square. The other options are not perfect squares.

“In general, for any number n, we write n x n = n², which is read as ‘n squared'”.
This notation is used consistently throughout the section on square numbers.

A pattern of perfect squares: “All these numbers end with 0, 1, 4, 5, 6 or 9. None of them end with 2, 3, 7 or 8”.
Therefore, if a number ends with 2, 3, 7 or 8, it is definitely not a square.

When discussing numbers with trailing zeros, the chapter highlights: “Can we say that squares can only have an even number of zeros at the end?”
The examples like 10² = 100 (two zeros) and 100² = 10000 (four zeros) support this observation.

The chapter demonstrates this pattern:
“1 = 1”, “1 + 3 = 4”, “1 + 3 + 5 = 9”, and so on, concluding, “We see that the sum of the first n odd numbers is n²”.
This is a fundamental property discussed.

“Also, we can find out whether a number is a perfect square by successively subtracting odd numbers”.
An example with 25 shows that when zero is obtained, it is a perfect square and the number of steps indicates its square root.

To find the 36ᵗʰ odd number, the question asks, “What is the nᵗʰ odd number?” and then provides the answer: “The nᵗʰ odd number is 2n – 1”.
This formula is used directly in the subsequent calculation.

The definition of a square root is provided: “We know that 7×7 = 49 or 7² = 49”.
It then explicitly states, “We call 7 the square root of 49”.

“Every perfect square has two integer square roots. One is positive and the other is negative”.
This is illustrated with examples like the square roots of 64 being +8 and -8.

It is clearly introduces the notation: “The square root of a number is denoted by √”.
An example is then given:
“Thus, √64 = ±8 and √100 = ±10”.

The story of Srinivasa Ramanujan and G. H. Hardy defines 1729 as “the smallest number that can be expressed as the sum of two cubes in two different ways”.
“And numbers that can be expressed as the sum of two cubes in two different ways are called taxicab numbers”. 

The chapter illustrates with 3375:
“3375 = 3 × 3 × 3 × 5 × 5 × 5”.
It then explains, “Can the factors be split into three identical groups? For 3375, we can form three groups of (3 × 5)”.
This leads to it being a perfect cube.

It is defines cube roots and provides examples: “Similarly, ∛27 = ∛(3³) = 3”.
This directly answers the question about the cube root of 27.

The “A Pinch of History” section states, “In ancient Sanskrit works the term varga was used both for the square figure or its area, as well as the square power..”.
This directly links ‘varga’ to the square power.

The historical section explains, “It is because, in ancient India, the Sanskrit word mula, meaning root of a plant, basis, cause, origin, etc., was used for the mathematical operations of taking roots”.
This highlights the origin of the term ‘root’.