RS Aggarwal Test: Real Numbers - 2 - Class 10 MCQ

5²ⁿ −2²ⁿ is of the form a²ⁿ − b²ⁿ which is divisible by (a – b).

25ⁿ − 4ⁿ ⇒ aⁿ − bⁿ
is always divisible by (a−b)
(25 − 4) = 21
∴ So factors are both 7 and 3.

Old NCERT

LCM of 50 and 48 = 1200
∴ 1200 sec = 20 min
Hence at 12.20 pm they will beep again for the first time.

Step 1: Find the highest common factor (HCF) of 576 and 448 by prime factorization.

576 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3

448 = 2 × 2 × 2 × 2 × 2 × 2 × 7

Common prime factors: six factors of 2 ⇒ HCF = 2 × 2 × 2 × 2 × 2 × 2 = 64

Step 2: Number of sections of boys = 576 divided by 64 = 9

Step 3: Number of sections of girls = 448 divided by 64 = 7

Step 4: Total sections = 9 + 7 = 16

Hence, the correct answer is 16.

Given that, LCM (2472, 1284, and n) is 2³ × 3² × 5 × 103 × 107

Let us express the numbers 2472, and 1284 as a product of prime numbers.

2472 = 2 × 2 × 2 × 3 × 103  (2³ × 3 × 103)

1284 = 2 × 2 × 3 × 107 (2² × 3 × 107)

HCF (2472, 1284)=  2 × 2 × 3 (2² × 3)

'n' should also have one of the factors as HCF, and another factor as the missing element of other numbers from the LCM (i.e., 3 × 5)

Therefore,

n =  2² × 3 × (3 × 5)

n = 2² × 3² × 5

n = 4 × 9 × 5

n = 180

Therefore, if the HCF of 2472 and 1284 and a third number 'n' is 12 and if their LCM is 2³ × 3² × 5 × 103 × 107, then the number 'n' is 180 (2² × 3² × 5)

The product of a non-zero rational number and an irrational number is always irrational.

Explanation:

• A rational number is a number that can be expressed as a fraction a/b ​, where a and b are integers and b≠0
• An irrational number cannot be expressed as a simple fraction and has a non-repeating, non-terminating decimal expansion.

When a non-zero rational number is multiplied by an irrational number, the product cannot be expressed as a simple fraction, and the non-terminating, non-repeating nature of the irrational number is preserved in the product, making the result irrational.

Answer: 1 (Always irrational)

To find the HCF (Highest Common Factor) of two numbers, knowing their LCM (Lowest Common Multiple), use the formula:

• The product of the HCF and LCM of two numbers equals the product of the numbers themselves.

For 77 and 99:

• Given: LCM of 77 and 99 is 693.
• Calculate the product: 77 × 99 = 7623.

Use the formula:

• HCF × 693 = 7623
• HCF = 7623 / 693
• Therefore: HCF = 11

Factor of m is x²y⁵ = y³(x²y²)

Factor of n is x³y² = x(x²y²)

Therefore HCF (m, n) is x²y².

To determine when the traffic lights will change simultaneously again, we need to find the least common multiple (LCM) of their change intervals: 48 seconds, 72 seconds, and 108 seconds.

Find the prime factors of each number:

48: 2⁴ × 3

72: 2³ × 3²

108: 2² × 3³

For the LCM, take the highest power of each prime:

2⁴ from 48

3³ from 108

Calculate the LCM: 2⁴ × 3³ = 432 seconds

Convert 432 seconds into minutes and seconds:

432 seconds = 7 minutes and 12 seconds

Starting from 8:20:00 hrs, add 7 minutes and 12 seconds:

The lights will change simultaneously again at 8:27:12 hrs.

The denominator 2⁶ x 5² is of the form 2^m x 5ⁿ, where m and n are non-negative integers. Hence, it is a terminating decimal expansion.

To get the greatest volume of each cask, we have to find the largest number which exactly divides 403, 434 and 465. That is nothing but the H.C.F of (403, 434, 465)

The HCF of 403. 434, 465 = 31 liters

Each cask must be of the volume 31 liters.

Req. No. of casks is

= (403/31) + (434/31) + (465/31)

= 13 + 14 + 15

= 42

Hence, the least possible number of casks of equal size required is 42.

It is given that,

a = x³y² = x × x × x × y × y

b = xy³ = x × y × y × y

The HCF (Highest Common Factor) of two or more numbers is the highest number among all the common factors of the given numbers.

As HCF is the product of the smallest power of each common prime factor involved in the numbers.

HCF of a and b = HCF (x³y², xy³)

= x × y × y

= xy²

Therefore,HCF (a, b) is xy²