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.

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

Let the three consecutive positive integers be n, n+1 and n+2.

Whenever a number is divided by 3, the remainder obtained is either 0,1 or 2.

Therefore, n=3p or 3p+1 or 3p+2, where p is some integer.

If n=3p, then n is divisible by 3.

If n=3p+1, then n+2=3p+1+2=3p+3=3(p+1) is divisible by 3.

If n=3p+2, then n+1=3p+2+1=3p+3=3(p+1) is divisible by 3.

So, we can say that one of the numbers among n,n+1 and n+2 is always divisible by 3 that is:

n(n+1)(n+2) is divisible by 3.

Similarly, whenever a number is divided by 2, the remainder obtained is either 0 or 1.

Therefore, n=2q or 2q+1, where q is some integer.

If n=2q, then n and n+2=2q+2=2(q+1) is divisible by 2.

If n=2q+1, then n+1=2q+1+1=2q+2=2(q+1) is divisible by 2.

So, we can say that one of the numbers among n, n+1 and n+2 is always divisible by 2.

Since, n(n+1)(n+2) is divisible by 2 and 3.

Hence, n(n+1)(n+2) is divisible by 6.

Taking different values of n we find that A and B are coprime

e.g. put n=1

we get A= 2(1)+13=15

B=1+7=8
∴ HCF = 1
 

put n=2

we get A= 2(2)+13=17

B=2+7=9
∴ HCF = 1

we have to  find the largest number which divide 615 and 963 leaving remainder 6 in each case.

so let us subtract 6 from 615 and 963

⇒ 615 − 6 = 609

⇒ 963 − 6 = 957

now lets find the HCF

⇒ prime factorisation of 609=29×3×7

prime factorisation of 957=29×3×11

now lets take out the common factors from both the cases

⇒ 29 and 3

x = 29 × 3 = 87

∴ 87 is the number which will divide 615 and 963 leaving remainder 6 in each case.

The number 576 can be factorised as,
576 = 2×2×2×2×2×2×3×3
The number 448 can be factorised as,
448=2×2×2×2×2×2×7
Write the common factors of the given numbers.
2×2×2×2×2×22×2×2×2×2×2
Multiply the common factors to determine the highest common factor (HCF) of the given numbers.
2×2×2×2×2×2 = 642×2×2×2×2×2 = 64
Since the highest common factor (HCF) of the given numbers is 64, this implies that each section will have 64 number of students.
Now, we need to find the number of sections formed.
Let us first find the number of sections formed by the total number of boys by dividing 576 by 64.
576/64 = 9
Now, find the number of sections formed by the total number of girls by dividing 448 by 64.
448/64=7
Thus, the total number of sections formed will be 9+7=16
Hence, option B is the correct answer.

The largest number by which x , y

divisible and gives the remainder a ,

and b is

the HCF of ( x - a ) and ( y - b)
According to the given problem ,

The largest number which divides

70 and 125 leaving remainders 5 and

8 respectively are

HCF of ( 70 - 5 ) = 65 and

( 125 - 8 ) = 117

65 = 5 × 13

117 = 3 × 3 × 13

HCF ( 65 , 117 ) = 13

Required number is 13.

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)

Let the two numbers be x and x+66.

Since, the LCM is 360, there must be 5,2 and 3 as the prime factors of the two numbers.

So, one pair is 24 & 90, because 90−24=66.

Also, 24=2×2×2×3

90=2×3×3×5

∴ LCM=2×2×2×3×3×5=360.

So, the numbers are 24 and 90.

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.

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