RS Aggarwal Test: Real Numbers - 2 - Class 10

Multiply by 5 in both numerator and denominator
We get,
(47 × 5)/1000 = 0.235

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
∴ 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=29×3×3

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.
 

HCF of 576 and 448 = 64

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.

Difference of 90 and 24 = 66
and LCM of 90 and 24 = 360
∴ Numbers are 90 and 24.

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.

The LCM of two numbers = 1200.

By formula, product of two numbers =HCF x LCM

Let the two numbers be y,z

From the details given in question, 

HCF x 1200 = y x z

The four options given in question are 600, 400, 500, and 200.

From the available options given, 500 cannot be the HCF since it cannot divide 1200