RD Sharma Test: Real Numbers - 2 - Class 10 MCQ

The HCF of two consecutive numbers is always 1.

Trick is HCF of (237-132), (132-62), (237-62)
= HCF of (70,105,175) = 35

We have two numbers such that their product is equal to -20/9.
So we have x*y=-20/9
One no. is given 4, so
x*4=-20/9
x=-20/9 = 4
x=-20/9 x 1/4=-5/9

As the number 2m will always be even, so if we add 1 to it then, the number will always be odd.

 Therefore,  will terminate after 6 places of decimals.

For the relation x = qy+r, 0 ⩽ r < y So, here r lies between 0 ⩽ r < 6. Hence r = 0, 1, 2, 3, 4, 5

The HCF of any two consecutive even numbers is always 2. The HCF (Highest Common Factor) of two or more numbers is the highest number among all the common factors of the given numbers.

For the largest number exactly divisible by 12, 15, 18 and 27, we need to find the LCM.


∴ LCM (12, 15, 18, 27) = 2 × 2 × 3 × 3 × 3 × 5 = 540

To check whether the greatest 4-digit number is divisible by 540, we must divide by 9999 by 540.

On dividing, 9999/540= 279 is remainder

So, 9999 – 279 = 9720 is the largest 4-digit number divisible by 540.

√9 is a rational number because √9 = 3 and 3 is a rational number.

We need to find the L.C.M of 40, 42 and 45 cm to get the required minimum distance.
40 = 2×2×2×5
42 = 2×3×7
45 = 3×3×5
L.C.M. = 2×3×5×2×2×3×7 = 2520

Comparing the denominators of both fractions, we have m = 5 and n = 3

693 = 3×3×7×7
567 = 3×3×3×3×7
441 = 3×3×7×11
Therefore H.C.F of 693, 567 and 441 is 63.

The LCM of two consecutive numbers is their product always.

Let the no.s be a & b,

GCD [ a , b ] = 13;
=> Let a = 13m and b = 13n

Now, LCM [ a , b ] = 78
=> 13m | 78
=> m | 6 and similarly, n | 6; --> Also, m does not divide n

.'. Possible values of (m,n) are :-> 
[ 1 , 6 ] , [ 2 , 3 ]
.'. Possible values for ( a , b ) are :-> [ 13 , 78 ] , [ 26 , 39 ]

√p is an irrational number because square root of every prime number is an irrational number.

Since 5 and 8 are the remainders of 70 and 125, respectively. Thus after subtracting these remainders from the numbers, we have the numbers
65 = (70 − 5), 117 = (125 − 8) which is divisible by the required number.
Now required number = H.C.F of (65,117)

A = 2n + 13
B = n + 7

1) If the value of n = 1, then

A = (2*1) + 13
= 2 + 13
= 15

B = 1 + 7
= 8

Now, A = 15 and B = 8
HCF of 15 and 8 is 1.

2) If the value of n =2 

A = (2*2) + 13
= 4 + 13 
= 17

B = 2 + 7 
= 9

A = 17 and B = 9
HCF of 17 and 9 is 1.

3) If the value of n 3

A = (2*3) + 13
= 6 + 13
= 19

B = 3 + 7
= 10

A = 19 and B = 10
HCF of 19 and 10 is 1

4) If the value of n = 4
A = (2*4) + 13
= 8 +13 
= 21

B = 4 + 7
= 11

A = 21 and B = 11

HCF of 21 and 11 is 1

5) If the value of n = 5
A = (2*5) + 13
= 10 + 13
= 23

B = 5 + 7
= 12

A = 23 and B = 12
HCF of 23 and 12 is 1

6) If the value of n = 6
A = (2*6) + 13
= 12 + 13
= 25

B = 6 + 7
= 13

A = 25 and B = 13
HCF of 25 and 13 is 1

7) If the value of n = 7
A = (2*7) + 13
14 + 13
= 27

B = 7 + 7
= 14

A = 27 and B = 14
HCF of 27 and 14 is 1

8) If the value of n = 8
A = (2*8) + 13
16 + 13
= 29

B = 8 + 7
= 15

A = 29 and B = 15
HCF of 29 and 15 is 1

9) If the value of n = 9
A = (2*9) + 13
18 + 13
= 31

B = 9 + 7
= 16

A = 31 and B = 16
HCF of 31 and 16 is 1

10) If the value of n = 10 
A = (2*10) + 13
20 + 13
= 33

B = 10 + 7
= 17

A = 33 and B = 17
HCF of 33 and 17 is 1

We have seen that whatever the value n has, the HCF of A and B (separate values they hold in each situation) is always 1. It means we can take any value of n and the HCF will be 1.

LCM = 45 HCF
LCM + HCF = 1150
45*HCF + HCF = 1150
46 *HCF = 1150
HCF = 25
Then,
LCM = 45 *25 = 1125.
Now, use the formula,
1st number * second Number = LCM * HCF
125 * second number = 1125 *25
Second Number = 225.

The difference of a rational and an irrational number is always an irrational number.

5²ⁿ −2²ⁿ is of the form a²ⁿ − b²ⁿ which is divisible by both (a + b) and (a – b).
So, 5²ⁿ − 2²ⁿ is divisible by both 7, 3.

Some rules associated with Multiplicative Inverse are discussed in following ways :- 
Rule 1 = If product of two Fractional Numbers is equal to 1, then each of the Fractional Numbers is the Multiplicative Inverse of other. 
Rule 2 = If the product of a Fractional Number and a Whole Number is equal to 1, then each is the Multiplicative Inverse of other. 
Rule 3 = Multiplicative Inverse of 1 is also 1. 
Rule 4 = Multiplicative Inverse of 0 (zero) does not exists 

As the denominator has factor 5³ × 2² and which is of the type 5^m × 2ⁿ, So this is a terminating decimal expansion.

The product of LCM and HCF of any two given natural numbers is equivalent to the product of the given numbers.

LCM × HCF = Product of the Numbers

Suppose A and B are two numbers, then.

LCM (A & B) × HCF (A & B) = A × B

Example: Prove that: LCM (9 & 12) × HCF (9 & 12) = Product of 9 and 12.
->LCM and HCF of 9 and 12:
9 = 3 × 3 = 3²
12 = 2 × 2 × 3 = 2² × 3
LCM of 9 and 12 = 2² × 3² = 4 × 9 = 36
HCF of 9 and 12 = 3
LCM (9 & 12) × HCF (9 & 12) = 36 × 3 = 108
Product of 9 and 12 = 9 × 12 = 108
Hence, LCM (9 & 12) × HCF (9 & 12) = 108 = 9 × 12