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 two consecutive even numbers is 2.

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