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


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25 Questions MCQ Test Mathematics (Maths) Class 10 - RD Sharma Test: Real Numbers - 2

RD Sharma Test: Real Numbers - 2 for Class 10 2024 is part of Mathematics (Maths) Class 10 preparation. The RD Sharma Test: Real Numbers - 2 questions and answers have been prepared according to the Class 10 exam syllabus.The RD Sharma Test: Real Numbers - 2 MCQs are made for Class 10 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for RD Sharma Test: Real Numbers - 2 below.
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RD Sharma Test: Real Numbers - 2 - Question 1

The HCF of two consecutive numbers is

Detailed Solution for RD Sharma Test: Real Numbers - 2 - Question 1

The HCF of two consecutive numbers is always 1.

RD Sharma Test: Real Numbers - 2 - Question 2

Find the largest number which divides 62,132,237 to leave the same reminder

Detailed Solution for RD Sharma Test: Real Numbers - 2 - Question 2

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

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RD Sharma Test: Real Numbers - 2 - Question 3

The product of two numbers is -20/9. If one of the numbers is 4, find the other. 

Detailed Solution for RD Sharma Test: Real Numbers - 2 - Question 3

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

RD Sharma Test: Real Numbers - 2 - Question 4

For some integer m, every odd integer is of the form

Detailed Solution for RD Sharma Test: Real Numbers - 2 - Question 4

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

RD Sharma Test: Real Numbers - 2 - Question 5

The decimal expansion of  terminates after:

Detailed Solution for RD Sharma Test: Real Numbers - 2 - Question 5

 Therefore,  will terminate after 6 places of decimals.

RD Sharma Test: Real Numbers - 2 - Question 6

If 112 = q×6+r, then the possible values of r are:

Detailed Solution for RD Sharma Test: Real Numbers - 2 - Question 6

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

RD Sharma Test: Real Numbers - 2 - Question 7

The HCF of two consecutive even numbers is

Detailed Solution for RD Sharma Test: Real Numbers - 2 - Question 7

The HCF of two consecutive even numbers is 2.

RD Sharma Test: Real Numbers - 2 - Question 8

The largest number of 4 digits exactly divisible by 12, 15, 18 and 27 is

Detailed Solution for RD Sharma Test: Real Numbers - 2 - Question 8

LCM (12, 15, 18, 27) = 540
Now, largest four digit number = 9999 
∴ 9999 ÷ 540 = 18 x 540 + 279 (Remainder = 279) Therefore, the largest number of 4 digits exactly divisible by 12, 15, 18 and 27 is 9999 – 279 = 9720

RD Sharma Test: Real Numbers - 2 - Question 9

Which of the following is a rational number?

Detailed Solution for RD Sharma Test: Real Numbers - 2 - Question 9

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

RD Sharma Test: Real Numbers - 2 - Question 10

On a morning walk, three persons step off together and their steps measure 40 cm, 42 cm and 45 cm, respectively. What is the minimum distance each should walk so that each can cover the same distance in complete steps?

Detailed Solution for RD Sharma Test: Real Numbers - 2 - Question 10

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

RD Sharma Test: Real Numbers - 2 - Question 11

Detailed Solution for RD Sharma Test: Real Numbers - 2 - Question 11

 

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

RD Sharma Test: Real Numbers - 2 - Question 12

The H.C.F of 441, 567 and 693 is

Detailed Solution for RD Sharma Test: Real Numbers - 2 - Question 12

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.

RD Sharma Test: Real Numbers - 2 - Question 13

The LCM of two consecutive numbers is

Detailed Solution for RD Sharma Test: Real Numbers - 2 - Question 13

The LCM of two consecutive numbers is their product always.

RD Sharma Test: Real Numbers - 2 - Question 14

Pairs of natural numbers whose least common multiple is 78 and the greatest common divisor is 13 are:      

Detailed Solution for RD Sharma Test: Real Numbers - 2 - Question 14

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 ]

RD Sharma Test: Real Numbers - 2 - Question 15

If ‘p’ is a prime number, then √p is

Detailed Solution for RD Sharma Test: Real Numbers - 2 - Question 15

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

RD Sharma Test: Real Numbers - 2 - Question 16

The largest number which divides 70 and 125, leaving remainders 5 and 8, respectively, is

Detailed Solution for RD Sharma Test: Real Numbers - 2 - Question 16

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)

RD Sharma Test: Real Numbers - 2 - Question 17

If 9x+2 = 240+9x, then the value of ‘x’ is

Detailed Solution for RD Sharma Test: Real Numbers - 2 - Question 17

RD Sharma Test: Real Numbers - 2 - Question 18

If A = 2n + 13, B = n + 7, where n is a natural number then HCF of A and B is:      

Detailed Solution for RD Sharma Test: Real Numbers - 2 - Question 18

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.

RD Sharma Test: Real Numbers - 2 - Question 19

The LCM of 24, 60 and 150 is

Detailed Solution for RD Sharma Test: Real Numbers - 2 - Question 19

RD Sharma Test: Real Numbers - 2 - Question 20

The LCM of two number is 45 times their HCF. If one of the numbers is 125 and the sum of HCF and LCM is 1150, the other number is:

Detailed Solution for RD Sharma Test: Real Numbers - 2 - Question 20

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.

RD Sharma Test: Real Numbers - 2 - Question 21

The difference of a rational and an irrational number is always

Detailed Solution for RD Sharma Test: Real Numbers - 2 - Question 21

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

RD Sharma Test: Real Numbers - 2 - Question 22

 If n is a rational number, then 52n − 22n is divisible by

Detailed Solution for RD Sharma Test: Real Numbers - 2 - Question 22

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.

RD Sharma Test: Real Numbers - 2 - Question 23

The multiplicative inverse of zero is

Detailed Solution for RD Sharma Test: Real Numbers - 2 - Question 23

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 

RD Sharma Test: Real Numbers - 2 - Question 24

will have

Detailed Solution for RD Sharma Test: Real Numbers - 2 - Question 24



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

RD Sharma Test: Real Numbers - 2 - Question 25

The relationship between HCF and LCM of two natural numbers is

Detailed Solution for RD Sharma Test: Real Numbers - 2 - Question 25

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

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