Real number class 10 Important Questions with solution 2021

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Practice following  “Real number questions for the upcoming 2021 board” and if you found any problem difficult to solve, see explanation. 

Problem:01  The LCM of the two numbers is 1200. Which of them cannot be their HCF?

a) 600

b) 500

c) 400

d) 200

Correct: (b) 500 


Solution: 

We know, HCF of any two numbers is the divisor of their LCM as well. Here 1200 is the LCM and it is divisible by 600,400 and 200. So, one of these numbers might be their HCF but 500 can’t be their HCF.


Problem:02 For some integer q, every odd integer is of the form 

a) q

b) q+1

c) 2q

d) 2q+1

Correct answer : (d) 2q+1 


Explanation : 

Using Euclid’s Division lemma a=bq+r, where 0≤r<b

Let b= 2 

∴ a=2q+r where 0≤r<2 i.e r=0,1 

Case:01 When b= 2 and r=0 

a=2q+r

a=2q+0

a=2q ,it is form of an even integer 

Case: 02 When b= 2 and r=1 

a=2q+r

a=2q+1 

a=2q+1 ,it is form of an odd integer 

So, Every odd integer is of the form 2q+1 


Problem : 03    n²-1 is divisible by 8, if n is

a) An integer

b) A natural number 

c) An odd integer

d) An even integer

Correct answer : (c) Odd integer 


Explanation : 

Using Euclid’s Division lemma a=bq+r, where 0≤r<b

Let b=4

∴  a=4q+r, where 0≤r<4 i.e 0,1,2,3 

a= 4q ,4q+1 ,4q+2 ,4q+3 

Case:01 a=4q 

n²-1 = (4q)²-1 =16q²-1 is not divisible by 4 

Case:02 a= 4q+1

n²-1 =(4q+1)²-1 = (4q)²+(1)²+2.4q -1 

= 16q²+1+8q-1 

= 16q²+8q 

= 8q(2q+1) is divisible by 8 

Similarly ,you can check with other two cases as well . 


Problem:04  If the HCF of 65 and 117 is expressible in the form 65m-117, then the value of m is

a) 4

b) 2

c) 1

d) 3

 

Correct answer: m=2 


Explanation : 

HCF of 65 and 117 is 13 

And ,according to question, it is expressible as 65m-117

∴ 65m-117=13

65m=130 

m=2 


Problem:05  What is a composite number?

 

Definition: A natural number having two prime factors or more than 2 factors is a composite number.For Example 4,6,8 etc 

 

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Problem:06  What is the HCF of the smallest composite number and the smallest prime number?

 

Answer:


Explanation : 

We know, smallest composite and prime numbers are 4 and 2 respectively 

∴ HCF of 4 and 2 is 2 


Problem:07  Find the HCF of 81 and 237 and express it as a linear combination of 81 and 237. Also, show that this combination is not unique.

Solution: We have 81 and 237 as two positive integers such that 237>81 

Using Euclid’s division lemma ,find the HCF (237,81) 

HCF (237,81)=3

HCF of 81 and 237

Now express it as linear combination of given two numbers 

3= 75 – 6×12

3=75 -(81-75)×12                               

3=75-81×12 + 75×12 

3=13×75 – 12×81                        

3=13×(237-81×2) – 12×81                   

3= 13×237 -26×81 -12×81

3= 237×13 – 81(26+12) 

3=237×13 – 81×38 

3=237x + 81y where x= 13 and y=-38 

Why this linear representation is not unique ?see 

We have 3=237×13 – 81×38  as linear expression 

Now Add and subtract 237×81 

3= 237×13 – 81×38 + 237×81  – 237×81 

3= (237×13 +237×81) -(81×38+237×81 ) 

3= 237×94  – 81(275) 

3= 237×94 + 81 ×(-275) 

Now we have different values for x and y i.e x=94 and  y=-275.That’s why linear representations are not unique


Problem:08 Find the largest number which divides 285 and 1249 leaving remainders 9 and 7 respectively.

 

Answer: 138 


Explanation : 

Since it is given that the required number when divides 285 and 1249 leaves 9 and 7 as remainder respectively.

∴ 285 -9=276 and 1249-7=1242 are completely divisible by the required number

So, the required number is the HCF of 276 and 1242 

hcf of 276 and 1242

Thus, Largest no. which divides 285 and 1249 leaving remainder 9 and 7 respectively is 138 

Problem:09 The length, breadth, and height of a room are 8 m 25 cm, 6m 75cm, and 4 m 50 cm, respectively. Determine the longest rod which can measure the three dimensions of the space exactly.

Real number class 10 important questions

Solution: 

Length of the room = 8 m 25 cm=825 cm 

Breadth of the room =6m 75cm = 675 cm 

Height of the room =4 m 50 cm=450 cm 

Required length of rod which can exactly measure three dimensions of room is HCF (825,675 and 450)cm=75 cm 


 Problem:10  Express the HCF of 468 and 222 as 468x +222 y where x, y are integers.

Solution : 

We have 468 and 222 as two positive integers such that 468>222 

Using Euclid’s division lemma, find the HCF of 468 and 222 

hcf of 222 and 468

HCF(468,222) = 6 

Now express it as linear combination of given numbers 

6= 222 – 24×9

6=222 -(468-222)×9

6=222 – 468×9+222×9

6= 222 + 222×9 – 468×9

6=222(10) – 468×9

6 = 222×10 + 468 ×(-9) 

6=222x + 468y   where x = 10 and y=-9 

Problem:11 Explain why 3×5×7+7 is a composite number.

Explanation  : 

A composite number is a natural number that has more than 2 factors or 2 prime factors. 

We have,

3×5×7+7

3×5×7(1+1) 

3×5×7×2 

It has 4 prime factors that’s why it is a composite number 

Problem:12 Find the smallest number which when increased by 17 is exactly divisible by both 520 and 468. 

Solution: 

The smallest number which is exactly divisible by 520 and 468 is the LCM (520,468) 

So, LCM (520,468) = 4680 

Hence Required smallest number which is divisible by 520 and 468 when increased by 17 =4680-17=4667 

Problem:13 A rectangular courtyard is 18m 72 cm long and 13 m 20 cm broad. It is to be paved with square tiles of the same size. Find the least possible number of such tiles.

Real number Important question class 10

 

Answer : Least possible number of needed tiles = 4290


 Solution : 

Length of the courtyard = 18m 72 cm = 1872 cm 

Breadth of the courtyard = 13 m 20 cm =1320 cm 

Since ,it is to be paved with square tiles of same size

Largest square tile which will fit in the courtyard =HCF(1872,1320) 

HCF(1872,1320) =24 cm 

Required number of tiles for the whole courtyard =\frac{Area of court yard}{Area of single square tile}

=\frac{1872×1320}{24×24}

=4290 tiles 

Check these stuff as well 

 

 

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