CBSE Class 10 Mathematics Real Numbers Assignment Set A

CBSE Class 10 Mathematics Real Numbers Assignment Set A. Students are advised to refer to the attached assignments and practise them regularly. This will help them to identify their weak areas and will help them to score better in examination. Parents should download and give the assignments to their children for practice.

REAL NUMBER
LEVEL WISE QUESTIONS AND SOLUTIONS

LEVEL- 1 

Question. Find HCF of 128 and 216.
Solution : Two numbers are : 216, 128
216 > 128
By using Euclid division lemma :
216 = 128 x 1 +88 ( R≠0 )
128 = 88 x 1 + 40 ( R ≠0 )
88 = 40 x 2 + 8 (R ≠ 0)
40 = 8 x 5 + 0
Here R = 0,
Divisor is 8, so HCF = 8.         

Question. Given HCF(306,657) = 9. Find LCM(306,657)
Solution : HCF = 9, Ist number = 306, IInd number = 657
We know that :
HCFxLCM = Products of two number
9 x LCM = 306 X 657
LCM = 306x657/9
LCM = 22338. 

Question. Prove that 3+2 √5 is an irrational number.
Solution : Let us suppose that 3+2 √5 is rational so, we can find co prime a and b
where a and b are integers and b≠0.
So ,3+2 √5 =a/b
2 √5 = a/b - 3
√5 = (a-3b)/2b
Since a and b both are integers
(a-3b)/2b is rational number, so, √5 is also rational number
But this contradicts the facts that √5 is irrational.
our supposition is wrong.
Hence 3+2 √5 is rational.

Question. Explain why 7×6×5×4×3×2×1+5 is a composite number.
Solution : we have
7x6x5x4x3x2x1+5
5(7x6x1x4x3x2x1+1)
5(1008+1)
5(1009)
5045
The of factors of 5045 are 1,5,1009,5045.
It has more than two factors, so it is a composite number

LEVEL -2

Question. Check whether 6n can end with the digit 0. For any natural number n.
Solution : Let us suppose that 6n ends with the digits 0 for some n ∈N
So,6ⁿ is divisible by 5
But prime factor of 6 are 2 and 3
prime factor of 6n are ( 2x3)ⁿ
It is clear that in prime factorization of 6n there is no place of 5.
So our supposition is wrong . Hence there exists no any natural number n for which 6n ends
with digit zero. 2

Question. Find the LCM and HCF of 91 and 26, also verify that HCF×LCM= product of two
Solution : Two number are 91 and 26
91 = 7x13
26 = 2x13
HCF = 13
LCM = 13x2x7 = 182
Verification
HCFXLCM = Products of two number
13x182 = 91x26
2366 2366 

Question. The HCF of two number is 4 and their LCM is 9696,if one number is 96,find the other number.
Solution : HCF = 4
LCM = 9696, Ist number = 96,
We know that HCFXLCM = Ist x IInd
4x 9696 = 96 x IInd
96
4x9696
= IIndIInd number = 404

Question. Find the LCM and HCF of 12 , 15 and 21 by prime factorization method.
Solution : numbers are 12, 15, 21
12 = 2x2x3
15 = 3x5
21 = 3x7
So, HCF = 3
LCM = 3X2X2X5X7 = 420

LEVEL -3

Question. Prove that √2 is irrational number.
Solution : Let us suppose that √2 is rational number, so it can be put in the form of p/q
q≠0, and p and q are co prime
√2 = p/q
squaring both side
2 = p²/q²
q² = p²/2 here p² is divisible by 2, so p is also divisible by 2.
Let p = 2r putting this value in above relation
q² = 4 r² /2
so, r² = q²/2 , here q² is divisible by 2, so q is also divisible by 2
from these two p and q both divisible by 2 but it is suppose p and q are co prime which is contradiction so, √2 is irrational.

Question. Show that the square of any positive integer is of the form 3m or 3m+1,for some integer m.
Solution : Let a be any positive integer then it is of the form 3q,3q+1or 3q+2
If a = 3q squaring both sides,
a² =9q²
= 3(3q²)
= 3m, where m = 3q²
If a = 3q+1 squaring both sides,
a² = (3q+1)²
= 9q²+1+2x3qx1
= 3(3q²+2q) +1
4 = 3m+1, where m = (3q²+2q)
Where m is also an integer, 
Hence , square of any positive integer is either of the form 3m or 3m+1 for some integer m.

Question. Find the LCM and HCF of 510 and 92 and verify the thatLCMxHCF = Product of two no.
Solution : Two number are 510 and 92
510 = 2x3x5x17
92 = 2x2x23
HCF = 2 1 Mark
LCM = 2X2X3X5X17X23 = 23460
Verification : LCM X HCF =23460X2 = 46920
Products of two numbers = 510x92 = 46920
Hence verified .

LEVEL -4

Question. Show that 3√2 is irrational.
Solution : Let 3√2 is rational.
Therefore x = 3√2
=> x/3 = √2 
=> x/3 is rational and √2 is also rational.
But this contradicts the fact that √2 is irrational.
=> 3√2 is irrational. 

Question. Explain why ( 7 x 11 x 13) + 13 is a composite number.
Solution : ( 7 X 11 X 13) +13
= 13 x ( 7 x 11 + 1 ) x 1
So , it is product of more than 2 factors. Hence it is a composite number. 

Question. Find the HCF of 616 and 32 using Euclid’s division algorithm.
Solution : Using Euclid’s division lemma,
a = bq + r, . 0 ≤ r < b
Step 1 : 616 = 32 x 19 + 8 
Step 2 : 32 = 8 x 4 + 0
HCF of 616 and 32 is 8.

Question. Use Euclid’s division algoritha to find the largest number which divides 957 and 1280 leaving remainder 5 in each case.
Solution : 957-5=952 and 1280-5= 1275,are completely divisible by required number.
Now find the HCF by Euclid division lemma,
1275 > 952 by apply division lemma
1275 = 952 x 1 + 323 ( since R≠ 0 )
952 = 323 x 2 + 306 ( since R≠ 0 )
323 = 306 x 1 + 17 ( since R ≠ 0)
306 = 17 x 18 + 0 here R = 0
Divisor in the last step is 17
HCF of 1275 and 952 is 17.
Hence required number is 17.

Question. Euclid’s division algorithm can be applied to :
(A) only positive integers
(B) only negative integers
(C) all integers
(D) all integers except 0
Answer. A

Question. If the HCF of 65 and 117 is expressible in the form 65m – 117, then the value of m is :
(A) 1
(B) 2
(C) 3
(D) 4
Answer. B

Question. The least number that is divisible by all the numbers from 1 to 10 (both inclusive) is :
(A) 10
(B) 100
(C) 504
(D) 2520
Answer. D

Question. 3.24636363... is :
(A) a terminating decimal number
(B) a non-terminating repeating decimal number
(C) a rational number
(D) both (B) and (C)
Answer. D

Question. The decimal expansion of the rational number 47/2².5 . will terminate after :
(A) one decimal place
(B) three decimal places
(C) two decimal places
(D) more than 3 decimal places
Answer. C

Question. Euclid’s division lemma states that if a and b are any two +ve integers, then there exist unique integers q and r such that :
(A) a = bq + r, 0 < r < b
(B) a = bq + r, 0 ≤ r ≤ b
(C) a = bq + r, 0 ≤ r < b
(D) a = bq + r, 0 < b < r
Answer. C

Question. The value of x in the factor tree is :
(A) 30
(B) 150
(C) 100
(D) 50
Answer. B

Question. Which of the following numbers has terminating decimal expansion ?
(A) 37/45
(B) 21/2³5⁶
(C) 17/49
(D) 89/2²3²
Answer. B

Question. How many prime factors are there in prime factorisation of 5005 ?
(A) 2
(B) 4
(C) 6
(D) 7
Answer. B

Question. If a, b are coprime, then a², b² are :
(A) Coprime
(B) Not coprime
(C) Odd numbers
(D) Even numbers
Answer. A

Question. Find three different rational numbers between 5/7 & 9/11 .
Answer. 56/77, 57/77, 58/77

Question. Given that HCF (2520, 6800) =40 and LCM (2520, 6800) = 252 x K. Find the value of K.
Answer. 1700

Question. Radius of a circular track is 63m. Two cyclists Surjeet and Jacob start together from the same position,at the same time and in the same direction with speeds 33m/min. and 44m/min. After how many minutes they meet again at the starting point.
Answer. 36 min

Please click the link below to download CBSE Class 10 Mathematics Real Numbers Assignment Set A