CBSE Class 10 Mathematics Real Numbers Assignment Set A

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

Vert Short Answer type Questions

Question. If two positive integers a and b are written as a = p³q² and b = pq³; p, q are prime numbers, then HCF (a, b) is:
Answer : HCF = pq²

Question. If p and q are two coprime numbers, then p³ and q³ are?
Answer : Coprime

Question. Find the LCM of smallest prime and the smallest odd composite natural number
Answer : LCM of 2 and 4 is 4

Question. Prime factorization of 120 is …
Answer : 2³ x 3 x 5

Question. 144 cartons of coke cans and 90 cartons of Pepsi cans aer : re to be stacked in a canteen. If each stack is of the same height and is to contain cartons of the same drink, what would be the greatest number of cartons each stack would have?
Answer : 18

Question. Explain why 3 × 5 × 7 + 7 is a composite number.
Answer : 112 is an even number and is therefore a composite number

Question. If n is an even prime number then, 2(7ⁿ + 8ⁿ) ends with?
Answer : 6

Question. Given that LCM (91, 26) = 182, then HCF (91, 26) is:
Answer : HCF = 13

Question. Determine the prime factorisation of 2057?
Answer : 2 × 5 × 11² × 17

Question. Can the number 4n, n being a natural number, end with the digit 0? Give reasons.
Answer : No

Question. If the HCF of 408 and 1032 is expressible in the form 1032 m -408x5 ,find m.
Answer : 2

Question. The length, breadth and height of a room are 825 cm, 675 cm and 450 cm respectively. Find the longest tape which can measure the three dimensions of the room exactly.
Answer : 75 cm

 

Short Answer type Questions

 

Question. Find the ratio between the LCM and HCF of 5, 15 and 20.
Answer : 5 = 5 × 1
15 = 3 × 5
20 = 2 × 2 × 5
∴ LCM of 5, 15, 20 = 60
and HCF = 5
∴ Ratio = 60 : 5 = 12 : 1

Question. Find HCF and LCM of 448, 1008 and 168 using fundamental theorem of arithmetic.
Answer : 448 = 2 × 2 × 2 × 2 × 2 × 2 × 7 = 26 × 7
1008 = 2 × 2 × 2 × 2 × 3 × 3 × 7 = 24 × 32 × 7
168 = 2 × 2 × 2 × 3 × 7 = 23 × 3 × 7
∴ HCF = 23 × 7 = 56
LCM = 26 × 7 × 32 = 4032

Question. Find the HCF and LCM of 96 and 404 using fundamental theorem of arithmetic.
Answer : 96 = 25 × 3
404 = 22 × 101
∴ HCF = 22 = 4
LCM = 25 × 3 × 101 = 9696

Question. In a seminar the number of participants in Mathematics, Physics and Biology are 336,  240 and96. Find the minimum number of rooms required if in each room same number of participants is to be seated and all of them being in the same subject.
Answer : 336 = 2 × 2 × 2 × 2 × 3 × 7; 240 = 2 × 2 × 2 × 2 × 3 × 5, 96 = 2 × 2 × 2 × 2 × 2 × 3
HCF of 336, 240, 96 = 48
Now number of rooms for participants in Mathematics = 336/48 = 7
Number of rooms for participants in Physics = 240/48 = 5
Number of rooms for participants in Biology = 96/48 = 2
Total no. of rooms = 14

Question. Find the HCF and LCM of 6, 72 and 120 using fundamental theorem of arithmetic.
Answer : 6 = 2 × 3
72 = 2 × 2 × 2 × 3 × 3
120 = 2 × 2 × 2 × 3 × 5
Common factors of 6, 72 and 120 are 2 and 3.
∴ HCF = 2 × 3 = 6
and LCM = 2 × 2 × 2 × 3 × 3 × 5
⇒ LCM = 360

Question. Can we have any n ∈ N, where 4n ends with the digit zero?
Answer : For unit’s digit to be 0, then 4n should have 2 and 5 as its prime factors, but 4n =
(22)n = 22n.
It does not contain 5 as one of its prime factors.
∴ 4n will not end with digit 0 for n ∈ N.

Question. Find the largest number that divides 2053 and 967 and leaves a remainder of 5 and 7 respectively.
Answer : Required number is HCF of 2053 – 5 and 967 – 7 = HCF of 2048 and 960 = 64

Question. Find the largest number which divides 615 and 963 leaving remainder 6 in each case.
Answer : According to the question, Number divides 615 – 6 and 963 – 6 i.e. 609 and 957
Required number = HCF of 609 and 957.
609 = 3 × 7 × 29
957 = 3 × 29 × 11
HCF of 609 and 957 = 3 × 29 = 87
Required number = 87

Question. Given that √3 is irrational, prove that (2 + 5√3) is an irrational number.
Answer : Let 2 + 5√3 be a rational number such that
2 + 5√3 = a, where a is a non-zero rational number.
⇒ 5 √3 = a - 2 ⇒ √3 = a - 2/5
Since 5 and 2 are integers and a is a rational number, therefore a-2/5 is a rational number
⇒ √3 is a rational number which contradicts the fact that √3 is an irrational number.
Therefore, our assumption is wrong.
Hence 2 + 5√3 is an irrational number

Question. Two numbers are in the ratio 21 : 17. If their HCF is 5, find the numbers.
Answer : Let numbers are 21x and 17x.
Now, common factor of 21x and 17x = x
Also HCF = 5
⇒ x = 5
∴ numbers are 21 × 5 and 17 × 5 i.e. 105 and 85.

Question. The HCF of two numbers is 29 and other two factors of their LCM are 16 and 19. Find
the larger of the two numbers.
Answer : HCF of the two numbers is 29.
∴ Numbers are 29 × a and 29 × b
where a and b are co-prime.
Now other two factors of the LCM are 16 and 19.
∴ LCM = 29 × 16 × 19
⇒ 29 × 16 × 19 = 29 × a × b
⇒ a = 16 and b = 19
So, larger of the two number is 29 × 19 = 551.

Question. Prove that √5 is and irrational number.
Answer : Let √5 is a rational number then we have √5 = p/q, where p and q are co-primes.
⇒ p = √5q
Squaring both sides, we get p² = 5q²
⇒ p² is divisible by 5 ⇒ p is also divisible by 5
So, assume p = 5m where m is any integer.
Squaring both sides, we get p² = 25m²
But p² = 5q²
Therefore, 5q² = 25m2 ⇒ q² = 5m2
⇒ q² is divisible by 5 ⇒ q is also divisible by 5
From above we conclude that p and q have one common factor i.e. 5 which
contradicts that p and q are co-primes.
Therefore, our assumption is wrong.
Hence, √5 is an irrational number.

Question. The length, breadth and height of a room are 8 m 25 cm, 6 m 75 cm and 4 m 50 cm respectively. Determine the length of the longest rod which can measure the three dimensions of the room exactly.
Answer : Length of the longest rod = HCF of 825 cm, 675 cm and 450 cm.
825 = 3 × 5 × 5 × 11
675 = 3 × 3 × 3 × 5 × 5
450 = 2 × 3 × 3 × 5 × 5
⇒ HCF = 75
Length of the longest rod = 75 cm

Question. Prove that √2 is and irrational number.
Answer : Let √2 is a rational number then we have √2 = p/q , where p and q are co-primes.
⇒ p = √2 q
Squaring both sides, we get p² = 2q²
⇒ p² is divisible by 2 ⇒ p is also divisible by 2
So, assume p = 2m where m is any integer.
Squaring both sides, we get p² = 4m2
But p² = 2q²
Therefore, 2q² = 4m2 ⇒ q² = 2m2
⇒ q² is divisible by 2 ⇒ q is also divisible by 2
From above we conclude that p and q have one common factor i.e. 2 which contradicts that p and q are co-primes.
Therefore, our assumption is wrong.
Hence, √2 is an irrational number.

Question. Given that √2 is irrational, prove that (5 + 3√2) is an irrational number.
Answer : Let 5 + 3 √2 be a rational number such that
5 + 3 √2 = a, where a is a non-zero rational number.
⇒ 3 √2 = a - 5 ⇒ √2 = a - 5 / 3
Since 5 and 3 are integers and a is a rational number, therefore a-5/3 is a rational number
⇒ √2 is a rational number which contradicts the fact that √2 is an irrational number.
Therefore, our assumption is wrong.
Hence 5 + 3 2 is an irrational number

Question. The LCM of two numbers is 14 times their HCF. The sum of LCM and HCF is 600. If one number is 280, then find the other number.
Answer : Let HCF = x
⇒ LCM = 14x
According to the question, x + 14x = 600
⇒ x = 40
Now 280 × other number = HCF × LCM = 40 × 560
⇒ Other number = 80

Question. Find the HCF and LCM of 426 and 576 using fundamental theorem of arithmetic.
Answer : 426 = 2 × 3 × 71
576 = 2 × 3 × 3 × 2 × 2 × 2 × 2 × 2
∴ HCF = 2 × 3 = 6
LCM = 426 x 576/6 = 40896

Question. Given that √3 is irrational, prove that (2 + √3) is an irrational number.
Answer : Let 2 + √3 be a rational number such that
2 + √3 = a, where a is a non-zero rational number.
⇒ √3 = a – 2
Since 2 is a integer and a is a rational number, therefore a – 2 is a rational number
⇒ √3 is a rational number which contradicts the fact that √3 is an irrational number.
Therefore, our assumption is wrong.
Hence 2 + √3 is an irrational number

Question. Two alarm clocks ring their alarms at regular intervals of 50 seconds and 48 seconds if they first beep together at 12 noon, at what time will they beep again for the first time?
Answer : 50 = 2 × 5 × 5, 48 = 2 × 2 × 2 × 2 × 3
∴ LCM of 50 and 48 = 2 × 2 × 2 × 2 × 3 × 5 × 5 = 1200
∴ 1200 sec = 20 min
Hence at 12.20 pm they will beep again for the first time.

Question. Find the HCF and LCM of 288, 360 and 384 by prime factorisation method.
Answer : Here 288 = 2⁵ × 3²
360 = 2⁵ × 3² × 5 and 384 = 2⁷ × 3
The HCF of 288, 360 and 384 is the product of their common prime factor with least exponents.
HCF of (288, 360 and 384) = 2³ × 3 = 24.
The LCM of 288, 360 and 384 is product of all prime factors with their highest exponents.
LCM of 288, 360 and 384 = 2⁷ × 3² × 5 = 5760

Question. Prove that √3 is and irrational number.
Answer : Let √3 is a rational number then we have √3 = p/q, where p and q are co-primes.
⇒ p = √3 q
Squaring both sides, we get p² = 3q²
⇒ p² is divisible by 3 ⇒ p is also divisible by 3
So, assume p = 3m where m is any integer.
Squaring both sides, we get p² = 9m²
But p² = 3q²
Therefore, 3q² = 9m² ⇒ q² = 3m²
⇒ q² is divisible by 3 ⇒ q is also divisible by 3
From above we conclude that p and q have one common factor i.e. 3 which
contradicts that p and q are co-primes.
Therefore, our assumption is wrong.
Hence, √3 is an irrational number.

Question. Three sets of physics, chemistry and mathematics books have to be stacked in such a way that all the books are stored topic wise and the number of books in each stack is the same. The number of physics books is 192, the number of chemistry books is 240 and the number of mathematics books is 168. Determine the number of stacks of physics, chemistry and mathematics books.
Answer : Here, we have to find the HCF of 192, 240 and 168 because the HCF will be the
largest number which divides 192, 240 and 168 exactly.
192 = 2⁶ × 3
240 = 2⁴ × 3 × 5
168 = 2³ × 3 × 7
Now, the HCF of 192, 240 and 168 is = 2³ × 3 = 24
There must be 24 books in each stack.
Number of stacks of physics books = 192/24 = 8
Number of stacks of chemistry books = 240/24 = 10
Number of stacks of mathematics books = 168/24 = 7

Question. Find HCF and LCM of 625, 1125 and 2125 using fundamental theorem of arithmetic.
Answer : 625 = 54
1125 = 32 × 53
2125 = 53 × 17
∴ HCF = 53 = 125
and LCM = 54 × 32 × 17 = 95625

Question. Prove that √7 is and irrational number.
Answer : Let √7 is a rational number then we have √7 = p/q, where p and q are co-primes.
⇒ p = √7 q
Squaring both sides, we get p² = 7q²
⇒ p² is divisible by 7 ⇒ p is also divisible by 7
So, assume p = 7m where m is any integer.
Squaring both sides, we get p² = 49m2
But p² = 7q²
Therefore, 7q² = 49m2 ⇒ q² = 7m2
⇒ q² is divisible by 7 ⇒ q is also divisible by 7
From above we conclude that p and q have one common factor i.e. 7 which contradicts that p and q are co-primes.
Therefore, our assumption is wrong.
Hence, √7 is an irrational number.

Question. There are 576 boys and 448 girls in a school that are to be divided into equal sections of either boys or girls alone. Find the total number of sections thus formed.
Answer : 576 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3
448 = 2 × 2 × 2 × 2 × 2 × 2 × 7
∴ HCF of 576 and 448 = 64
∴ Number of sections = 576/64 + 448/64 = 9 + 7 = 16

Question. A forester wants to plant 66 apple trees, 88 banana trees and 110 mango trees in equal rows (in terms of number of trees). Also he wants to make distinct rows of trees (i.e., only one type of trees in one row). Find the number of minimum rows required.
Answer : 66 = 2 × 3 × 11; 88 = 2 × 2 × 2 × 11; 110 = 2 × 5 × 11
HCF of 66, 88 and 110 = 22
Number of trees in each row = 22
Number of rows = 66/22 + 88/22 + 110/22 = 3 + 4 + 5 = 12

Question. 4 Bells toll together at 9.00 am. They toll after 7, 8, 11 and 12 seconds respectively.
How many times will they toll together again in the next 3 hours?
Answer : 7 = 7 × 1
8 = 2 × 2 × 2
11 = 11 × 1
12 = 2 × 2 × 3
∴ LCM of 7, 8, 11, 12 = 2 × 2 × 2 × 3 × 7 × 11 = 1848
∴ Bells will toll together after every 1848 sec.
∴ In next 3 hrs, number of times the bells will toll together = 3 x 3600/1848 = 5.84
= 5 times.

Question. On a morning walk three persons step off together and their steps measure 40 cm, 42 cm, 45cm, what is the minimum distance each should walk so that each can cover the same distance incomplete steps?
Answer : Minimum distance = LCM of 40, 42 and 45
Now 40 = 23 × 5
42 = 2 × 3 × 7
45 = 32 × 5
⇒ LCM of 40, 42 and 45 = 23 × 32 × 5 × 7 = 2520
They should walk 2520 cm or 25.20 m to cover the distance in complete steps.

1. Use Euclid’s division algorithm to find the HCF of :
(i) 56 and 814
(ii) 6265 and 76254
Answer : (i) 2 (ii) 179

2. Find the HCF and LCM of following using Fundamental Theorem of Arithmetic method.
(i) 426 and 576
(ii) 625, 1125 and 2125
Answer : 2. (i) 6,40896 (ii) 125, 95625

3. Prove that √3 is an irrational number.

4. Prove that √5 is irrational number.

5. Prove that 5 √2 is irrational.

6. Prove that 2 √3 is irrational.

7. Can we have any n ∈ N , where 7n ends with the digit zero.
Answer : No

8. Without actually performing the long division, state whether the following rational number will have a terminating decimal expansion or non - terminating decimal expansion :
(i) 77/210
(ii) 15/1600
Answer : (i) Non-terminating (ii) Terminating

9. An army contingent of 616 members is to march behind and army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?
Answer : 8 columns

10. There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point ?
Answer : 36 minutes

11. Write a rational number between √2 and √3.
Answer : 3/2