CBSE Class 10 Mathematics Real Numbers Worksheet Set E

Question. Let x be the greatest number by which if we divide 366, 513 and 324, then in each case the remainder is the same. The sum of digit of x is:   
(A) 3
(B) 4
(C) 5
(D) 7

Answer: A

Question. If a number m is divided by 5 leaves a remainder 2, while another number n is divided by 5 leaves a remainder 4, then the remainder when (m + n) is divided by 5 is:     
(A) 1
(B) 2
(C) 3
(D) 4

Answer: A

Question. Among the numbers 2²⁵⁰ ,3²⁰⁰ ,4¹⁵⁰ and 5¹⁰⁰ the greatest is   
(A) 2²⁵⁰
(B) 3²⁰⁰ 
(C) 4¹⁵⁰ 
(D) 5¹⁰⁰ 

Answer: B

Question. The sum of the digit in (10⁴ⁿ²+8 + 1) , where ‘n’ is a positive integer, is……….   
(A) 4
(B) 4n
(C) 2+2n
(D) 4n2

Answer: A

Question. The HCF of two expressions p and q is 1. Their LCM is:     
(A) p + q
(B) p - q
(C) pq
(D) 1/pq

Answer: C

Question. The ascending order of  √2, ³√4, ⁴√6 is:     
(A) √2, ³√4, ⁴√6
(B) √2, ⁴√6, ³√4
(C) ³√4, √2, ⁴√6
(D) ⁴√6, ³√4, √2

Answer: B

Question. How many pairs of positive integers x, y exist such that H.C.F of x, y = 35 and the sum of x & y = 1085?   
(A) 10
(B) 2
(C) 3
(D) 15

Answer: D

Question. How many pairs of positive integers x, y exist such that H.C.F (x, y) + L.C.M (x, y) = 91? 
(A) 8
(B) 2
(C) 6
(D) None of these

Answer: A

Question. The smallest number which when increased by 17 is exactly divisible by both 520 and 468 is:     
(A) 4697
(B) 4663
(C) 4656
(D) 4680

Answer: D

Question. What is the smallest number which when increased by 5 is completely divisible by 8, 11 and 24?   
(A) 259
(B) 355
(C) 255
(D) None of these

Answer: A

Question. The sum of 18 consecutive positive integers is a perfect square. The smallest possible value of this sum is…   
(A) 169
(B) 225
(C) 289
(D) 361

Answer: B

Question. When the decimal point of a certain positive decimal number is moved two places to the right, the new number is two times the sum of the original number and the reciprocal of the original number. The product of 42 and the original number is:   
(A) 3
(B) 6
(C) 7
(D) 14

Answer: B

Question. Find the least number that when divided by 16, 18 and 20 leaves a remainder of 4 in each case, but is completely divisible by 7.     
(A) 2800
(B) 2882
(C) 2884
(D) None of these

Answer: C

Question. What least number should be subtracted from 1856 so that remainder when divided by 7, 12 and 16 is 4? 
(A) 170
(B) 172
(C) 174
(D) None of these

Answer: B

Question. The L. C. M. of two numbers is 48. The numbers are in the ratio 2 : 3. The sum of the numbers is. 
(A) 40
(B) 42
(C) 44
(D) None of these

Answer: A

Question. L. C. M. of two prime numbers x & y (x > y) is 161. The value of 3y – x is.     
(A) -1
(B) -2
(C) -3
(D) None of these

Answer: B

Question. Find the greatest number that will divide 43, 91 & 183 so as to leave the same remainder in each case.       
(A) 4
(B) 3
(C) 2
(D) None of these

Answer: A

Question. A number when divided by 342 gives a remainder 47. When the same number is divided by 19,what would be the remainder?
(A 9
(B) 8
(C) 18
(D) 11

Answer: A

Question. If xyz + xy + xz + yz + x + y + z = 384where x, y, z are positive integers, then the value of x, y , z is:   
(A) 23
(B) 18
(C) 20
(D) 35

Answer: C

Question. The H.C.F of the polynomials 9(x + a)^p (x – b)^q (x + c)^r and 12(x + a) ^p+3 (x- b)^{q – 3} (x + c)^{r + 2} is 3(x+ a)⁶ (x –b)⁶ (x + c)⁶, then the value of p + q – r is.   
(A) 10
(B) 11
(C) 9
(D) None of these

Answer: C

Question. If the H.C.F of 8x³ya and 12x^by² in 4x^ay^b, then find the maximum value of a + b.   
(A) 6
(B) 4
(C) 8
(D) None of these

Answer: B

Question. How many integers (a, b) exist such that the product of a, b and H.C.F (a, b) = 1080.   
(A) 12
(B) 9
(C) 10
(D) None of these

Answer: B

Question. Without actually performing the long division, find if 987/10500 will have terminating or non-terminating repeating decimal expansion. Give reason for your answer.
Sol. Terminating decimal expansion, because 987/10500 = 47/500 and 500 = 22 ´53.

Question. The values of remainder r, when a positive integer a is divided by 3 are 0 and 1 only. Is this statement true or false? Justify your answer.
Sol. No. According to Euclid’s division lemma
             a = 3q + r, where 0 £ r< 3
and r is an integer. Therefore, the values of r can be 0, 1 or 2.

Question. Using prime factorisation method, find the HCF and LCM of 30, 72 and 432. Also show that HCF × LCM ¹ Product of the three numbers.
Sol. Given numbers = 30, 72, 432
                         30 = 2 × 3 × 5
                         72 = 2³ × 3²
                       432 = 2⁴ × 3³
Here, 21 and 31 are the smallest powers of the common factors 2 and 3 respectively.
So, HCF (30, 72, 432) = 2¹ X 3¹ = 2 X 3 = 6
Again, 24 , 33 and 51 are the greatest powers of the prime factors 2, 3 and 5 respectively.
So, LCM (30, 72, 432) = 2⁴ X 3³ X 5¹ = 2160
                HCF × LCM = 6 × 2160 = 12960
    Product of numbers = 30 × 72 × 432 = 933120
Therefore, HCF × LCM ≠ Product of the numbers.

Question Find the LCM and HCF of 12, 15 and 21 by applying the prime factorisation method. [NCERT]
Sol. The prime factors of 12, 15 and 21 are
                12 = 2² X 3,15 = 3 X 5    and        21 = 3 X 7
Therefore, the HCF of these integers is 3
          2² , 3¹ , 5¹ and 7¹ are the greatest powers involved in the prime factors of 12, 15 and 21.
So, LCM (12,15, 21) = 2² X 3¹ X 5¹ X 7¹ = 420.

Question. 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 from 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? 
Sol. To find the time after which they meet again at the starting point, we have to find
LCM of 18 and 12 minutes. We have
                                     18 = 2 X 3²
                               and 12 = 2² X 3
Therefore, LCM of 18 and 12 = 2  X 3² = 36
So, they will meet again at the starting point after 36 minutes.

Question. Find the LCM and HCF of the following pairs of integers and verify that LCM ´ HCF = product of the two numbers.
(i) 26 and 91                                                          (ii) 198 and 144
Sol. (i) We have
                                           26 = 2 X13
                                           91 = 7 X13
Thus,                    LCM (26, 91) = 2 X 7 X 13 =182
HCF                             (26, 91) =13
Now, LCM (26, 91) X HCF (26, 91) =182 X13 = 2366
and Product of the two numbers = 26 X 91 = 2366
Hence,                     LCM X HCF = Product of two numbers.
(ii)                144 = 2⁴ X 3²
                    198 = 2 X 3² X11
∴  LCM (198,144) = 24 X 3² X 11 = 1584
   HCF (198,144) = 2 X 3² =18
Now, LCM (198, 144) × HCF (198,144) =1584 X 18 = 28512
and product of 198 and 144 = 28512
Thus, product of LCM (198, 144) and HCF (198, 144) = Product of 198 and 144.

Question. Write whether the square of any positive integer can be of the form 3m +2, where m is a natural number. Justify your answer.
Sol. No, because any positive integer can be written as 3q, 3q +1,3q +2, therefore, square will be 9q² = 3m,
9q² +6q +1 = 3 (3q² +2q) +1 = 3m+1, 9q² +12q + 4 = 3(3q² + 4q +1) + 1 = 3m+1.

Question. Can two numbers have 18 as their HCF and 380 as their LCM? Give reason.
Sol. No, because HCF (18) does not divide LCM (380).

Question. The product of two consecutive integers is divisible by 2. Is this statement true or false? Give reason.
Sol. True, because n(n +1) will always be even, as one out of the n or (n +1) must be even.

Question. “The product of three consecutive positive integers is divisible by 6”. Is this statement true or false? Justify your answer.
Sol. True, because n(n +1)(n +2) will always be divisible by 6, as at least one of the factors will be divisible by 2 and at least one of the factors will be divisible by 3.

Question. A rational number in its decimal expansion is 1.7351. What can you say about the prime factors of q when this number is expressed in the form p/q ? Give reason.
Sol. As 1.7351 is a terminating decimal number, so q must be of the form 2m 5n, where m,n are natural numbers.
Real Numbers

Question Write a rational number between √3 and √5 .
Sol. 1. Write a rational number between √3 and √5 .
Sol.  A rational number between √3 and √5 is √3.24 = 1.8 = 18/10=9/5

Question. Explain why 3 × 5 ×7 +7 is a composite number.
Sol. 3 × 5 ×7 + 7= 7 (3 × 5 +1) = 7 × 16, which has more than two factors.

Question. Can the number 4n, n being a natural number, end with the digit 0? Give reason.
Sol. If 4n ends with 0, then it must have 5 as a factor. But, (4)n = (22)n = 22n i.e., the only prime factor of 4n is 2. Also, we know from the fundamental theorem of arithmetic that the prime factorisation of each number is unique.
∴ 4n^∗ can never end with 0.

Question. Prove that √7 is irrational.
Sol. Let us assume, to the contrary, that √7 is rational.
Then, there exist co-prime positive integers a and b such that
√7 = a/b, b ≠ 0
So, a = 7 b
Squaring on both sides, we have
    a² = 7b² …(i)
⇒ 7 divides a2 ⇒ 7 divides a So, we can write
     a = 7c, (where c is any integer)
Putting the value of a = 7c in (i), we have
49c² = 7b²⇒ 7c² = b²
It means 7 divides b2 and so 7 divides b.
So, 7 is a common factor of both a and b which is a contradiction.
So, our assumption that √7 is rational is wrong.
Hence, we conclude that √7 is irrational.

Type A: Problems Based on Euclid’s Division Algorithm
Question. Use Euclid’s division algorithm to find the HCF of:
(i) 960 and 432                            (ii) 4052 and 12576. 
Sol. (i) Since 960 > 432, we apply the division lemma to 960 and 432.
We have
                            960 = 432 X 2 + 96
Since the remainder 96 ≠ 0, so we apply the division lemma to 432 and 96.
We have                432 = 96 ´ 4 + 48
Again remainder 48 ≠ 0, so we again apply division lemma to 96 and 48.
We have                  96 = 48 X 2 + 0
The remainder has now become zero. So our procedure stops.
Since the divisor at this stage is 48.
Hence, HCF of 225 and 135 is 48.
i.e.,     HCF (960, 432) = 48
(ii) Since 12576 > 4052, we apply the division lemma to 12576 and 4052, to get
                                 12576 = 4052 X 3 + 420
Since the remainder 420 ¹ 0, we apply the division lemma to 4052 and 420, to get
                                 4052 = 420 X 9 + 272
We consider the new divisor 420 and the new remainder 272, and apply the division lemma to get
                                 420 = 272 X 1 +148
We consider the new divisor 272 and the new remainder 148, and apply the division lemma to get
                                 272 =148 X1 +124
We consider the new divisor 148 and the new remainder 124, and apply the division lemma to get
                                 148 =124 X1 + 24
We consider the new divisor 124 and the new remainder 24, and apply the division lemma to get
                                 124 = 24 X 5 + 4
We consider the new divisor 24 and the new remainder 4, and apply the division lemma to get
                                 24 = 4 X 6 + 0
The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 4,
the HCF of 12576 and 4052 is 4.

Question. Find the largest positive integer that will divide 398, 436 and 542 leaving remainders 7, 11 and 15 respectively.
Sol. It is given that on dividing 398 by the required number, there is a remainder of 7. This means tha 398 – 7 = 391 is exactly divisible by the required number. In other words, required number is a factor of 391.
Similarly, required positive integer is a factor of 436 -11 = 425 and 542 -15 = 527.
Clearly, required number is the HCF of 391, 425 and 527.
Using the factor tree, we get the prime factorisations of 391, 425 and 527 as follows :
391 =17 X 23,           425 = 5² X 17            and 527 =17 X 31
∴ HCF of 391, 425 and 527 is 17.
Hence, required number = 17.

Question. Show that any positive odd integer is of the form 6q +1, or 6q + 3, or 6q + 5, where q is some integer.
Sol. Let a be any positive integer and b = 6. Then, by Euclid’s algorithm, a = 6q + r , for some integer q ³ 0
and 0 £ r < 6.
Mathematics X: Term – I
i.e., the possible remainders are 0, 1, 2, 3, 4, 5.
Thus, a can be of the form 6q, or 6q +1, or 6q + 2, or 6q + 3, or 6q + 4, or 6q + 5, where q is some quotient.
Since a is odd integer, so a cannot be of the form 6q, or 6q + 2, or 6q + 4, (since they are even).
Thus, a is of the form 6q +1, or 6q + 3, or 6q + 5, where q is some integer.
Hence, any odd positive integer is of the form 6q +1 or 6q + 3 or 6q + 5, where q is some integer.

Question. An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the  ximum number of columns in which they can march?
Sol. For the maximum number of columns, we have to find the HCF of 616 and 32.
Now, since 616 > 32, we apply division lemma to 616 and 32.
We have,        616 = 32 X 19 + 8
Here, remainder 8 ≠ 0. So, we again apply division lemma to 32 and 8.
We have,         32 = 8 X 4 + 0
Here, remainder is zero. So, HCF (616, 32) = 8
Hence, maximum number of columns is 8.

Question. Check whether 6n can end with the digit 0 for any natural number n.
Sol. If the number 6n, for any n, were to end with the digit zero, then it would be divisible by 5. That is, the prime factorisation of 6n would contain the prime 5. This is not possible because 6ⁿ = (2 X 3)ⁿ = 2ⁿ X 3ⁿ so the primes in factorisation of 6n are 2 and 3. So the uniqueness of the Fundamental Theorem of Arithmetic guarantees that there are no other primes except 2 and 3 in the factorisation of 6ⁿ . So there is no natural number n for which 6ⁿ ends with digit zero.

Question. Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m +1 for some integer m. 
Sol. Let a be any positive integer, then it is of the form 3q, 3q +1 or 3q + 2. Now, we have to show that the square of these numbers can be rewritten in the form of 3m or 3m + 1.
Here, on squaring, we have
     (3q)² = 9q² = 3(3q2) = 3m,                                                                  where m = 3q²
         (3q +1)² = 9q² + 6q +1 = 3 (3q² + 2q) +1 = 3m +1,                         where m = 3q² + 2q
and,  (3q + 2)² = 9q² +12q + 4 = (9q² +12q + 3) +1
                       = 3 (3q² + 4q +1) +1 = 3m +1,                                          where m = 3q² + 4q +1.
Hence, square of any positive integer is either of the form 3m or 3m +1.

Question. Show that one and only one out of n, n +2, n + 4 is divisible by 3, where n is any positive integer.
Sol. Let q be the quotient and r be the remainder when n is divided by 3.
Therefore, n = 3q + r, where r = 0, 1, 2
⇒              n = 3q            or n = 3q+ 1 or n = 3q + 2.
Case (i) if n = 3q, then n is divisible by 3.
Case (ii) if n = 3q +1 then n +2 = 3q +3= 3(q +1), which is divisible by 3 and n + 4 = 3q +5, which is not divisible by 3.
So, only (n +2) is divisible by 3.
Case (iii) if n = 3q +2, then n +2 = 3q + 4, which is not divisible by 3 and (n + 4) = 3q +6 = 3(q +2), which is divisible by 3.
So, only (n + 4) is divisible by 3.
Hence one and only one out of n, (n +2), (n + 4) is divisible by 3.

Question. An army contingent of 616 members is to march behind an 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?
Solution. Clearly, the maximum number of columns in which members can march is the HCF of 616 and 32. So,
let us find the HCF of 616 and 32 by Euclid’s division algorithm.
Since, 616 > 32. Let a = 616, b = 32
Applying Euclid’s division algorithm, we get
616 = 32 × 19 + 8
Since the remainder is non-zero, so we apply the Euclid’s division algorithm to 32 and 8, to get
32 = 8 × 4 + 0
We observe that the remainder is zero at this stage.
∴ The divisor at this stage i.e. 8 is the HCF of 616 and 32.
Hence, the required maximum number of columns is 8.

Question. Use Eculid’s division demma, to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.
Solution. Let x be any positive integer. Then, it is of the form 3q or, 3q + 1 or, 3q + 2.
So, we have the following cases :
Case I : When x = 3q.
then, x³ = (3q)³ = 27q³ = 9 (3q3) = 9m, where m = 3q3.
Case II : When x = 3q + 1
then, x³ = (3q + 1)3
= 27q³ + 27q2 + 9q + 1
= 9 q (3q² + 3q + 1) + 1
= 9^m + 1, where m = q (3q² + 3q + 1)
Case III. When x = 3q + 2
then, x³ = (3q + 2)³
= 27 q³ + 54q2 + 36q + 8
= 9q (3q² + 6q + 4) + 8
= 9 m + 8, where m = q (3q2 + 6q + 4)
Hence, x³ is either of the form 9 m or 9 m + 1 or, 9 m + 8.

Question. The following real numbers have decimal expression as given below. In each case decide whether
they are rational or not. If they are rational, and of the form p/q, what can you say about the prime factors of q?
(i) 43.123456789 (ii) 0.120120012000120000.... (iii) 43.123456789
Solution. (i) 43.123456789 is a rational number of the form p/q and q is of the form 2^m × 5ⁿ.
(ii) 0.120120012000120000 .... is not a rational number since it has non-recurring and non-terminating decimal and q is not of the form 2m × 5n.
(iii) 43.123456789 is a rational number having recurring and non-terminating decimal expansion since q is not of the form 2^m × 5ⁿ.

Question. Use Eculid’s division algorithm to find the HCF of 135 and 225.
Solution. Since 225 > 135. Let a = 225, b = 135.
Applying Eculid’s division algorithm, we get
225 = 135 × 1 + 90
Since the remainder 90 ≠ 0. So, we apply the euclid’s division
algorithm to 135 and 90, to get
135 = 90 × 1 + 45
Again, since the remainder is non-zero, so we apply the euclid’s division algorithm to 90 and 45, to get
90 = 45 × 2 + 0
We observe that remainder at this stage is zero. ∴ the divisor at this stage i.e. 45 is the HCF of 135 and 225.
Symbolically we write HCF (135, 225) = 45 Ans.

Question. Prove that √2 is an irrational number.
Solution. Let us assume, to the contrary, that √2 is rational. So, we can find integers r and s (≠ 0) such that √2 = r/s . Suppose r and s have a common factor other than 1. Then, we divide by the common factor to get  √2 = a/b , where a and b are co-prime.
So, b√2 = a.
Squaring on both sides, we get 2b² = a2. Therefore, 2 divides a2 ⇒ 2 divides a.
So, we can write a = 2c for some integer c.
Substituting for a, we get 2b² = 4c², or b2 = 2c². This means that 2 divides b², and so 2 divides b.
Therefore, a and b have atleast 2 as a common factor. But this contradicts the fact that a and b
have no common factors other than 1.
This contradiction has arisen because of our incorrect assumption that √2 is rational.
So, we conclude that √2 is irrational. Hence proved.

Question. Prove that 3+2√ 5 is irrational.
Solution. Let us assume on the contrary that 3 + 2 √5 is rational. Then there exists co-prime integers a and b such that,
3 + 2√5 
⇒ 2√5 = a/b -3
⇒ √5 = a - 3b/2b
⇒ √5 is rational. [∴ a, b are integers, ∴ a - 3b/2b is a rational]
This contradicts the fact that √5 is irrational. So, our supposition is incorrect. 
Hence, 3 + 2√5 is an irrational number. 

Question. Without actually performing the long division, state whether the following rational number will have a terminating decimal expansion or a non-terminating repeating decimal expansion :
(i) 13/3125
(ii) 17/8
(iii) 64/455
(iv) 15/1600
(v) 29/343
Solution. (i) 13/3125 = 13 /55 = 13/20 x 55
So, the denominator is of the form 2m × 5n. Hence, it has terminating decimal expansion.
(ii) 17/8 = 17/23 = 17/23x50
So, the denominator is of the form 2m × 5n. Hence, it has terminating decimal expansion.
(iii) 64/455 = 64/5 x 7 x 13
Clearly, the denominator is not of the form 2m × 5n. Hence, it has non-terminating repeating decimal expansion.
(iv) 15/1600 = 3/320 = 3/26 x 51
So, the denominator is of the form 2m × 5n. Hence, it has terminating decimal expansion.
(v) 29/343 = 29/35
Clearly, the denominator is not of the form 2m × 5n. Hence, it has non-terminating repeating decimal expansion.

Question. Find the HCF of 96 and 404 by prime factorisation method. Hence, find their LCM.
Solution. We have, 96 = 25 × 3 and 404 = 22 × 101
∴ HCF = 22 = 4.
Now, HCF × LCM = 96 × 404
∴ LCM = 96 x 404 / HCF = 96 x 404 /4 = 96 x 101 = 9696 Ans.

Question. Check whether 6n can end with the digit 0 for any natural number n.
Solution. We have, 6n = (2 × 3)n = 2ⁿ × 3ⁿ. Therefore, prime factorisation of 6n does not contain 5 as a factor.
Hence, 6ⁿ can never end with the digit 0 for any natural number n.