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Q1. State whether the following real numbers are rational or not:       (i)              (ii)  47.5369821               (iii)  3.2010010001……

Solution

(i)  rational        (ii)  rational                      (iii)  irrational

Q2. Find the HCF of 75 and 243 by Euclid’s division lemma. 

• 1) 1
• 2) 3
• 3) 6
• 4) 18

Solution

243 = 75 × 3 + 18 75 = 18 × 4 + 3 18 = 3 × 6 + 0 Hence, HCF of 75 and 243 is 3.

Q3. Give an example and show that the sum of two irrational may be an irrational number  but their product may not  always be an irrational number.

Solution

Consider two numbers  and Addition= Product=10 So we see that  the addition  is an irrational number but the product is a rational number.

Q4. Which of the following is rational?

• 1)
• 2)
• 3)
• 4)

Solution

Since, , it is rational.

Q5. For any two positive integers a and b, there exist unique integers q and r such that a = bq + r, 0 ≤ r < b. If b = 4 then which is not the value of r?

• 1) 1
• 2) 2
• 3) 4
• 4) 3

Solution

Given, a = bq + r, 0 ≤ r < b and b = 4 So, r can take values 0, 1, 2, 3. Thus, r ≠ 4.

Q6. If H.C.F. of 65 and 117 is expressible in the form of 65 m - 117, then the value of m is :

• 1) 2
• 2) 3
• 3) 1
• 4) 4

Solution

65 = 13 × 5 and 117 = 13 × 9 So, HCF (65, 117) = 13 Because HCF of 65 and 117 is 13,which is expressible in the form 65 m - 117 Thus we get 65 m - 117 = 13 65 m = 130m = 2

Q7. A rational number between and  is:

• 1) 1.4
• 2) 1.75
• 3) 1.8
• 4) 1.45

Solution

= 1.414… = 1.732… Amongst the given alternatives, 1.45 is a rational number between and .

Q8. Find HCF of 592 and 252 by using Euclid’s Division Algorithm.

• 1) 76
• 2) 6
• 3) 4
• 4) 12

Solution

In the given pair, 592 > 252, thus let 592 = a and 252 = b. Now by applying the Euclid’s division algorithm a = bq + r, we get 592 = 252 × 2 + 88 252 = 88 × 2 + 76 88 = 76 × 1 + 12 76 = 12 × 6 + 4  12 = 4 × 3 + 0 Since in the above equation we get r = 0; therefore, 4is the HCF of the given pair 592 and 252. 

Q9. Without actually performing division state whether the number will have a terminating decimal representation or not.

Solution

A rational number will have a terminating decimal representation only if the denominator can be expressed  in terms of prime numbers 2 and 5. We see that, 343=7 x 7 x 7So according to the condition given above, the denominator of 29/343 can not be expressed fully in terms of 2 and 5. Hence, the number can't have a terminating decimal representation.

Q10. Given that HCF (26 , 91) = 13, then LCM of (26 , 91) is :

• 1) 2366
• 2) 91
• 3) 364
• 4) 182

Solution

HCF(26, 91) = 13 and let the LCM(26, 91) = x We know LCM  HCF = 26 9113 x = 26 91 x =

Q11. Is 7 6 5 4 3 2 1 + 5 a composite number? Justify your answer.

Solution

7 6 5 4 3 2 1 + 5 = 5040 + 5 = 5045 It has more than two prime factors. Thus, the given number is a composite number.

Q12. The value of x in the factor tree is:

• 1) 150
• 2) 100
• 3) 50
• 4) 30

Solution

To find the value of x, we will have to go backwards. 2 × 3 = 6 Further, 6 × 5 =  30 Then 30 × 5 = 150 Thus, x = 150

Q13. The length, breadth and height of a room is 6m 80 cm, 5m 10 cm and 3 m 40 cm, respectively. Find the longest tape which can measure the dimensions of the room exactly. 

• 1) 1m 60 cm
• 2) 2m 40 cm
• 3) 2m 70 cm
• 4) 1m 70 cm

Solution

6m 80 cm = 680 cm 5m 10 cm = 510 cm 3m 40 cm = 340 cm We have to find HCF of 680, 510 and 340. HCF of 680 and 510, 680 = 510 × 1 + 170 510 = 170 × 3 + 0 ∴ HCF (680, 510) = 170 HCF of 170 and 340, 340 = 170 × 2 + 0  ∴ HCF (170, 340) = 170 ∴ HCF (680, 510, 340) = 170 170 cm = 1m 70 cm

Q14. Prove that tan1° tan2° tan3°… tan89° = 1. OR Prove that    

Solution

tan1° tan2° tan3° … tan89° = tan1° tan2° tan3°… tan44° tan45° tan46°… tan88° tan89° = tan1° tan89° tan2° tan88° ………. tan 44° tan46° tan45°                           (½) = tan1° tan(90° - 1°) tan2° tan(90° - 2°) … tan 44° tan(90° - 44°) tan45°   (½) = tan1° cot1° tan2° cot2° … tan 44° cot44° tan45°                                     (½) = 1                                                                                                          (½) OR                                                (½)                                                                                    (½)                                                                                                    (½) = tan A ∴                                                                  (½)    

Q15. What is the probability that a leap year selected at random will have 53 Sundays? OR What is the probability that a number selected from the first 25 natural numbers is a prime number?   

Solution

A leap year contains 366 days, i.e. 52 weeks and 2 additional days. The possibility for these two additional days are   Sample space = {(Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday)} n(S) = 7                                                                                                        (1) Thus, there are seven possibilities out of which two are favourable for getting the 53^rd Sunday.  A = {(Sunday, Monday), (Saturday, Sunday)}                                                   (1) n(A) = 2 P(A) = n(A) / n(S) = 2/7                                                                                 (1) OR  S = {1, 2, 3, 4, 5, … 25}  n(S) = 25                                                                                                        (1)   Let A denote the event of getting a prime number.  A = {2, 3, 5, 7, 11, 13, 17, 19, 23}                                                                   (1)  n(A) = 9  P(A) = n(A)/n(S) = 9/25                                                                                    (1) 

Q16. Euclid’s division lemma can be used to find

• 1) Prime factors of numbers
• 2) HCF of numbers
• 3) Product of numbers
• 4) LCM of numbers

Solution

Euclid’s division lemma can be used to find HCF of numbers.

Q17. Use Euclid's division algorithm to find the HCF of 10224 and 9648.

Solution

Here 10224> 9648 By Euclid Division Algorithm, 10224 = 9648 1 + 576 9648 = 576 16 + 432 576 = 432 1 + 144 432 = 144 3 + 0 HCF (10224, 9648) = 144

Q18. Solve :  

Solution

                                   (1)                               (1) x = −4, 3                                                                              (1) 

Q19. Find the H.C.F. and L.C.M. of 6, 72 and 120 using the prime factorisation method.

Solution

Q20. Prove that is irrational.

Solution

is an irrational while is rational and an irrational number can never be equal to a rational number. Thus our assumption is wrong. Hence, is irrational.

Q21. Which of the following is a non-terminating, repeating decimal?

• 1)
• 2)
• 3)
• 4)

Solution

The denominator of the rational number is not of the form, Thus, is a non-terminating, repeating decimal.

Q22. Prove that 13 + 25 is irrational.

Solution

Let us assume, on contrary that 13 + 25 is rational. 13 + 25 = , where a and b are integers and b ≠ 0. a and b are integers, is rational. is rational. But this contradicts the fact that is irrational. This contradiction has arisen because of our incorrect assumption that 13 + 25 is rational. Hence, 13 + 25 is irrational.

Q23. Find the ratio in which the line 2x + y - 4 = 0 divides the line segment joining a(2, −2) and B(3, 7). OR Find the value of k for which the area formed by the triangle with vertices A(k, 2k), B(−2, 6) and C(3, 1) is 5 square units. 

Solution

Let the required ratio be k:1.  Let P(x, y) be the point on the given line dividing AB in the ratio k:1. Then   and  (1) This point P(x, y) lies on the line 2x + y - 4 = 0.                                                      (1) 6k + 4 + 7k - 2 = 4k + 4 9k = 2 k = 2/9                                                                                        (1) Hence, the required ratio is 2:9.                                                    (1)    OR    Vertices of the given triangle ABC are A(k, 2k), B(−2, 6) and C(3, 1).  x₁ = k, y₁ = 2k, x₂ = −2, y₂ = 6 and x₃ = 3, y₃ = 1   Area of triangle ABC = = = =                                                       (2) According to the question,                                                       (2) 

Q24. Given that HCF (2520, 6600) = 120, LCM (2520, 6600) = 252k, then the value of k is:

• 1) 625
• 2) 165
• 3) 500
• 4) 550

Solution

We know that the product of two numbers is equal to the product of their LCM and HCF.

Q25. is 

• 1) Irrational number
• 2) Rational number
• 3) Prime number
• 4) Negative number

Solution

 = 18 = …. form

Q26. Which of the following rational number has a terminating decimal expansion?

• 1)
• 2)
• 3)
• 4)

Solution

We know that a rational number will have a terminating decimal expansion if q is of the form 2ⁿ5^m, where n, m are non-negative integers. Thus, has a terminating decimal expansion.

Q27. Show that any positive even integer is of the form 6m, 6m + 2 or 6m + 4, where m is some integer.

Solution

Let a and b be any positive integers, a = bm + r, 0 r <b Let b = 6. Then, r = 0,1,2,3,4,5 a = 6m + r, where r = 0, 1, 2, 3, 4, 5 When r = 0, a = 6m even When r = 1a = 6m + 1odd When r = 2a = 6m + 2even When r = 3a = 6m + 3odd When r = 4a = 6m + 4even When r = 5a = 6m + 5odd Hence, all positive even integers are of the form 6m, 6m + 2 or 6m + 4.

Q28. 119² - 111² is:

• 1) An odd prime number
• 2) Composite number
• 3) An odd composite number
• 4) Prime number

Solution

119² - 111² = (119 + 111) (119 - 111) = 230 8 (Even number) Hence, 119² - 111² is a composite number, since a composite number has more than 2 factors other than itself and 1.

Q29. Find HCF of 1288 and 575 by using Euclid’s Division Algorithm.

• 1) 138
• 2) 6
• 3) 23
• 4) 4

Solution

In the given pair, 1288 > 575, thus let 1288 = a and 575 = b. Now by applying the Euclid’s division algorithm a = bq + r, we get 1288 = 575 × 2 + 138 575 = 138 × 4 + 23 138= 23 × 6 + 0 Since in the above equation we get r = 0; Therefore, 23 is the HCF of the given pair 1288 and 575.

Q30. Find HCF of 35, 56 and 91.

• 1) 21
• 2) 14
• 3) 35
• 4) 7

Solution

First find the HCF of 35 and 56.  56 = 35 × 1 + 21  35 = 21 × 1 + 14  21 = 14 × 1 + 7 14 = 7 × 2 + 0 So, HCF (35, 56) = 7  Now, HCF (7, 91)  91 = 7 × 13 + 0  So, HCF (7, 91) = 7  Therefore, HCF (35, 56, 91) = 7 Hence, HCF of 35, 36, 91 is 7. 

Q31. Find HCF of 963 and 657 by using Euclid’s Division Algorithm.

• 1) 45
• 2) 36
• 3) 4
• 4) 9

Solution

In the given pair, 963 > 657, thus let 963 = a and 657 = b. Now by applying Euclid’s division algorithm a = bq + r, we get 963 = 657 × 1 + 306 657 = 306 × 2 + 45 306 = 45 × 6 + 36 45 = 36 × 1 + 9  36 = 9 × 4 + 0 Since in the above equation we get r = 0; therefore, 9 is the HCF of the given pair 963 and 657.

Q32. Find HCF of 24, 15 and 36.

• 1) 3
• 2) 9
• 3) 6
• 4) 1

Solution

First find the HCF of 24 and 15.  24 = 15 × 1 + 9  15 = 9 × 1 + 6  9 = 6 × 1 + 3 So, HCF (24, 15) = 3  Now, HCF (3, 36)  36 = 3 × 13 + 0  So, HCF (3, 36) = 3  Therefore, HCF (24, 15, 36) = 3 

Q33. Euclid's division lemma states that if a and b are two positive integers, then there exist unique integers q and r such that:

• 1) a = bq + r, 0 r b
• 2) a = bq + r, 0 < r < b
• 3) a = bq + r, 0 < b r
• 4) a = bq + r, 0 r < b

Solution

Recall Euclid's division lemma. Euclid's division lemma states that if a and b are two positive integers, then there exist unique integers q and r such that: a = bq + r, 0 r <b

Q34. Prove that 7 - 2 is irrational.

Solution

Suppose 7 - 2 is rational. 7 - , where q 0, p and q are co-prime But is irrational while is rational and an irrational number can never be equal to a rational number. Thus, our assumption is wrong and hence 7 - is irrational.

Q35. If the HCF of 85 and 153 is expressible in the form 85n - 153, then value of n is :

• 1) 3
• 2) 1
• 3) 2
• 4) 4

Solution

HCF of 85 and 153 : 153 = 85 x 1 + 68 85 = 68 x 1 + 17 68 = 17 x 4 + 0 HCF of (85, 153) = 17 It is given that the HCF is expressible as 85n - 153. 85n - 153 = 17 85n = 17 + 153 = 170 n =  = 2

Q36. The decimal expansion of will terminate after how many places of decimal?

• 1) 4
• 2) 1
• 3) 3
• 4) 2

Solution

Thus, the decimal expansion of the given rational number will terminate after 3 places of decimal.

Q37. prove that is irrational number.                     

Solution

 Let us assume, to the contrary that is rational                 =  or              --- (1) Therefore 2 divides  and 2 divides p.  So we can  write  p = 2r for some integer r Substituting this value of p = 2r in (1), we get                                     This means 2 divides to  or 2 divides q also.  Therefore, p and q have at least 2 as a common factor.  But this contradict the fact that p and q have no common factor other than 1. This contradiction arises because of our wrong assumption that  is a rational number.  So we conclude that   is irrational number.

Q38. 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

Let x be the maximum number of columns is which the two groups can march, then x is HCF of 616 and 32. Here 616 > 32 By Euclid division algorithm, 616 = 32 × 19 + 8 32   = 8 × 4 + 0 Thus, HCF (616, 32) = 8 Hence the maximum number of columns in which they can march is 8.

Q39. Use Euclid's division lemma to show that cube of any positive integer is either of form 9q, 9q + 1, or 9q + 8 for some integer q.

Solution

Let a be any positive integer and b = 3 a = 3m + r, where  and  Therefore, every number can be represented as these three forms. There are three cases. Case 1: When a = 3m, Where q is an integer such that q = Case 2: When a = 3m + 1, a³ = (3m +1)³ a³ = 27m³ + 27m² + 9m + 1 a³ = 9(3m³ + 3m² + m) + 1 a³ = 9q + 1 Where q is an integer such that q = (3m³ + 3m² + m) Case 3: When a = 3m + 2, a³ = (3m + 2)³ a³ = 27m³ + 54m² + 36m + 8 a³ = 9(3m³ + 6m² + 4m) + 8 a³ = 9q + 8 Where q is an integer such that q = (3m³ + 6m² + 4m) Hence, the cube of any positive integer is of the form 9q, 9q + 1, or 9q + 8.

Q40. Find the largest number that will divide 398, 436 and 542 leaving remainder 7, 11 and 15 respectively.

Solution

Find the largest number that will divide 398, 436 and 542 leaving remainder 7, 11 and 15 respectively. Clearly, the required number is the H.C.F of 398 - 7 = 391, 436 - 11 = 425 and 542 - 15 = 527. From 391, 425 and 527, we will take 391 and 425. Here 425 > 391 425 = 391 × 1 +  34 391 =  34  × 11 +  17 34   =   17 × 11 +  0 Now the remainder is zero. Now we will find the HCF of 527 and 17. 527 = 17 × 31 + 0 Now the remainder is zero. Hence largest number i.e the required number is 17.

Q41. Find HCF of 506 and 1155 by using Euclid’s Division Algorithm.

• 1) 2
• 2) 1
• 3) 11
• 4) 5

Solution

In the given pair, 1155 > 506, thus let 1155 = a and 506 = b. Now by applying the Euclid’s division algorithm A = bq + r, we get 1155 = 506 × 2 + 143 506 = 143 × 3 + 77 143 = 77 × 1 + 66 77 = 66 × 1 + 11  66 = 11 × 6 + 0 Since in the above equation we get r = 0; therefore, 11 is the HCF of the given pair 1155 and 506.

Q42. State why  is a non-terminating repeating decimal.

Solution

6²  × 5³  = (2 × 3)² × 5³  = 2² × 3² × 5³ The prime factorization of the denominator contains a prime factor 3 other   than 2 and 5. Therefore the decimal expansion of a given rational number is non - terminating repeating.

Q43. To prove  is an irrational number, which of the following methods is used? 

• 1) Indirect method
• 2) None of these
• 3) Contradictory method
• 4) Direct method

Solution

To prove  is an irrational number, the contradictory method is used.

Q44. H.C.F. of two consecutive even numbers is:

• 1) 1
• 2) 0
• 3) 4
• 4) 2

Solution

We know that 2 is a factor of all the even numbers. Consider an even number, say 2q. This number will have its next consecutive even number as 2q+2  Further, this number will have its consecutive even number are 2q + 2 + 2= 2q + 4 2q + 2 can be written as 2(q+1), while 2q + 4 can be written as 2(q+2).  So, the only common factor is 2. The HCF of two consecutive even numbers is 2.

Q45. If two positive integers A and B can be expressed as A = ab² and B = a³b, where a and b are prime numbers, then LCM (A, B) is:

• 1) a⁴b³
• 2) ab
• 3) a²b²
• 4) a³b²

Solution

A = ab² and B = a³b LCM (A, B) = a³b²

Q46. Find the HCF of 117 and 45 by Euclid’s division lemma.

• 1) 8
• 2) 9
• 3) 7
• 4) 10

Solution

117 = 45 × 2 + 27 45 = 27 × 1 + 18 27 = 18 × 1 + 9 18 = 9 × 2 + 0 So, the HCF of 117 and 45 by Euclid’s division lemma is 9.

Q47. Show that 4ⁿ can never end with the digit zero for any natural number n.

Solution

Hence 4ⁿ can never end with the digit zero for any n N. We have 4ⁿ, where n = 1, 2, 3, 4, ... If n = 1 then 4ⁿ = 4¹ = 4 If n = 2 then 4ⁿ = 4² = 16   and so on. If a number ends with zero then it is divisible by 5. Here, 4 and 16 are not divisible by 5. Therefore, 4ⁿ can never end with zero fro any natural number n.

Q48. Show that every positive even integer is of the form 2q and that every positive odd integer is of the form 2q + 1, where q is some integer.

Solution

By Euclid's lemma, a = bq + r, 0 r < b, a, b, q are integers. Let b = 2 a = 2q when r = 0, this is an even number and a = 2q + 1, when r = 1, which is an odd number.

Q49. Find the H.C.F. and L.C.M. of 6, 72 and 120 using the prime factorisation method.

Solution

6    = 2 x 3 72  = 2 x 2 x 2 x 3 x 3 120= 2 x 2 x 2 x 3 x 5   Marking the factors common to all the three numbers, we see that the HCF = 2 X 3=6 LCM =

Q50. D is a point on the side BC of ΔABC such that ∠ADC = ∠BAC. Prove that CA² = CB × CD. OR  Prove that the line segment joining the midpoints of any two sides of a triangle is parallel to the third side. 

Solution

                                                                                                                        (1) In ΔABC and ΔDAC, ∠ADC = ∠BAC and ∠C = ∠C According to the AA-criterion of similarity, ΔABC ≅ ΔDAC                                                                                                    (1)                                                                                    (1) OR                                                                                                                 (1)    Given: Triangle ABC in which D and E are the midpoints of AB and AC, respectively.  To prove: DE is parallel to BC.                                                                    (1) Proof : Since D and E are the midpoints of AB and AC, respectively. We have AD = DB and AE = EC. ∴ Hence, by the converse of Thale’s theorem, DE is parallel to BC.                     (1) 

Q51. Prove that is irrational.

Solution

Let us suppose that is rational. Then, where a and b are integers and b 0 8 + 2 = 8 + 4 = Since, a, b, 8 and 4 are integers, is rational.   is a rational number. But we know that is irrational. irrational = rational, which is a contradiction Hence, is an irrational.

Q52. Prove that is irrational.

Solution

Suppose is rational. and HCF (p, q) = 1 5q² = p².... (1) 5 is a factor of p², therefore, 5 is a factor of p. Now, suppose p = 5 m From (1), 5q² = 25 m² q² = 5 m² 5 is a factor of q² and therefore 5 is a factor of q. Thus, 5 is a common factor of p and q, which contradicts to our assumption of HCF (p, q) = 1 Hence, is irrational.

Q53. Find HCF of 155 and 1385 by using Euclid’s Division Algorithm.

• 1) 25
• 2) 10
• 3) 15
• 4) 5

Solution

In the given pair, 1385 > 155, thus let 1385 = a and 155 = b. Now by applying the Euclid’s division algorithm a = bq + r, we get 1385 = 155 × 8 + 145 155 = 145 × 1 + 10 145 = 10 × 14 + 5 10 = 10 × 2 + 0 Since in the above equation we get r = 0; therefore, 5is the HCF of the given pair 1385 and 155. 

Q54. Prove that is irrational.

Solution

Let be a rational number. Then,, Where a and b are co-prime and b 0 As a and b are integers, is a rational number is rational This is a contradiction as is irrational. is irrational.

Q55. Find the HCF of 876 and 255 by Euclid’s division lemma.

• 1) 5
• 2) 37
• 3) 111
• 4) 3

Solution

876 = 225 × 3 + 111 225 = 111 × 2 + 3 111 = 3 × 37 + 0 Hence, HCF of 876 and 255 is 3.

Q56. The radius and height of a cylinder are in the ratio 2:7. If the curved surface area of the cylinder is 352 sq. cm, find its radius. 

Solution

Let the radius be 2x and height be 7x.  Curved surface area = 352 cm²  2πrh = 352     (½)   88x² = 352           (1) x = 2  Radius = 2 × 2 = 4 cm      (½) 

Q57. Prove that 5 + is an irrational number. OR Prove that is an irrational number.

Solution

Let 5 + 7 be a rational number. Then, 5 + 7 = 7 = = Since, p and q are integers, is a rational number. Therefore, is also a rational number. This is a contradiction as we know that is irrational. Out assumption is wrong. Hence, 5 + 7 is irrational.   OR   Let be a rational number. Let = , where q 0, p and q are integers and are co-prime. 7 q² = p² Thus, 7 divides p² and hence 7 divides p. Let p = 7m 7q² = 49 m² q² = 7m² 7 divides q² 7 divides q 7 divides p and q both. This is a contradiction as p and q are co-prime. Hence, is an irrational number.

Q58. Find the HCF of 70 and 30 by Euclid’s division lemma. 

• 1) 10
• 2) 20
• 3) 5
• 4) 2

Solution

70 = 30 × 2 + 10 30 = 10 × 3 + 0 Hence, HCF of 70 and 30 is 10.

Q59. Find the HCF of 135 and 225 by Euclid’s division lemma.

• 1) 90
• 2) 45
• 3) 10
• 4) 30

Solution

225 = 135 × 1 + 90 135 = 90 × 1 + 45 90 = 45 × 2 + 0 Hence, HCF of 135 and 225 is 45.

Q60. Find HCF of 84, 90 and 120.

• 1) 7
• 2) 6
• 3) 14
• 4) 20

Solution

First find the HCF of 84 and 90.  90 = 84 × 1 + 6 84 = 6 × 14 + 0 So, HCF (90, 84) = 6 Now, HCF (6, 120)  120 = 6 × 20 + 0  So, HCF (6, 120) = 6  Therefore, HCF (84, 90, 120) = 6 

Q61. Find the HCF of 105 and 120 by Euclid’s division lemma. 

• 1) 25
• 2) 1
• 3) 7
• 4) 15

Solution

120 = 105 × 1 + 15 105 = 15 × 7 + 0 Hence, HCF of 105 and 120 is 15.

Q62. For the decimal number , the rational number is:

• 1)
• 2)
• 3)
• 4)

Solution

Let x =  Then, x = 0.7777…………(1) Here, the number of digits recurring is only 1, so we multiply both sides of the equation by 10. …….(2) Subtracting (1) from (2), we get: 9x = 7

Q63. If d is the HCF of 45 and 27, find x, y satisfying d = 27x + 45y.

Solution

Since 45 > 27 By applying Euclid's division Lemma to 27 and 45, we get, 45 = 27 1 + 18 27 = 18 1 + 9 18 = 9 2 + 0   The remainder has now become zero, so we will stop the procedure.   Since the divisor at this stage is 9. HCF (45, 27) = 9 9 = 27 - 18    = 27 - (45 - 27 1)    = 27 - 45 + 27    = 54 - 45    = 27 2 - 45 1 Comparing this with d = 27x + 45y, we get, x = 2, y = -1

Q64. If x = 2³ 3 5², y = 2² 3², then HCF (x, y) is :

• 1) 36
• 2) 12
• 3) 6
• 4) 18

Solution

Q65. Find the HCF and LCM of 336 and 54. Verify that HCF LCM = Product of the two number.

Solution

336 = 2 2 2 2 3 7 54 = 2 3 3 3 HCF (336, 54) = 2 3 = 6 LCM (336, 54) = 2⁴ 3³ 7 = 3024 Verification: HCF LCM = 6 3024 = 18144 Product of numbers = 336 54 = 18144

Q66. A man observes a bird in the east flying at an angle of elevation 30°. After 2 minutes, he observes that the bird has flown west at an angle of elevation 60°. If the bird flies in a straight line all along at a height h√3, find the speed of the bird. OR A bridge across a river makes an angle of 45° with the river bank. If the length of the bridge across the river is 150 m, what is the width of the river?   

Solution

  In triangle BPD, tan 60˚ = In triangle APD, tan 30˚ = DA = √3 √3 h = 3h BD + DA = h + 3h = 4h                                                   (1) The distance travelled by the bird in 2 minutes = 4h m  Speed = 4h/2 metre per minute                                     (½)   OR     In the right-angled triangle ABC,   sin 45˚ =                                                                (½)                                                                            (½)                                                                      (1)  Hence, the width of the river is  m.                              (1) 

Q67. Show that 2√3 is irrational. 

Solution

If possible, let 2√3 be rational. Let its simplest form be 2√3 = a/b, where a and b are positive integers having no common factor other than 1. Then,   Since a and 2b are non-zero integers,  is rational.                            (1) Thus, from (i), it follows that √3 is rational. This contradicts the fact that √3 is irrational. The contradiction arises by assuming that 2√3 is rational. Hence, 2√3 is irrational.   (1)

Q68. Find the H.C.F. and L.C.M. of 6, 72 and 120 using the prime factorisation method.

Solution

Q69. The areas of two similar triangles ABC and PQR are in the ratio 9:16. If BC = 4.5 cm, find the length of QR.

Solution

Since ∆ABC  ~ ∆PQR, According to the question,      (½)   QR² = 36  QR = 6  (½) 

Q70. show that  is irrational.

Solution

Let = a rational number                     rearranging this equation  Since 12 and  are rational number which mean  is also a rational number. But this contradicts the fact that  is an irrational because we already learnt that square roots of prime numbers are irrational. So, we conclude that  is irrational.

Q71. A proven statement used for proving another statement is called

• 1) Procedure
• 2) Walkthrough
• 3) Lemma
• 4) Algorithm

Solution

A proven statement used for proving another statement is called a lemma. 

Q72. Prove that  is an irrational number.

Solution

Suppose  is not an irrational number, then it must be a rational number. So let =q where q is a rational number. )= = = So here we have an equation with its   left side is a rational number(as the set of rational numbers is closed under subtraction and division i.e the subtraction and division of two rational numbers is always a rational number)and  so its right side must also be a rational number.  is a rational number Which is a   contradiction. So our basic assumption   must be wrong. =q where q is a rational number,is wrong  is an irrational number. Hence proved.

Q73. Show that  is irrational.

Solution

Let us assume that is a rational i.e. Or,   Here  and 3 both are rational and we know that the difference of two rational is always rational. this is a contradiction to our assumption. hence, is not rational, i.e. it is an irrational number

Q74. Find the HCF of 1000 and 1125.

• 1) 100
• 2) 150
• 3) 75
• 4) 125

Solution

1125 = 1000 × 1 + 125 1000 = 125 × 8 + 0 Hence, HCF of 1000 and 1125 is 125.

Q75. Find the value of k for which x = 2 is a solution of the equation kx2 + 2x - 3 = 0.

Solution

It is given that 2 is a solution of the quadratic equation kx² + 2x - 3 = 0. Hence, k × 2² + 2 × 2 - 3 = 0   (½) 4k = −1 k = −1/4   (½) 

Q76. Find the HCF and LCM of 72 and 120 using prime factorization.

Solution

Q77. Given that HCF(306,657) = 9 ,find LCM(306,657).

Solution

Q78. The expression  is :

• 1) an irrational number
• 2) a natural number
• 3) a whole number
• 4) a rational number

Solution

Q79. A series of well-defined steps which give a procedure for solving a type of problem is called

• 1) Lemma
• 2) Procedure
• 3) Algorithm
• 4) Walkthrough

Solution

A series of well-defined steps which give a procedure for solving a type of problem is called an algorithm.

Q80. Using Euclid Division Lemma, prove that for any positive integer n, n³ - n is divisible by 6.

Solution

Any positive integer is of the form 6m, 6m + 1, 6m + 2, 6m + 3, 6m + 4, 6m + 5 for some positive integer n. When n = 6m, n³ - n = (6m)³ - 6m          = 216m³ - 6m           = 6m(36m² - 1)           = 6q,  where q = m(36m² -1) n³ - n is divisible by 6  When n = 6m + 1, n³ - n = n(n² - 1)          = n (n - 1) (n + 1)           = (6m + 1) (6m) (6m + 2)          = 6m(6m + 1) (6m + 2)          = 6q,     where q = m(6m + 1) (6m + 2) n³ - n is divisible by 6 When n = 6m + 2, n³ - n = n (n - 1) (n + 1)            = (6m + 2) (6m + 1) (6m + 3)          = (6m + 1) (36 m² + 30m + 6)          = 6m (36 m² + 30m + 6) + 1 (36m² + 30m + 6)          = 6[m (36m² + 30m + 6)] + 6 (6m² + 5m + 1)          = 6p + 6q,    where p = m (36m² + 30m + 6) and q = 6m² + 5m + 1 n³ - n is divisible by 6 When n = 6m + 3 n³ - n = (6m + 3)³ - (6m + 3)            = (6m + 3) [(6m + 3)² - 1]          = 6m [6m + 3)² - 1] + 3 [(6m + 3)² - 1]          =6 [m [(6m + 3)² - 1] + 3 [36m² + 36m + 8]          =6 [m [(6m + 3)² - 1] + 6 [18m² + 18m + 4]          = 6p + 3q,   where p = m[(6m + 3)² - 1] and q = 18m² + 18m + 4 n³ - n is divisible by 6 When n = 6m + 4 n³ - n = (6m + 4)³ - (6m + 4)          = (6m + 4) [(6m + 4)² - 1]          = 6m [(6m + 4)² - 1] + 4 [(6m + 4)² - 1]          = 6m [(6m + 4)² - 1] + 4 [36m² + 48m + 16 - 1]          = 6m [(6m + 4)² - 1] + 12 [12m² + 16m + 5]          = 6p + 6q,   where p = m [(6m + 4)² - 1]  and q = 2 (12 m² + 16m + 5) n³ - n is divisible by 6 When n = 6m + 5 n³ - n = (6m + 5) [(6m + 5)² - 1]          = 6m [(6m + 5)² - 1] + 5 [(6m + 5)² - 1]          = 6m [(6m + 5)² - 1] + 5 [36m² + 60m + 24]          = 6p + 30q          = 6(p + 5q),  where p = m [(6m + 5)² - 1] and  q = 6m² + 10m + 4 n³ - n is divisible by 6 Hence, n³ - n is divisible by 6, for any +ve integer n.

Q81. Show that the square of any positive integer cannot be of the form 5q + 2 or 5q + 3 for any integer q.

Solution

Let a be any positive integer. By Euclid's division lemma, a = bm + r where b = 5 ⇒ a = 5m + r So, r can be any of 0, 1, 2, 3, 4 ∴  a = 5m when r = 0,    a = 5m + 1 when r = 1,    a = 5m + 2 when r = 2,    a = 5m + 3 when r = 3,    a = 5m + 4 when r = 4 Case I : a = 5m ⇒ a² = (5m)² = 25m² ⇒ a² = 5(5m²) = 5q, where q = 5m² Case II : a = 5m + 1 ⇒ a² = (5m + 1)² = 25m² + 10 m + 1 ⇒ a² = 5 (5m² + 2m) + 1 = 5q + 1, where q = 5m² + 2m Case III : a = 5m + 2 ⇒ a² = (5m + 2)² ⇒ a² = 25m² + 20m + 4 ⇒ a² = 5 (5m² + 4m) + 4 ⇒ a² = 5q + 4 where q = 5m² + 4m Case IV: a = 5m + 3 ⇒ a² = (5m + 3)² = 25m² + 30m + 9 ⇒ a² = 5 (5m² + 6m + 1) + 4 ⇒ a² = 5q + 4 where q = 5m² + 6m + 1 Case V: a = 5m + 4 ⇒ a² = (5m + 4)² = 25m² + 40m + 16 ⇒ a² = 5 (5m² + 8m + 3) + 1 ⇒ a² = 5q + 1 where q = 5m² + 8m + 3 From all these cases, it is clear that square of any positive integer cannot be of the form 5q + 2 or 5q + 3.

Q82. The 6^th term of an AP is −10 and its 10^th term is −26. Determine the 15^th term of the AP.

Solution

In the given AP, let the first term = a and the common difference = d. T_n = a + (n - 1)d T₆ = a + (6 - 1)d and T₁₀ = a + (10 - 1)d T₆ = a + 5d and T₁₀ = a + 9d                                                      (1) −10 = a + 5d … (i) −26 = a + 9d … (ii) On subtracting (i) from (ii), we get 4d = −16, then d = −4 On substituting d = −4 in (i), we get a + 5 (−4) = −10                                                                        (1) a = 10 Thus, a = 10 and d = −4 15^th term = T₁₅  = a + 14d  = 10 + 14 × (−4)  = −46                                                                                            (2)  

Q83. Prove that is irrational.

Solution

Let us assume that = Where p and q are co-prime integers, q 0 Since, p and q are integers, RHS is rational. So, LHS = is also rational. This contradicts the fact that is irrational. Thus, our assumption was wrong. Hence, is irrational.

Q84. From a point P, 10 cm away from the centre of a circle, a tangent PT of length 8 cm is drawn. Find the radius of the circle.  

Solution

According to the question, OP = 10 cm PT = 8 cm PT is tangent at T and OT is the radius through T. OT is perpendicular to PT. In the right-angled triangle OTP, OP² = OT² + PT² (by Pythagoras theorem)  (½) OT² = OP² - PT² OT =      (1)  Hence, the radius of the circle is 6 cm.       (½)  

Q85. A sweet seller has 867 kaju barfis and 255 badam barfies. She wants to stack them in such a way that each stack has the same number and they take up the least area of the tray. What is the number of that can be placed in each stack for this purpose?

Solution

Q86. There is a circular path around a sports field whose circumference is 5280 m. A cycle at the speed 330m/min of the field B at the speed of 220 m/min and C at the speed of 165 m/min. If A,B and C all starts from the same point, then after how many minutes they will be together at the same starting point?

Solution

Q87. If d is the HCF of 45 and 27, find x, y satisfying d = 27x + 45y.

Solution

Here 45 > 27 By Applying Euclid's division Lemma to 27 and 45 we get, 45 = 27  × 1 + 18 27 = 18 × 1 + 9 18 = 9 × 2 + 0 HCF(27, 45) = 9 9 = 27 - 18 × 1    = 27 - (45 - 27 )×1    = 27 - 45 × 1 + 27 ×1    = 27 - 45 + 27    = 54 - 45    = 27 ×2 - 45 ×1 Comparing this with d = 27x + 45y, we get, x = 2, y = -1, d = 9

Q88. There is a circular path around a sports field. Sonia takes 18 minute to drive one round of the field while Ravi takes 12 minute for the same. Suppose they both start at the same point and at the same time and go in same direction. After how many minutes will they meet at the starting point.

Solution

 Required number of minutes is the L.C.M of 18 and 12. We have LCM is Product of the greatest power of each prime factor involved in the numbers.        L.C.M of 18 and 12 = Hence Sonia and Ravi will meet after 36 min.

Q89. Draw a circle of radius 4 cm. Take a point P outside the circle. Without using the centre of the circle, draw two tangents to the circle from point P. 

Solution

Steps of construction: Draw a line segment 4 cm. Take a point P outside the circle and draw a secant PAB, intersecting the circle at A and B. Produce AP to C such that AP = CP. Draw a semi-circle with CB as diameter. Draw PD perpendicular to CB, intersecting the semi-circle at D. With P as centre and PD as radius, draw arcs to intersect the given circle at T and M. Join PT and PM. Then PT and PM are the required tangents.       NOTE: Correct construction of circle = 1 mark   Correct construction of tangents = 3 marks  

Q90. Find the LCM and HCF of 15, 18, 45 by the prime factorisation method.

Solution

15 = 3 5, 18 = 2 3², 45 = 3² 5 HCF = 3 LCM = 2 3² 5 = 90

Q91. In the figure below, triangle ABC is circumscribed touching the circle at P, Q, R. If AP = 4 cm, BP = 6 cm, AC = 12 cm and BC = x cm, find x.   

Solution

Since the lengths of tangents drawn from an external point to a circle are equal, AC = 12 AR + RC = 12 4 + RC = 12 RC = 12 - 4 = 8 cm CR = CQ = 8 cm BP = BQ = 6 cm   (½) BC = x BQ + CQ = x 8 + 6 = x x = 14 cm   (½)  

Q92.

Solution

Q93. Is 523 a prime number? Justify your answer.

Solution

523 lies between 484 and 529  Prime numbers less than 23 are 2,3,5,7,11,13,17,19Since 523 is not divisible by any of these numbers Hence,  523 is a prime number

Q94. Prove that 263 is a prime number

Solution

256 < 263 < 289   16²  < 263  < 17²   Prime numbers that are less than 17 are 2,3,5,7,11,13   Since 263 is not divisible by any of these numbers, so 263 is a prime number.

Q95. Show that a number of the form n³ - n is always divisible by 6, where n is any positive integer.

Solution

Q96. If the polynomial f(x) = x⁴ - 6x³ + 16x² - 26x + 10 is divided by another polynomial x² - 2x + k, the remainder comes out to be x + a. Find k and a.

Solution

By the division algorithm, Dividend = Divisor × Quotient + Remainder Dividend - Remainder = Divisor × Quotient Dividend - Remainder is always divisible by the divisor. It is given that f(x) = x⁴ - 6x³ + 16x² - 26x + 10 when divided by x² - 2x + k leaves x + a as remainder.   (1) f(x) - (x + a) = x⁴ - 6x³ + 16x² - 26x + 10 - a is exactly divisible by x² - 2x + k. Let us now divide x⁴ - 6x³ + 16x² - 26x + 10 - a by x² - 2x + k                                                              (1) 

Q97. Which of the following rational number has non-terminating and repeating decimal expansion?

• 1)
• 2)
• 3)
• 4)

Solution

Rational number will be terminating decimal if the prime factorisation of q is of the form 2^m 5ⁿ. The denominator of the rational number is not of the form 2^m 5ⁿ. Hence, it has non-terminating and repeating decimal expansion.

Q98.

Solution

Q99. Find a relation between x and y such that the point (x, y) is equidistant from the points (7, 1) and (3, 5). 

Solution

Let P(x, y) be equidistant from the points A(7, 1) and B(3, 5).  According to the question,  AP = BP  AP² = BP²  (x - 7)² + (y - 1)² = (x - 3)² + (y - 5)²                               (1)  x² - 14x + 49 + y² - 2y + 1 = x² - 6x + 9 + y² - 10y + 25  −14x + 6x - 2y + 10y + 50 - 34 = 0  −8x + 8y + 16 = 0                                                            (1)  −8x + 8y = −16  x - y = 2 which is the required relation.                                (1) 

Q100. Find the greatest number of four digits which is divisible by 5,12 and 36.

Solution

The greatest number of four digits is 9999. Any number which is divisible by all of 5, 12 and 36 has to be a least common multiple of 5, 12 and 36.... So, L.C.M. (5, 12, 36) =2 x 2 x 3 x 3 x 5=180...Now, 9999 is not divisible by 180 So the required number has to be less than 9999.On dividing 9999 by 180 we get Quotient=55 and Remainder=99.So the required number is 9999-99=9900.  So the greatest number of four digits which is divisible by 5,12, 36 is 9900

Q101. If sec θ + tan θ = x, show that sin θ =   OR If sec x = 5/4, evaluate     

Solution

sec θ + tan θ = x                                                                       (1)                                                                  (1)                                                                    (1) By componendo-dividendo,                                                                (1) OR sec x = Consider triangle ABC, right angled at B. Hypotenuse = AC = 5 units, Base = AB = 4 units and ∠BAC = x  (1)                                              (1)   AC² = AB² + BC²  5² = 4² + BC²  BC² = 25 - 16 = 9  BC = 3  tan x = BC/AB = ¾                                                             (1)                                        (1)  

Q102.

Solution

Q103. A rectangular lawn, 75 m by 60 m, has two roads, each 4 m wide, running through the middle of the lawn, one parallel to the length and the other parallel to the breadth, as shown in the figure. Find the cost of gravelling the roads at Rs. 4.50 per m².   

Solution

Area of the road ABCD = 75 × 4 = 300 m² Area of the road EFGH = 60 × 4 = 240 m²  (1) Area PQRS is common to both roads. Area PQRS = 4 × 4 = 16 m²  (1) Area of the roads to be gravelled = 300 + 240 - 16 = 524 m²  (1)  

Q104. In a morning walk three persons steps off together. Their steps measure 80 cm, 85 cm and 90 cm respectively. What is the minimum distance each should walk so that they can cover the distance in complete steps?

Solution

Required distance will be the L.C.M of 80cm, 85cm and 90cm. Prime factorisation of 80, 85 and 90 are given by 80 = 2 × 2 × 2 × 2 × 5 = 2⁴ × 5 85 = 5 × 17 90 = 2 × 3 × 3 × 5 = 2 × 3² × 5 LCM(80, 85, 90) = 2⁴ × 3² × 5 × 17 = 12240 cm Therefore the minimum  distance thet should walk to cover the distance is 122m and 40cm.

Q105. Two persons jog together from the same point and in the same direction. One takes 12 minutes to jog one round while the other takes 18 minutes. After how many minutes will they meet first together at the starting point?

Solution

The two persons will be first together at the starting point after an interval of time which is the LCM of the times in which each of them make one complete round. So, the required time will be the LCM of 12 and 18 minutes. Prime factorisation of 12 and 18 are: 12 = 2 2 3 18 = 2 3 3 LCM (12, 18) = 2 2 3 3 = 36 Thus, the two persons will together meet for the first time after 36 minutes.

Q106. Without performing the long division whether the following rational numbers will have a terminating or a non-terminating repeating decimal expansion: (i)   (ii)   (iii)   (iv)    (v)   (vi)   (vii)

Solution

(i) Non terminating repeating (ii) Terminating  (iii) Non terminating repeating  (iv) Non terminating repeating (v) Non terminating repeating (vi) Non terminating repeating (vii) Terminating

Q107. Prove that is irrational.

Solution

Suppose is rational. Squaring both sides, we get, is irrational while is rational and an irrational number can never be equal to a rational number. Thus our assumption is wrong. Hence, is irrational.

Q108. 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

Let x be the maximum number of columns in which the two groups can march. x is HCF of 616 and 32. By Euclid's division algorithm, 616 = 32 19 + 8 32 = 8 4 + 0 HCF (616, 32) = 8 Hence, the required maximum number of columns in which they can march is 8.

Q109. A solid sphere 6 cm in diameter is formed into a tube 10 cm in external diameter and 4 cm in length. Find the thickness of the tube.

Solution

Diameter of the sphere = 6 cm Radius of the sphere = 3 cm Volume =                                                               (1) As the external diameter of the tube is 10 cm, External radius = 5 cm Internal radius = 5 - x, where x cm = thickness the tube And the length of the tube = 4 cm                                                                (1) Volume of the material of the tube = External volume - Internal volume = π × 5² × 4 - π(5 - x)² × 4 = 4π [5² - (5 - x)²] = 4π(10 - x)x cc                                                                                           (1) According to the question, 36π = 4π(10 - x)x 10x - x² = 9 x² - 10x + 9 = 0 (x - 1)(x - 9) = 0 x = 1 or x = 9 which is neglected as the thickness of the tube cannot be more than its external radius. Thickness of the tube = 1 cm                                                             (1) 

Q110. If two triangles are equiangular, prove that the ratio of the corresponding sides is same as the ratio of the corresponding altitudes.

Solution

  Given: Two triangles ABC and DEF in which ∠A = ∠D, ∠B = ∠E, ∠C = ∠F and AL perpendicular to BC and DM perpendicular to EFb To prove:                                                                                               (1) Proof: Since equiangular triangles are similar. ∆ABC ∼ ∆DEF … (i)          (1) In ∆ALB and ∆DME,∠ALB = ∠DME = 90˚ ∠B = ∠E∆ALB ∼ ∆DME  A-A criterion            (1)  … (ii)  from (i) and (ii)                                                   (1)  

Q111. A bag contains 5 red balls, 8 white balls, 4 green balls and 7 black balls. If one ball is drawn at random, find the probability that it is  Black  Red  Not green  OR  A jar contains 54 marbles each of which is blue, green or white. The probability of selecting a blue marble at random from the jar is 1/3, and the probability of selecting a green marble at random is 4/9. How many white marbles does the jar contain?   

Solution

Total number of balls in the bag = 5 + 8 + 4 + 7 = 24  Total number of elementary events = 24  There are 7 black balls in the bag Favourable number of elementary events = 7 P(getting a black ball) = 7/24                               (1)  There are 5 red balls in the bag Favourable number of elementary events = 5 P(getting a red ball) = 5/24                                  (1)  There are 5 + 8 + 7 = 20 balls which are not green. Favourable number of elementary events = 20 P(not getting a green ball) = 20/24 = 5/6               (1)    OR Let there be b blue, g green and w white marbles in the jar. Then  b + g + w = 54 … (i)  P(selecting a blue marble) = b/54    The probability of getting a blue marble is 1/3.  1/3 = b/54  b = 18                                                           (1)   P(selecting a green marble) = 4/9  g/54 = 4/9  g = 24                                                            (1)  Put b = 18 and g = 24 in (i)  18 + 24 + w = 54  w = 12   Hence, the jar contains 12 white marbles.           (1) 

Q112. Why  is a non terminating rational number?

Solution

120 = 2³x 3 x 5 Because the prime factorization of denominator contains a prime factor 3 other 2 and 5. so _{is a non- terminating rational number}

Q113. Use Euclid's algorithm to find H. C. F of 135 and 225.

Solution

Q114. Given that the sum and the difference of the H.C.F and L.C.M are 96 and 48 respectively. Find the product of the numbers.

Solution

Q115. If the mean of the following frequency distribution is 54, then what is the value of p?  Class  0-20  20-40  40-60  60-80  80-100  Frequency  7  p  10  9  13      OR   Find the median daily wages from the following frequency distribution:   Daily wages in Rs. 100-150 150-200 200-250 250-300 300-350 Number of labourers 6 3 5 20 10  

Solution

                                                                                                                                                                               (1)     Class interval Frequency Mid-value x_i f_i x_i 0−20 7 10 70 20−40 p 30 30p 40−60 10 50 500 60−80 9 70 630 80−100 13 90 1170   ∑f_i = 39 + p   ∑f_ix_i = 2370 + 30p                                                                                                                                            (1)                                                                                                                                                             (1)   OR                                                                                                                                                                           (1)   Class Frequency f_i Cumulative frequency 100−150 6 6 150−200 3 9 200−250 5 14 250−300 20 34 300−350 10 44   N = ∑f_i = 44      N = 44  N/2 = 22 The cumulative frequency just greater than 22 is 34, and the corresponding class is 250−300. Median class is 250−300. l = 250, h = 30, f = 20, c = 14, N/2 = 22  (1) Median = l +   = 250 + = 270    Hence, the median wages are Rs. 270.  (1)  

Q116. Find the H.C.F. between 24 and 404 and express it as a linear combination of the two numbers.

Solution

Since 404 > 24 By using Euclid's division lemma, we get 404 = 24 × 16 + 20  24   = 20 × 1 + 4 20   =  4 × 5 + 0 The remainder at this stage is zero. So the divisor in the last step i.e 4 is the HCF. Therefore HCF(24,404) = 4 Now working steps backward, we get 4 = 24 - (20 × 1)    = 24 - {[404 - (24 × 16)]}    = (24 × 17) - (404 × 1)    Hence,  4 = (24 × 17) - (404 × 1) is a linear combination of 24 and 404. 

Q117. Find the value of k for which the equation kx² - 6x - 2 = 0 has real roots. OR Solve: 2x² - x + 1/8 = 0

Solution

kx² - 6x - 2 = 0  Comparing with ax² + bx + c = 0 a = k, b = −6, c = −2     (½) b² - 4ac = (−6)² - 4 × k × (−2)  = 36 + 8k      (½) The given equation has real roots if D ≥ 0 (½) 36 + 8k ≥ 0 k ≥ −36/8 k ≥ −9/2     (½)  OR  2x² - x + 1/8 = 0      (1) 4x - 1 = 0  x = ¼      (½) Hence, each of the two roots of the given equation is x = 1/4.    (½)  

Q118. Solve the following system of equations for x and y.   

Solution

  Put  and                               (½) 5u + v = 2 … (i) 6u - 3v = 1 … (ii) Multiplying (i) by 3 and adding (ii) to it, 21u = 7 u = 7/21 = 1/3                                                            (1) Putting u = 1/3 in (i), 5/3 + v = 2 v = 2 - 5/3 = 1/3 u = 1/3                                                                          (½) x - 1 = 3 x = 4 v = 1/3                                                                             (½) y - 2 = 3 y = 5                                                                               (½) 

Q119. Show that one and only one out of n, n+3 n+6 or n+9 is divisible by 4.

Solution

Any positive integer is of the form 4q, 4q + 1, 4q + 2, 4q + 3 for some integer q. Case (I): When n = 4q, then n is divisible by 4 n + 3 = 4q + 3, leaves a remainder of 3 when divided by 4 n + 6 = 4q + 6, leaves a remainder of 2 when divided by 4 n + 9 = 4q + 9, leaves a remainder 1 when divided by 4 Case (II): When n = 4q + 1, then it leaves a remainder 1 when divided by 4 n + 3 = 4q + 4  is divisible by 4. n + 6 = 4q + 7, leaves a remainder 3 when divided by 4 n + 9 = 4q + 10, leaves a remainder 2 when divided by 4 Case (III): When n = 4q + 2, then n + 6 = 4q + 8 is divisible by 4 n + 3 = 4q + 5, leaves a remainder of 1 when divided by 4 n + 6 = 4q + 8  is divisible by 4, n + 9 = 4q + 11, leaves a remainder 3 when divided by 4 Case (IV): When n = 4q + 3, leaves a remainder of 3 when divided by 4 n + 3 = 4q + 6, leaves a remainder of 2 when divided by 4 n + 6 = 4q + 9, leaves a remainder 1 when divided by 4 n + 9 = 4q + 12  is divisible by 4.   Thus, we see that in each case only one of the number n, n + 3, n + 9 is divisible by 4. Hence proved.

Q120. If α, β are the zeroes of the quadratic polynomial f(x) = x² - px + q, then find the values of α² + β² and . 

Solution

Since α and β are the zeroes of a polynomial f(x) = x² - px + q,  and                                   (1)                               (1)                                                             (1) 

Q121. Show that the square of any positive odd integer is of the form 8m + 1, for some integer m.

Solution

Let a be any positive integer and b = 4. Then, by Euclid's algorithm a = 4q + r for some integer q 0 and 0 r < 4 Thus, r = 0, 1, 2, 3 Since, a is an odd integer, so a = 4q + 1 or 4q + 3 Case I: When a = 4q + 1 Squaring both sides, we have, a² = (4q + 1)² a² = 16q² + 1 + 8q     = 8(2q² + q) + 1      = 8m + 1, where m = 2q² + q Case II: When a = 4q + 3 Squaring both sides, we have, a² = (4q +3)²     = 16q² + 9 + 24q     = 16 q² + 24q + 8 + 1      = 8(2q² + 3q + 1) +1      = 8m +1 where m = 2q² + 3q + 1 Hence, a is of the form 8m + 1 for some integer m.

Q122. Prove that is an irrational. OR Prove that - 7 is an irrational.

Solution

Let be a rational number Let = where p and q are co-prime.   So, = p 11q² = p² Thus, 11 divides p². Hence 11 divides p.… (1) Let p = 11c 11q² = 121 c² Or q² = 11c² This implies that 11 divides q². Hence 11 divides q.… (2) From (1) and (2), we can say that: p and q have a common factor 11 which contradicts our assumption. Hence, is irrational.   OR Let - 7 be rational. where p and q are co-prime = Since p and q are integers, is rational. is rational But we know that is irrational Our assumption is wrong. Hence 2 - 7 is irrational.  

Q123. A kite is flying at a height of 75 m from the level ground, attached to a string inclined at 60° to the horizontal. Find the length of the string, assuming that there is no slack in it.

Solution

Let the length of the string be l m.      (½)    (½)  

Q124. Show by an example that the sum and product of two irrational numbers is not always  an irrational number.

Solution

Consider two numbers A= and B= We  see that A*B== -11, which is a rational number. A+B=+=2, which also is a rational number. So we see  that , the sum and product of two irrational numbers is not always  an irrational number.  

Q125. Show that, any positive integer is of form 3q, 3q + 1 or 3q + 2, where q is some integer.

Solution

Euclid's Division Lemma : For any two positive integers a and b, there exists two unique integers q and r such that a = bq + r, 0 r < b. If we take b = 3, the possible values of r will be 0, 1 and 2 Hence, either a = 3q or a = 3q + 1 or a = 3q + 2.

Q126. The marks obtained in a class test by 30 students of a class are as follows: Marks obtained  Number of students  More than or equal to 5  30  More than or equal to 10  28  More than or equal to 15  16  More than or equal to 20  14  More than or equal to 25  10  More than or equal to 30  7  More than or equal to 35  3    Draw less than type and more than type ogive curves for the given data and hence find the median.   OR The mode of the following frequency distribution is 55 and the modal class is 45-60. Find the value of p and q.  Class Interval  0-15  15-30  30-45  45-60  60-75  75-90  Total  Frequency  6  7  p  15  10  q  51   

Solution

More than method:    Marks obtained  Number of students  More than or equal to 5  30  More than or equal to 10  28  More than or equal to 15  16  More than or equal to 20  14  More than or equal to 25  10  More than or equal to 30  7  More than or equal to 35  3                                                                                                                                                                                          (1)  Thus, we plot the points (5, 30), (10, 28), (15, 16), (20, 14), (25, 10), (30, 7), (35, 3).    Less than method:    Marks obtained  No. of students   (Frequency)  Marks less than  Cumulative frequency  5-10  30 - 28 = 2  10  2  10-15  28 - 16 = 12  15  14  15-20  16 - 14 = 2  20  16  20-25  14 - 10 = 4  25  20  25-30  10 - 7 = 3  30  23  30-35  7 - 3 = 4  35  27  35-40  3 - 0 = 3  40  30                                                                                                                                                                                       (1)    We plot points (10, 2), (15, 14), (20, 16), (25, 20), (30, 23), (35, 27) and (40, 30) to get the ‘less than’ curve.    Median = 17. 5 (x-coordinate of intersection of both curves)                                                                                       (2) OR  We have 38 + p + q = 51  p + q = 51 − 38  p + q = 13 … (1)   Mode = 55 (Given)  ∴ Modal class is 45 − 60  = 45, f₀ = p, f₁ = 15, f₂ = 10, h = 15                                                                                                                        (1)  55 = 45 +   10 =   200 − 10p = 225 − 15p                                                                                                                                            (1)  5p = 25  p = 5   From (1), we have  q = 13 − 5 = 8                                                                                                                                                        (1) 

Q127. Show that 8ⁿ cannot end with the digit zero for any natural number n.

Solution

We know 8ⁿ = (2³)ⁿ = 2³ⁿ If 8ⁿ end with zero then 10 is factor of 8ⁿ. 8ⁿ = 2³ⁿ = (5)(2) 5 is factor of 2, which is a contradiction So, our assumption is wrong. Hence 8ⁿ cannot end with zero.

Q128. Show that any positive odd integer is of the form 4q + 1 or 4q + 3 where q is a positive.

Solution

Let 'a' any odd positive integer and b = 4 Using Euclid Division Lemma, a = 4q + r, where  and 0 r < 4 a = 4q or 4q + 1 or 4q + 2 or 4q + 3 a = 4q + 1 or 4q + 3 Any odd integer is of the form 4q + 1 or 4q + 3

Q129. Without actually performing the long division, state whether the following number will have a terminating decimal or a non-terminating repeating decimal expansion. (i) _(ii) (iii)  

Solution

(i)                           Here, the prime factorisation of denominator is of the form.           Hence, the decimal expansion of the given rational number is terminating.                                (ii)                          Here, the prime factorisation of denominator is not of the form .            So, the decimal expansion of the given rational number is non - terminating repeating.         _(iii)                   Here, the denominator is not of the form_.          Hence, the decimal expansion of the given rational number is non - terminating repeating.

Q130. Find the LCM of 72, 80 and 120 using the fundamental theorem of arithmetic.

Solution

72 = 2³ 3², 80 = 2⁴ 5, 120 = 2³ 3 5 LCM (72, 80, 120) = 2⁴ 3² 5 = 720

Q131. Show that the square of any positive integer is of the form 4q or 4q + 1 for some integer q.

Solution

Applying Euclid's division algorithm with a and b = 4. a = 4Q + r, 0 r < 4 When r = 0, a = 4Q a² = 16Q² = 4(4Q²) = 4q, where q = 4Q²   When r = 1, a = 4Q + 1 a² = (4Q + 1)² = 16Q² + 1 + 8Q = 4Q(4Q + 2) + 1 = 4q + 1, where q = Q(4Q + 2)   When r = 2, a = 4Q + 2 a² = (4Q + 2)² = 16Q² + 4 + 16Q = 4(4Q² + 4Q + 1) = 4q, where q = 4Q² + 4Q + 1   When r = 3, a = 4Q + 3 a² = (4Q + 3)² = 16Q² + 9 + 24Q = 4(4Q² + 6Q + 2) + 1 = 4q + 1, where q = 4Q² + 6Q + 2   Hence, the square of any positive integer is of the form 4q or 4q + 1 for some integer q.

Q132. A die is thrown at once. What is the probability of getting a number other than 4? OR One card is drawn at random from a well-shuffled deck of 52 cards. What is the probability that the card drawn is a face card? 

Solution

In a single throw of a die, all possible outcomes are 1, 2, 3, 4, 5, 6. Total number of possible outcomes = 6 Let E be the event of getting a number other than 4. Then, the favourable outcomes are 1, 2, 3, 5, 6. Number of favourable outcomes = 5    (½) P(getting a number other than 4) = P(E) = 5/6 (½) OR     Total number of possible outcomes = 52   We know that kings, queens and jacks are face cards.   Number of face cards = 12   Let E be the event of getting a face card.   The number of favourable outcomes = 12 (½)   P(getting a face card) = 12/52 = 3/13  (½)    

Q133. Find without actual division whether the 66/180 rational number is a terminating or non-terminating repeating decimal. OR Show that any number of the form 6ⁿ, n ∊ N can never end with digit 0. 

Solution

  30 = 2 × 3 × 5 ≠ 2^m × 5ⁿ   (½) So, the given rational number is a non-terminating repeating decimal.  (½)   OR    If 6ⁿ ends with zero, then it must have 5 as a factor. But, we know that the only prime factors of 6ⁿ are 2 and 3.  (½) Using the fundamental theorem of arithmetic that the prime factorisation of each number is unique. 6ⁿ can never end with zero. (½)  

Q134. Find LCM and HCF of 120 and 144 by fundamental theorem of Arithmetic.

Solution

120  = 2³ 3 5   144 = 2⁴ 3² LCM = 2⁴ 3² 5 = 720 HCF = 2³ 3 = 24

Q135. Find the HCF and LCM of 42, 72, and 120 using prime factorization

Solution

Q136. The measure of the angle of depression of the bottom of a building on a level ground from the top of a tower 50 m high is 60˚. How far is the building from the tower? 

Solution

    (1) Let AB be the tower and CD be the building.  AB = 50 m  Angle of depression ∠CBE = 60°  ∠BCA = 60°  In right-angled triangle CAB,  tan 60° = AB/AC         (1)