Contents

## GSEB Solutions for Class 10 mathematics – Euclid’s Algorithm and Real Numbers (English Medium)

### Exercise-1.1

**Question 1:**

Prove that 16 divides n^{4} + 4n^{2} + 11, if n is an odd integer.

**Solution :
**Here n is an odd integer for some k ∊ Z.

Then n = 4k + 1 or n = 4k – 1.

**Case 1:**

For n = 4k + 1

n

^{4}+ 4n

^{2}+ 11 = (4k + 1)

^{4}+ 4(4k + 1)

^{2}+ 11

= (4k)

^{4}+ 4(4k)

^{3}+ 6(4k)

^{2}+ 4(4k) + 1 + 4(16k

^{2}+ 8k+1) + 11

= 256k

^{4}+ 256k

^{3}+ 96k

^{2}+ 16k + 1 + 64k

^{2}+ 32k + 4 + 11

=256k

^{4}+ 256k

^{3}+ 160k

^{2}+ 48k + 16

= 16(16k

^{4}+ 16k

^{3}+ 10k

^{2}+ 3k + 1).

Thus the above step is divisible by 16.

**Case2:**

For n = 4k – 1,

n

^{4}+ 4n

^{2}+ 11 = (4k – 1)

^{4}+ 4(4k – 1)

^{2}+ 11

= (4k)

^{4}– 4(4k)

^{3}+ 6(4k)

^{2}– 4(4k) + 1 + 4(16k

^{2}– 8k+1) + 11

= 256k

^{4}– 256k

^{3}+ 96k

^{2}– 16k + 1 + 64k

^{2}– 32k + 4 + 11

= 256k

^{4}– 256k

^{3}+ 160k

^{2}– 48k + 16

= 16(16k

^{4}– 16k

^{3}+ 10k

^{2}– 3k + 1).

Thus the above step is divisible by 16.

Thus in any case n

^{4}+ 4n

^{2 }+ 11 is divisible by 16.

**Question 2:**

Prove that if n is positive even integer, then 24 divides n(n+1)(n+2).

**Solution :****
**Here n is a positive even integer for some m ∊ N and

n = 2m.

Then n(n + 1)(n + 2) = 2m(2m + 1)(2m + 2)

= 4m(m + 1)(2m + 1) ….. (i)

**Case 1: If m = 1**

m(m + 1)(2m + 1) = 1(1 + 1)(2(1) + 1) = 2 × 3 = 6, which is divisible by 6.

∴ m(m + 1)(2m + 1) is divisible by 6.

∴ 4m(m + 1)(2m + 1) is divisible by 24.

Hence n(n + 1)(n + 2) is divisible by 24. (∵ by (1))

**Case2: If m = 2**

m(m + 1)(2m + 1) = 2(2 + 1)(2(2) + 1) = 2 × 3 × 5 = 30, which is divisible by 6.

∴ m(m + 1)(2m + 1) is divisible by 6.

∴ 4m(m + 1)(2m + 1) is divisible by 24.

**Case 3: If m≥3**

Here m and m + 1 being consecutive integers, one of them will always be even and other will be odd.

∴ m(m + 1)(2m + 1) is always divisible by 2.

Also m(m ≥ 3) is a positive integer, so for some k ∊ N, m = 3k or m = 3k + 1 or m = 3k + 2

- For m = 3k

m(m + 1)(2m + 1) = 3k(3k + 1)(2(3k) + 1)

= 3[k(3k + 1)(6k + 1)]

This is divisible by 3. - For m = 3k + 1

m(m + 1)(2m + 1) = (3k + 1)(3k + 1 + 1)(2(3k + 1) + 1)

= [(3k + 1)(3k + 2)(6k + 3)]

= 3[(3k + 1)(3k + 2)(2k + 1)]

This is also divisible by 3. - For m = 3k + 2

m(m + 1)(2m + 1) = (3k + 2)(3k + 2+1)(2(3k + 2)+ 1)

= [(3k + 2)(3k + 3)(6k + 5)]

= 3[(k + 1)(3k + 2)(6k + 5)]

This is also divisible by 3.

Hence in any case, m(m + 1)(2m + 1) is divisible by 3 and 2.

As 2 and 3 are mutually prime numbers

m(m + 1)(2m + 1) is divisible by 6.

∴ n(n + 1)(n + 2) is divisible by 24. (∵ by (1))

Thus in any case n(n + 1)(n + 2)is divisible by 24.

**Question 3:**

Prove that if either of 2a + 3b and 9a + 5b is divisible by 17, so is the other. a, b ∈ N

(Hint :4(2a + 3b) + 9a + 5b = 17a + 17b)

**Solution :**

Let 2a + 3b be divisible by 17.

For some integer m, 2a + 3b = 17m —- (i)

Now, 9a + 5b = 17a + 17b – 4(2a + 3b) [∵ Using the hint]

= 17(a + b) – 4(17m) (∵ by (i))

= 17(a + b – 4m)

= 17k where k = a + b – 4m is an integer.

∴ 9a + 5b is divisible by 17.

Similarly, by considering 9a + 5b divisible by 17, we can show 2a + 3b is divisible by 17.

**Question 4:**

Prove that every natural number can be written in the form 5k or 5k ± 1 or 5k ± 2, k ∈ N ∪ {0}.

**Solution :****
**From Euclid’s division lemma, for every natural number a, if we take b = 5, then unique non-negative integers k and m can be obtained such that a = 5k + m, 0 ≤ m ≤ 5

When 0 ≤ m ≤ 5 i.e. m = 0, 1,2, 3 or 4

∴ We can say that either a = 5k or a = 5k + 1 or

a = 5k + 2 or a = 5k + 3 or a = 5k + 4, where k∊ N υ {0}.

Now,

a = 5k + 3

= 5k + 5 – 2

= 5(k + 1) – 2

= 5k’ – 2, where k’ = k + 1, k ∊ N υ {0}.

= 5k- 2, k ∊ N υ {0} (taking k instead of k’)

∴ a = 5k + 3 can be written as a = 5k – 2 and

a= 5k + 4

= 5k + 5 – 1

= 5(k + 1) – 1

= 5k’ – 1, where k’ = k + 1, k ∊ N υ {0}.

= 5k – 1, k ∊ N υ {0} (∵taking k instead of k’)

∴ a = 5k + 4 can be written as a = 5k – 1. Every natural number can be written in the form of 5k, 5k ± 1or 5k ± 2, where k ∊ N υ {0}.

**Question 5:**

Prove that if 6 has no common factor with n, n^{2} – 1 is divisible by 6.

**Solution :****
**Here, there is no common factor between n and 6.

2 and 3 are prime factors of 6.∴ n is not divisible by any of 2 and 3.

Let n be an integer that is not divisible by 2.

∴ It can be written in the form of n = 2k + 1; k ∊ Z.

n

^{2}– 1 = (2k + 1)

^{2}– 1

= 4k

^{2}+ 4k + 1 – 1

= 4k

^{2}+ 4k

= 2(2k

^{2}+ 2k)

∴ It is divisible by 2. — (i)

Also n is not divisible by 3.

∴ It can be written in the form of n = 3k ± 1; where k ∊ Z.

n

^{2}– 1 =(3k ± 1)

^{2}– 1

= 9k

^{2}± 6k + 1 – 1

= 9k

^{2}± 6k

= 3(3k

^{2}± 2k)

∴ It is divisible by 3. — (ii)

By (i) and (ii), 2 and 3 being mutually prime integers, n

^{2 }– 1 is divisible by 6.

**Question 6:**

Prove that product of four consecutive positive integers is divisible by 24.

**Solution :
**Suppose n(n + 1)(n + 2)(n + 3) is the product of four consecutive integers.

Here we have to find the product of four consecutive integers, such that one of them will be divisible by 4.

Of the remaining three, one must be even and one must be divisible by 3.

⇒ n(n + 1)(n + 2)(n + 3) is divisible by (4 × 2 = 8)

Also n(n + 1)(n + 2)(n + 3) is divisible by 3.

Also 3 and 8 are mutually prime integers.

∴ n(n + 1)(n + 2)(n + 3) is divisible by both 3 and 8.

∴ n(n + 1)(n + 2)(n + 3) is divisible by 24.

### Exercise-1.2

**Question 1:**

Find g.c.d. (1) 144, 233 (2) 765, 65 (3) 10211, 2517

**Solution :**

- 144, 233

Here 233 > 144233 = 144 × 1 + 89

144 = 89 × 1 + 55

89 = 55 × 1 + 34

55 = 34 × 1 + 21

34 = 21 × 1 + 13

21 = 13 × 1 + 8

13 = 8 × 1 + 5

8 = 5 × 1 + 3

5 = 3 × 1 + 2

3 = 2 × 1 + 1

2 = 1 × 2 + 0

The last non-zero divisor is 1.

∴ g.c.d.(144, 233) = 1 - 765, 65

Here 765 > 65765 = 65 × 11 + 50

65 = 50 × 1 + 15

50 = 15 × 3 + 5

15 = 5 × 3 + 0

The last non-zero divisor is 5.

∴ g.c.d.(144, 233) = 5 - 10211, 2517

Here 10211> 251710211 = 2517 × 4 + 143

2517 = 143 × 17 + 86

143 = 86 × 1 + 57

86 = 57 × 1 + 29

57 = 29 × 1 + 28

29 = 28 × 1 + 1

28 = 1 × 28 + 0

The last non-zero divisor is 1.

∴ g.c.d.(10211, 2517) = 1

**Question 2:**

Find g.c.d. of 736 and 85 by using Euclid’s algorithm.

**Solution :****
**Here 736 > 85,

736 = 85 × 8 + 56

85 = 56 × 1 + 29

56 = 29 × 1 + 27

29 = 27 × 1 + 2

27 = 2 × 13 + 1

2 = 1 × 2 + 0

The last non-zero divisor is 1.

∴ g.c.d.(736, 85) = 1

**Question 3:**

Prove g.c.d. (a – b, a + b) = 1 or 2, if g.c.d. (a, b) = 1

**Solution :****
**Suppose, g.c.d.(a – b, a + b) = d

⇒ d | a – b and d | a + b

⇒ a – b = md —- (1)

Also, a + b = nd … where m, n ∊ N —-(2)

On adding and subtracting equations (2) and (1) respectively,

2a = (m + n)d and 2b = (n – m)d —-(3)

Now, g.c.d.(a, b) = 1

∴ 2 × g.c.d.(a, b) = 2

∴ g.c.d.(2a, 2b) = 2

∴ g.c.d[(m + n)d, (n – m)d] = 2

Using values of 2a and 2b from (3) we get,

d × g.c.d(m + n, n – m) = 2 = 2 × 1 = 1 × 2

∴ As 1 and 2 are mutually prime numbers ,

d = 2 or d = 1

∴ g.c.d.(a – b, a + b) is either d = 1 or d = 2

∴ g.c.d.(a – b, a + b) = 1 or 2

**Question 4:**

Using the fact that g.c.d. (a, b) l.c.m. (a, b) = ab, find l.c.m. (115, 25).

**Solution :****
**First we find g.c.d. of 115 and 25.

Here, being 115 > 25,

115 = 25 × 4 + 15

25 = 15 × 1 + 10

15 = 10 × 1 + 5

10 = 5 × 2 + 0

The last non-zero number is 5.

g.c.d.(115, 25) = 5

Subs. a = 115 and b = 25 and g.c.d.(115, 25) = 5 in

g.c.d.(a, b) × l.c.m.(a, b) = ab

5 × l.c.m.(115, 25) = (115)(25)

Thus, l.c.m(115, 25) = 575

### Exercise-1.3

**Question 1:**

Express as a product of primes:

(1) 7007 (2) 7500 (3) 10101 (4) 15422

**Solution :**

**Question 2: **

Find g.c.d. and l.c.m. using the fundamental theorem of arithmetic:

- 250 and 336
- 4000 and 25
- 225 and 145
- 175 and 1001

**Question 2(1):**

**Solution :**

**Question 2(2):**

**Solution :**

**Question 2(3):**

**Solution :**

**Question 2(4):**

**Solution :**

**Question 3:**

Find g.c.d. and l.c.m. :

- 15, 21, 35
- 40, 60, 80
- 49, 42, 91

**Question 3(1):**

**Solution :**

**Question 3(2):**

**Solution :**

**Question 3(3):**

**Solution :**

**Question 4:**

Prove that following numbers are irrational :

- √5
- √15
- √3 + 1
- √5 + √7
- 5√2

**Solution :**

**Question 5:**

Find l.c.m. (105, 91) using g.c.d. (a, b) l.c.m. (a, b) = ab

**Solution :**

First find the g.c.d. of 105 and 91 by Euclid’s algorithm,

Here 105 > 91,

105 = 91 × 1 + 14

91 = 14 × 16 + 7

14 = 7 × 2 + 0

Last non-negative remainder is 7.

∴ g.c.d.(105, 91) = 7.

Subs. a = 105, b = 91 and g.c.d.(105, 91) = 7 in the formula we get,

7 × l.c.m.(105, 91) = 105 × 91

**Question 6:**

Prove √3 + √2 + 1 is irrational.

**Solution :**

**Question 7:**

Using (√7 + √3)(√7 – √3) = 4 and the fact that (√7 + √3) is irrational prove that √7 – √3 is irrational.

**Solution :**

**Question 8:**

Two buses start from the same spot for the same circular root. One is a BRTS bus returning in 35 minutes. The other is a regular express bus taking 42 minutes to return. After how many minutes will they meet again at the same initial spot?

**Solution :
**The BRT bus returns to the initial spot in 35 minutes. Thus, it returns to the initial spot in time intervals which are multiples of 35 minutes.

The other express bus returns to the initial spot in 42 minutes. Thus, it returns to the initial spot in time intervals which are multiples of 42 minutes.

∴ Both buses meet again at the initial spot at the time of l.c.m. of 35 and 42.

∴ 42 = 2 × 3 × 7

∴ l.c.m.(35, 42)=2 × 3 × 5 × 7 = 210 minutes.

Thus, after 210 minutes both buses will meet again at the initial spot.

### Exercise-1.4

**Question 1:**

**Question 1(1):**

**Solution :**

**Question 1(2):**

**Solution :**

**Question 1(3):**

**Solution :**

**Question 1(4):**

**Solution :**

**Question 1(5):**

**Solution :**

**Question 1(6):**

**Solution :**

**Question 1(7):**

**Solution :
**

**Question 1(8):**

**Solution :**

**Question 1(9):**

**Solution :
**

**Question 1(10):**

**Solution :
**

**Question 2:**

Following real numbers are expressed in decimal form. Find whether they are rational or not. If rational, express them in the form . Comment on factors of q :

- 0.01001000100001…..
- 5.123456789
- 0.090909… = 0.
- 2.03
- 0.9999…. = 0.
- 5.781
- 2.312
- 0.12345

**Question 2(1):**

**Solution :
**Decimal expression 0.01001000100001…. is non-terminating and non-recurring. Hence it an irrational number.

**Question 2(2):**

**Solution :
**

**Question 2(3):**

**Solution :**

**Question 2(4):**

**Solution :
**0.090909… is non-terminating and recurring, so it is a rational number.

Let x = 0.090909 —(i)

⇒ 100x = 9.0909… —(ii)

Subtracting eq. (1) from (2),

99x = 9

**Question 2(5):**

**Solution :**

**Question 2(6):**

**Solution :**

**Question 2(7):**

**Solution :**

**Question 2(8):**

**Solution :**

**Question 2(9):**

**Solution :**

**Question 2(10):**

**Solution :**

### Exercise-1.5

**Question 1:**

Find the square roots of following surds :

- 5 + 2√6
- 9 + 2√14
- 2 – √3
- a + √a
^{2}– b^{2} - 7 + √48
- 6 + 4√2
- 5 + √21
- 8 – 3√7

**Question 1(1):**

**Solution :
**

**Question 1(2):**

**Solution :
**

**Question 1(3):**

**Solution :**

**Question 1(4):**

**Solution :
**

**Question 1(5):**

**Solution :**

**Question 1(6):**

**Solution :
**

**Question 1(7):**

**Solution :
**

**Question 1(8):**

**Solution :**

**Question 2:**

Simplify :

**Solution :**

### Exercise-1

Find g.c.d. and l.c.m. (1 to 10) :

- 25, 35
- 105, 125
- 220, 132
- 3125, 625
- 15625, 35
- 15, 25, 35
- 18, 12, 16
- 16, 24, 36
- 35, 25, 63
- 112, 128, 144

**Question 1:**

**Solution :**

**Question 2:**

**Solution :
**

**Question 3:**

**Solution :
**

**Question 4:**

**Solution :
**

**Question 5:**

**Solution :**

**Question 6:**

**Solution :**

**Question 7:**

**Solution :
**

**Question 8:**

**Solution :**

**Question 9:**

**Solution :**

**Question 10:**

**Solution :**

**Question :**

prove following numbers are irrational. (11 to 20) :

**Question 11:**

**Solution :**

**Question 12:**

**Solution :**

**Question 13:**

**Solution :**

**Question 14:**

**Solution :
**

**Question 15:**

**Solution :
**

**Question 16:**

**Solution :
**

**Question 17:**

**Solution :**

**Question 18:**

**Solution :**

**Question 19:**

**Solution :**

**Question :**

Which of the following numbers have terminating decimal expansion and why? (21 to 25)

**Question 20:**

**Solution :**

**Question 21:**

**Solution :**

**Question 22:**

**Solution :**

**Question 23:**

**Solution :**

**Question 24:**

**Solution :**

**Question :**

Find the square root of the following in the form of a binomial surd (26 to 30) :

26. 12 +

27. 8 +

28. 2 +

29. 14 +

30. n +

**Question 25:**

**Solution :**

**Question 26:**

**Solution :**

**Question 27:**

**Solution :**

**Question 28:**

**Solution :**

**Question 29:**

**Solution :**

**Question 30:**

**Solution :**

**Question :**

Simplify (31 to 32) :

31.

32.

**Question 31:**

**Solution :**

**Question 32:**

**Solution :
**

**Question 33:**

Find the largest number dividing 230 and 142 and leaving remainders 5 and 7 respectively.

**Solution :****
**Suppose d is the greatest number such that when 230 and 142 are divided by d, the remainders are respectively 5 and 7.

∴ For some integers p

_{1}and p

_{2},

230 = dp

_{1 }+ 5 and 142 = dp

_{2 }+ 7

∴ dp

_{1 }= 230 – 5 = 225 and dp

_{2 }= 142 – 7 = 135

But, d is the greatest number. So, d becomes the greatest common divisor of 225 and 135.

∴ d is theg.c.d. of 225 and 135.

Now, 225 = 5

^{2 }× 3

^{2}and 135 = 5 × 3

^{3 }∴ g.c.d.(225, 135) = 5 × 3

^{2 }= 45

Thus, 45 is the greatest number which leaves remainders 5 and 7 respectively when it divides 230 and 142.

**Question 34:**

Find the largest number dividing 110, 62, 92 and leaving remainders 5, 6 and 1 respectively.

**Solution :
**Suppose, d is the greatest number such that when 110, 62 and 92 are divided by d, the remainders are 5, 6 and 1 respectively.

∴ For some integers p

_{1 }, p

_{2}and p

_{3},

110 = dp

_{1 }+ 5, 62 = dp

_{2 }+ 6, 92 = dp

_{3 }+ 1

∴ dp

_{1 }= 110 – 5 = 105, dp

_{2 }= 62 – 6 = 56, dp

_{3 }= 92 – 1 = 91

But, d is the largest number. So d becomes the greatest common divisor of 105, 56 and 91.

∴ d is the g.c.d. of 105, 56 and 91.

Now, 105 = 5 × 7 × 3, 56 = 2

^{3 }× 7 and 91 = 13 × 7

∴g.c.d.(105, 56, 91) = 7

Thus, 7 is the greatest number which leaves remainders 5, 6 and 1 respectively when it divides 110, 62 and 92.

**Question 35:**

The length and the breadth and the height of a room are 735 cm, 625 cm and 415 cm. Find the length og the largest scale measuring instrument which can measure all the three dimensions.

**Solution :
**To find the length of the largest scale measuring instrument which can measure all the three dimensions, we find theg.c.d. of their measures.

735 = 5 × 3 × 7

^{2 }625 = 5

^{4}and

415 = 5 × 83

∴ g.c.d.(735, 625, 415) = 5

Thus, the length of the largest scale measuring instrument is 5 cm which can measure all the three dimensions of the room.

**Question 36:**

A milk man has 150 liters of milk of higher fat and 240 litres of milk of lower fat. He wants to pack the milk in tins of equal capacity. What should be the capacity of each tin ?

**Solution :
**We need to pack a 150 litres milk of high fat and 240 liters milk of low fat.

For packing milk in tins of equal capacity we find the g.c.d. of the quantity of high fat milk and the quantity of low fat milk.

Now, 150 = 5

^{2 }× 3 × 2 and 240 = 5 × 2 × 3 = 30

Thus, the capacity of each tin should be 30 litres.

**Question 37:**

Find the smallest number which decreased by 15 is a multiple of 125 and 225

**Solution :
**The smallest number which is a multiple of 125 and 225 is the l.c.m. of 125 and 225.

But it is given that this number is obtained by subtracting 15.

∴ Required number = (Smallest positive integer – 15) = l.c.m.(125, 225)

∴ Required smallest positive integer = l.c.m.(125, 225) + 15 … (1)

Now, 125 = 5

^{3}and 225 = 5

^{2 }× 3

^{2 }∴ l.c.m.(125, 225) = 5

^{3 }× 3

^{2 }= 1125

Putting this value in result (1),

Required smallest positive integer = 1125 + 15 = 1140

**Question 38:**

Find the smallest number of six digits divisible by 18, 24 and 30

**Solution :
**

**Question 39:**

Prove if 3 | (a^{2} + b^{2}) then 3 | a and 3 | b, a ∈ N, b ∈ N.

**Solution :
**Case 1: First we prove that if 3|a then 3|b also follows.

Here, as 3| (a

^{2 }+ b

^{2}), a

^{2 }+ b

^{2 }= 3k, k ∈ N and being 3|a, a = 3m, m ∈ N.

Now, putting a = 3m in a

^{2 }+ b

^{2 }= 3k,

(3m)

^{2 }+ b

^{2 }= 3k

b

^{2}= 3k – 9m

^{2 }∴ b

^{2 }= 3(k – 3m

^{2 })

∴ b

^{2}is divisible by 3.

∴ b is divisible by 3.

(∵ By Lemma 1, if a

^{2 }(a ∈ N) is divisible by some prime number p then p|a)

Also b ∈ N and 3 is a prime number.

∴ 3|b

Thus if 3|a, then 3|b follows.

Hence, 3|a and 3|b are obtained.

Case 2: If 3|b, then according to case 1, we can prove that 3|a.

∴ If 3|b, then it implies that 3|a.

Hence 3|b and 3|a are obtained.

Case 3:

Suppose, 3∤a and 3∤b

Here, 3∤a i.e., a is not divisible by 3 and 3∤b i.e., b is not divisible by 3.

From version (4) of division algorithm, we know that,

if a is not divisible by 3 than for some p ∈ N, a = 3p ± 1 and b is not divisible by 3 so for some q∈ N, b = 3q ± 1.

Now, a

^{2 }+ b

^{2 }= (3p ± 1)

^{2 }+ (3q ± 1)

^{2 }= 9p

^{2 }± 6p + 1 + 9q

^{2 }± 6q + 1

= 3(3p

^{2 }± 3q

^{2 }± 2p ± 2q) + 2

∴ If we divide a

^{2 }+ b

^{2}by 3, then remainder is 2.

∴ a

^{2 }+ b

^{2}is not divisible by 3.

∴ 3|a

^{2 }+ b

^{2}is not divisible by 3.

But 3|a

^{2}+b

^{2}is given.

∴ Our supposition 3∤a and 3∤b is wrong.

Hence, 3|a and 3|b are obtained.

Thus, in any case, it is proved that,

If 3|a

^{2 }+ b

^{2}, then 3|a and 3|b.

**Question 40:**

Prove n^{4 }+ 4 is a composite number for n > 1

**Solution :
**n

^{4 }+ 4 = n

^{4 }+ 4n

^{2 }+ 4 – 4n

^{2 }= (n

^{2 }+ 2)

^{2 }– ( 2n)

^{2 }=(n

^{2 }– 2n + 2)( n

^{2 }+ 2n + 2)

=[(n – 1)

^{2 }+ 1][(n + 1)

^{2 }+ 1] … (1)

Now, n > 1 so n – 1 > 0 are distinct positive integers.

Also, n – 1 ≠ n + 1

∴ n – 1 and n + 1 are distinct positive integers.

∴ (n – 1)

^{2 }+ 1 and (n + 1)

^{2 }+ 1 are distinct positive integers.

Thus, from (1), n

^{4}+ 4 has two distinct positive integer factors [(n – 1)

^{2 }+ 1] and [(n + 1)

^{2 }+ 1].

∴ n

^{4 }+ 4 is a composite number for n > 1.

**Question 41:**

In a morning walk a man, a woman and a child step off together. Their steps measure 90 cm, 80 cm and 60 cm. What is the minimum distance each should walk to cover the distance in complete steps ?

**Solution :
**A man, woman and a child step off together.

∴ The measure of the steps of the man, woman and are 90 cm, 80 cm and 60 cm respectively.

If they walk the distance of l.c.m. of 90cm, 80 cm and 60 cm, then they cover the distance in complete steps.

90 = 2 × 3

^{2 }× 5, 80 = 2

^{4 }× 5 and 60 = 2

^{2 }× 3 × 5

∴ l.c.m.(90, 80, 60)= 2

^{4 }× 3

^{2 }× 5 =

**720**

Thus, the man, woman and child have to cover at least a distance of 720 cm to cover the distance in complete steps.

**Question 42**

Find the number nearest to 24001 and between 24001 and 25000 divisible by 16, 24, 40.

**Solution :
**

**Question 43 :**

Select a proper option (a), (b), (c) or (d) from given options :

**Question 43(1):**

Product of any four consecutive positive integers is divisible by……

**Solution :****
**c. 24

We have learnt that the product of four consecutive positive integers is divisible by 24.

**Question 43(2):**

+ 3 is …..

**Solution :
**

**Question 43(3):**

If g.c.d. of two numbers is 8 and their product is 384, then their l.c.m. is ……

**Solution :
**

**Question 43(4):**

If l.c.m. of two numbers (greater than 1) is the product of them, then their g.c.d. is …..

**Solution :
**

**Question 43(5):**

If P_{1} and p_{2} are distinct primes, their g.c.d. is …..

**Solution :
**d. 1

p

_{1}and p

_{2}are distinct prime numbers.

∴ There is no common factor other than 1 between them.

∴ g.c.d. of them is 1.

**Question 43(6):**

p, q, r are distinct primes, their l.c.m. is ……

**Solution :
**

**Question 43(7):**

g.c.d. (15, 24, 40) = ….

**Solution :
**b. 1

15 = 5

^{1 }× 3

^{1}, 24 = 2

^{3 }× 3

^{1}, 40 = 2

^{3 }× 5

^{1},

∴ g.c.d.(15, 24, 40) = 1

**Question 43(8):**

l.c.m. (15, 24, 40) = ….

**Solution :
**c. 120

15 = 5

^{1 }× 3

^{1}, 24 = 2

^{3 }× 3

^{1}, 40 = 2

^{3 }× 5

^{1},

∴ l.c.m.(15, 24, 40) = 2

^{3 }× 5

^{1 }× 3

^{1 }= 120

**Question 43(9):**

0.02222…. is a …..

**Solution :
**

**Question 43(10):**

= …..

**Solution :
**

**Question 43(11):**

= …..

**Solution :
**

**Question 43(12):**

g.c.d. (136, 221, 391) = …..

**Solution :
**b. 17

136 = 2

^{3 }× 17, 221 = 13 × 17, 391= 23 × 17,

∴ g.c.d.(136, 221, 391) = 17

**Question 43(13):**

l.c.m. (136, 221, 391) = …..

**Solution :
**a. 40664

136 = 2

^{3 }× 17, 221 = 13 × 17, 391 = 23 × 17

∴g.c.d. (136, 221, 391)= 2

^{3}×17×13×23 = 40664

**Question 43(14):**

If g.c.d. ( a, b) = 8, l.c.m. (a, b) = 64 and a > b then a = ……

**Solution :
**

**Question 43(15):**

If g.c.d. ( a, b) = 1, then g.c.d. (a – b, a + b)= ……

**Solution :
**a. 1 or 2

Suppose, g.c.d.(a – b, a + b) = d

∴ d | a – b and d | a + b

∴ a – b = md —- (1)

And a + b = nd where m, n ∈ N —-(2)

Now on adding and subtracting equations (2) and (1) respectively.

2a = (m + n)d and 2b = (n – m)d —-(3)

Now, g.c.d.(a, b) = 1

∴ 2 × g.c.d.(a, b) = 2

∴ g.c.d.(2a, 2b) = 2

∴ g.c.d((m + n)d, (n – m)d) = 2

Using values of 2a and 2b from (3)

d × g.c.d(m + n, n – m) = 2 = 2 × 1 = 1 × 2

∴ As 1 and 2 are mutually prime numbers, d = 2 or d = 1

∴ g.c.d.(a – b, a + b) is either d = 1 or d = 2

∴ g.c.d.(a – b, a + b) = 1 or 2

**Question 43(16):**

If n > 1, n^{4}+ 4 is … n ∈ N

**Solution :
**b. a composite integer

n

^{4 }+ 4 = n

^{4 }+ 4n

^{2 }+ 4 – 4n

^{2 }= (n

^{2 }+ 2)

^{2 }– (2n)

^{2 }=(n

^{2 }– 2n + 2)(n

^{2 }+ 2n + 2)

=[(n – 1)

^{2 }+ 1][(n + 1)

^{2 }+ 1] … … (1)

Now, n > 1 so n – 1 > 0 are distinct positive integers.

Also, n – 1 ≠ n + 1

∴ n – 1 and n + 1 are distinct positive integers.

∴ (n – 1)

^{2 }+ 1 and (n + 1)

^{2 }+ 1 are distinct positive integers.

Thus, from (1), n

^{4 }+ 4 has two distinct positive integer factors

[(n – 1)

^{2 }+ 1] and [(n + 1)

^{2 }+ 1].

∴n

^{4 }+ 4 is a composite number for n > 1.

**Question 43(17):**

If g.c.d. ( a, b) = 18, l.c.m. (a, b) ≠….

**Solution :
**c. 48

48 is not divisible by 18.

**Question 43(18):**

has …… digits after decimal point.

**Solution :
**

**Question 43(19):**

The decimal expansion of will terminate after …. digits.

**Solution :
**

**Question 43(20):**

5^{n }(n ∈ N) ends with ……

**Solution :
**b. 5

Any power of 5 always ends with 5.

**Question 43(21):**

2^{m}5^{n }(m, n ∈ N) ends with ……

**Solution :
**a. 0

2

^{m }5

^{n }= 2

^{m-1 }× 5

^{n-1 }× 2 × 5 = 2

^{m-1 }× 5

^{n-1 }× 10

(∵ m, n ∈ N so m – 1, n – 1 ∈ N ∪ {0})

∴ 2

^{m }5

^{n}ends with 0.

**Question 43(22):**

represents …….

**Solution :
**

**Question 43(23):**

(5k + 1)^{2} leaves reminder …… on dividing by 5.

**Solution :
**d. 1

(5k + 1)

^{2 }= 25k

^{2 }+ 10k + 1 = 5(5k

^{2 }+ 2k) + 1

∴ (5k + 1)

^{2}is divided by 5 and the remainder is 1.

**Question 43(24):**

On division by 6, a^{2} cannot leave remainder…..(a ∈ N)

**Solution :
**c. 5

From Euclid’s division lemma, for any positive integer a ∈ N,

a = 6m + r where, 0 ≤ r < 6 and m ∈ N

∴ a = 6m, a = 6m + 1, a = 6m + 2, a = 6m + 3, a = 6m + 4 or a = 6m + 5

Hence if a = 6m,

then a

^{2 }= 6

^{2}m

^{2 }= 36m

^{2 }= 6 × (6m

^{2}) + 0

If a = 6m + 1, then a

^{2 }= (6m + 1)

^{2 }= 36m

^{2 }+ 12m + 1

= 6(6m

^{2 }+ 2m) + 1

If a = 6m + 2, then a

^{2 }= (6m + 2)

^{2 }= 36m

^{2 }+ 24m + 4

= 6(6m

^{2 }+ 4m) + 4

If a = 6m + 3, then

a

^{2 }= (6m + 3)

^{2 }= 36m

^{2 }+ 36m + 9

= 6(6m

^{2 }+ 6m + 1) + 3

If a = 6m + 4, then

a

^{2 }= (6m + 4)

^{2 }= 36m

^{2 }+ 48k + 16

= 6(6m

^{2 }+ 8m + 2) + 4

If a = 6m + 5, then

a

^{2 }= (6m + 5)

^{2 }= 36m

^{2 }+ 60m + 25

= 6(6m

^{2 }+ 10m + 4) + 1

Thus in any possibility, if a

^{2}is divided by 6, then the remainder is 0, 1, 3 or 4 and not 5.

**Question 43(25):**

Product of three consecutive integers is divisible by …..

**Solution :
**c. 6

For any two consecutive integers one is even and the other is odd.

∴ In the product of three consecutive integers at least one of them is even.

∴ It is divisible by 2.

Also, in any three consecutive integers one of them is a multiple of 3.

Also, 3 and 2 are mutually prime numbers. So product of three consecutive integers is divisible by 6.

bhuvana says

good!

Mehul Jirwal says

Useful to score good marks

Mehul Jirwal says

Useful to complete syllabuse

pathan faizan says

You help me so much my freindthanks so much☺☺☺☺✋✋✋✋✋so much thanks for you you help me so much my brother