TN Samacheer Kalvi 10th Maths Solutions Chapter 2 Numbers and Sequences Ex 2.3

பத்தாம் வகுப்பிற்கு  கணித பாடத்தில் கற்பிக்கும் ஆசிரியர்களும்  மற்றும் மாணவ  மாணவியர்களுக்கும்  அன்பான வணக்கம்..., 

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 Chapter - 1 : Numbers & sequences EXERCISE : 2.3   அனைத்து கணக்குகளும் தனித்தனியாக தொகுக்கப்பட்டு விடைகளுடன் solved செய்து கீழே கொடுக்கப்பட்டுள்ளது.

 

       ஆசிரியர்களுக்கு பாடம் நடத்தும் வேலையில் ஏதேனும் சந்தேகங்கள் இருந்தால் கீழே கொடுக்கப்பட்டுள்ள கணக்குகளை refer செய்து கொள்ளலாம்.Chapter -1 பயிற்சி கணக்குகள் அனைத்தும் தனித்தனியாக தொகுக்கப்பட்டு கீழே கொடுக்கப்பட்டுள்ளது.

 மேலும் மாணவர்களுக்கும் சந்தேகங்கள் இருந்தால் கீழே கொடுக்கப்பட்டுள்ள கணக்குகளை பார்த்து சந்தேகங்களை சரி பார்த்துக் கொள்ளலாம்.

Question 1.

Find the least positive value of x such that

(i) 71 = x (mod 8)

Answer:

71 = 7 (mod 8)
∴ The value of x = 7

(ii) 78 + x = 3 (mod 5)
78 + x – 3 = 5n (n is any integer)
75 + x = 5n
(Let us take x = 5)
75 + 5 = 80 (80 is a multiple of 5)
∴ The least value of x is 5

(iii) 89 = (x + 3) (mod 4)
89 – (x + 3) = 4n
(n may be any integer)
89 – x – 3 = 4n
89 – x = 4n
86 – x is a multiple of 4
(84 is a multiple of 4)
86 – 2 = 4n
84 = 4n
The value of x is 2

(iv) 96 = x7 (mod 5)
96 – x7 = 5n (n may be any integer)
672 – x = 35n (multiple of 35 is 665)
672 – 7 = 665
∴ The value of x = 7

(v) 5x = 4 (mod 6)
5x – 4 = 6n (n may be any integer)
5x = 6n + 4
x = 6n+45
Substitute the value of n as 1, 6, 11, 16 …. as n values in x = 6n+45 which is divisible by 5.
2, 8, 14, 20,…………
The least positive value is 2.

Question 2.

If x is congruent to 13 modulo 17 then 7x -3 is congruent to which number modulo 17?

Solution:

x ≡ 13 (mod 17)
Let p be the required number …………. (1)
7x – 3 ≡ p (mod 17) ………….. (2)
From (1),
x – 13 = 17n for some integer M.
x – 13 is a multiple of 17.
x must be 30.
∴ 30 – 13 = 17
which is a multiple of 17.
From (2),
7 × 30 – 3 ≡ p (mod 17)
210 – 3 ≡ p (mod 17)
207 ≡ p (mod 17)
207 ≡ 3 (mod 17)
∴ P ≡ 3

Question 3.

Solve 5x ≡ 4 (mod 6)

5x – 4 = 6n (n may be any integer)
5x = 6n + 4
x = 6n+45
The value of n 1, 6, 11, 16 ……..
∴ The value of x is 2, 8, 14, 20 …………..

Question 4.

Solve 3x – 2 = 0 (mod 11)

Answer:

Given 3x – 2 = 0(mod 11)
3x – 2 = 11n (n may be any integer)
3x = 2 + 11n
x = 11n+23
Substitute the value of n = 2, 5, 8, 11 ….
When n ≡ 2 ⇒ x = 22+23 = 243 = 8
When n = 5 ⇒ x = 55+23 = 573 = 19
When n = 8 ⇒ x = 88+23 = 903 = 30
When n = 11 ⇒ x = 121+23 = 1233 = 41
∴ The value of x is 8, 19, 30,41

Question 5.

What is the time 100 hours after 7 a.m.?

Answer:

100 ≡ x (mod 12) Note: In a clock every 12 hours
100 ≡ 4 (mod 12) the numbers repeats.

The time repeat after 7 am is 7 + 4 = 11 o’ clock (or) 11 am.

Question 6.

What is time 15 hours before 11 p.m.?

Solution:

15 ≡ x (mod 12)
15 – x = 12n
15 – x is a multiple of 12 x must be 3.

∴ The time 15 hrs before 11 O’clock is 11 – 3 = 8 O’ clock i.e. 8 p.m

Question 7.

Today is Tuesday. My uncle will come after 45 days. In which day my uncle will be coming?

Answer:

Number of days in a week = 7
45 ≡ x (mod 7)
45 ≡ 3 (mod 7)

The value of x must be 3.
Three days after tuesday is friday uncle will come on friday.

Question 8.

Prove that 2n + 6 × 9n is always divisible by 7 for any positive integer n.

Solution:

21 + 6 × 91 = 2 + 54 = 56 is divisible by 7
When n = k,
2k + 6 × 9k = 7 m [where m is a scalar]
⇒ 6 × 9k = 7 m – 2k …………. (1)
Let us prove for n = k + 1
Consider 2k+1 + 6 × 9k+1 = 2k+1 + 6 × 9k × 9
= 2k+1 + (7m – 2k)9 (using (1))
= 2k+1 + 63m – 9.2k = 63m + 2k.21 – 9.2k
= 63m – 2k (9 – 2) = 63m – 7.2k
= 7 (9m – 2k) which is divisible by 7
∴ 2n + 6 × 9n is divisible by 7 for any positive integer n

Question 9.

Find the remainder when 281 is divided by 17?

Answer:

281 ≡ x(mod 17)
240 × 240 × 21 ≡ x(mod 17)
(24)10 × (24)10 × 21 ≡ x(mod 17)
(16)10 × (16)10 × 21 ≡ x(mod 17)
(162)5 × (162)5 × 21 ≡ x(mod 17)
= 1 × 1 × 2 (mod 17)
[(16)2 = 256 = 1 (mod 17)]
= 2 (mod 17)
281 = 2(mod 17)
∴ x = 2
The remainder is 2

Question 10.

The duration of flight travel from Chennai to London through British Airlines is approximately 11 hours. The airplane begins its journey on Sunday at 23:30 hours. If the time at Chennai is four and half hours ahead to that of London’s time, then find the time at London, when will the flight lands at London Airport?

Answer:

Duration of the flight time = 11 hours
(Chennai to London)
Starting time on Sunday = 23 : 30 hour

Time difference is 4 12 horns ahead to london
The time to reach London airport = (10.30 – 4.30)
= 6 am
The first reach the london airport next day (monday) at 6 am