Ex 2.10 | Aster Classes

Tamilnadu, Samacheer Kalvi, 10th, Maths, Solutions, Chapter 2, Numbers and Sequences, unit, exercise 2,

Question 1.
Prove that n2 – n divisible by 2 for every positive integer n.
Answer:
We know that any positive integer is of the form 2q or 2q + 1 for some integer q.
Case 1: When n = 2 q
n2 – n = (2q)2 – 2q = 4q2 – 2q
= 2q (2q – 1)
In n2 – n = 2r
2r = 2q(2q – 1)
r = q(2q + 1)
n2 – n is divisible by 2

Case 2: When n = 2q + 1
n2 – n = (2q + 1)2 – (2q + 1)
= 4q2 + 1 + 4q – 2q – 1 = 4q2 + 2q
= 2q (2q + 1)
If n2 – n = 2r
r = q (2q + 1)
∴ n2 – n is divisible by 2 for every positive integer “n”


Question 2.
A milk man has 175 litres of cow’s milk and 105 litres of buffalow’s milk. He wishes to sell the milk by filling the two types of milk in cans of equal capacity. Calculate the following
(i) Capacity of a can
(ii) Number of cans of cow’s milk
(iii) Number of cans of buffalow’s milk.
Answer:
175 litres of cow’s milk.
105 litres of goat’s milk.
H.C.F of 175 & 105 by using Euclid’s division algorithm.
175 = 105 × 1 + 70, the remainder 70 ≠ 0
Samacheer Kalvi 10th Maths Guide Chapter 2 Numbers and Sequences Unit Exercise 2 1
Again using division algorithm,

105 = 70 × 1 + 35, the remainder 35 ≠ 0
Again using division algorithm.
70 = 35 × 2 + 0, the remainder is 0.
∴ 35 is the H.C.F of 175 & 105.
(i) ∴ The milk man’s milk can’s capacity is 35 litres.
(ii) No. of cow’s milk obtained = 17535 = 5 cans
(iii) No. of buffalow’s milk obtained = 10535= 3 cans


Question 3.
When the positive integers a, b and c are divided by 13 the respective remainders are 9,7 and 10. Find the remainder when a + 2b + 3c is divided by 13.
Answer:
Given the positive integer are a, b and c
a = 13q + 9 (divided by 13 leaves remainder 9)
b = 13q + 7
c = 13q + 10
a + 2b + 3c = 13q + 9 + 2(13q + 7) + 3 (13q + 10)
= 13q + 9 + 26q + 14 + 39q + 30
= 78q + 53
When compare with a = 3q + r
= (13 × 6) q + 53
The remainder is 53

Question 4.
Show that 107 is of the form 4q +3 for any integer q.
Solution:
107 = 4 × 26 + 3. This is of the form a = bq + r.
Hence it is proved.


Question 5.
If (m + 1)th term of an A.P. is twice the (n + 1)th term, then prove that (3m + 1)th term is twice the (m + n + 1)th term.
Answer:
tn = a + (n – 1)d
Given tm+1 = 2 tn+1
a + (m + 1 – 1)d = 2[a + (n + 1 – 1)d]
a + md = 2(a + nd) ⇒ a + md =2a + 2nd
md – 2nd = a
d(m – 2n) = a ….(1)
To Prove t(3m + 1) = 2(tm+n+1)
L.H.S. = t3m+1
= a + (3m + 1 – 1)d
= a + 3md
= d(m – 2n) + 3md (from 1)
= md – 2nd + 3md
= 4md – 2nd
= 2d (2m – n)
R.H.S. = 2(tm+n+1)
= 2 [a + (m + n + 1 – 1) d]
= 2 [a + (m + n)d]
= 2 [d (m – 2n) + md + nd)] (from 1)
= 2 [dm – 2nd + md + nd]
= 2 [2 md – nd]
= 2d (2m – n)
R.H.S = L.H.S
∴ t(3m+1) = 2 t(m+n+1)
Hence it is proved.


Question 6.
Find the 12th term from the last term of the A.P -2, -4, -6,… -100.
Answer:
The given A.P is -2, -4, -6, …. 100
d = -4 – (-2) = -4 + 2 = – 2
Finding the 12 term from the last term
a = -100, d = 2 (taking from the last term)
n = 12
tn = a + (n – 1)d
t12 = – 100 + 11 (2)
= -100 + 22
= -78
∴ The 12th term of the A.P from the last term is – 78


Question 7.
Two A.P’s have the same common difference. The first term of one A.P is 2 and that of the other is 7. Show that the difference between their 10th terms is the same as the difference between their 21st terms, which is the same as the difference between any two corresponding terms.
Solution:
Let the two A.Ps be
AP1 = a1, a1 + d, a1 + 2d,…
AP2 = a2, a2 + d, a2 + 2d,…
In AP1 we have a1 = 2
In AP2 we have a2 = 7
t10 in AP1 = a1 + 9d = 2 + 9d ………….. (1)
t10 in AP2 = a2 + 9d = 7 + 9d …………… (2)
The difference between their 10th terms
= (1) – (2) = 2 + 9d – 7 – 9d
= -5 ………….. (I)
t21 m AP1 = a1 + 20d = 2 + 20d …………. (3)
t21 in AP2 = a2 + 20d = 7 + 20d ………… (4)
The difference between their 21 st terms is
(3) – (4)
= 2 + 20d – 7 – 20d
= -5 ……………. (II)
I = II
Hence it is Proved.


Question 8.
A man saved ₹16500 in ten years. In each year after the first he saved ₹100 more than he did in the preceding year. How much did he save in the first year?
Answer:
Amount of saving in ten years = ₹ 16500
S10 = 16500, d= 100
Sn = n2 [2a + (n – 1)d]
S10 = 102 [2a + 9d]
16500 = 102 [2a + 900] = 5(2a + 900)
16500 = 10a + 4500 ⇒ 16500 – 4500 = 10a
12000 = 10a
a = 1200010 = 1200
Amount saved in the first year = ₹ 1200


Question 9.
Find the G.P. in which the 2nd term is 6–√and the 6th term is 9 6–√.
Answer:
2nd term of the G.P = 6–√
t2 = 6–√
[tn = a rn-1]
a.r = 6–√ ….(1)
6th term of the G.P. = 9 6–√
a. r5 = 96–√ ……..(2)


Question 10.
The value of a motorcycle depreciates at a rate of 15% per year. What will be the value of the motorcycle 3 year hence, which is now purchased for ₹45,000?
Solution:
a = ₹45000
Depreciation = 15% for ₹45000
= 45000 × 15100
d = ₹6750 since it is depreciation
d = -6750
At the end of 1st year its value = ₹45000 – ₹6750
= ₹38250,
Again depreciation = 38250 × 15100 = 5737.50
At the end of 2nd year its value
= ₹38250 – ₹5737.50 = 32512.50
Again depreciation = 32512.50 × 15100 = 4876.88
At the end of the 3rd year its value
= 32512.50 – 4876.88 = 27635.63
∴ The value of the automobile at the 3rdyear
= ₹ 27636


Tamilnadu, Samacheer Kalvi, 10th, Maths, Solutions, Chapter 2, Numbers and Sequences, Ex 2.10,

Question 1.
Euclid’s division lemma states that for positive integers a and b, there exist unique integers q and r such that a = bq + r, where r must satisfy ………………….
(1) 1 < r < b
(2) 0 < r < b
(3) 0 < r < 6
(4) 0 < r < b
Ans.
(3) 0 < r < b


Question 2.
Using Euclid’s division lemma, if the cube of any positive integer is divided by 9 then the possible remainders are ………………….
(1) 0, 1, 8
(2) 1, 4, 8
(3) 0, 1, 3
(4) 1, 3, 5
Answer:
(1) 0, 1, 8
Hint: Let the +ve integer be 1, 2, 3, 4 …………
13 = 1 when it is divided by 9 the remainder is 1.
23 = 8 when it is divided by 9 the remainder is 8.
33 = 27 when it is divided by 9 the remainder is 0.
43 = 64 when it is divided by 9 the remainder is 1.
53 = 125 when it is divided by 9 the remainder is 8.
The remainder 0, 1, 8 is repeated.


Question 3.
If the H.C.F of 65 and 117 is expressible in the form of 65m – 117 , then the value of m is
(1) 4
(2) 2
(3) 1
(4) 3
Answer:
(2) 2
Hint:
H.C.F. of 65 and 117
117 = 65 × 1 + 52
65 = 52 × 1 + 13
52 = 13 × 4 + 0
∴ 13 is the H.C.F. of 65 and 117.
65m – 117 = 65 × 2 – 117
130 – 117 = 13
∴ m = 2


Question 4.
The sum of the exponents of the prime factors in the prime factorization of 1729 is …………………….
(1) 1
(2) 2
(3) 3
(4) 4
Answer:
(3) 3
Hint: 1729 = 7 × 13 × 19
Sum of the exponents = 1 + 1 + 1
= 3


Question 5.
The least number that is divisible by all the numbers from 1 to 10 (both inclusive) is
(1) 2025
(2) 5220
(3) 5025
(4) 2520
Answer:
(4) 2520
Hint:

L.C.M. = 23 × 32 × 5 × 7
= 8 × 9 × 5 × 7
= 2520


Question 6.
74k ≡ ______ (mod 100)
(1) 1
(2) 2
(3) 3
(4) 4
Answer:
(1) 1
Hint:
74k ≡______ (mod 100)
y4k ≡ y4 × 1 = 1 (mod 100)


Question 7.
Given F1 = 1 , F2 = 3 and Fn = Fn-1 + Fn-2then F5 is ………….
(1) 3
(2) 5
(3) 8
(4) 11
Answer:
(4) 11
Hint:
Fn = Fn-1 + Fn-2
F3 = F2 + F1 = 3 + 1 = 4
F4 = F3 + F2 = 4 + 3 = 7
F5 = F4 + F3 = 7 + 4 = 11


Question 8.
The first term of an arithmetic progression is unity and the common difference is 4. Which of the following will be a term of this A.P
(1) 4551
(2) 10091
(3) 7881
(4) 13531
Answer:
(3) 7881
Hint:
t1 = 1
d = 4
tn = a + (n – 1)d
= 1 + 4n – 4
4n – 3 = 4551
4n = 4554
n = will be a fraction
It is not possible.
4n – 3 = 10091
4n = 10091 + 3 = 10094
n = a fraction
4n – 3 = 7881
4n = 7881 + 3 = 7884
n = 78844, n is a whole number.
4n – 3 = 13531
4n = 13531 – 3 = 13534
n is a fraction.
∴ 7881 will be 1971st term of A.P.


Question 9.
If 6 times of 6th term of an A.P is equal to 7 times the 7th term, then the 13th term of the A.P. is ………..
(1) 0
(2) 6
(3) 7
(4) 13
Answer:
(1) 0
Hint:
6 t6 = 7 t7
6(a + 5d) = 7 (a + 6d) ⇒ 6a + 30d = 7a + 42d
30 d – 42 d = 7a – 6a ⇒ -12d = a
t13 = a + 12d (12d = -a)
= a – a = 0


Question 10.
An A.P consists of 31 terms. If its 16th term is m, then the sum of all the terms of this A.P. is
(1) 16 m
(2) 62 m
(3) 31 m
(4) 312 m
Answer:
(3) 31 m
Hint:
t16 = m
S31 = 312 (2a + 30d)
= 312 (2(a + 15d))
(∵ t16 = a + 15d)
= 31(t16) = 31m


Question 11.
In an A.P., the first term is 1 and the common difference is 4. How many terms of the A.P must be taken for their sum to be equal to 120?
(1) 6
(2) 7
(3) 8
(4) 9
Answer:
(3) 8
Here a = 1, d = 4, Sn = 120
Sn = n2[2a + (n – 1)d]
120 = n2 [2 + (n – 1)4] = n2 [2 + 4n – 4)]
= n2 [4n – 2)] = n2 × 2 (2n – 1)
120 = 2n2 – n
∴ 2n2 – n – 120 = 0 ⇒ 2n2 – 16n + 15n – 120 = 0
2n(n – 8) + 15 (n – 8) = 0 ⇒ (n – 8) (2n + 15) = 0
n = 8 or n = −152 (omitted)
∴ n = 8


Question 12.
A = 265 and B = 264 + 263 + 262 …. + 20 which of the following is true?
(1) B is 264 more than A
(2) A and B are equal
(3) B is larger than A by 1
(4) A is larger than B by 1
Answer:
(4) A is larger than B by
A = 265
B = 264+63 + 262 + …….. + 20
= 2
= 1 + 22 + 22 + ……. + 264
a = 1, r = 2, n = 65 it is in G.P.
S65 = 1 (265 – 1) = 265 – 1
A = 265 is larger than B


Question 13.
The next term of the sequence 316,18,112,118 is ………..
(1) 124
(2) 127
(3) 23
(4) 181
Answer:
(2) 127
Hint:
316,18,112,118
a = 316, r = 18 ÷ 316 = 18 × 163 = 23
The next term is = 118 × 23 = 127


Question 14.
If the sequence t1,t2,t3 … are in A.P. then the sequence t6,t12,t18 … is
(1) a Geometric Progression
(2) an Arithmetic Progression
(3) neither an Arithmetic Progression nor a Geometric Progression
(4) a constant sequence
Answer:
(2) an Arithmetic Progression
Hint:
If t1, t2, t3, … is 1, 2, 3, …
If t6 = 6, t12 = 12, t18 = 18 then 6, 12, 18 … is an arithmetic progression


Question 15.
The value of (13 + 23 + 33 + ……. + 153) – (1 + 2 + 3 + …….. + 15) is …………….
(1) 14400
(2) 14200
(3) 14280
(4) 14520
Answer:
(3) 14280
Hint:

1202 – 120 = 120(120 – 1)
120 × 119 = 14280


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