Numbers and Sequences Ex 2.10-10th Std Maths-Book Back Question And Answer
Multiple choice questions:
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
Answer
(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-2 then 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
