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Maths 3 marks

T2 CH1 Numbers · Tamil Nadu Board STD 6 (Tamil - English Medium). 25 questions.

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25 questions · timed · auto-graded

Question 13 Marks
The sum of three prime numbers is 80. The difference of two of them is 4. Find the numbers
Answer
Given the sum of three prime numbers is 80
The numbers will be one or two-digit prime numbers. Also one of them is 2
Sum of the remaining 2 numbers = 78 ...[∵ one number must be even]
Also, their difference = 4 ....(Given) [Otherwise sum of three odd numbers is odd]
The numbers will be 37 and 41
The required numbers are 2, 37, 41
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Question 23 Marks
For which of the numbers, from n = 2 to 8, is 2n − 1 a prime?
Answer
n 2n − 1 Result Prime/Not Prime
2 2 × 2 − 1 = 4 − 1 3 Prime
3 2 × 3 − 1 = 6 − 1 5 Prime
4 2 × 4 − 1 = 8 − 1 7 Prime
5 2 × 5 − 1 = 10 − 1 9 Not prime
6 2 × 6 − 1 = 12 − 1 11 Prime
7 2 × 7 − 1 = 14 − 1 13 Prime
8 2 × 8 − 1 = 16 − 1 15 Not prime
For n = 2, 3, 4, 6 and 7 it is prime.
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Question 33 Marks
Every even number greater than $2$ can be expressed as the sum of two prime numbers. Verify this statement for every even number upto $16$
Answer
Even number greater then 2 upto 16 are $4,6,8,10,12,14$ and $16$
$ 4=2+2$
$6=3+3$
$8=3+5$
$10=3+7 \text { or } 5+5$
$12=5+7$
$14=7+7 \text { or } 3+11$
$16=5+11 \text { or } 3+13 $
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Question 43 Marks
Wilson, Mathan and Guna can complete one round of a circular track in 10, 15 and 20 minutes respectively. If they start together at 7 a.m from the starting point, at what time will they meet together again at the starting point?
Answer
This is an LCM related problem
So, we need to find the LCM of 10, 15 and 20
5 10, 15, 20
2 2, 3, 4
1, 3, 2
LCM = 5 × 2 × 1 × 3 × 2 = 60 min
They will meet together again after 60 minutes.
ie. 7 a.m + 60 minutes = 8 a.m
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Question 53 Marks
In an apartment consisting of 108 floors, two lifts A and B starting from the ground floor, stop at every $3^{\text {rd }}$ and $5^{\text {th }}$ floors respectively. On which floors, will both of them stop together?
Answer
Multiples of 3 are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, 99, 102, 105
Multiples of 5 are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, 100, 105
Common multiples of 3 and 5 are 15, 30, 45, 60, 75, 90 and 105
Both the lifts will stop at floors 15, 30, 45, 60, 75, 90 and 105
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Question 63 Marks
Malarvizhi, Karthiga and Anjali are friends and natives of the same village. They work at different places. Malarvizhi comes to her home once in 5 days. Similarly, Karthiga and Anjali come to their homes once in 6 days and 10 days respectively. Assuming that they met each other on the $1^{\text {st }}$ of October, when will all the three meet again?
Answer
Find the $\operatorname{LCM}(5,6,10)$
$\operatorname{LCM}(15,25,30)=5 \times 6=30$
They meet again after 30 days
$\therefore$ They met on $1^{\text {st }}$ October
They will meet again on $31^{\text {st }}$ October
5 15, 25, 30
3 3, 5, 6
5 1, 5, 2
2 1, 1, 2
  1, 1, 1
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Question 73 Marks
Find the smallest number which is exactly divisible by all the numbers from 1 to 9
Answer
To find the smallest number we have to find the LCM (1, 2, 3, 4, 5, 6, 7, 8, 9)
LCM is 2 × 3 × 2 × 5 × 7 × 2 × 3 = 2520
The required number is 2520
2 1, 2, 3, 4, 5, 6, 7, 8, 9
3 1, 1, 3, 2, 5, 3, 7, 4, 9
2 1, 1, 1, 2, 5, 1, 7, 4, 3
5 1, 1, 1, 1, 5, 1, 7, 2, 3
7 1, 1, 1, 1, 1, 1, 7, 2, 3
2 1, 1, 1, 1, 1, 1, 1, 2, 3
3 1, 1, 1, 1, 1, 1, 1, 1, 3
1, 1, 1, 1, 1, 1, 1, 1, 1
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Question 83 Marks
Find the LCM set of numbers using prime factorisation method.
$30, 40, 60$
Answer
$ 30,40,60$
$30=3 \times 2 \times 5$
$40=2 \times 2 \times 2 \times 5$
$60=2 \times 3 \times 2 \times 5 $
Product of highest powers of the common factors $=3 \times 2^3 \times 5=120$ $\operatorname{LCM}(30,40,60)=120$
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Question 93 Marks
Find the LCM set of numbers using prime factorisation method.
$15, 25, 75$
Answer
$ 15,25,75$
$15=5 \times 3$
$25=5 \times 5$
$75=5 \times 5 \times 3 $
Product of the highest powers of the common factors $=3 \times 5^2$
$ =3 \times 25=75$
$\operatorname{LCM}(15,25,75)=75 $
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Question 103 Marks
Find the LCM set of numbers using prime factorisation method.
$14, 42$
Answer
$ 14,42$
$14=2 \times 7$
$42=2 \times 21$
$=2 \times 3 \times 7 $
Product of common factors $=2 \times 7$
Product of other factor $=3$
$\operatorname{LCM}(14,42)=$ Product of common factors $\times$ Product of other factors
$ =2 \times 7 \times 3$
$=42 $
$\operatorname{LCM}(14,42)=42$
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Question 113 Marks
Find the LCM set of numbers using prime factorisation method.
$8, 12$
Answer
$ 8,12$
$8=2 \times 4=2 \times 2 \times 2$
$12=2 \times 6=2 \times 2 \times 3 $
Product of common factors $=2 \times 2=4$
Product of other factors $=2 \times 3=6$
LCM $=$ Product of common factors $\times$ Product of other factors
$=4 \times 6=24$
$\operatorname{LCM}(8,12)=24$.
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Question 123 Marks
Find the LCM set of numbers using prime factorisation method.
$10, 15$
Answer
$ 10,15$
$10=2 \times 5$
$15=3 \times 5 $
Product of common factors $=5$
Product of other factors $=2 \times 3=6$
$\operatorname{LCM}(10,15)=$ Product of common factors $\times$ Product of other factors
$ =5 \times 6$
$=30 $
$\operatorname{LCM}(10,15)=30$
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Question 133 Marks
Find the HCF set of numbers using prime factorisation method.
27, 45, 81
Answer
27, 45, 81
Prime factorisation of 27 = 3 × 3 × 3
Prime factorisation of 45 = 3 × 3 × 5
Prime factorisation of 81 = 3 × 3 × 3 × 3
Common factors of 27, 45, 81 = 3 × 3 = 9
HCF (27, 45, 81) = 9
3 27
3 9
3 3
1
3 45
3 15
5 5
1
3 81
3 27
3 9
3 3
1
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Question 143 Marks
Find the HCF set of numbers using prime factorisation method.
45, 55, 95
Answer
45, 55, 95
Prime factorisation of 45 = 3 × 3 × 5
Prime factorisation of 55 = 5 × 11
Prime factorisation of 95 = 5 × 19
Common factors of 45, 55, 95 = 5
HCF (45, 55, 95) = 5
3 45
3 15
5 5
1
5 55
11 11
1
5 95
19 19
1
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Question 153 Marks
Find the HCF set of numbers using prime factorisation method.
84, 120
Answer
84, 120
Prime factorisation of 84 = 2 × 2 × 3 × 7
Prime factorisation of 120 = 2 × 2 × 2 × 3 × 5
Common factors of 84 and 120 = 2 × 2 × 3
HCF (84, 120) = 12
2 84
2 42
3 21
7 7
4
2 120
2 60
2 30
3 15
5 5
1
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Question 163 Marks
Find the HCF set of numbers using prime factorisation method.
18, 24
Answer
18, 24
Prime factorisation of 18 = 2 × 3 × 3
Prime factorisation of 24 = 2 × 2 × 2 × 3
Common factors of 18 and 24 = 2 × 3 = 6
HCF (18, 24) = 6
2 18
3 9
3 3
1
2 24
2 12
2 6
3 3
1
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Question 173 Marks
Find the HCF set of numbers using prime factorisation method.
51, 85
Answer
51, 85
Prime factorisation of 51 = 3 × 17
Prime factorisation of 85 = 5 × 17
Common factors of 51 and 85 = 17
HCF (51, 85) = 17
3 51
17 17
1
5 85
17 17
1
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Question 183 Marks
Find the HCF set of numbers using prime factorisation method.
61, 76
Answer
61, 76
Prime factorisation of 61 = 1 × 61
Prime factorisation of 76 = 2 × 2 × 19 × 1
Common factors of 61 and 76 = 1
HCF (61, 76) = 1
61 61
1
2 76
2 38
19 19
1
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Question 193 Marks
The digits of the prime number 13 can be reversed to get another prime number 31. Find if any such pairs exist upto 100
Answer
The required pairs are
(i) 17 and 71
(ii) 37 and 73
(iii) 79 and 97
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Question 203 Marks
If there are 143 math books to be arranged in equal numbers in all the stacks, then find the number of books in each stack and also the number of stacks.
Answer
Total number of books = 143
Factorizing 143 = 11 × 13
11143
1313
1
A number of stacks and number of books in each stack maybe 11, 13 or 13, 11.
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Question 213 Marks
Find the prime factorisation number by factor tree method and division method.
999
Answer
999
∴ 999 = 3 × 3 × 3 × 37
Also
3999
3333
3111
3737
1

999 = 3 × 333
= 3 × 3 × 111
= 3 × 3 × 3 × 37
∴ 999 = 3 × 3 × 3 × 37
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Question 223 Marks
Find the prime factorisation number by factor tree method and division method.
420
Answer
420

∴ 420 = 2 × 2 × 3 × 5 × 7
Also
2420
2210
3105
535
77
77
1

420 = 2 × 210
= 2 × 2 × 105
= 2 × 2 × 3 × 35
= 2 × 2 × 3 × 5 × 7
∴ 420 = 2 × 2 × 3 × 5 × 7
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Question 233 Marks
Find the prime factorisation number by factor tree method and division method.
198
Answer
198

Also
2198
399
333
1111
1

198 = 2 × 99
= 2 × 3 × 33
= 2 × 3 × 3 × 11
∴ 198 = 2 × 3 × 3 × 11
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Question 243 Marks
Find the prime factorisation number by factor tree method and division method.
144
Answer
144
144

∴ 144 = 2 × 2 × 2 × 2 × 3 × 3
Also
2144
272
236
236
218
39
33
3
1

144 = 2 × 72 = 2 × 2 × 36
= 2 × 2 × 2 × 18 = 2 × 2 × 2 × 2 × 9
= 2 × 2 × 2 × 2 × 3 × 3
∴ 144 = 2 × 2 × 2 × 2 × 3 × 3
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Question 253 Marks
Find the prime factorisation number by factor tree method and division method.
128
Answer
128

∴ 128 = 2 × 2 × 2 × 2 × 2 × 2 × 2
Also

2128
264
232
216
216
28
24
24
22
1


128 = 2 × 64 = 2 × 2 × 32
= 2 × 2 × 2 × 16
= 2 × 2 × 2 × 2 × 8
128 = 2 × 2 × 2 × 2 × 2 × 4
∴ 128 = 2 × 2 × 2 × 2 × 2 × 2 × 2

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