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Exercise 1.3

6th Maths : Term 2 Unit 1 : Numbers : Exercise 1.3 : Miscellaneous Practice Problems, Challenge Problems, Text Book Back Exercises Questions with Answers, Solution

Exercise 1.3

Miscellaneous Practice Problems

1. 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.

4 = 2 + 2

6 = 3 + 3

8 = 3 + 5

10 = 3 + 7 (or) 5 + 5

12 = 5 + 7

14 = 7 + 7 (or) 3 + 11

16 = 5+ 11 (or) 3 + 13

2. Is 173 a prime? Why?

173, is a prime number. Because it has only two factors 173 = 1 × 173.

3. For which of the numbers, from n = 2 to 8, is 2n − 1 a prime?

n = 2, 3, 4, 5, 6, 7, 8

2n − 1

n = 2, 2 × 2 − 1 = 4 − 1 = 3

n = 3, 2 × 3 – l = 6 – l = 5

n = 4, 2 × 4 – 1 = 8 – 1 = 7

n = 5, 2 × 5 − 1 = 10 − 1 = 9 composite number

n = 6, 2 × 6 − 1 = 12 − 1 = 11

n = 7, 2 × 7 − 1 = 14 − 1 = 13

n = 8, 2 × 8 − 1 = 16 − 1 = 15 composite number

2n −1 is prime for the numbers 2, 3, 4, 6 and 7.

a) A number is divisible by 9, if it is divisible by 3.

b) A number is divisible by 6, if it is divisible by 12.

a) A number is divisible by 9, if it is divisible by 3.

False − The number 6 is divisible by 3.

But it is not divisible by 9.

b) A number is divisible by 6, if it is divisible by 12.

True − The number 24 is divisible by 12

It is divisible by 6 also.

5. Find A as required:

(i) The greatest 2 digit number 9A is divisible by 2.

(ii) The least number 567A is divisible by 3.

(iii) The greatest 3 digit number 9A6 is divisible by 6.

(iv) The number A08 is divisible by 4 and 9.

(v) The number 225A85 is divisible by 11.

(i) The greatest 2 digit number 9A is divisible by 2.

The greatest 2 digit number divisible by 2 is 98. A = 8

(ii) The least number 567A is divisible by 3.

The least number divisible by 3 is 5670. A = 0

(iii) The greatest 3 digit number 9A6 is divisible by 6.

The greatest 3 digit number divisible by 2 is 996. A = 9

(iv) The number A08 is divisible by 4 and 9.

The number divisible by 4 and 9 is 108. A = 1

(v) The number 225A85 is divisible by 11.

The number divisible by 11 is 225885. A = 8

6. Numbers divisible by 4 and 6 are divisible by 24. Verify this statement and support your answer with an example.

False. The number divisible by 4 and 6 is 12. But it is not divisible by 24.

7. The sum of any two successive odd numbers is always divisible by 4. Justify this statement with an example.

True. Two successive odd numbers 17, 19

Sum of this numbers = 17 + 19 = 36, It is divisible by 4.

8. Find the length of the longest rope that can be used to measure exactly the ropes of length 1m 20cm, 3m 60cm and 4m.

Length of ropes = 1 m 20cm, 3m 60cm and 4m.

H.C.F. must be found.

Length of three ropes must be in same units.

120 cm, 360 cm and 400 cm.

H.C.F. = 4 × 10 = 40 cm

The length of the longest rope that can be used to measure exactly = 40cm.

Challenge Problems

9. The sum of three prime numbers is 80. The difference of two of them is 4. Find the numbers.

The difference of two numbers 4

The two numbers 37, 41

41 – 37 = 4.

Sum of three prime numbers = 80

2 + 37 + 41 = 80

The three prime numbers 2, 37 and 41.

10. Find the sum of all the prime numbers between 10 and 20 and check whether that sum is divisible by all the single digit numbers.

The prime numbers between 10 and 20

11,13,17,19

Sum of these numbers 11 + 13 + 17 + 19 = 60

60, It is divisible by 1

It is divisible by 2

It is divisible by 3

It is divisible by 4

It is divisible by 5

It is divisible by 6

11. Find the smallest number which is exactly divisible by all the numbers from 1 to 9.

The smallest number which is exactly divisible by all the numbers from 1 to 9 is 2520.

12. The product of any three consecutive numbers is always divisible by 6. Justify this statement with an example.

The product of any three consecutive numbers is always divisible by 6.

Yes.

2 × 3 × 4 = 24, It is divisible by 6

3 × 4 × 5 = 60, It is divisible by 6

4 × 5 × 6 = 120, It is divisible by 6

13. 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 1st of October, when will all the three meet again?

To find the again meeting days of Malarvizhi, Karthiga and Anjali calculate L.C.M.

L.C.M. = 5 × 2 × 3 = 30

Once in 30 days the three will meet again.

14. In an apartment consisting of 108 floors, two lifts A & B starting from the ground floor, stop at every 3rd and 5th floors respectively. On which floors, will both of them stop together?

To find on which floors, both of the lifts stop together calculate L.C.M.

L.C.M. of 3 and 5 = 3 × 5 = 15

Both the lifts will stop together at floors 15, 30, 45, 60, 75, 90 and 105.

15. The product of 2 two digit numbers is 300 and their HCF is 5. What are the numbers?

The product of 2 two digit numbers = 300

H.C.F. = 5

The product of two numbers = H.C.F × L.C.M

300 = 5 × L.C.M.

L.C.M =  300/5 = 60

L.C.M = 60 = 2 × 2 × 3 × 5

The two numbers = 15 and 20

16. Find whether the number 564872 is divisible by 88. (use of the test of divisibility rule for 8 and 11 will help )

The number 564872 is divisible by 88.

It must be divisible by 8 and 11.

564 872 If the last three digits divisible by 8

It will divisible by 8

872/8  = 109 remainder 0.

It is divisible by 8.

5 6 4 8 7 2

Sum of alternate digits

5 + 4 + 7 = 16

6 + 8 + 2 = 16

Difference = 0

It is divisible by 11.

So, This number 564872 is divisible by 88.

17. 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?

To calculate the meeting time of Wilson, Mathan and Guna find L.C.M.

L.C.M. = 5 × 2 × 2 × 3 = 60 minutes

= 1 hour.

After 60 minutes, at 8 a.m they will meet together again at the starting point.

Exercise 1.3

1. 4 = 2 + 2; 6 = 3 + 3; 8 = 3 + 5;

10 = 3 + 7 (or) 5 + 5; 12 = 5 + 7;

14 = 7 + 7 (or) 3 + 11;

16 = 5 + 11 (or) 3 + 13

2. Yes, because it has only two factors.

3. For n = 2, 3, 4, 6 and 7

4. a) False b) True

5. i) 8 ii) 0 iii) 9 iv) 1 v) 8

6. False. 12 is divisible by both 4 and 6 but not by 24

7. True. 17+19 =36 is divisible by 4

8. 40 cm

Challenge problems

9. 2, 37, 41

10. 11, 13, 17, 19; The sum 60 is divisible by 1, 2, 3, 4, 5 and 6

11. 2520

12. Yes. 2 × 3 × 4 = 24 is divisible by 6.

13. Once in 30 days

14. The lifts will stop at floors 15, 30, 45, 60, 75, 90 and 105

15. (15, 20)

16. Yes. Since it is divisible by both 8 and 11 and hence by 88

17. After 60 minutes, at 8 a.m

Two numbers are said to be amicable numbers if the sum of the factors of one number (except the number itself) gives the other number.

The numbers 220 and 284 are amicable, since the sum of the factors of 220 (except 220) i.e., 1 + 2 + 4 + 5 + 10 + 11 + 20 + 22 + 44 + 55 + 110 = 284 and the sum of the factors of (except 284) i.e., 1 + 2 + 4 + 71 + 142 = 220.

Check whether 1184 and 1210 are amicable numbers.

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6th Maths : Term 2 Unit 1 : Numbers : Exercise 1.3 | Questions with Answers, Solution | Numbers | Term 2 Chapter 1 | 6th Maths