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.

** **

**4.
Explain your answer with reason for the following statements.**

**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 9** 8**.
A = 8

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

The least number divisible by 3 is 567** 0**. A = 0

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

The greatest 3 digit number divisible by 2 is 9** 9**6.
A = 9

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

The number divisible by 4 and 9 is ** 1**08. A = 1

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

The number divisible by 11 is 225** 8**85. 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 20 cm, 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 1 ^{st} 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.**

** **

__ANSWERS: __

__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.

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

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