# Q&A - Ask Doubts and Get Answers

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Q. Are only big animals facing extinction?

No, It is not necessary that only big animal is going to be extinct.  Ex- Dodo is not that big but is extinct now.

Animal life is also affected by deforestation. How?

Animal life is also affected by deforestation because wild animals lose their shelter and food amount is decreased. So, they start migrating for the search of the food, in this process they might get killed by other animals or hunters. In this way, their life will be affected by deforestation.

Q How does deforestation reduce rainfall on the one hand and lead to floods on the other?

Deforestation means cutting down the trees. And we know that trees play an essential role in the water cycle as it contributes moisture to the air which brings rainfall. So if no. of trees decreases the process of transpiration and water cycle get disturbed. Hence rainfall will be reduced. Reduced rain will reduce the amount of water absorption in the soil. This soil will be dry and leads to...

Q. What is the purpose of making national parks, wildlife sanctuaries and biosphere reserves?

The main purpose of making national parks, wildlife sanctuaries and biosphere reserves is to protect the animals from an external threat like prevent from hunting and other danger. And also to maintain their natural habitat as well as their own environment.

Q1. If the division N ÷ 2 leaves a remainder of 1, what might be the one’s digit of N? (N is odd; so its one’s digit is odd. Therefore, the one’s digit must be 1, 3, 5, 7 or 9.)

The detailed solution for the above-written question is as follows N is odd; so it's unit digit is odd. Therefore, the unit digit must be 1, 3, 5, 7 or 9.

Q1. If the division N ÷ 5 leaves a remainder of 3, what might be the ones digit of N? (The one’s digit, when divided by 5, must leave a remainder of 3. So the one’s digit must be either 3 or 8.)

The detailed solution for the above-written problem is as follows, The unit digit, when divided by 5, must be leaving a remainder of 3. So the unit digit must be either 3 or 8.

Q2.

(iii) From (i) and (ii) above, can you say that a number will be divisible by 11 if the difference between the  sum of digits at its odd places and that of digits at the even places is divisible by 11?

Yes,  A number will always be divisible by 11 if the difference between the sum of digits at its odd places and that of digits at the even places is divisible by 11. So, for instance, 2728 has the alternating sum of digits 2-7+2-8 = -11. Here -11 is divisible by 11, so is 2728. Similarly, for 31415, the alternating sum of digits is 3-1+4-1+5 = 10. This would not divisible by 11, so neither is 31415.

Q2.

(ii) Write a 4-digit number $abcd$ as $1000a + 100b + 10c + d$
$\\= (1001a + 99b + 11c) - (a - b + c - d)\\ = 11(91a + 9b + c) + [(b + d) - (a + c)]$
If the number $abcd$ is divisible by 11, then what can you say about $[(b + d) - (a + c)]$ ?

If the number abcd is divisible by 11 then [ (b + d) - (a + c) ] also must be divisible by 11. let the number be 1089 here a = 1, b = 0, c = 8 and d = 9 1089 = 1000*1 + 100*0 + 10*8 + 9          = (1001*1 + 99*0 + 11*8) + [(0 + 9) - (1 + 8)]          = 11(91*1 + 9*0 + 8) + [ 9 - 9 ] here  [ (b + d) - (a + c) ] = [9 - 9 ] = 0 which is divisible by 11. hence If the number abcd is divisible by 11...

Q2.

(i) Write a 3-digit number $abc$ as $100a + 10b + c$
$\\= 99a + 11b + (a - b + c)\\ = 11(9a + b) + (a - b + c)$
If the number $abc$ is divisible by 11, then what can you say about
$(a - b + c)$

Is it necessary that $(a - b + c)$  should be divisible by 11?

let the number abc be 132 Here a = 1, b = 3 and c = 2 132= 100*1 + 10*3 + 2 = 99 + 11*3 + (1 - 3 + 2)     = 11(9*1+3) + (1 - 3 + 2 )  if number is divisible by 11 then (a - b + c )  must be divisible by 11. as in above case of number 132 the a - b + c = 1 -3 + 2 = 0 whichis divisible by 11. Hence we conclude ( a - b + c ) should be divisible by 11 if abc is divisible by 11.

Q1. You have seen that a number 450 is divisible by 10. It is also divisible by 2 and 5 which are factors of 10. Similarly, a number 135 is divisible 9. It is also divisible by 3 which is a factor of 9.

Can you say that if a number is divisible by any number m, then it will also be divisible by each of the factors of m?

Yes, it has been prooved that if a number is divisible by any number m, then it will also be divisible by each of the factors of m. Let's Assume n is divisible by m, and m is divisible by k.This means n=pm and m = qk where all are integers Now, n = p(qk) =( pq)k which means n is divisible by k. Hence a number is divisible by any number m, then it will also be divisible by each of the factors of m.

Check the divisibility of the following numbers by 3.

5. 927

Any number will be divisible by 3 only if the sum of all the digits in that number will be divisible by 3. Sum of digits of number 927 = 9 + 2 + 7 = 18 which is divisible by 3.  Hence we conclude number 108 is divisible by 3.

Check the divisibility of the following numbers by 3.

Q4. 432

Any number will be divisible by 3 only if the sum of all the digits in that number will be divisible by 3. Sum of digits of number 432 = 4 + 3 + 2 = 9 which is divisible by 3.  Hence we conclude number 432 is divisible by 3.

Check the divisibility of the following numbers by 3.

Q3. 294

Any number will be divisible by 3 only if the sum of all the digits in that number will be divisible by 3. Sum of digits of number 294= 2 + 9 + 4 = 15 which is divisible by 3.  Hence we conclude number 294 is divisible by 3.

Check the divisibility of the following numbers by 3.

Q2. 616

Any number will be divisible by 3 only if the sum of all the digits in that number will be divisible by 3. Sum of digits of number 616 = 6 + 1 + 6 = 13 which is not divisible by 3.  Hence we conclude number 616 is not divisible by 3.

Check the divisibility of the following numbers by 3.

Q1. 108

Any number will be divisible by 3 only if the sum of all the digits in that number will be divisible by 3. Sum of digits of number 108 = 1 + 0 +8 = 9 which is divisible by 3.  Hence we conclude number 108 is divisible by 3.

Check the divisibility of the following numbers by 9.

Q5. 927

Any number will be divisible by 9 only if the sum of all the digits in that number will be divisible by 9. Sum of digits of 927 = 9 + 2 + 7 = 18 which is divisible by 18. Hence we conclude number 927 is divisible by 9.

Check the divisibility of the following numbers by 9.

Q4. 432

Any number will be divisible by 9 only if the sum of all the digits in that number will be divisible by 9. Sum of digit 432 = 4 + 3 + 2 = 9 which is divisible by 9. Hence we conclude 432 is divisible by 9.

Check the divisibility of the following numbers by 9.

Q3. 294

Any number will be divisible by 9 only if the sum of all the digits in that number will be divisible by 9. Sum of digits of 294 = 2 + 9 + 4 = 15 which is not divisible by 9. Hence we conclude 294 is not divisible by 9.

Check the divisibility of the following numbers by 9.

2. 616

Any number will be divisible by 9 only if the sum of all the digits in that number will be divisible by 9. Sum of digits of 616 = 6 + 1 + 6 = 13 which is not divisible by 9, Hence we conclude 616 is not divisible by by 9.

Check the divisibility of the following numbers by 9.

1. 108

Any number will be divisible by 9 only if the sum of all the digits in that number will be divisible by 9. Sum of digit of 108 = 1 + 0 + 8 = 9 which is divisible by 9 i.e (9/9 = 1). hence we conclude 108 is divisible by 9.
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