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We know that the sum of digits of the number is divisible by 9 then the number is divisible by 9.

Sum of the digits is 15, which is not divisible by 9

∴The given number 64302 is not divisible by 9

We know that the sum of digits of the number is divisible by 9 then the number is divisible by 9.

Sum of the digits is 27, which is divisible by 9

∴The given number 89361 is divisible by 9

We know that the sum of digits of the number is divisible by 9 then the number is divisible by 9.

Sum of the digits is 30, which is not divisible by 9

∴The given number 14799 is not divisible by 9

We know that the sum of digits of the number is divisible by 9 then the number is divisible by 9.

Sum of the digits is 36, which is divisible by 9

∴The given number 66888 is divisible by 9

We know that the sum of digits of the number is divisible by 9 then the number is divisible by 9.

Sum of the digits is 9, which is divisible by 9

∴The given number 30006 is divisible by 9

We know that the sum of digits of the number is divisible by 9 then the number is divisible by 9.

Sum of the digits is 15, which is not divisible by 9

∴The given number 33333 is not divisible by 9

We know that if the number formed by last two digits of a given number is divisible by 4 then the whole number is divisible by 4.

The number formed by the last two digits is 34, which is not divisible by 4

∴The given number 134 is not divisible by 4

We know that if the number formed by last two digits of a given number is divisible by 4 then the whole number is divisible by 4.

The number formed by the last two digits is 18, which is not divisible by 4

∴The given number 618 is not divisible by 4

We know that if the number formed by last two digits of a given number is divisible by 4 then the whole number is divisible by 4.

The number formed by the last two digits is 28, which is divisible by 4

∴The given number 3928 is divisible by 4

We know that if the number formed by last two digits of a given number is divisible by 4 then the whole number is divisible by 4.

The number formed by the last two digits is 76, which is divisible by 4

∴The given number 50176 is divisible by 4

We know that if the number formed by last two digits of a given number is divisible by 4 then the whole number is divisible by 4.

The number formed by the last two digits is 92, which is divisible by 4

∴The given number 39392 is divisible by 4

We know that if the number formed by last two digits of a given number is divisible by 4 then the whole number is divisible by 4.

The number formed by the last two digits is 94, which is not divisible by 4

∴The given number 56794 is not divisible by 4

We know that if the number formed by last two digits of a given number is divisible by 4 then the whole number is divisible by 4.

The number formed by the last two digits is 02, which is not divisible by 4

∴The given number 86102 is not divisible by 4

We know that if the number formed by last two digits of a given number is divisible by 4 then the whole number is divisible by 4.

The number formed by the last two digits is 66, which is not divisible by 4

∴The given number 66666 is not divisible by 4

We know that if the number formed by last two digits of a given number is divisible by 4 then the whole number is divisible by 4.

The number formed by the last two digits is 18, which is not divisible by 4

∴The given number 99918 is not divisible by 4

We know that if the number formed by last two digits of a given number is divisible by 4 then the whole number is divisible by 4.

The number formed by the last two digits is 36, which is divisible by 4

∴The given number 77736 is divisible by 4

we know that if the number formed by last three digits of a given number is divisible by 8 then the whole number is divisible by 8.

The number formed by the last three digits is 132, which is not divisible by 8

∴The given number 6132 is not divisible by 8

we know that if the number formed by last three digits of a given number is divisible by 8 then the whole number is divisible by 8.

The number formed by the last three digits is 304, which is divisible by 8

∴The given number 7304 is divisible by 8

Let the given statement P(n) be defined as P(n) : 1 + 3 + 5 +...+ (2n – 1) = n², for n ∈ N. Note that P(1) is true, since

P(1) : 1 = 1²

Assume that P(k) is true for some k ∈ N, i.e.,

P(k) : 1 + 3 + 5 + ... + (2k – 1) = k²

Now, to prove that P(k + 1) is true, we have

1 + 3 + 5 + ... + (2k – 1) + (2k + 1)

= k² + (2k + 1)

= k² + 2k + 1 = (k + 1)²

Thus, P(k + 1) is true, whenever P(k) is true.

Hence, by the Principle of Mathematical Induction, P(n) is true for all n ∈ N.

We know that if the number formed by last three digits of a given number is divisible by 8 then the whole number is divisible by 8.

The number formed by the last three digits is 312, which is divisible by 8

∴The given number 59312 is divisible by 8

We know that if the number formed by last three digits of a given number is divisible by 8 then the whole number is divisible by 8.

The number formed by the last three digits is 664, which is divisible by 8

∴The given number 66664 is divisible by 8

We know that if the number formed by last three digits of a given number is divisible by 8 then the whole number is divisible by 8.

The number formed by the last three digits is 444, which is not divisible by 8

∴The given number 44444 is not divisible by 8

we know that if the number formed by last three digits of a given number is divisible by 8 then the whole number is divisible by 8.

The number formed by the last three digits is 360, which is divisible by 8

∴The given number 154360 is divisible by 8

We know that if the number formed by last three digits of a given number is divisible by 8 then the whole number is divisible by 8.

The number formed by the last three digits is 818, which is not divisible by 8

∴The given number 998818 is not divisible by 8

we know that if the number formed by last three digits of a given number is divisible by 8 then the whole number is divisible by 8.

The number formed by the last three digits is 472, which is divisible by 8

∴The given number 265472 is divisible by 8

we know that if the number formed by last three digits of a given number is divisible by 8 then the whole number is divisible by 8.

The number formed by the last three digits is 162, which is not divisible by 8

∴The given number 7350162 is not divisible by 8

We know that in the unit place A + 7 = 3

A = 3-7 = -4, which is not possible

Since A is greater than 10, where, 1 is carried over to tens place

∴ A + 7 = 13

A = 13-7 = 6

Now in tens place

5 + 8 + 1 = B where 1 is carried over

∴ B = 14

Here, 1 is carried over to hundred’s place

∴ C = 1

We know that in the unit place 6 + A = 3

A = 3-6 = -3, which is not possible

Since A is greater than 10, where, 1 is carried over to tens place

∴ 6 + A = 13

A = 13-6 = 7

Now in tens place

B + 9 + 1 = 7 where 1 is carried over

B = 7-10 = -3, which is not possible

Since B is greater than 10, where, 1 is carried over to hundred’s place

B + 9 + 1 = 10+7

B = 17-10 = 7

∴ B = 7

Now in hundred’s place

C + 6 + 1 = 1 where, 1 is carried over

C= 1-7 = -6 which is not possible

Since C is greater than 10, where, 1 is carried over to hundred’s place

C + 6 + 1 = 1 + 10

C = 11 -7 = 4

∴ C = 4

We know that in the unit place A + A + A = A

3A = A, which is not possible

Since A is greater than 10, where, 1 is carried over to tens place

A + A + A = A + 10

3A = A + 10

3A-A = 10

2A=10

∴A=10/2 = 5

Now in tens place

∴B = 1 where 1 is carried over

Here, in units place,

6 + A = 3

Ie. A = - 3, which is not possible.

Hence A is greater than 10,where, 1 carry is given to tens place.

∴ A + 6 = 13

∴ A = 7

Now, in tens place,

B + 9 + 1 = 7 … as 1 is carried

∴ B = -3, which is not possible.

Hence B is greater than 10,where, 1 carry is given to hundreds place.

∴ B + 9 + 1 = 17

∴ B = 7

Now, in hundreds place,

C + 6 + 1 = 1 … as 1 is carried

Ie. C = -6, which is not possible.

Hence C is greater than 10, where, 1 carry is given to thousands place.

∴ C + 6 + 1 = 11

∴ C = 4

Here, in units place, 

A + A +A = A Ie. 3A = A, 

ie. 3 = 1 which is wrong. 

Hence A is greater than 10, where, 1 carry is given to tens place. 

∴ A + A + A = A + 10 

∴ 3A = A + 10 

∴ 2A = 10 

∴ A = 5 

Now, in tens place, 

B = 1 … as 1 is carried

In tens place

6 – A = 3

Which implies that A ≤ 3

Now in units place

A – B = 7

This involves borrowing

∴ in tens place

6 – A -1 = 3

∴ A= 2

Now in units place

A + 10 – B = 7, where borrowing is involved

2 + 10 –B = 7

12 – B = 7

12-7 = B

∴ B= 5

Here, we can observe that B × A = B

i.e. A = 1

Here,

First digit = B+1

Thus, 1 will be carried from 1+B² and becomes (B+1) (B² -9) B.

∴ C = B² -1

Now, all B, B+1 and B² -9 are one digit number.

This condition is satisfied for B=3 or B=4.For B< 3, B² -9 will be negative.

For B>3, B² -9 will become a two digit number.

For B=3, C = 32 - 9 = 9-9 = 0

For B = 4, C = 42 -9 = 16-9 = 7

Hence, A=1, B=3, C = 0 or A=1, B=4, C = 7

Here, (B × 3) = B

Here, B can be either 0 or 5, which satisfies above condition.

If B is 5, then 1 will be carried,

then, A×3+1 = A will not be possible for any number

∴ B = 0

Also, A×3=A is possible for either 0 or 5.

If we take A=0, then all number will become 0, which is not possible

∴ A= 5

So, 1 will be carried.

∴ C = 1

(i) 524618

We know that if sum of digits of a number is divisible by 9, then the number is divisible by 9.

Here, sum of digits is 26, which is not divisible by 9.

Hence, 524618 is not divisible by 9.

(ii) 7345845

We know that if sum of digits of a number is divisible by 9, then the number is divisible by 9.

Here, sum of digits is 36, which is divisible by 9.

Hence, 7345845 is divisible by 9.

(iii) 8987148

We know that if sum of digits of a number is divisible by 9, then the number is divisible by 9.

Here, sum of digits is 45, which is divisible by 9.

Hence, 8987148 is divisible by 9.

In unit place

5 – A = 9, this involves borrowing

i.e. 10 + 5 –A = 9

15 – A = 9

15-9 = A

∴ A= 6

Now in tens place, as 1 is borrowed from hundred’s place as well as lent

B – 5 + 10 -1 = 8

B = 8-4 = 4

∴ B= 4

Now in hundred’s place as 1 is lent

C – 2 -1 = 2

∴ C= 5

Here, in units place,

A – 8 = 3

This implies that 1 is borrowed.

11 – 8 = 3

∴ A = 1

Now in tens place,Then, 7−B=9

1 is borrowed from 7.

∴ 16−B=9

∴ B = 7

Now in hundreds place,

5−C=2

But 1 has been borrowed from 5.

∴ 4 - C = 2

∴ C=2

∴ A=1, B=7 and C=2

Now in units place,

5 - A = 9

This states that borrowing is involved.

Ie. 10 + 5 - A = 9

∴ A = 6

Now in tens place, as 1 is borrowed from hundreds place and also lent,

B – 5 +10 - 1 = 8

B = 4

Now in hundred place, as 1 is lent,

C – 2 – 1 = 2

C = 5

Here, in tens place,

6 – 3 = 3

Which implies that maximum value of A is 3

Ie. A ≤ 3

Now in units place,

A - B = 7

This states that borrowing is involved.

∴ in tens place,

6 – A – 1 = 3

∴ A = 2

Now in units place,

A + 10 – B = 7 …as borrowing is involved

∴ 12 – B = 7

∴ B = 5

Here, (B × 3) = B

Wherein B can be either 0 or 5, which satisfies the above condition

If B is 5, then 1 can be carried

A × 3 + 1 = A, this not be possible for any number

∴ B= 0

Also, A × 3 = A is possible for either 0 0r 5

If we consider A=0, then all the numbers will become 0 which is not possible

∴ A= 5

So 1 will be carried over

∴ C= 1

If more amount of pulses, the cost will be more. Therefore it’s directly proportional.

Let us consider the cost be x, 882/14 = x/22

14 × x = 882 × 22

14x = 19404

x = 19404/14

x = 1386

∴ ₹1386 is the cost

Here, B×A= A i.e. A=1

Now first digit (B+1)

Where 1 can be carried from 1+B² and becomes (B+1)(B²-9)B

∴ C= B²-1

Now B, B+1 and B²-9 are one single digit

This condition is satisfied for B=3 or B=4

For B<3, B²-9 is negative

For B >3, B²-9 will become two digit number

For B=3, C= 3²– 9 = 9-9 = 0

For B=4, C= 4²-9 = 16-9 = 7

∴ A=1, B= 3, C= 0

A=1, B=4, C=7

We know that the sum of the digits of a number is divisible by 3 then the whole number is divisible by 3

5+x+6 = multiple of 3

11+x= 0, 3, 6, 9….

11+x= 12

∴ x=12-11 = 1

The least value of x is 1

We know that the sum of the digits of a number is divisible by 3 then the whole number is divisible by 3

6+4+y+8 = multiple of 3

18+y= 0, 3, 6, 9….

18+y= 18

∴ y=18-18 = 0

The least value of y is 0

We know that the sum of the digits of a number is divisible by 9 then the whole number is divisible by 9

7+x+8 = multiple of 9

15+x= 0, 9, 18, 27….

15+x= 18

∴ x=18-15 = 3

The least value of x is 3

We know that the sum of the digits of a number is divisible by 9 then the whole number is divisible by 9

3+7+y+4 = multiple of 9

14+y= 0, 9, 18, 27….

14+y= 18

∴ y=18-14 = 4

The least value of y is 4

Given A = N x N

* is a binary operation on A defined by 

(a, b) * (c, d) = (a + c, b + d) 

(i) Commutativity: Let (a, b), (c, d) ∈ N x N 

Then (a, b) * (c, d) = (a + c, b + d) = (c + a, d + b) 

(∵ a, b, c, d ∈ N, a + c = c + a and b + d = d + c) 

= (c, d) * b 

Hence, (a, b) * (c, d) = (c, d) * (a, b)

∴ * is commutative. 

(ii) Associativity: let (a, b), (b, c), (c, d) 

Then [(a, b) * (c, d)] * (e, f) = (a + c, b + d) * (e, f) = ((a + c) + e, (b + d) + f) 

= {a + (c + e), b + (d + f)] (∵ set N is associative) 

= (a, b) * (c + e, d + f) = (a, b) * {(c, d) * (e, f)} 

Hence, [(a, b) * (c, d)] * (e, f) = (a, b) * {(c, d) * (e, f)} 

∴ * is associative. 

(iii) Let (x, y) be identity element for ∀ on A, 

Then (a, b) * (x, y) = (a, b) 

⇒ (a + x, b + y) = (a, b) 

⇒ a + x = a, b + y = b 

⇒ x = 0, y = 0 

But (0, 0) ∉ A 

∴ For *, there is no identity element.

Given f : R → R and g : R → R defined by f (x) = sinx and g(x) = 5x²

∴ gof(x) = g [f(x)] = g (sinx) = 5 (sinx)² = 5 sin²x

1. The gravitational force that acts on the mass is called the weight. 

2. The gravitational force is always inversely proportional to the distance. 

3. So as the distance increases between the earth and the object, the gravitational force decreases. 

4. The gravitational force by which the earth attracts an object towards its centre is called weight. 

5. The weight at the sea-level is higher than at higher altitude, as the object is at a greater distance at higher altitudes compared to sealevel.

Mass is a scalar quantity because it can be completely expressed by its magnitude alone, so it is scalar quantity.