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**In each of the questions 1 to 17, out of the four options, only one is correct. Write the correct answer.Q.1. Generalised form of a four-digit number abdc is(a) 1000 a + 100 b + 10 c + d(b) 1000 a + 100 c + 10 b + d(c) 1000 a + 100 b + 10 d + c(d) a × b × c × dAns: **c

Solution:

We know that, the numbers are expressed as the sum of the product of it digits with the respective place value. So the generalised form of abdc is 1000 a + 100 b + 10 d + c

(a) x + y

(b) 10x + y

(c) 10x – y

(d) 10y + x

Ans:

Solution:

The numbers are expressed as the sum of the product of it digits with the respective place value. So the generalised form of xy is 10x+y

(a) abc

(b) abco

(c) aobc

(d) aboc

Ans: c

Solution:

The usual form to represent the aobc is 1000xa+100xo+ 10xb+1xc, which is equal 1000a+10b+c

(a) 9

(b) 11

(c) 18

(d) 33

Ans: c

Solution:

The general form of abc is 100a+10b+c

The genera form of cba is

Now (abc- cba) = (100a+10b+c ) – (100c+10b+a) =99a-99c

= 99(a – c)

Now, abc – cba is divisible by 99, because 99 is the factor of abc – cba

So, all the numbers which are the factors of 99 will also be divisible by abc-cba

Here, 9, 11 and 33 are the factors of 99. But 18 is not a factor of 99. Hence abc- cba is not divisible by 18.

(a) 11

(b) 33

(c) 37

(d) 74

Ans: c

Solution:

It is known that, three numbers which can be formed by the using the digits x, y and z are xyz, yzx and zxy.

The general form of xyz = 100x + 10y + z.

The general form of yzx = 100y + 10z + x.

The general form of zxy = 100z + 10x + y.

Add the three numbers, we will get

xyz + yzx + zxy = (100x + 10y + z) + (100y + 10z + x) + (100z + 10x + y).

xyz + yzx + zxy = 100x + 10x + x + 100y + 10y + y + 100z + 10z + z

xyz + yzx + zxy = 111x + 111y + 111z.

Now, simplify the expression, we will get

xyz + yzx + zxy = 111(x + y + z).

Now, it is clear that 111 is the common factor. So,

xyz + yzx + zxy is divisible by 111

Also xyz + yzx + zxy will be divisible by factors of 111.

From the given options, 11, 33, and 74 are not the factors of 11 whereas 37 is the factor of 111.

(a) 0 and 2

(b) 2 and 5

(c) 0 and 5

(d) 7

Ans: c

Solution:

It means that the number aabb is divisible by 5. By using the divisibility test of 5, it must be 0 or 5.

(a) a + b + c

(b) 3

(c) 37

(d) 9

Ans:

Solution:

= abc+bca+cab = 111(a+b+c)

Hence, abc+bca+cab is divisible by 111 and also it is divisible by the factors of 111.

Here, 3 and 7 are the factors of 111, and a+b+c is also a factor of 111(a+b+c).

But 9 is not the factor of 111.

(a) 0

(b) 1

(c) 4

(d) 5

Ans: b

Solution:

It means that

(4+b) – (a+5) =0, 11, 22, …it becomes

b – a – 1 = 0

Hence, b – a = 1

(a) 9

(b) 90

(c) 10

(d) 11

Solution:

Then the given number is abc – a – b – c =100a+10b+c – a – b- c

By simplifying the above expression, we will get

abc – a – b – c = 9(11a+b)

Hence, then number abc – a – b – c is divisible by 9.

(a) 7 only

(b) 11 only

(c) 13 only

(d) 1001

Ans: d

**Solution: **From the given question, the number should be of the form abcabc

So the general form of abcabc is 1000000a+100000b+1000c+100a+10b+c

Now, abcabc = is 1000000a+100000b+1000c+100a+10b+c

By simplifying the above expression, we will get

abcabc = 1001(100a+10b+c)

Hence, the six digit number should be divisible by 1001.**Q.11. If the sum of digits of a number is divisible by three, then the number is always divisible by(a) 2(b) 3(c) 6(d) 9Ans: **b

Solution:

(a) an odd multiple of 3

(b) odd multiple of 6

(c) even multiple of 3

(d) even multiple of 9

**Ans: **a

**Solution: **Given that, the sum of the digits xyz is given as x+y+z is 6, where z is an odd integer.

The divisibility test of 3, the number xyz is divisible by z. Since the last digit is an odd digit, then xyz is an odd multiple of 3.**Q.13. If 5 A + B 3 = 65, then the value of A and B is(a) A = 2, B = 3(b) A = 3, B = 2(c) A = 2, B = 1(d) A = 1, B = 2Ans: **c

**Solution:**

In the 1’s column A +3 = 5

When A is added with 3, it gives 5

Since A is a digit, it should be between 0 and 9

When you substitute A = 2, you will get 2+3 = 5

Similarly, repeat the process for 10’s column

Then we will get B= 1

Therefore A= 2, and B = 1**Q.14. If A 3 + 8 B = 150, then the value of A + B is(a) 13(b) 12(c) 17(d) 15Ans: aSolution:** From the question,

In 1’s place, we have to add 3+B = 0

Since B is a digit, the value of B should between 0 and 9

When you substitute the value of B = 7, then you will get

3+7 = 10

So here the unit digit is 0.

Now carry 1 to the 10’ s place,

Repeat the same process for 10’s place, you will get the value of A

So, A = 6

Therefore, A+B = 6+7 = 13

(a) 3

(b) 6

(c) 7

(d) 9

Ans: c

Solution:

Here, A is a number whose multiplication gives the two digit number having 9 in its 1’s place.

So, the product of 7 and 7 gives 49, where 1’s place is 9.

(a) –2

(b) 2

(c) –3

(d) 3

Ans:

Solution:

So, AB = B

Therefore A = 1

Now substitute A = 1 in the given equation

6B = A8

6B = 18

B = 18/6 = 3

Now, A – B = 1 -3 = -2

(a) 913462

(b) 114345

(c) 135792

(d) 3572406

Ans: b

**Solution: **If a number is divisible is 99, it should be divisible by both 9 and 11.

From the given option, option (b) 114345 is divisible by both 9 and 11.

All other numbers are either divisible by 9 or 11,but not both 9 and 11.

In questions 18 to 33, fill in the blanks to make the statements true.**Q.18. 3134673 is divisible by 3 and ______.****Ans:** 9

Adding all the digits in the given number, we get 27, which is divisible by 3 and 9.**Q.19. 20×3 is a multiple of 3 if the digit x is ______ or ______ or ______.Ans:** 1 or 4 or 7

It is known that, when a number is a multiple of 3, if the sum of digits in the number is a multiple of 3

By equating the closet multiple of 3 such as 6, 9, and 12, we get the digit

Either the digit x should be 1 or 4 or 7.

Ans:

It is known that, if a number is a multiple of 9, then the sum of digits of a number is divisible by 9.

From, the given equation, we can write it as:

3+x+5 = x + 8

It means that x+8 is divisible by 8, if x+8 is equal to 9,

Then the value of x is 1.

Ans:

Let “a” and “b” be the digits, then X = ab

By reversing the digits, then take Y =ba

Therefore sum of digits in X is 10a+b

Sum of digits in Y is 10b+a

By adding X and Y, we get

11a+11b = 11(a+b)

Therefore, the number is divisible by 11.

Ans:

Let “a” and “b” be the digits, then X = ab

By reversing the digits, then take Y =ba

Therefore sum of digits in X is 10a+b

Sum of digits in Y is 10b+a

By subtracting X and Y, we get

9a – 9a = 9(ab)

Therefore, the number is divisible by 9.

Ans:

Let “a”, “b” and “c” be the digits, then X = abc

By reversing the digits, then take Y =cba

Therefore sum of digits in X is 100a+10b+c

Sum of digits in Y is 100c+10b+a

By subtracting X and Y, we get

99(x-z) = (9)(11)(x-z)

Therefore, the number is divisible by 9 and 11.

, then A = ______ and B = ______.

Ans:

By adding the given digit according to the place value,

20+B + 10 A +B = 80 +A

9A+2B = 60

9A = 60 -2B

As, both A and B are whole numbers, 60-2B should be divisible by 9

So, the value of B should be 3

Substitute B = 3 in 9A + 2B = 60

Simplifying this, we will get

A = 6.

, then A = ___________ and B = ___________

Ans:

From the question, it is given that the units place contains a 6. So, the product of B and B should give a number with 6 in units place. It is possible when B = 4 or 6.

When B = 4,

(10A + B) x (B) = 96

It becomes (10A + 4) x 4 = 96

40A + 16 = 96,

By simplifying the equation, it gives A = 2.

Similarly, when B = 6,

(10A + B)x(B) = 96

By simplifying, it becomes (10A + 6) x 6 = 96

60A + 36 = 96, gives A = 1.

Hence, A = 2 B = 4 or A = 1 B = 6 (Both are feasible solutions)

, then B = ______.

Ans:

From the question, it is clear that the value of B should be greater than 4 and less than 9.

If the B value is less than 4, the solution should have only two digits. If the B value is greater than 9, the solution should have more than 3 digit.

So, the value of B ranges from 4 to 9. The 10’s place when multiplied by B gives 49

Hence, B x B = 49

Therefore, B = 7

Ans:

We know that, when a number is a multiple of 9, then the sum of digits in the number is divisible by 9

So, by adding the digits, we get:

X+9 = 9

Hence, x = 0

Ans:

Using the divisibility rule of 11, if abcd is divisible by 11, then

a-b + c-d = 0

By rearranging the terms, we get

a+ c = b+ d, or each should be zero

Ans:

Assume that the four digit number abcd

Here, the digits at odd places are a and c and in even places are b and d

Hence, (a+c) – (b+d) = 0

It follows that, (a+c) – (b+d) is divisible by 11, when (a+c) – (b+d) is not equal to 0.

Ans:

From the direct consequence from the divisibility rule of 11, we say that (a+c)- b is either 0 or a multiple of 11.

Ans:

Using the place values of the numbers, we can write it as:

10 + A = 3A

10 = 2A

A = 5

Ans:

From the given equations, B can take the values between 4 and 9

When you take B = 4, 4×4 = 16

Then, A=1 and B = 6

Ans:

Assume that, the number is initially 10t+u or tu, but after adding 1 to the unit place, tu gets shifted to the one unit higher. It means that 100t+10u+1 or tu1.

State whether the statements given in questions 34 to 44 are true (T) or false (F):

Ans:

If a number with two digits and the number in the unit place is even, then the number is said to be an even number and it is divisible by 2 (using the divisibility test of 2)

Ans:

If a number with three digits and if the unit digit is an even number, then the number is said to be an even number. A number is divisible by 5, only if the unit digit is either 0 or 5 (divisibility test of 5)

Ans:

Assume that 2463 is a four digit number which is equal to abcd.

We know that 24 is divisible by 4, but 2463 is not divisible by 4.

a + b + c is a multiple of 3.

Ans:

In case, if a number is divisible by 6, it is also divisible by 2 and 3.

If c is an even number, then the sum of digits is divisible by 3.

Ans:

The given number is of the form 3N+2

It can be written as (multiples of 3) + 2

When the given number 3N+2 is divide by 3, we will get the remainder 2. For example take 11.

11 can be written as (3×3) + 2

Ans:

The given number is of the form 7N+1

It can be written as (multiples of 7) + 1

When the given number 7N+1 is divide by 7, we will get the remainder 1. For example take 22.

22 can be written as (7×3) + 1

Ans:

Let a= 27, and b = 9

Here, when 27 is divisible by 9, we will get 3

Now, consider the factor of 9 = 1, 3 and 9

In this case, 27 is divisible by the factors such as 1 and 3. Hence the given statement is true.

Ans:

From the given question, the value of B should be either 3 or 8.

If you take the value of B is 3, it should be equal to 19.

Hence, the value of B is 8.

We know that, the value of A should between 0 and 9.

Hence, A x 4 = 19- 3 = 16

A = 16/4

A = 4

Therefore A= 4 and B= 8

Hence, A+B = 4+8 = 12

Ans:

From the given number, B+C is either 2 or the two digit number that gives the unit digit as 2.

It is given that B ≠ 0, C ≠ 0,

If B or C = 5 or 7, then A should be 3, then it becomes A+B+C = 14

Or else, if B= C = 6, and A = 2, we will get A+B+C = 14

Ans:

If 213 x 27 is divisible by 9, then the sum of the digits of a number is a multiple of 9.

2+1+3+x+2+7 = 15+x

Then 15x must be any multiple of 9 such as 9, 18, 27 …

Now assume that,

15+x = 18

X = 18 -15

X = 3

Ans:

Given that, when N is divided by 5, it leaves the remainder 5. (i.e) N = 5n+3 where n= 0, 1, 2, 3, …

Similarly, when N is divided by 2, it leaves the remainder 0. So N is an even Number. (Using divisibility test rule of 2).

But in N = 5n+3, the second term is odd.

So, 5n is an odd number.

When you substitute n = 1, 3, 5 … in 5n+3, we will get 8, 18, 28 …

Now, if we divide N by 10, it should be written as

N = 10 n+8

So, when N is divided by 10, it always leaves the remainder 8.

Ans:

By using the divisibility test of 8, if a number is divisible by 8, then the last three digit of a number should be divisible by 8.

In the given question, the last three digit is 6a2.

The value of “a” varies from 0 to 9.

If a = 0, then 6a2 = 602 is not divisible by 8.

If a = 1, then 612 = 612 is not divisible by 8.

If a = 2, then 6a2 = 622 is not divisible by 8.

If a = 3, then 6a2 = 632 is divisible by 8.

Hence, the least value of “a” is 3.

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