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Page No 408:
Question 1:
In the questions out of the four options, only one is correct.
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 × d
Answer:
In generalised form, we express a number as the sum of the products of its digits with their respective place values.
abdc is written in generalised form as 1000a + 100b + 10d + c.
i.e. abdc = 1000a + 100b + 10d + c
Hence, the correct answer is option C.
Page No 409:
Question 2:
In the questions out of the four options, only one is correct.
Generalised form of a two-digit number xy is
(a) x + y
(b) 10x + y
(c) 10x – y
(d) 10y + x
Answer:
In generalised form, xy can be written as the sum of the products of its digits with their respective place values.
i.e. xy = 10x + y
Hence, the correct answer is option B.
Page No 409:
Question 3:
In the questions out of the four options, only one is correct.
The usual form of 1000a + 10b + c is
(a) abc
(b) abc0
(c) a0bc
(d) ab0c
Answer:
Given expanded form of a number is 1000a + 10b + c.
We can write it as 1000 × a + 100 × 0 + 10 × b + c
= a0bc
Hence, the correct answer is option C.
Page No 409:
Question 4:
In the questions out of the four options, only one is correct.
Let abc be a three-digit number. Then abc – cba is not divisible by
(a) 9
(b) 11
(c) 18
(d) 33
Answer:
Given, abc is a three-digit number.
Then, abc = 100a + 10b + c
and cba = 100c + 10b + a
Then, abc – cba is divisible by 99.
Thus, abc – cba is divisible by 9, 11, 33, but it is not divisible by 18.
Hence, the correct answer is option C.
Page No 409:
Question 5:
In the questions out of the four options, only one is correct.
The sum of all the numbers formed by the digits x, y and z of the number xyz is divisible by
(a) 11
(b) 33
(c) 37
(d) 74
Answer:
We know that, the sum of three-digit numbers taken in cyclic order can be written as xyz + yzx + zxy.
Thus, xyz + yzx + zxy is divisible by 37, but not divisible by 11, 33 and 74.
Hence, the correct answer is option C.
Page No 409:
Question 6:
In the questions out of the four options, only one is correct.
A four-digit number aabb is divisible by 55. Then possible value(s) of b is/are
(a) 0 and 2
(b) 2 and 5
(c) 0 and 5
(d) 7
Answer:
Given that, aabb is divisible by 55.
Then, it is also divisible by 5.
Now, if a number is divisible by 5, then its unit digit is either 0 or 5.
Thus, the possible values of b are 0 and 5.
Hence, the correct answer is option C.
Page No 409:
Question 7:
In the questions out of the four options, only one is correct.
Let abc be a three digit number. Then abc + bca + cab is not divisible by
(a) a + b + c
(b) 3
(c) 37
(d) 9
Answer:
Given that abc be a three digit number.
Thus, the sum is divisible by 3, 37 and (a + b + c), but not divisible by 9.
Hence, the correct answer is option D.
Page No 409:
Question 8:
In the questions out of the four options, only one is correct.
A four-digit number 4ab5 is divisible by 55. Then the value of b – a is
(a) 0
(b) 1
(c) 4
(d) 5
Answer:
Given, a four-digit number 4ab5 is divisible by 55.
Since 55 = 5 ×11
Then, it is also divisible by 11.
The difference of the sum of its digits in odd places and the sum of its digits in even places is either 0 or multiple of 11.
i.e. (4 + b) – (a + 5) is 0 or a multiple of 11
So, 4 + b – a – 5 = 11n, where n = {0, 1, 2, 3....}
⇒ b – a = 1 + 11n, where n = {0, 1, 2, 3....}
Thus, the possible values of b – a is 1, 12, 23...
Hence, the correct answer is option B.
Page No 409:
Question 9:
In the questions out of the four options, only one is correct.
If abc is a three digit number, then the number abc – a – b – c is divisible by
(a) 9
(b) 90
(c) 10
(d) 11
Answer:
We have, abc = 100a + 10b + c
Thus, the given number abc – a – b – c is divisible by 9.
Hence, the correct answer is option A.
Page No 409:
Question 10:
In the questions out of the four options, only one is correct.
A six-digit number is formed by repeating a three-digit number. For example 256256, 678678, etc. Any number of this form is divisible by
(a) 7 only
(b) 11 only
(c) 13 only
(d) 1001
Answer:
Let the six digit number be abcabc.
Thus, it is divisible by 1001.
Since 1001 = 7 × 11 × 13, therefore, abcabc is also divisible by all the three numbers, i.e., 7, 11, 13 and not by only one of these.
Hence, the correct answer is option D.
Page No 409:
Question 11:
In the questions out of the four options, only one is correct.
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) 9
Answer:
We know that, if the sum of digits of a number is divisible by three, then the number must be divisible by 3.
The remainder obtained by dividing the number by 3 is same as the remainder obtained by dividing the sum of its digits by 3.
Hence, the correct answer is option B.
Page No 409:
Question 12:
In the questions out of the four options, only one is correct.
If x + y + z = 6 and z is an odd digit, then the three-digit number xyz is
(a) an odd multiple of 3
(b) odd multiple of 6
(c) even multiple of 3
(d) even multiple of 9
Answer:
We have, x + y + z = 6 and z is an odd digit.
Since, sum of the digits is divisible by 6, it will also be divisible by 2 and 3 but the unit digit is odd, so it is divisible by 3 only.
Thus, the number is an odd multiple of 3.
Hence, the correct answer is option A.
Page No 410:
Question 13:
In the questions out of the four options, only one is correct.
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 = 2
Answer:
We have,
A + 3 is a equal to 5.
So, either A + 3 is 5 or it is 15.
∴ A + 3 = 5
⇒ A = 2
And, 5 + B = 6
⇒ B = 1
Thus, A = 2 and B = 1.
Hence, the correct answer is option C.
Page No 410:
Question 14:
In the questions out of the four options, only one is correct.
If A 3 + 8 B = 150, then the value of A + B is
(a) 13
(b) 12
(c) 17
(d) 15
Answer:
We have,
Here, 3 + B = 0, so 3 + B is a two-digit number whose unit’s digit is zero.
∴ 3 + B = 10
⇒ B = 7
Now, considering ten’s column, A + 8 + 1 = 15
⇒ A + 9 = 15
⇒ A = 6
Thus, A + B = 6 + 7 = 13.
Hence, the correct answer is option A.
Page No 410:
Question 15:
In the questions out of the four options, only one is correct.
If 5 A × A = 399, then the value of A is
(a) 3
(b) 6
(c) 7
(d) 9
Answer:
We have, 5A × A = 399
Here, A × A is the number 9 or a number whose unit’s digit is 9.
Thus, the number whose product with itself produces a two-digit number having its unit’s digit as 9 is 7.
i.e. A × A = 49
⇒ A = 7
Now, 5 × A + 4 = 39
⇒ 5 × 7 + 4 = 39
So, A satisfies the product.
Thus, the value of A is 7.
Hence, the correct answer is option C.
Page No 410:
Question 16:
In the questions out of the four options, only one is correct.
If 6 A × B = A 8 B, then the value of A – B is
(a) –2
(b) 2
(c) –3
(d) 3
Answer:
Given, 6A × B = A 8 B
Here, LHS = 6A × B and RHS = A 8 B
Let us assume, A = 1 and B = 3
Then, LHS = 61 × 3 = 183 and RHS = 183 Thus, our assumption is true.
∴ A – B = 1 – 3 = –2
Hence, the correct answer is option A.
Page No 410:
Question 17:
In the questions out of the four options, only one is correct.
Which of the following numbers is divisible by 99
(a) 913462
(b) 114345
(c) 135792
(d) 3572406
Answer:
A number is divisible by 99 means it is divisible by 9 and 11 both.
Now, going through the options, we observe that the number 114345 is divisible by 9 and 11 both as the sum of digits of the number (1 + 1 + 4 + 3 + 4 + 5 = 18) is divisible by 9 and sum of digits at odd places (1 + 4 + 4 = 9) = sum of digits at even places (1 + 3 + 5 = 9).
Hence, the correct answer is option B.
Page No 410:
Question 18:
Fill in the blanks to make the statements true.
3134673 is divisible by 3 and ______.
Answer:
Sum of the digits, 3 + 1 + 3 + 4 + 6 + 7 + 3 = 27 is divisible by both 3 and 9.
Hence, 3134673 is divisible by 3 and 9.
Page No 410:
Question 19:
Fill in the blanks to make the statements true.
20x3 is a multiple of 3 if the digit x is ______ or ______ or ______.
Answer:
We know that, if a number is a multiple of 3, then the sum of its digits is again a multiple of 3, i.e. 2 + 0 + x + 3 is a multiple of 3.
⇒ x + 5 = 0, 3, 6, 9, 12, 15...
But, x is a digit of the number 20x3.
x can take values 0, 1, 2, 3, ..., 9.
⇒ x + 5 = 6 or 9 or 12 (for any other value of x + 5, x will not be greater than 0 and single digit)
⇒ x = 1 or 4 or 7
Hence, 20x3 is a multiple of 3 if the digit x is 1 or 4 or 7.
Page No 410:
Question 20:
Fill in the blanks to make the statements true.
3x5 is divisible by 9 if the digit x is __________.
Answer:
Since, the number 3x5 is divisible by 9, then the sum of its digits is also divisible by 9. i.e. 3 + x + 5 is divisible by 9.
⇒ x + 8 can take values 9, 18, 27,….
⇒ x = 1, 10, 19, ...
But x is a digit of the number 3x5, so x = 1.
Hence, 3x5 is divisible by 9 if the digit x is 1.
Page No 410:
Question 21:
Fill in the blanks to make the statements true.
The sum of a two-digit number and the number obtained by reversing the digits is always divisible by __________.
Answer:
Let ab be any two-digit number, then the number obtained by reversing its digits is ba.
Now,
Thus, ab + ba is always divisible by 11 and (a + b).
Hence, the sum of a two-digit number and the number obtained by reversing the digits is always divisible by 11 and (a + b).
Page No 410:
Question 22:
Fill in the blanks to make the statements true.
The difference of a two-digit number and the number obtained by reversing its digits is always divisible by ___________.
Answer:
Let ab be any two-digit number, then we have
Thus, ab – ba is always divisible by 9 and (a – b).
Hence, the difference of a two-digit number and the number obtained by reversing its digits is always divisible by 9 and (a – b).
Page No 410:
Question 23:
Fill in the blanks to make the statements true.
The difference of three-digit number and the number obtained by putting the digits in reverse order is always divisible by 9 and ___________.
Answer:
Let abc be a three-digit number, then we have
Thus, abc – cba is always divisible by 9, 11 and (a – c).
Hence, the difference of three-digit number and the number obtained by putting the digits in reverse order is always divisible by 9 and 11 and (a – c).
Page No 410:
Question 24:
Fill in the blanks to make the statements true.
If then A = ______ and B = ______.
Answer:
We have,
B is a digit, thus can take values from 0 to 9.
For B = 0, A = 0 which is not true.
For B = 1, A = 2 which is not true.
For B = 2, A = 4 which is not true.
For B = 3, A = 6 which is true.
Thus, A = 6 and B = 3.
Hence, if then A = 6 and B = 3.
Page No 410:
Question 25:
Fill in the blanks to make the statements true.
If then A = ______ and B = ______.
Answer:
We have,
Here, unit place of B × B is 6.
Thus, B can take values either 4 or 6.
For B = 4,
A × B + 1 = 9
⇒ A × 4 = 8
⇒ A = 2
Hence, if then A = 2 and B = 4.
Page No 410:
Question 26:
Fill in the blanks to make the statements true.
If then B = _______.
Answer:
We have,
Here, B is a digit, thus can take values from 0 to 9.
∴ 1 × B = B and B × B = 49
⇒ B = 7
Hence, if then B = 7.
Page No 411:
Question 27:
Fill in the blanks to make the statements true.
1x35 is divisible by 9 if x = _______.
Answer:
We know that, if a number is a multiple of 9, then sum of digits is also divisible by 9.
So, 1 + x + 3 + 5 is divisible by 9.
⇒ 9 + x = 0, 9, 18, ...
⇒ x = 0 (because x is a digit)
Hence, 1x35 is divisible by 9 if x = 0.
Page No 411:
Question 28:
Fill in the blanks to make the statements true.
A four-digit number abcd is divisible by 11, if d + b = _______ or _____.
Answer:
We know that, a number is divisible by 11, if the difference between the sum of digits at odd places and at even places is either 0 or a multiple of 11.
Thus, abcd is divisible by 11, if (d + b) − (a + c) = 0, 11, 22, 33,…
⇒ d + b = a + c or d + b = (11 + a + c)
Hence, a four-digit number abcd is divisible by 11, if d + b = a + c or 11 + a + c.
Page No 411:
Question 29:
Fill in the blanks to make the statements true.
A number is divisible by 11 if the differences between the sum of digits at its odd places and that of digits at the even places is either 0 or divisible by ______.
Answer:
By divisibility test of 11, we know that, a number is divisible by 11, if the sum of digits at odd places and even places is either 0 or divisible by 11.
Page No 411:
Question 30:
Fill in the blanks to make the statements true.
If a 3-digit number abc is divisible by 11, then ______ is either 0 or multiple of 11.
Answer:
Since, abc is divisible by 11, the difference of the sum of its digits at odd places and that of even places is either zero or multiple of 11.
Thus, (a + c) – b is either zero or multiple of 11.
Page No 411:
Question 31:
Fill in the blanks to make the statements true.
If A × 3 = 1A, then A = ______.
Answer:
We have,
Here, A × 3 is a two-digit number whose unit digit is A.
Since A is a digit, this can take values from 0 to 9.
Thus, A = 5 satisfies .
Hence, A × 3 = 1A, then A = 5.
Page No 411:
Question 32:
Fill in the blanks to make the statements true.
If B × B = AB, then either A = 2, B = 5 or A = ______, B = ______.
Answer:
We have,
Here, B × B is a two-digit number, whose unit digit is B.
Thus, B is either 5 or 6.
If B = 5, then A = 2.
If B = 6, then A = 3.
Hence, if B × B = AB, then either A = 2, B = 5 or A = 3, B = 6.
Page No 411:
Question 33:
Fill in the blanks to make the statements true.
If the digit 1 is placed after a 2-digit number whose tens is t and ones digit is u, the new number is ______.
Answer:
Given, a two-digit number whose ones digit is u and tens digit is t.
If the digit 1 is placed after this number, then the new number will be 100t + 10u + 1.
Page No 411:
Question 34:
State whether the statements given in questions are true (T) or false (F):
A two-digit number ab is always divisible by 2 if b is an even number.
Answer:
True
By the divisibility test of 2, we know that a number is divisible by 2, if its unit’s digit is even.
Thus, for a two-digit number ab is always divisible by 2 if b is an even number.
Page No 411:
Question 35:
State whether the statements given in questions are true (T) or false (F):
A three-digit number abc is divisible by 5 if c is an even number.
Answer:
False
By divisibility test of 5, we know that if a number is divisible by 5, then its one’s digit will be either 0 or 5.
Thus, the numbers ending with the digits 0 or 5 are divisible by 5.
Page No 411:
Question 36:
State whether the statements given in questions are true (T) or false (F):
A four-digit number abcd is divisible by 4 if ab is divisible by 4.
Answer:
False
We know that, if a number is divisible by 4, then the number formed by its digits in unit’s and ten’s place is divisible by 4.
Thus, a four-digit number abcd is divisible by 4, if cd is divisible by 4.
Page No 411:
Question 37:
State whether the statements given in questions are true (T) or false (F):
A three-digit number abc is divisible by 6 if c is an even number and a + b + c is a multiple of 3.
Answer:
True
If a number is divisible by 6, then it is divisible by both 2 and 3.
Since, abc is divisible by 6, it is also divisible by 2 and 3.
Therefore, c is an even number and the sum of digits is divisible by 3, i.e. multiple of 3.
Page No 411:
Question 38:
State whether the statements given in questions are true (T) or false (F):
Number of the form 3N + 2 will leave remainder 2 when divided by 3.
Answer:
True
Let x = 3N + 2.
⇒ x = (a multiple of 3) + 2
i.e. x is a number which is 2 more than a multiple of 3
i.e. x is a number, which when divided by 3, leaves the remainder 2.
Page No 411:
Question 39:
State whether the statements given in questions are true (T) or false (F):
Number 7N + 1 will leave remainder 1 when divided by 7.
Answer:
True
As 7N is completely divisble by 7.
Thus, when 7N + 1 is divisible by 7 the remainder will be 1.
Page No 411:
Question 40:
State whether the statements given in questions are true (T) or false (F):
If a number a is divisible by b, then it must be divisible by each factor of b.
Answer:
True
Given that, a is divisible by b.
Let b = xy, where x and y are prime factors.
Since, a is divisible by b, a is a multiple of b
i.e. a = mb
⇒ a = mxy
⇒ a = cy = dx, where c = mx, d = my
⇒ a is a multiple of x as well as y.
Hence, a is divisible by each factor of b.
Page No 411:
Question 41:
State whether the statements given in questions are true (T) or false (F):
If AB × 4 = 192, then A + B = 7.
Answer:
False
We have,
Here, B × 4 is a two-digit number whose unit's digits is 2.
Thus, B is either 3 or 8.
If B = 3, then A × 4 + 1 ≠ 19.
If B = 8, then A × 4 + 3 = 19 for A = 4
∴ A + B = 4 + 8 = 12
Page No 411:
Question 42:
State whether the statements given in questions are true (T) or false (F):
If AB + 7C = 102, where B ≠ 0, C ≠ 0, then A + B + C = 14.
Answer:
True
We have,
Here, B + C is either 2 or a two-digit number whose unit digit is 2.
If B = C = 1, then A = 3 and A + B + C = 3 + 1 + 1 = 5 ≠ 14
If B = 5,C = 7, then A = 2 and A + B + C = 2 + 5 + 7 = 14.
Page No 411:
Question 43:
State whether the statements given in questions are true (T) or false (F):
If 213x27 is divisible by 9, then the value of x is 0.
Answer:
False
Given, 213x27 is divisible by 9, so sum of its digits is also divisible by 9.
⇒ 2 + 1 + 3 + x + 2 + 7 is either 0, 9, 18, 27, 36,…
⇒ x + 15 = 0, 9, 18, 27, 36,…
⇒ x + 15 = 18 [x is a digit of a number]
⇒ x = 3
Page No 411:
Question 44:
State whether the statements given in questions are true (T) or false (F):
If N ÷ 5 leaves remainder 3 and N ÷ 2 leaves remainder 0, then N ÷ 10 leaves remainder 4.
Answer:
False
If N ÷ 2 leaves the remainder 0, then N must be an even number.
And, if N ÷ 5 leaves the remainder 3, then
N = 5n + 3, n = 0, 1, 2, 3, ...
For N to be even, 5n must be odd.
⇒ n = 1, 3, 5, ...
⇒ N = 8, 18, 28, ...
⇒ N = 10n + 8, n = 0, 1, 2, 3, ...
Thus, N ÷ 10 leaves remainder 8 always.
Page No 411:
Question 45:
Solve the following :
Find the least value that must be given to number a so that the number 91876a2 is divisible by 8.
Answer:
Given that, 91876a2 is divisible by 8.
Since, a number is divisible by 8, then the number formed by the last 3 digits is divisible by 8.
So, 6a2 is divisible by 8.
Here, a can take values from 0 to 9.
For a = 0; 602 is not divisible by 8.
For a = 1; 612 is not divisible by 8.
For a = 2; 622 is not divisible by 8.
For a = 3; 632 is divisible by 8.
Hence, the minimum value of a is 3 to make 91876a2 divisible by 8.
Page No 412:
Question 46:
Solve the following:
If where Q – P = 3, then find the values of P and Q.
Answer:
We have, and Q − P = 3.
Here, the unit's place of P × P is 6.
So, P is either 4 or 6.
If P = 4, then Q = 5.
∴ Q − P = 5 − 4 = 1 ≠ 3
If P = 6, then Q = 9.
∴ Q − P = 9 − 6 = 3
Hence, P = 6 and Q = 9.
Page No 412:
Question 47:
Solve the following :
If 1AB + CCA = 697 and there is no carry-over in addition, find the value of A + B + C.
Answer:
We have,
Since there is no carry-over in addition.
∴ 1 + C = 6
⇒ C = 5
∴ A + C = 9
⇒ A + 5 = 9
⇒ A = 4
∴ B + A = 7
⇒ B + 4 = 7
⇒ B = 3
∴ A + B + C = 4 + 3 + 5 = 12
Hence, the value of A + B + C is 12.
Page No 412:
Question 48:
Solve the following :
A five-digit number AABAA is divisible by 33. Write all the numbers of this form.
Answer:
Given, a number of the form AABAA is divisible by 33.
Then, it is also divisible by 3 and 11.
Since, AABAA is divisible by 3, sum its digits is also divisible by 3.
∴ A + A + B + A + A = 0, 3, 6, 9, …
⇒ 4A + B = 0, 3, 6, 9, ... .....(1)
Now, the given number is also divisible by 11, therefore (A + A + B) – (A + A) = 0, 11, 22,…
⇒ B = 0, 11, 22,… .....(2)
⇒ B = 0 [∵ B is a digit of the given number]
From (1) and (2), we get
⇒ 4A = 0, 3, 6, 9, ...
⇒ 4A = 12 or 24 or 36
⇒ A = 3, 6, 9 [∵ A is a digit of the given number]
Hence, the required numbers are 33033, 66066 and 99099.
Page No 412:
Question 49:
Solve the following :
Find the value of the letters in each of the following questions.
Answer:
We have,
Here, A + A = Z
Since, the sum of the second column is a two-digit number.
The possible values of A are 5 to 9.
Thus only possible value of A is 9.
⇒ Z = 8 and X = 1
Page No 412:
Question 50:
Solve the following :
Answer:
We have,
Here, 5 + A is a two-digit number whose one's digit is 3.
∴ 5 + A = 13
⇒ A = 8
For second column,
BC = 8 + 4 + 1
⇒ BC = 13
⇒B = 1 and C = 3
Hence, A = 8, B = 1 and C = 3.
Page No 412:
Question 51:
Solve the following :
Answer:
We have,
Here, 6 + A cannot be equal to 2.
∴ 6 + A = 12
⇒ A = 6
For second column,
B + 8 + 1 = CA
⇒ B + 9 = C6
⇒ B = 7 and C = 1
Hence, A = 6, B = 7 and C = 1.
Page No 412:
Question 52:
Solve the following :
Answer:
We have,
Here, A + A is a number whose one's digit is 2.
⇒ A + A = 12
⇒ A = 6
Also, B + B + 1 = B
⇒ B = 9
Also, 1 + A + 1 = 8
which is satisfied by A = 6
Hence, A = 6 and B = 9.
Page No 412:
Question 53:
Solve the following :
Answer:
We have,
Here, A + A = 0
⇒ A = 0 or 5
For A = 0, B + B cannot be 3.
For A = 5, A + A = 10.
∴ B + B + 1 = 3
⇒ B = 1
But C + C cannot be 15.
Thus, B + B + 1 = 13
⇒ 2B = 12
⇒ B = 6
∴ C + C + 1 = 15
⇒ C = 7
Hence, A = 5, B = 6 and C = 7.
Page No 412:
Question 54:
Solve the following :
Answer:
We have,
Here, A + A = 8
⇒ A = 4
For second column,
A + A = A is not satisfied by A = 4
So, A + A is a two-digit number.
∴ A + A = 18
⇒ A = 9
For second column,
A + A + 1 = 19 is satisfied by A = 9.
∴ B + B + 1 = 3
⇒ B = 1
Hence, A = 9 and B = 1.
Page No 412:
Question 55:
Solve the following :
Answer:
We have,
Here, B + B = 8
⇒ B = 4 or 9
Since, in second column 1 + A = 0 and A is a digit. Thus,
1 + A + 1 = 10
⇒ A = 8 and B = 9
Now, A + 1 = B also satisfied by A = 8 and B = 9.
Hence, A = 8 and B = 9.
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Question 56:
Solve the following:
Answer:
We have,
Here, B × 6 is a two-digit number with unit digit 8.
⇒ B = 3 or 8
For B = 3,
A × 6 + 1 = C6,
which is not satisfied by any value of A and C.
For B = 8,
A × 6 + 4 = C6
⇒ A = 7 and C = 4
Hence, A = 7, B = 8 and C = 4.
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Question 57:
Solve the following:
Answer:
We have,
Here, B × B is a two-digit number whose unit digit is B.
⇒ B = 1 or 5
Also, AB × AB = 6AB .....(1)
The square of a two-digit number is a three-digit number.
So A can be 1, 2 or 3.
For A = 1, 2, 3 and B = 1, (1) is not satisfied.
For A = 1, B = 5, (1) is not satisfied.
For A = 2, B = 5, (1) is satisfied.
As, 25 × 25 = 625
Hence, A = 2 and B = 5.
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Question 58:
Solve the following:
and B – A = 1
Answer:
Given: and B – A = 1
Here, AA × A is a 3-digit number, whose unit digit is B.
Thus, A can be 4 to 9.
Since the tens digit is A itself, so A cannot be 4, 5, 6, 7, 8.
Hence, A = 9, B = 1 and C = 8.
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Question 59:
Solve the following:
Answer:
We have,
Here, B − 7 = 5
As 12 − 7 = 5
⇒ B = 2
∴ A − B − 1 = 4
⇒ A − 2 − 1 = 4
⇒ A − 3 = 4
⇒ A = 7
Hence, A = 7 and B = 2.
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Question 60:
Solve the following:
Answer:
We have,
Here, C − 5 = 8
⇒ C = 13
Thus, C = 3 (because C is a digit)
In second column, B − C − 1 = 8
⇒ B − 3 − 1 = 8
⇒ B = 12
Thus, B = 2 (because B is a digit)
In third column, A − B − 1 = 4
⇒ A − 2 − 1 = 4
⇒ A = 7
In forth column, 8 − A = D
⇒ 8 − 7 = D
⇒ D = 1
Hence, A = 7, B = 2, C = 3, D = 1.
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Question 61:
Solve the following :
If 2A7 ÷ A = 33, then find the value of A.
Answer:
We observe that, A × 3 can never be a single digit number 2, so A × 3 must be a two-digit number, whose ten’s digit is 2 and unit’s digit is the number less than or equal to A.
Therefore, the value of A can be 9, as the values of A from 1 to 8 does not satisfied.
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Question 62:
Solve the following :
212x5 is a multiple of 3 and 11. Find the value of x.
Answer:
Since, 212x5 is a multiple of 3.
2 +1 + 2 + x + 5 = 0, 3, 6, 9, 12, 15, 18, ...
⇒ 10 + x = 0, 3, 6, 9, 12, 15, 18, ...
⇒ x = 2, 5, 8 … .....(1)
Again, 212x5 is a multiple of 11, (2 + 2 + 5) − (1 + x) = 0, 11, 22, 33, ...
⇒ 8 − x = 0, 11, 22, …
⇒ x = 8 …..(2)
From Eqs. (1) and (2), we have
x = 8
Hence, the value of x is 8.
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Question 63:
Solve the following :
Find the value of k where 31k2 is divisible by 6.
Answer:
Given, 31k2 is divisible by 6.
Then, it is also divisible by both 2 and 3.
Now, 31k2 is divisible by 3, the sum of its digits is a multiple of 3.
i.e. 3 + 1 + k + 2 = 0, 3, 6, 9, 12, …
⇒ k + 6 = 0, 3, 6, 9, 12, ...
⇒ k = 0 or 3, 6, 9, ...
Hence, the value of k is 0 or 3, 6, 9, ... .
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Question 64:
Solve the following :
1y3y6 is divisible by 11. Find the value of y.
Answer:
It is given that, 1y3y6 is divisible by 11.
Then, we have (1 + 3 + 6) − (y + y) = 0, 11, 22, …
⇒ 10 − 2y = 0, 11, 22, …
⇒ 10 − 2y = 0 [other values give a negative number]
⇒ 2y = 10
⇒ y = 5
Hence, the value of y is 5.
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Question 65:
Solve the following :
756x is a multiple of 11, find the value of x.
Answer:
Given that, 756x is a multiple of 11.
Since, 756x is divisible by 11, then (7 + 6) – (5 + x) is a multiple of 11,
i.e. 8 – x = 0,11,22,…
⇒ 8 – x = 0
⇒ x = 8
Hence, the value of x is 8.
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Question 66:
Solve the following :
A three-digit number 2a3 is added to the number 326 to give a three-digit number 5b9 which is divisible by 9. Find the value of b – a.
Answer:
Given that .
Here, 3 + 6 = 9.
And, 2 + 3 = 5.
Thus, a + 2 = b
⇒ b − a = 2
Hence, the value of b − a is 2.
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Question 67:
Solve the following :
Let E = 3, B = 7 and A = 4. Find the other digits in the sum .
Answer:
We have, E = 3, B = 7 and A = 4.
Here, 3 + L = S
⇒ S − L = 3 .....(1)
Also, S + L = 3 .....(2)
From (1) and (2), we get
S = 3 and L = 0
In third column, 4 + 4 = M
⇒ M = 8
In forth column, 7 + 7 = G4
⇒ G4 = 14
⇒ G = 1
Hence, G = 1, M = 8, S = 3 and L = 0.
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Question 68:
Solve the following :
Let D = 3, L = 7 and A = 8. Find the other digits in the sum
Answer:
Given that, D = 3, L = 7 and A = 8.
Thus, .
In first column,
3 + S + 8, which is a two-digit number whose unit’s digit is 7.
⇒ S = 6
Now, in second column, 8 + 8 + 1 = 16 +1 = 7 [1 is carry forward]
In third column, M + 1 is a 2-digit number, therefore M must be 9.
Then, M + 1 = 9 + 1 = 10 = BU
⇒ B = 1, U = 0
Hence, S = 6, M = 9, B = 1 and U = 0.
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Question 69:
Solve the following :
If from a two-digit number, we subtract the number formed by reversing its digits then the result so obtained is a perfect cube. How many such numbers are possible? Write all of them.
Answer:
Let ab be any two-digit number.
Then, the digit formed by reversing its digits is ba.
Further, since ab − ba is a perfect cube and is a multiple of 9.
Thus, the possible value of a − b is 3.
i.e. a = b + 3
Here, b can take values from 0 to 6 (because a is a single digit).
Thus, possible numbers are as follows:
For b = 0, a = 3, i.e. ab = 30
For b = 1, a = 4, i.e. ab = 41
For b = 2, a = 5, i.e. ab = 52
For b = 3, a = 6, i.e. ab = 63
For b = 4, a = 7, i.e. ab = 74
For b = 5, a = 8, i.e. ab = 85
For b = 6, a = 9, i.e. ab = 96
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Question 70:
Solve the following :
Work out the following multiplication.
Use the result to answer the following questions.
(a) What will be 12345679 × 45?
(b) What will be 12345679 × 63?
(c) By what number should 12345679 be multiplied to get 888888888?
(d) By what number should 12345679 be multiplied to get 999999999?
Answer:
We have, .
(a)
Here, sum of digits of 45 is 9.
The product consists of digit 5 as unit's digit of 9 × 5 is 5.
Thus, .
(b)
Here, sum of the digits of the multiplier 63 is 9 and the unit's digit of the products of 3 × 9 is 7.
Thus, .
(c) The number in the product is 8.
The number 12345679 should be multiplied by 72 as the sum of digits is 9 and 2 × 9 = 18.
(d) The number in the product is 9.
The number 12345679 should be multiplied by 81 as the sum of digits is 9 and 1 × 9 = 9.
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Question 71:
Solve the following :
Find the value of the letters in each of the following:
(i)
(ii)
Answer:
(i) We have,
Here, unit digit of Q × 6 is Q.
Thus, the possible values of Q are 2, 4, 6, and 8.
For Q = 2, P × 6 + 1 cannot be 22.
For Q = 4, P × 6 + 2 = 44
⇒ P × 6 = 42
⇒ P = 7
Hence, P = 7 and Q = 4.
(ii) We have,
Here, M + 1 = 8
⇒ M = 7
And, L + 7 = 1
⇒ L = 4
And, 2 + L + 1= M
⇒ 2 + 4 + 1 = 7 = M
Hence, L = 4 and M = 7.
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Question 72:
Solve the following :
If 148101B095 is divisible by 33, find the value of B.
Answer:
Given that the number 148101B095 is divisible by 33.
Therefore it is also divisible by both 3 and 11.
Now, the number is divisible by 3, its sum of digits is a multiple of 3.
⇒ 1 + 4 + 8 + 1 + 0 + 1 + B + 0+ 9 + 5 is a multiple of 3.
⇒ 29 + B = 0, 3, 6, 9, …
⇒ B = 1, 4, 7, ... .....(1)
Also, the number is divisible by 11, therefore
(1 + 8 + 0 + B + 9) − (4 + 1 + 1 + 0 + 5) = 0, 11, 22, …
⇒ (18 + B) − 11 = 0, 11, 22, ...
⇒ B + 7 = 0, 11, 22, ...
⇒ B + 7 = 11, 22, ...
⇒ B = 4, 15 .....(2)
From (1) and (2), we get B = 4.
Hence, the value of B is 4.
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Question 73:
Solve the following :
If 123123A4 is divisible by 11, find the value of A.
Answer:
Given, 123123A4 is divisible by 11.
Tthen we have (1 + 3 + 2 + A) – (2 + 1 + 3 + 4) is a multiple of 11.
i.e. (6 + A) − 10 = 0, 11, 22, ..
⇒ A − 4 = 0, 11, 22, …
⇒ A − 4 = 0 [A is a digit of the given number]
⇒ A = 4
Hence, the value of A is 4.
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Question 74:
Solve the following :
If 56x32y is divisible by 18, find the least value of y.
Answer:
Given that, the number 56x32y is divisible by 18.
Thus, it is divisible by 2 as well as 3.
Now, the number is divisible by 2, its unit’s digit must be an even number that is 0, 2, 4, 6, 8.
Therefore, the least value of y is 0.
Again, the number is divisible by 3 also, sum of its digits is a multiple of 3. i.e. 5 + 6 + x + 3 + 2 + y is a multiple of 3.
⇒ 16 + x + y = 0, 3, 6, 9,…
⇒ 16 + x = 18 (because least value of y is 0)
⇒ x = 2, which is the least value of x.
Hence, the least value of y is 0.
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