Math Ncert Exemplar 2019 Solutions for Class 8 Maths Chapter 13 Playing With Numbers are provided here with simple step-by-step explanations. These solutions for Playing With Numbers are extremely popular among class 8 students for Maths Playing With Numbers Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Math Ncert Exemplar 2019 Book of class 8 Maths Chapter 13 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Math Ncert Exemplar 2019 Solutions. All Math Ncert Exemplar 2019 Solutions for class 8 Maths are prepared by experts and are 100% accurate.
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.
Page No 412:
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.
Page No 412:
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.
Page No 412:
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.
Page No 412:
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.
Page No 412:
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.
Page No 412:
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.
Page No 412:
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.
Page No 412:
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, ... .
Page No 412:
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.
Page No 412:
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.
Page No 412:
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.
Page No 413:
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.
Page No 413:
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.
Page No 413:
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
Page No 413:
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.
Page No 413:
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.
Page No 413:
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.
Page No 413:
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.
Page No 413:
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.
View NCERT Solutions for all chapters of Class 8