RD Sharma Solutions for Class 8 Math Chapter 5 - Playing with Numbers (Part-1) Class 8 Notes | EduRev

RD Sharma Solutions for Class 8 Mathematics

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Class 8 : RD Sharma Solutions for Class 8 Math Chapter 5 - Playing with Numbers (Part-1) Class 8 Notes | EduRev

The document RD Sharma Solutions for Class 8 Math Chapter 5 - Playing with Numbers (Part-1) Class 8 Notes | EduRev is a part of the Class 8 Course RD Sharma Solutions for Class 8 Mathematics.
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Question 1: Without performing actual addition and division write the quotient when the sum of 69 and 96 is divided by
(i) 11
(ii) 15

Answer 1: (i) Clearly, 69 and 96 are two numbers such that one can be obtained be reversing the digits of the other. Therefore, when the sum of 69 and 96 is divided by 11, we get 15 (sum of the digits) as quotient.
(ii) Clearly, 69 and 96 are two numbers such that one can be obtained be reversing the digits of the other. Therefore, when the sum of 69 and 96 is divided by 15 (sum of the digits), we obtain 11 as quotient. 

Question 2: Without performing actual computations, find the quotient when 94 − 49 is divided by
(i) 9
(ii) 5

Answer 2:  (i) We know that when abba  is divided by 9, the quotient is ab.Therefore, when (9449) is divided by 9, the quotient is (94 = 5).(ii) We know that when abba is divided by (ab), the quotient is 9.Therefore, when (9449) is divided by (94=5), the quotient is 9.Question 3: If sum of the number 985 and two other numbers obtained by arranging the digits of 985 in cyclic order is divided by 111, 22 and 37 respectively. Find the quotient in each case. 

Answer 3: The sum of (985+859+598) when divided by:(i) 111Quotient = (9+8+5)=22(ii) 22, i.e, (9+8+5)Quotient = 111(iii) 37 (= 111/3)Quotient = 3(9+8+5)=66

Question 4: Find the quotient when the difference of 985 and 958 is divided by 9.

Answer 4: If abc-acb is divided by 9, the quotient is (b-c).If (985-958) is divided by 9, quotient=(8-5)=3 

Question 5: Given that the number ¯¯¯¯¯¯¯¯¯¯    35α6435α64¯ is divisible by 3, where α is a digit, what are the possible values of α?

Answer 5:  It is given that¯¯¯¯¯¯¯¯¯¯¯       35a64 is a multiple of 3. (3+5+a+6+4) is a multiple of 3. (a+18) is a multiple of 3. (a+18)=0, 3, 6, 9, 12, 15, 18, 21...But a is a digit of number ¯¯¯¯¯¯¯¯¯¯¯¯       35a64  . So, a can take value 0, 1, 2, 3, 4...9.a+18=18 a=0a+18=21 a=3a+18=24 a=6a+18=27 a=9 a=0, 3, 6, 9

Question 6: If x is a digit such that the number ¯¯¯¯¯¯¯¯¯¯       18x7118x71¯ is divisible by 3, find possible values of x.

Answer 6: It is given that ¯¯¯¯¯¯¯¯¯¯      18x71 is a multiple of 3.(1+8+x+7+1) is a multiple of 3.(17+x) is a multiple of 3.17+x=0, 3, 6, 9, 12, 15, 18, 21...But x is a digit. So, x can take values 0,1,2,3,4...9.17+x=18 x=117+x=21 x=417+x=24 x=7x=1,4,7Question 7: If x is a digit of the number ¯¯¯¯¯¯¯¯¯¯¯¯        66784x66784x¯ such that it is divisible by 9, find possible values of x. 

Answer 7: It is given that ¯¯¯¯¯¯¯¯¯¯¯¯¯      66784x is a multiple of 9.Therefore, (6+6+7+8+4+x) is a multiple of 9.And,(31+x) is a multiple of 9. Possible values of (31+x) are 0, 9, 18, 27, 36, 45,...But x is a digit. So, x can only take value 0, 1, 2, 3, 4,...9.31+x=36 x=3631x=5

Question 8: Given that the number ¯¯¯¯¯¯¯¯¯¯       67y1967y19¯ is divisible by 9, where y is a digit, what are the possible values of y?

Answer 8: It is given that ¯¯¯¯¯¯¯¯¯¯      67y19 is a multiple of 9. (6+7+y+1+9) is a multiple of 9. (23+y) is a multiple of 9. 23+y=0, 9, 18, 27, 36...But x is a digit. So, x can take values 0,1,2,3,4...9.23+y=27y=4 Question 9: If  ¯¯¯3x23x2¯ is a multiple of 11, where x is a digit, what is the value of x?

Answer 9: Sum of the digits at odd places = 3+2= 5Sum of the digit at even place = x Sum of the digit at even placeSum of the digits at odd places=(x5) (x5) must be multiple by 11. Possible values of (x5) are 0, 11, 22, 33...But x is a digit; therefore x must be  0, 1 ,2, 3...9. x5=0 x=5

Question 10: If  ¯¯¯¯¯¯¯¯    98215x298215x2¯ is a number with x as its tens digit such that is is divisible by 4. Find all possible values of x.

Answer 10:  A natural number is divisible by 4 if the number formed by its digits in units and tens places is divisible by 4.RD Sharma Solutions for Class 8 Math Chapter 5 - Playing with Numbers (Part-1) Class 8 Notes | EduRev will be divisible by 4 if RD Sharma Solutions for Class 8 Math Chapter 5 - Playing with Numbers (Part-1) Class 8 Notes | EduRev is divisible by 4. 

RD Sharma Solutions for Class 8 Math Chapter 5 - Playing with Numbers (Part-1) Class 8 Notes | EduRev=10x + 2x is a digit; therefore possible values of x are 0, 1, 2, 3...9.¯¯¯¯¯   x2 =2, 12, 22, 32, 42, 52, 62, 72, 82, 92The numbers that are divisible by 4 are 12, 32, 52, 72, 92.Therefore, the values of x are 1, 3, 5, 7, 9.

Question 11: If x denotes the digit at hundreds place of the number ¯¯¯¯¯¯  67x1967x19¯ such that the number is divisible by 11. Find all possible values of x.

Answer 11 : A number is divisible by 11, if the difference of the sum of its digits at odd places and the sum of its digits at even places is either 0 or a multiple of 11.Sum of digits at odd places−Sum of digits at even places=(6 + x + 9)(7 + 1)=(15 + x)8=x + 7 x + 7 =11  x = 4

Question 12: Find the remainder when 981547 is divided by 5. Do this without doing actual division. 

Answer 12: If a natural number is divided by 5, it has the same remainder when its unit digit is divided by 5.Here, the unit digit of 981547 is 7. When 7 is divided by 5, remainder is 2.Therefore, remainder will be 2 when 981547 is divided by 5.  

Question 13: Find the remainder when 51439786 is divided by 3. Do this without performing actual division.

Answer 13: Sum of the digits of the number 51439786 = 5 + 1 + 4 + 3 + 9 + 7 + 8 + 6 = 43

The remainder of 51439786, when divided by 3, is the same as the remainder when the sum of the digits is divided 3.

When 43 is divided by 3, remainder is 1.

Therefore, when 51439786 is divided by 3, remainder will be 1. 

Question 14: Find the remainder, without performing actual division, when 798 is divided by 11.

Answer 14: 798 =A multiple of 11 + (Sum of its digits at odd places − Sum of its digits at even places)798 =A multiple of 11 + (7 + 8  9)798 =A multiple of 11 + (15  9)798 =A multiple of 11 + 6Therefore, the remainder is 6.

Question 15: Without performing actual division, find the remainder when 928174653 is divided by 11.

Answer 15: 928174653= A multiple of 11 + (Sum of its digits at odd places −Sum of its digits at even places)

928174653= A multiple of 11 + {(9 + 8 + 7 + 6 + 3) − (2 + 1 + 4 + 5)}928174653= A multiple of 11 + (33  12)928174653= A multiple of 11 + 21928174653= A multiple of 11 + (11 × 1 + 10)928174653= A multiple of 11 + 10Therefore, the remainder is 10.

Question 16: Given an example of a number which is divisible by
(i) 2 but not by 4.
(ii) 3 but not by 6.
(iii) 4 but not by 8.
(iv) both 4 and 8 but not by 32.

Answer 16: (i) 10
Every number with the structure (4n + 2) is an example of a number that is divisible by 2 but not by 4.
(ii) 15
Every number with the structure (6n + 3) is an example of a number that is divisible by 3 but not by 6.
(iii) 28
Every number with the structure (8n + 4) is an example of a number that is divisible by 4 but not by 8.
(iv) 8
Every number with the  structure (32n + 8), (32n + 16) or (32n + 24) is an example of a number that is divisible by 4 and 8 but not by 32. 

Question 17: Which of the following statements are true?
(i) If a number is divisible by 3, it must be divisible by 9.
(ii) If a number is divisible by 9, it must be divisible by 3.
(iii) If a number is divisible by 4, it must be divisible by 8.
(iv) If a number is divisible by 8, it must be divisible by 4.
(v) A number is divisible by 18, if it is divisible by both 3 and 6.
(vi) If a number is divisible by both 9 and 10, it must be divisible by 90.
(vii) If a number exactly divides the sum of two numbers, it must exactly divide the numbers separately.
(viii) If a number divides three numbers exactly, it must divide their sum exactly.
(ix) If two numbers are co-prime, at least one of them must be a prime number.
(x) The sum of two consecutive odd numbers is always divisible by 4. 

Answer 17: (i) False
Every number with the structures (9n + 3) or (9n + 6) is divisible by 3 but not by 9. Example: 3, 6, 12 etc.
(ii) True
(iii) False
Every number with the structure (8n + 4) is divisible by 4 but not by 8. Example: 4, 12, 20 etc.
(iv) True
(v) False
Example: 24 is divisible by both 3 and 6 but it is not divisible by 18.
(vi) True
(vii) False
Example: 5 divides 10, which is a sum of 3 and 7. However, it neither divides 3 nor 7.
(viii) True
(ix) False
Example: 4 and 9 are co-prime numbers but both are composite numbers too.
(x) True 

Question 18: Solve each of the following Cryptarithms:

RD Sharma Solutions for Class 8 Math Chapter 5 - Playing with Numbers (Part-1) Class 8 Notes | EduRev

Answer 18:Two possible values of A are:(i) If 7 + B  9 3 + A = 9  A = 6But if A = 6, 7 + B must be larger than 9. Hence, it is impossible.(ii) If 7 + B  9 1 + 3 + A = 9 A = 5If A = 5 and 7 + B = 5, B must be 8 A = 5, B = 8

Question 19: Solve each of the following Cryptarithm:

RD Sharma Solutions for Class 8 Math Chapter 5 - Playing with Numbers (Part-1) Class 8 Notes | EduRev

Answer 19: Two possibilities of A are:(i) If B + 7  9, A = 6But clearly, if A = 6, B + 7  9; it is impossible(ii) If B + 7  9, A = 5 and B + 7 = 5 Clearly, B = 8 A = 5, B = 8

Question 20: Solve each of the following Cryptarithm:

RD Sharma Solutions for Class 8 Math Chapter 5 - Playing with Numbers (Part-1) Class 8 Notes | EduRev

Answer 20: If 1 + B = 0   Surely, B = 9If 1 + A + 1 = 9  Surely, A = 7

Question 21: Solve each of the following Cryptarithm:

RD Sharma Solutions for Class 8 Math Chapter 5 - Playing with Numbers (Part-1) Class 8 Notes | EduRev

Answer 21: B + 1 = 8, B = 7A + B = 1, A + 7 = 1, A = 4So, A = 4, B = 7

Question 22: Solve each of the following Cryptarithm:

RD Sharma Solutions for Class 8 Math Chapter 5 - Playing with Numbers (Part-1) Class 8 Notes | EduRev

Answer 22: A + B = 9  as the sum of two digits can never be 192 + A = 0, A must be 8A + B = 9, 8 + B = 9, B = 1So, A = 8, B = 1

Question 23: Solve each of the following Cryptarithm:

RD Sharma Solutions for Class 8 Math Chapter 5 - Playing with Numbers (Part-1) Class 8 Notes | EduRev

Answer 23:  If A + B = 8, A + B  9 is possible only if A = B = 9But from 7 + B = A, A = B = 9 is impossibleSurely, A + B = 8, A + B  9So, A + 7 = 9, Surely A = 27 + B = A, 7 + B = 2,  B = 5So, A = 2, B = 5

Question 24: Solve each of the following Cryptarithm: 

RD Sharma Solutions for Class 8 Math Chapter 5 - Playing with Numbers (Part-1) Class 8 Notes | EduRev

Answer 24: 0 is the only unit digit number, which gives the same 0 at the unit digit when multiplied by 4. So, the possible value of B is 0.Similarly, for A also, 0 is the only possible digit.But then A, B and C will all be 0.And if A, B and C become 0, these numbers cannot be of two−digit or three−digit.Therefore, both will become a onedigit number.Thus, there is no solution possible.

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