Selina Textbook Solutions: Playing with Numbers | Mathematics Class 8 ICSE PDF Download

 Page 1


5. Playing with Number 
EXERCISE 5(A) 
Question 1. 
Write the quotient when the sum of 73 and 37 is divided by 
(i) 11 
(ii) 10 
Solution: 
 
Question 2. 
Write the quotient when the sum of 94 and 49 is divided by 
(i) 11 
(ii) 13 
Solution: 
Page 2


5. Playing with Number 
EXERCISE 5(A) 
Question 1. 
Write the quotient when the sum of 73 and 37 is divided by 
(i) 11 
(ii) 10 
Solution: 
 
Question 2. 
Write the quotient when the sum of 94 and 49 is divided by 
(i) 11 
(ii) 13 
Solution: 
 
 
Question 3. 
Find the quotient when 73 – 37 is divided by 
(i) 9 
(ii) 4 
Solution: 
 
Question 4. 
Find the quotient when 94 – 49 is divided by 
Page 3


5. Playing with Number 
EXERCISE 5(A) 
Question 1. 
Write the quotient when the sum of 73 and 37 is divided by 
(i) 11 
(ii) 10 
Solution: 
 
Question 2. 
Write the quotient when the sum of 94 and 49 is divided by 
(i) 11 
(ii) 13 
Solution: 
 
 
Question 3. 
Find the quotient when 73 – 37 is divided by 
(i) 9 
(ii) 4 
Solution: 
 
Question 4. 
Find the quotient when 94 – 49 is divided by 
(i) 9 
(ii) 5 
Solution: 
 
Question 6. 
If a = 6, show that abc = bac. 
Solution: 
Given : a = 6 
To show : abc = bac 
Proof: abc = 100a + 106 + c …….(i) 
(By using property 3) 
bac = 1006 + 10a + c —(ii) 
(By using property 3) 
Question 5. 
Show that 527 + 752 + 275 is exactly divisible by 14. 
Solution: 
Property : 
abc = 100a + 106 + c ………(i) 
bca = 1006 + 10c + a ……..(ii) 
and cab = 100c + 10a + b ……….(iii) 
Adding, (i), (ii) and (iii), we get abc + bca + cab = 111a + 111b + 111c = 111(a + b + c) = 
3 x 37(a + b + c) 
Now, let us try this method on 
527 + 752 + 275 to check is it exactly divisible by 14 
Here, a = 5, 6 = 2, c = 7 
527 + 752 + 275 = 3 x 37(5 + 2 + 7) = 3 x 37 x 14 
Hence, it shows that 527 + 752 + 275 is exactly divisible by 14 
Page 4


5. Playing with Number 
EXERCISE 5(A) 
Question 1. 
Write the quotient when the sum of 73 and 37 is divided by 
(i) 11 
(ii) 10 
Solution: 
 
Question 2. 
Write the quotient when the sum of 94 and 49 is divided by 
(i) 11 
(ii) 13 
Solution: 
 
 
Question 3. 
Find the quotient when 73 – 37 is divided by 
(i) 9 
(ii) 4 
Solution: 
 
Question 4. 
Find the quotient when 94 – 49 is divided by 
(i) 9 
(ii) 5 
Solution: 
 
Question 6. 
If a = 6, show that abc = bac. 
Solution: 
Given : a = 6 
To show : abc = bac 
Proof: abc = 100a + 106 + c …….(i) 
(By using property 3) 
bac = 1006 + 10a + c —(ii) 
(By using property 3) 
Question 5. 
Show that 527 + 752 + 275 is exactly divisible by 14. 
Solution: 
Property : 
abc = 100a + 106 + c ………(i) 
bca = 1006 + 10c + a ……..(ii) 
and cab = 100c + 10a + b ……….(iii) 
Adding, (i), (ii) and (iii), we get abc + bca + cab = 111a + 111b + 111c = 111(a + b + c) = 
3 x 37(a + b + c) 
Now, let us try this method on 
527 + 752 + 275 to check is it exactly divisible by 14 
Here, a = 5, 6 = 2, c = 7 
527 + 752 + 275 = 3 x 37(5 + 2 + 7) = 3 x 37 x 14 
Hence, it shows that 527 + 752 + 275 is exactly divisible by 14 
Since, a = 6 
Substitute the value of a = 6 in equation (i) and (ii), we get 
abc = 1006 + 106 + c ………(iii) 
bac = 1006 + 106 + c ………(iv) 
Subtracting (iv) from (iii) abc – bac = 0 
abc = bac 
Hence proved. 
Question 7. 
If a > c; show that abc – cba = 99(a – c). 
Solution: 
Given, a > c 
To show : abc – cba = 99(a – c) 
Proof: abc = 100a + 10b + c ……….(i) 
(By using property 3) 
cba = 100c + 10b + a ………..(ii) 
(By using property 3) 
Subtracting, equation (ii) from (i), we get 
abc – cba = 100a + c – 100c – a 
abc – cba = 99a – 99c 
abc – cba = 99(a – c) 
Hence proved. 
Question 8. 
If c > a; show that cba – abc = 99(c – a). 
Solution: 
Given : c > a 
To show : cba – abc = 99(c – a) 
Proof: 
cba = 100c + 106 + a ……….(i) 
(By using property 3) 
abc = 100a + 106 + c ………(ii) 
(By using property 3) 
Subtracting (ii) from (i) 
cba – abc= 100c+ 106 + a- 100a- 106-c 
=> cba – abc = 99c – 99a 
=> cba – abc = 99(c – a) 
Hence proved. 
Question 9. 
If a = c, show that cba – abc = 0. 
Solution: 
Given : a = c 
To show : cba – abc = 0 
Proof: 
Page 5


5. Playing with Number 
EXERCISE 5(A) 
Question 1. 
Write the quotient when the sum of 73 and 37 is divided by 
(i) 11 
(ii) 10 
Solution: 
 
Question 2. 
Write the quotient when the sum of 94 and 49 is divided by 
(i) 11 
(ii) 13 
Solution: 
 
 
Question 3. 
Find the quotient when 73 – 37 is divided by 
(i) 9 
(ii) 4 
Solution: 
 
Question 4. 
Find the quotient when 94 – 49 is divided by 
(i) 9 
(ii) 5 
Solution: 
 
Question 6. 
If a = 6, show that abc = bac. 
Solution: 
Given : a = 6 
To show : abc = bac 
Proof: abc = 100a + 106 + c …….(i) 
(By using property 3) 
bac = 1006 + 10a + c —(ii) 
(By using property 3) 
Question 5. 
Show that 527 + 752 + 275 is exactly divisible by 14. 
Solution: 
Property : 
abc = 100a + 106 + c ………(i) 
bca = 1006 + 10c + a ……..(ii) 
and cab = 100c + 10a + b ……….(iii) 
Adding, (i), (ii) and (iii), we get abc + bca + cab = 111a + 111b + 111c = 111(a + b + c) = 
3 x 37(a + b + c) 
Now, let us try this method on 
527 + 752 + 275 to check is it exactly divisible by 14 
Here, a = 5, 6 = 2, c = 7 
527 + 752 + 275 = 3 x 37(5 + 2 + 7) = 3 x 37 x 14 
Hence, it shows that 527 + 752 + 275 is exactly divisible by 14 
Since, a = 6 
Substitute the value of a = 6 in equation (i) and (ii), we get 
abc = 1006 + 106 + c ………(iii) 
bac = 1006 + 106 + c ………(iv) 
Subtracting (iv) from (iii) abc – bac = 0 
abc = bac 
Hence proved. 
Question 7. 
If a > c; show that abc – cba = 99(a – c). 
Solution: 
Given, a > c 
To show : abc – cba = 99(a – c) 
Proof: abc = 100a + 10b + c ……….(i) 
(By using property 3) 
cba = 100c + 10b + a ………..(ii) 
(By using property 3) 
Subtracting, equation (ii) from (i), we get 
abc – cba = 100a + c – 100c – a 
abc – cba = 99a – 99c 
abc – cba = 99(a – c) 
Hence proved. 
Question 8. 
If c > a; show that cba – abc = 99(c – a). 
Solution: 
Given : c > a 
To show : cba – abc = 99(c – a) 
Proof: 
cba = 100c + 106 + a ……….(i) 
(By using property 3) 
abc = 100a + 106 + c ………(ii) 
(By using property 3) 
Subtracting (ii) from (i) 
cba – abc= 100c+ 106 + a- 100a- 106-c 
=> cba – abc = 99c – 99a 
=> cba – abc = 99(c – a) 
Hence proved. 
Question 9. 
If a = c, show that cba – abc = 0. 
Solution: 
Given : a = c 
To show : cba – abc = 0 
Proof: 
cba = 100c + 106 + a …………(i) 
(By using property 3) 
abc = 100a + 106 + c …………(ii) 
(By using property 3) 
Since, a = c, 
Substitute the value of a = c in equation (i) and (ii), we get 
cba = 100c + 10b + c ……….(iii) 
abc = 100c + 10b + c …………(iv) 
Subtracting (iv) from (iii), we get 
cba – abc – 100c + 106 + c – 100c – 106 – c 
=> cba – abc = 0 
=> cba = abc 
Hence proved. 
Question 10. 
Show that 954 – 459 is exactly divisible by 99. 
Solution: 
To show : 954 – 459 is exactly divisible by 3 99, where a = 9, b = 5, c = 4 
abc = 100a + 10b + c 
=> 954 = 100 x 9 + 10 x 5 + 4 
=> 954 = 900 + 50 + 4 ………(i) 
and 459 = 100 x 4+ 10 x 5 + 9 
=> 459 = 400 + 50 + 9 ……..(ii) 
Subtracting (ii) from (i), we get 
954 – 459 = 900 + 50 + 4 – 400 – 50 – 9 
=> 954 – 459 = 500 – 5 
=> 954 – 459 = 495 
=> 954 – 459 = 99 x 5 
Hence, 954 – 459 is exactly divisible by 99 
Hence proved.