ML Aggarwal CBSE Solutions Class 8 Math 5th Chapter Playing with Numbers Exercise 5.1
(1) (i) 89 = 10 x 8 + 9
(ii) 408 = 100 x 4 + 10 x 0 + 8
(iii) 369 = 100 x 3 + 10 x 6 + 9
(2) (i) Given number 23
Number obtained by reversing the digit = 32
Therefore, sum = 23 + 32 = 55 = 11 x 5
(i) When sum is divided by 11,
Quotient is 5,
(ii) Sum of digits = 2 + 3 = 5
Therefore, when sum 55 is divided by 5,
Quotient is 11.
(3) Given number = 83
Number obtained by reversing the digit is 38.
Therefore, difference = 83 – 38 = 45 = 5 x 9
(i) When sum is divided by 9,
Then the quotient will be ‘5’
(ii) Difference digit = 8 – 3 = 5
When difference ‘45’ is divided by ‘5’ then the quotient will be ‘9’
(4) Given number = a b c
Changing the order of digit = b c a
Changing the order of digit = c a b
Therefore, a b c = 100a + 10b + c
bca = 100a + 10c + a
Therefore, c a b = 100c + 10a + b
On adding number, we get —
abc + bca + cab = (100a + 10b + c) + (100b + 10c + a) + (100c + 10a + b)
= 100a + 10a + a + 100b + b + 100c + 10c + c
= 111a + 111b + 111c
= 111 x (a + b + c) = 3 x 37 x (a + b + c)
(i) When divided by 111, the quotient will be (a + b + c)
(ii) When divided by (a + b + c), the quotient will be 111
(iii) When divided by ‘37’, the quotient will be 3 x (a + b + c)
(iii) When divided by ‘3’, the quotient 37 x (a + b + c)
(5) Given number = 943
Reversing number = 349
Therefore difference = 943 – 349
= 594
= 99 x 6
(i) When dividing by 99
The equitant will be 6.
(ii) When dividing by ‘6’
the equitant will be ‘99’
(6) let,
One digit = a
Another digit = b
Original number = 10a + b
Reversing the digits number obtained = 10b + a
According to que——
(1) a + b = 11 ——(i)
(2) (10a + b) – (10b + a) = 9
We know that –
(10a + b) – (10b + a) = 9
Or, 10a + b – 10b – a = 9
Or, 9a – 9b = 9
Or, 9(a – b) = 9
Or, a – b = 1 —— (ii)
Now, (i) & (ii) are equating.
We get,
a = b = 11
a – b = 1
_________
2a = 12
Or, a = 12/2
= 6
Therefore, putting the value of a in eq. (ii) we get,
a – b = 1
Or, a – 1 = b
Or, b = 6 – 1
Therefore, a = 6
a = 5
Therefore, the original number = 65.
= 5
(7) Let, one digit = a
Another digit = b
Therefore, original number = 10a + b
Number obtained by reversing number = 10b + a
According to question——–
(10a + b) – (10b + a) = 27
Or, 10a + b – 10b – a = 27
Or, 9a – 9b = 27
Or, 9(a – b) = 27
Or, a – b = 27/9
= 3
Therefore, the difference between the digits of ‘2’ digit number = a – b = 3
(8) Let,
One digit = a
Another digit = b
Therefore, original number = 10a + b
Number obtained by reversing number = 10b + a
According to question—
(10a + b) + (10b + a) = 44
Or, 10a + b + 10b + a = 44
Or, 11a + 11b = 44
Or, 11(a + b) = 44
Or, a + b = 44/11
= 4
(9) Let,
1st digit = a
Last digit = C
Middle digit = b = 0
Therefore, original number = 100a + b + c
Given, a + c = 11 —- (i)
Reversing digit number obtain = 100c + b + a
According to question —
(100c + b + a) – (100a + b + c) = 495
Or, (100c + 0 + a) – (100a + 0 + c) = 495
Or, 100c + a – 100a – c = 495
Or, 99c – 99a = 495
Or, 99(c – a) = 495
Or, c – a = 495/99
= 5 ——(ii)
Equating (i) & (ii) we get –
C + a = 11
C – a = 5
_______
2c = 16
Therefore, a + c = 11
Or, a = 11 – c
Or, a = 11 – 8
Or, a = 3
Therefore, a = 3
C = 8
Therefore, original number = 308.
(10) Ratio = 1:2:3
Let, common factor = x
Therefore, units digits = x
Ten’s digit = 2x
Hundreds digit = 3x
Therefore, Original number [(100 X 3x) + (100 x 2x)+ x]
= (300x + 20x + x) = 321x
Therefore, reversing digit number obtained = [(100 X x) + (10 X 2x) + 3x]
= 123x
According to question —-
321x – 123x = 594
Or, 198x = 594
Or, x = 594/198 = 3
Therefore, units digit = 3
Ten’s digit = 2 x 3 = 6
Hundred’s digit = 3 x 3= 9
Therefore, original number = 963.
(11) Let,
Hundred’s unit (a) = x
Ten’s unit (b) = x – 1
Unit’s unit (c) = x + 1
Original number => a + b + c
= (100 X x) + 10(x – 1) + x +1
= 100x + 10x – 10 + x + 1
= 111x – 9
2nd reversing digit obtained number => c + a + b
= [100 (x +1) + 10 X x + (x – 1)]
= 100x + 100 + 10x + x – 1
= 111x + 99
3rd, reversing digit obtained number => b + c + a
= [100 (x – 1) + 10 ( x + 1) + x]
= 100x – 100 + 10x + 10 + x
= 111x – 90
According to question—–
(111x – 9) + (111x + 99) + (111x – 90) = 2664
Or, (111x + 111x + 111x) – (9 – 99 +90) = 2664
Or, 333x – (-90 + 90) = 2662
Or, 333x = 2664
Or, x = 2664/333
Or, x = 8
Ten’s units = 8 – 1 = 7 Therefore, original number = 879.
Unit’s unit = 8 + 1 = 9