Karnataka Class 8 Math Solution Chapter Playing with Numbers by Experience Math Teacher. Karnataka Board Class 8 Math Part 1 PDF Chapter 1 Exercise 1.1, Exercise 1.2 Solution.
Karnataka Class 8 Math Solution Chapter Playing with Numbers
Part 1 Chapter 1 Exercise 1.1
Find the values of the letters in each of the following and give reasons for the steps involved.
(1) 3A + 25 =B2
Solution: This has two letter A and B whose values are to be found. From ‘A+5’ we get as one’s place for the addition. For this to happen, the digit A should be ‘7’. After adding ‘7+5’ we get ‘1’ as a tens digit to carry forward. Then adding 3+2+1; we get ‘6’ as B. So the puzzle can be solved as shown below.
37 + 25 = 62 That is A = 7, and B = 6
(2) 4A + 98 =CB3
Solution: This has three letter A, B and C whose values are to be found. From ‘A+8’ we get ‘3’ as ones place for the addition, for this to happen, the digit A should be ‘5’. After adding ‘8+5’ we get 1 as a tens digit to carry forward. Then adding ‘4+9+1’ we get 14. After that 1 will be carry forward o hundredth place, hence the digit ‘C’ should be ‘1’
So, the puzzle can be solved as 45 + O98 =143
That is A = 5, B = 4 and C = 1
(3) 1A X A =9A
Solution: These is just one letter A whose value we have to find since the ones digit of AXA is A, it must be that A=0, A=5 or A=6 Now look at 1XA, 1XA must get added a number to get 9 after carrying the number we get ones place after multiplying supposed number. If we consider A as O then there would be no carry forward. If we take 5 then 5X5=25 two will be carry forward then ‘1X5+2’=7, it is not the required number. If we take A as 6, then ‘1X6+3’=9 will be the ten’s number. So, the ‘A’ should be ‘6’. So the puzzle can be solved as shown below.
16 X 6 = 96 That is, A=6
(4) AB + 37 =6A
Solution: This has two letter A and B whose values are to be found. Look at the place column. There is A+3=6 and tens place is the end of the addition. So, A should be 2 or 3, because there is possibility of any forward to tens column after addition. So, it we take A as 3 the ‘B+7=3’ for this happen B must be 6 to get ‘3’ as ones digit, then I will be carry. A+3+1=6 (3+3+1≠6 is it 7). Then A=3 is not possible. So, consider A=2, then ‘B+7=A’ so, B must be 5 to get 2 as ones digit.
Then (A+3+1=6) (2+3+1=6) it satisfies the value for A. The puzzle can be solved as shown below.
25 + 37= 62 That is A=2 and B=5
(5) AB X 3 =CAB
Solution: This has three letters A, band C whose values are to be found. Since the ones digit of BX3 is B, it must be B=O or B=5.
Now look at A. If B=1 or A=2 then ABX3 would at most be equal to 19X3=57 or 29X3=87 but the product is exceed than two digits. So, we cannot have A=1, or A=2.
It ‘B=0, and A=3 then, ABX3 would
be 30X3=90, or B=0 and A=4 then ABX3 would
be 40X3=120 or B=0 and A=5 than ABX3 would
be 150. Putting A=5 and B=0 satisfies the condition for given equation.
So the answer is A=5, B=0 and C=1
The puzzle can be solved as shown below
50 X 3 =150
(6) AB X 5 = CAB
Solution: This has three letters A, B and C whose values are to be found. Since the ones digit of BX5 is B, it must be that B=0 or B=5
Now look at A if A=1, 2, 3, or 5 then ABX5 could be if we took B=0, then 10X5=50, 20X5=100, 30X5=150, 40X5=200, 50X5=250.
After observing all possibilities above A=5 and B=0 works out correctly since 50X5=250
So the answer is A=5, B=0 and C=2
The puzzle can be solved as shown below.
50 X 5 = 250
(7) AB X 6 =BBB
Solution: This has two letters A and B whose values are to be found. Since the ones digit of BX6 is B, it must be that B=0 or B=2, 4, 6, or 8.
Now look at A if A=0, then ABX6 would at most be equal to 08X6=48 and 48 is less than three digit, so we cannot have A=0.
If B=2 and A=1 the ABX6 would be 12X6=72 its not possible. Now putting B=2 and A=2, 3, 4, and 5 we get ABX6 such that 22X6=132, 32X6=192, 42X6=252, and 52X6=312. It doesn’t satisfies the values of A and B.
Now putting A=1, 2, 3, 4, 5, 6 and 7 and B=4 we get 24X6=144, 34X6=204, 44X6=264, 54X6=324, 64X6=264 and 74X6=444.
Hence A=7 and B=4 is the answer for the puzzle so the puzzle can be solved as shown below.
74 X 6 =444
(8) A1 + 1B
Solution: From 1+B=0, we get ‘0’ that is a number whose ones digit is ‘0’. For this to happen the digit B should be 9. Then B should be 9. So from A+1=B is A+1=9, by solving it we get A=9-1=8
So, the digit A should be 8 but there is carry forward from ones column is 1. So B must be 8-1=7. Hence B=9 and A=7.
So the puzzle can be solved as shown below.
71 + 19 =90
(9) 2AB + AB1
Solution: From B+1=8 we get B=8-1=7, then B must be 7. If B=7, from A+B=1; is A+7=1 then we get ‘1’ as ones place after adding A into 7. For this to happen the digit B must be 4. If A=4 then 7+4=11 and I will be carry forward in hundredth column. Hence 2+4+1=7 our above possibilities are correct.
So the correct answer is A=4 and B=7 and above puzzle can be solved as shown below.
247 + 471 =718
(10) 12A + 6AB =A09
Solution: From ones column if A+B=9 then A and B must be less than. Let’s move to tens column. If A+2=0 for this to happen A must be 8 since after addition we should get 0 as a ones digit. Then A must be 8. If A is 8 then A+B=9 and B=9-8=1, then B must be 1. Let check hundredths column 1+6=7 and 1 is carry forward from tenth’s column so 1+6+1=8 and our all possibilities are correct.
So the correct answer is A=8, B=1 and above puzzle can be solved as shown below.
128 + 681 = 809
Part 1 Chapter 1 Exercise 1.2
(1) If 21y 5 is a multiple of 9, where y is a digit, what is the value of y?
Solution: If 21y 5 is a multiple of 9 then 21y 5 must be divisible by 9.
The sum of the digits of 21y 5 is 2+1+5+Y = 8+y
Since 21y 5 is divisible by 9 sum of its digit i.e. 8+y should be divisible by 9.
This is possible when 8+y=9 or 18
But since y is a digit, therefore, 8+y=9 i.e, y=1.
(2) If 31z 5 is multiple of 9, where z is digit, what is the value of z?
Solution: If 31z 5 is multiple of 9 then 31z 5 must be divisible by 9.
Since 31z 5 is divisible by 9, sum of its digits is 3+1+5+2= 9+Z should be divisible by 9.
This is possible when 9+Z=9 or 18…..
But Z is a digit, therefore 9+Z=9 or 9+Z=18
Z=9-9 or 9+9=18
Z=0 Z=9.
I.e. Z=0 or 9.
You will find that there are two answers for the last problem. Why is this so?
= Yes there are two answer for the value of Z. because we have 9+Z=9 or 9+Z=18 and Z is single digit. Since 0 and 9 both satisfies the value of Z. Hence for Z we have to values/answers.
(3) If 24x is a multiple of 3, where x is a digit, what is the value of x?
(Since 24x is a multiple of 3 its. Sum of digits 6+x is a multiple of 3, so 6+x is one of these numbers 0, 3, 6, 9, 12, 15, 18 ……….. but since x is a digit, it can only be that 6+x=6 or 9 or 12 or 15. Therefore x=0 or 3, or 6 or 9. Thus, x can be have any of four different values)
Solution: If 24x is a multiple of 3 then 24x must be divisible by 3.
From the divisibility of 3 the sum of its number i.e. 2+4+x=6+x must be divisible by 3. Since x is a digit, it can only be 0, 3, 6 or 9. I.e. x=0, or x=3, or x=6, or x=9.
(4) If 31z 5 is a multiple of 3. Where z is a digit, what might be the values of z?
Solution: If 31z5 is a multiple of 3. Then 31z5 must be divisible by 3.
From the divisibility of 3 the sum of its digits i.e. 3+1+5+z=9+z must be divisible by 3. Since z is digit it can be. 0, 3, 6, or 9.
I.e. z=0, or z=3, or z=6, or z=9.