- In the given addition, find the value of the letters and give reason.
4 A
+ 3 4
———-
B 3
Answer:
When we add A and 4, we are getting number 1 that is it gives number 1 in onesplace. This is possible only when A is number 9. Therefore, adding A that is 9 to 4, it gives 13. 1 will be carry for the next step.
In next step, the addition is 1 + 4 + 3 = 8. So the addition is as given below.
4 9
+ 3 4
————
8 3
Therefore, B is 8.
So, A and B are 4 and 8 respectively.
- In the given addition, find the value of the letters and also give reason.
5 X
+ 8 7
———-
Y Z 2
Answer:
The addition of X and 7 gives 2 that is a no. whose ones place is 2. It is possible only when digit X is 5. So, the addition of 5 and 7 gives 12. 1 will be carry for the next step.
1 + 5 + 8 = 14
Hence, the addition is as given below.
5 5
+ 8 7
—————
1 4 2
The value of X, Y and Z is 5, 1 and 4 respectively.
- In the given multiplication, find the value of the letters and give reason.
2 Y
× Y
—————–
1 2 Y
Answer:
The multiplication of a number with itself gives a number whose ones placeis that number itself. This occurs when that number is 1, 5 or 6.
Y = 1
The multiplication will be 21 × 1 = 21. Here, there is hundreds place as well. So, Y = 1 is not possible.
Y = 5
The multiplication will be 25 × 5 = 125. Here, tens as well as hundreds place match. So, Y = 5 is the correct answer.
- In the given addition, find the value of the letters and also give reason.
X Y
+ 4 6
————–
7 X
Answer:
The addition of X and 3 is giving 7. There can be two cases.
(i)No carry
The value of X will be 3 so we get 7 when we add 3 and 4 that is 3 + 4 = 7. Now consider the first step, Y + 6 = 3 so the value of Y has to be 7. Then we get 3 in ones place. But the value X is single digit so it is not possible.
(ii) With carry
The value of X will be 2 as 1 + 2 + 4 = 7. Now consider the first step where Y is added to 6 to give 2 in ones place. For that the value of Y will be 6 + Y = 12. Therefore, the value of Y is 6.
2 6
+ 4 6
——————
7 2
Hence, the value of X and Y is 2 and 6 respectively.
- In the given multiplication, find the value of the letters and give reason.
X Y
× 3
————-
Z X Y
Answer:
When 3is multiplied with Y it gives a number whose onesplace is Y again. So, Y must be 5 or 0.
Let Y = 5
First step: 5 × 3 = 15
1 will be carried forward. Therefore, (X × 3) + 1 = ZX. This is not possible for any number.
Therefore, value of Y has to be 0 only.
If Y = 0, then there will be no carry. So we get X × 3 = ZX.
When a number is multiplied with 3, its ones placeshould be the number itself. That is possible only for X = 0 or 5. But X cannot be 0 as it has to be two digit numbers. Therefore, the value of X is 5. Thus we get the following
5 0
× 3
—————
15 0
The value of X, Y and Z is 5, 0 and 1 respectively.
- In the given multiplication, find the value of the letters and also give reason.
X Y
× 5
————
Z X Y
Answer:
When 5 is multiplied with Y it gives a number whose ones placeis Y again. So, Y must be 5 or 0.
Let Y = 5
First step: 5 × Y = 5 × 5 = 25
2 will be carried forward. Therefore, (X × 5) + 2 = ZX. This is possible for number X = 2 or 7.
The multiplication is as given below.
2 5 7 5
× 5 × 5
———- ———-
1 2 5 3 7 5
Let Y = 0
First step: 5 × Y = 5
5 × 0 = 0
There will not be any carry in this case.
In the next step, 5 × X = ZX
This can happen only when the value of X is 5 or 0.
However, X cannot be 0 as XY is two digit numbers. Therefore, the value of X is 5.
5 0
× 5
———–
2 5 0
Therefore, there three possible values of X, Y and Z.
(i) 2, 5 and 1 respectively
(ii) 5, 0 and 2 respectively
(iii) 7, 5 and 3 respectively
- In the given multiplication, find the value of the letters and also give reason.
X Y
× 6
———-
Y YY
Answer:
When 6 is multiplied with Y, it gives a number whose ones placeis Y. It is possible only if Y = 0, 2, 4, 6 or 8.
Y = 0;
The product will be 0 in this case so it is not possible.
Y = 2;
Y × 6 = 12 and 1 will be carried forward for the next step.
6X + 1 = YY = 22. Then integer value of X is not possible.
Y = 6;
Y × 6 = 36 and 3 will be carried forward for the next step.
6X + 3 = YY = 66. Then integer value of X is not possible.
Y = 8;
Y × 6 = 48 and 4 will be carried forward for the next step.
6X + 4 = YY = 88.
6X = 84.
X = 14
But X is single digit number.
Then value of X is not possible.
Y = 4;
Y × 6 = 24 and 2 will be carried forward for the next step.
6X + 2 = YY = 44.
6X = 42.
X = 7
The multiplication is given below
7 4
× 6
———-
4 4 4
Thus integer value of X and Y is 7 and 4 respectively.
- In the given addition, find the value of the letters and also give reason.
X 1
+ 1 Y
————
Y 0
Answer:
When 1 is added to Y, it gives 0 that is a number whose ones place is 0. This is possible when digit Y is 9.
So the addition of 1 and Y will be 10 so 1 will be carried forward for the next step.
In the next step,
1+X+1=9
Therefore, X is 7.
1 + 7 + 1 = 9 = Y
Hence, the addition is as given below.
7 1
+ 1 9
———–
9 0
Thus value of X and Y is 7 and 9 respectively.
- In the given addition, find the value of the letters and also give reason.
2 X Y
+ X Y 1
—————-
Y 1 8
Answer:
When 1 is added to Y, it gives 8 that is a number whose ones place is 8. This is possible when digit Y is 7.
In the next step, X + Y = 1. Therefore, the value of X is 4.
4 + 7 = 11 and 1 will be carried forward for the next step.
In the next step,
1+ 2 + X = Y
1+ 2 + 4 = 7
Hence, the addition is as given below.
2 4 7
+ 4 7 1
————–
7 1 8
Thus value of X and Y is 4 and 7 respectively.
- In the given addition, find the value of the letters and also give reason.
1 2 X
+ 6X Y
—————-
X 0 9
Answer:
When X is added to Y, it gives 9 that is a number whose ones place is 9.Sum can be 9 only as summation of two single digits cannot be 19. So no carry generated.
In the next step, X + 2 = 0
It is possible if X = 8.
Therefore, 2 + 8 = 10 and 1 will be carried forward for the next step.
1 + 1 + 6 = 8. Therefore, value of X = 8.
When X is added to Y, it gives 9.
X + Y = 9
8 + Y = 9
Therefore, value of Y = 1
1 2 8
+ 6 8 1
—————–
8 0 9
Thus value of X and Y is 8 and 1 respectively.
