Students can refer to and download RD Sharma Solutions Class 8 Maths Chapter 5 Playing with Numbers from the links provided below. Subject experts have solved the RD Sharma Class 8 Solutions to help students grasp the concepts easily and thus boost their confidence level, which plays an essential role in their examinations. Hence, students whose intention is to score high in the exams are advised to practise RD Sharma Class 8 Solutions.
In Exercise 5.3, we will discuss problems based on the addition and multiplication of cryptarithms (cryptarithms are puzzles, which are operated on various operations and numbers, in which letters take the place of digits, and one has to find out which letter represents which digit).
RD Sharma Solutions for Class 8 Maths Chapter 5 Playing with Numbers Exercise 5.3
Access Answers to RD Sharma Solutions for Class 8 Maths Exercise 5.3 Chapter 5 – Playing with Numbers
Solve each of the following cryptarithms:
1.
Solution:
Firstly let us solve for unit’s place,
7 + B = A
And for ten’s place,
3 + A = 9
This means that A = 6 and B = -1, which is not possible.
So, there should be one carry in ten’s place which means 7 + B >9
Now solving for ten’s place with one carry,
3 +A +1 = 9
A = 9-1-3 = 5
For unit’s place, subtracting 10 as one carry is given to ten’s place,
7 + B – 10= 5
B = 5+10-7 = 8
∴A = 5 and B = 8
2.
Solution:
Firstly let us solve for unit’s place,
B + 7 = A
And for ten’s place,
A + 3 = 9
This means that A = 6 and B = -1, which is not possible.
So, there should be one carry in ten’s place, which means B + 7 >9
Now solving for ten’s place with one carry,
A +3 +1 = 9
A = 9-4 = 5
For unit’s place, subtracting 10 as one carry is given to ten’s place,
B + 7 – 10= 5
B = 5+10-7 = 8
∴ A = 5 and B = 8
3.
Solution:
Firstly let us solve for unit’s place,
1 + B = 0
This means that B = -1, which is not possible.
So, there should be one carry in ten’s place,
A + 1 +1 = B —- (1)
For unit’s place, we need to subtract 10 as one carry is given in ten’s place,
1 + B – 10 = 0
B = 10-1 = 9
Substituting B = 9 in (1),
A + 1 + 1 = 9
A = 9-1-1 = 7
∴ A = 7 and B = 9
4.
Solution:
Firstly let us solve for unit’s place,
B + 1 = 8
B = 7
Now let us solve for ten’s place,
A + B = 1
A + 7 = 1
A = -6, which is not possible.
Hence, A + B > 9
We know that now there should be one carry in hundred’s place, and so we need to subtract 10 from ten’s place,
i.e., A + B – 10 = 1
A + 7 = 11
A = 11-7 = 4
Now, to check whether our values of A and B are correct, we should solve for the hundred’s place.
2 + A + 1 = B
2 + 4 + 1 = 7
7 = 7
i.e., RHS = LHS
∴ A = 4 and B = 7
5.
Solution:
Firstly let us solve for unit’s place,
A + B = 9 —- (1)
With this condition, we know that sum of 2 digits can be greater than 18.
So, there is no need to carry one from ten’s place.
Now let us solve for ten’s place,
2 + A = 0
This means A = -2, which is never possible
Hence, 2 + A > 9
Now, there should be one carry in hundred’s place, and hence we need to subtract 10 from ten’s place,
i.e., 2 + A – 10 = 0
A = 10-2 = 8
Now, substituting A=8 in 1,
A + B = 9
8 + B = 9
B = 9 – 8
B = 1
∴ A = 8 and B = 1
6.
Solution:
Firstly let us solve for unit’s place,
We have two conditions here, 7 + B ≤ 9 and 7 + B > 9
For 7 + B ≤ 9
7 + B = A
A – B = 7 —- (1)
Now let us solve for ten’s place,
B + A = 8 —- (2)
Solving 1 and 2 simultaneously,
2A = 15 which means A = 7.5 which is not possible
So, our condition 7 + B ≤ 9 is wrong.
∴ 7 + B > 9 is the correct condition
Hence, there should be one carry in ten’s place and subtracting 10 from the unit’s place,
7 + B – 10 = A
B – A = 3 —- (3)
For ten’s place,
B + A + 1 = 8
B + A = 8-1
B + A = 7 —- (4)
Solving (3) and (4) simultaneously,
2B = 10
B = 10/2 = 5
Substituting the value of B in equation 4
B + A = 7
5 + A = 7
A = 7-5 = 2
∴ B = 5 and A = 2
7. Show that the Cryptarithm does not have any solution.
Solution:
If B is multiplied by 4 then only 0 satisfies the above condition.
So, for the unit place to satisfy the above condition, we should have B = 0.
Similarly, for ten’s place, only 0 satisfies the above condition.
But AB cannot be 00 as 00 is not a two-digit number.
So, A and B cannot be equal to 0
∴ there is no solution satisfying the condition
.
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