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Q.1. Find the values of the letters in each of the following and give reasons for the steps involved:
(i)
Since we know that,
12 – 5 = 7 = A
So, B = 8
Hence, A = 7 and B = 8
(ii)
Since we know that,
9 + 4 = 13,
∴ A = 4
B = 1 + 5 + 7 = 3
C = 1
Hence, A = 4, B = 3, C = 1
(iii)
Since we know that,
5 + 7 = 12, ∴ A = 7
1 + 2 + 7 = 10
1 + 4 + 2 = 7 = A
Hence, A = 7
(iv)
Since we know that,
A = 4 or 9
A ≠ 4 as A + A = A, 4 + 4 ≠ 4
A = 9
B = 1
Hence A = 9, B = 1
(v)
Since we know that,
5 + 7 = 12
∴ A = 5
B = 4, C = 6
Hence, A = 5, B = 4, C = 6
(vi)
Since we know that,
B = 3 or 8
If B = 8
1 + 1 + 7 = 9 then A = 7
C = 4 – 2 = 2
7 + 1 = 8 = B
Hence, A = 7, B = 8, C = 2
(vii)
Since we know that,
A = 7 (∵ 5 + 7 =12)
1 + 4 + 2 = 7(A)
∴ B = 2
1 + 2 + 5 = 8, C = 5
Hence, A = 7, B = 2, C = 5
(viii)
Since we know that,
B – 6 = 7
B = 3
A – 1 – B = 4
A – 1 – 3 = 4, A = 4 + 4 = 8
Hence, A = 8, B = 3
(ix)
Since we know that,
A = 1 or 6 or 5 as 1 × 1 = 1
or 6 × 6 = 6
or 5 × 5 = 5
Taking A = 5
B = 8
Hence, A = 5, B = 8
(x)
Since we know that,
B × B = B
B = 1, 6, 5
If B = 5, and A = 2,
then 25 × 25 = 625
Hence, A = 2 and B = 5
(xi)
Since we know that,
A = 2 or 8
Let A = 2
Hence, A = 2
Q.2. Fill in the numbers from 1 to 6 (without repetition) so that each side of the magic triangle adds up to 12.
Given:
Numbers 1 to 6 without repetition:
The sum from each side = 12
There can be much more solutions such as
Q.3. Complete the magic square given alongside using number 0, 1, 2, 3, ……., 15 (only once), so that sum along each row, column and diagonal is 30.
Given:
In the magic square, the use of number 0, 1, 2, 3… 15
So, only once the sum along each row, column and diagonal is 30.
Some numbers are already filled.
3 + 8 + 4 = 15 + 15 = 30
3 + 12 = 5 + 10 = 30
3 + 14 + 0 = 17 + 13 = 30
0 + 6 = 9 + 15 = 30
0 + 7 + 12 = 19 + 11 =30
15 + 1 + 12 = 28 + 2 = 30
8 + 6 + 11 = 25 + 5 = 30
13 + 6 + 1 = 20 + 10 = 30
14 + 5 + 2 = 21 + 9 = 30
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