Question

MAC + MAAR = JOCKO, find the value of 3A + 2M + 2C(C=2)​

Answer 1

Answer:

31

Step-by-step explanation:

Here J is carry, J=1 when J=1, O=0 with carry 1 and M=9 C+R=O à 0 with carry 1. So, C=2 and R=8 M+A=C à 2 with carry 1, A=3, A+A+1= K, 3+3+1=K=7, 932+9338=10270 so, finally A = 3, M = 9, C = 2, = 3A + 2M + 2C = 9 + 18 + 4 = 31

Answer 2

Question :- MAC + MAAR = JOCKO, find the value of 3A + 2M + 2C(C=2) ?

Solution :-

Conclusion 1 ) 3 digit number + 4 digit number = 5 digit number . Therefore, J = 1 . (From Carry.)

Conclusion 2) Now put value of J and C and take some example ,

MA2

MAAR.

-------------

1O2KO.

-------------

J = 1

Ex 1):- 642 Ex 2):- 652

6448 9558

---------- ----------

7090 10210

We can conclude that, in order to get J = 1 , value of M must be 9 .

Therefore, M = 9, and than , O = 0 . (9 + 1 carry) = 10

Conclusion 3) :- Putting these value now we get,

9A2

9AAR.

-------------

102K0.

-------------

→ 2 + R = 0 (1 carry)

→ R = 10 - 2 = 8 .

Conclusion 4) :- Putting value of R = 8 also Now,

9A2

9AA8.

-------------

102K0.

-------------

→ 1(carry) + A + A = K

→ 9 + A = 2 (1 carry)

So,

→ A = 12 - 9 = 3.

Therefore,

→ 1 + 3 + 3 = 7 = K .

Conclusion 5) :- Putting all values Now, we get,

932

932 9338.

-------------

10270.

-------------

Therefore :-

• M = 9
• A = 3
• C = 2
• R = 8
• J = 1
• O = 0
• K = 7

Hence,

→ 3A + 2M + 2C

→ 3*3 + 2*9 + 2*2

→ 9 + 18 + 4

→ 31. (Ans.)

________________________

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