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Q.

In the product of BA×B3=57A, what are the respective positional values of B and A?



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a

6, 7

b

5, 2

c

7, 4

d

2, 5 

answer is D.

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Detailed Solution

Concept- We will first investigate the likelihood of an A value in order to answer this query. The value of B will then be determined by plugging the potential value of A into the equation above. The error will then be examined after that. There is another way we can accomplish this, which we will go over in more detail later on.
BA×B3=57A is the given equation.
When we multiply BA with B3, the product of the multiplication of the unit places of both terms on the left hand side must have A in its unit place. In this case, A is on the unit place of the product side, or we can say on the right hand side.
This implies that the product value of 3 times A must include A in the unit location.
Two choices emerge if we check the possibility. i.e. A must be either 0 or 5.
BA×B3=57A
When we multiply the equation by 10, we obtain(10B+A) (10B+3) =(57×10) +A 
(10B+A) (10B+3) =570+A 
100B2+10(AB) +30B+3A=570+A.
100B2+10(A×B) +30B+2A=570……………..(1)
For A=0
Putting A=0 in equation 1 we get,
100b2+30B=570
Dividing 10 on each side we get,
10b2+3B=57………..(2) No B satisfies this as
3B=7For A=5 Putting A=5 in equation 1 we get,
100B2+80B+10=570Taking 10 common from the equation (or dividing the whole equation by 10) we get-
10B+8B+1=57 10B2+8B=56
Dividing the by 2 we get,
⇒5B2+4B=28=20+8 B=2Hence, the correct answer is Option 4.
 ExamType: CBSE
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