AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

Q.

In the product of $BA \times B 3 = 57 A$ , what are the respective positional values of B and A?

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

a

6, 7

b

5, 2

c

7, 4

d

2, 5

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

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 \times B 3 = 57 A

is the given equation.
When we multiply

BA

with

B 3,

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 \times B 3 = 57 A

When we multiply the equation by 10, we obtain

(10 B + A) (10 B + 3) = (57 \times 10) + A

(10 B + A) (10 B + 3) = 570 + A

{100 B}^{2} + 10 (AB) + 30 B + 3 A = 570 + A .

100 B^{2} + 10 (A \times B) + 30 B + 2 A = 570

……………..(1)
For

A = 0

Putting A=0 in equation 1 we get,
100

b^{2}

+30B=570
Dividing 10 on each side we get,

10 b^{2} + 3 B = 57 \dots \dots \dots . . (2)

No B satisfies this as

3 B = 7 For A = 5

Putting A=5 in equation 1 we get,

100 B^{2} + 80 B + 10 = 570

Taking 10 common from the equation (or dividing the whole equation by 10) we get-

10 B + 8 B + 1 = 57

10 B^{2} + 8 B = 56

Dividing the by 2 we get,

\Rightarrow 5 B^{2} + 4 B = 28 = 20 + 8 B = 2

Hence, the correct answer is Option 4.
ExamType: CBSE

Watch 3-min video & get full concept clarity

courses

No courses found

Related Blogs

Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Get Expert Academic Guidance – Connect with a Counselor Today!

Related Blogs

Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

best study material, now at your finger tips!

live classes
progress tracking
24x7 mentored guidance
study plan analysis

download the app

gplay

mentor

Download the App

gplay

personalised 1:1 online tutoring