A mathematician trying to cross a street happened to witness a bank robbery. When the police questioned him, he stated that the number plate of the van in which the thieves escaped had its last four digits as follows:
The first digit is 5. The last digit is the square of the second digit. The third digit is twice the second digit. Also, he noticed the sum of digits to be “9”. What is the quadratic equation that the police need to frame to find the $2^{n d}$ digit, given that ‘b’ is the second digit?

$b^{3} - 9 = 0$

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$b^{2} + 3 b - 4 = 0$

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$4 b + 5 = 9$

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$b^{2} + 2 b - 3 = 0$

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Solution

The correct option is B
$b^{2} + 3 b - 4 = 0$

Given that 2nd digit to be ‘b’, the registration number would be ‘5 b 2b $b^{2}$ ’.
The sum of the digits is $b^{2} + 2 b + b + 5$ which is given to be equal to ‘9’.
$\Rightarrow b^{2} + 3 b + 5 = 9$
Therefore the quadratic equation is
$\Rightarrow b^{2} + 3 b - 4 = 0$

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Similar questions

A mathematician trying to cross a street happened to witness a bank robbery. When the police questioned him, he stated that the number plate of the van in which the thieves escaped had its last four digits as follows:
The first digit is 5. The last digit is the square of the second digit. The third digit is twice the second digit. Also, he noticed the sum of digits to be “9”. What is the quadratic equation that the police need to frame to find the $2^{n d}$ digit, given that $b$ is the second digit?

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If in each number the third digit becomes the first digit, the first digit becomes the second digit and the second digit becomes the third digit, which of the following will be the first digit of the second highest number?

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The correct option is B b2+3b−4=0 Given that 2nd digit to be ‘b’, the registration number would be ‘5 b 2b b2’. The sum of the digits is b2+2b+b+5 which is given to be equal to ‘9’. ⇒b2+3b+5=9 Therefore the quadratic equation is ⇒b2+3b−4=0