Question

$\color{red}{\Huge{\mathbb{QUESTION}}}$
A student got twice as many sums as he got right. If he attempted 48 sums in all. How many did he solved correctly?

(Don't give irrelevant answers otherwise on the spot it will be reported)​

Answer 1

Answer:

$\huge\underbrace\mathcal\red{ANSWER}$...

Hi friend here is ur answer ✈

Let x is the value of sums he solved right.

Let 2x is the value of sums he solved wrong.

Total Questions = 48

So, let's carry on

x + 2x = 48

3x = 48

\frac{48}{3} = 16

3

48

=16

Answer = 16 Question are correct insted of 48.

{\bigstar}\large{\boxed{\sf{\pink{16 \: is \: the \: correct \: answer}}}}★

16isthecorrectanswer

Hope it's helpful :)

Thank you guys :)

Let x is the value of sums he solved right.

Let 2x is the value of sums he solved wrong.

Total Questions = 48

So, let's carry on

x + 2x = 48

3x = 48

\frac{48}{3} = 16

3

48

=16

Answer = 16 Question are correct insted of 48.

{\bigstar}\large{\boxed{\sf{\pink{16 \: is \: the \: correct \: answer}}}}★

16isthecorrectanswer

Hope it's helpful :)

Thank you guys :)

Answer 2

$\huge\colorbox{pink}{Answer}$

Suppose the boy got x sums right and 2x sums wrong. Then, x + 2x = 48 3x = 48 x = 16.