A chemical supply company has 60 liters of a 40% HNO3...

[BTGmoderatorLU]

A chemical supply company has 60 liters of a 40% HNO3 solution. How many liters of pure undiluted HNO3 must the chemist add so that the resultant solution is a 50% solution?

A. 12
B. 15
C. 20
D. 24
E. 30

The OA is A.

I get the solution in the following way,

60 liters of a 40% HNO3 solution means HNO3 = 24 liters in 60 liters of the solution.

Now, let x be the pure HNO3 added.

As per question,

24 + x = 50% of (60 + x)

or x = 12. Option A.

Experts, any suggestion about it? Thanks in advance.

[deloitte247]

$$60\ litres\ of\ 40\%\ HNO_3\ solution=60\cdot40\%$$
$$=60\cdot\frac{40}{100}=60\cdot0.4=24\ litres\ in\ 60litres$$
$$Let\ pure\ HNO_3\ added\ be\ =\ x$$
$$Therefore,\ 24+x=50\%\ of\ 60+x$$
$$24+x=\frac{50}{100}\cdot\left(60+x\right)$$
$$24+x=\frac{5}{10}\cdot\left(60+x\right)$$
$$24+x=\frac{\left(300+5x\right)}{10}$$
$$On\ cross\ multiplication,\ we\ have$$
$$10\left(24+x\right)=300+5x$$
$$240+10x=300+5x$$
$$10x-5x=300-240$$
$$5x=60$$
$$x=\frac{60}{5}=12$$
$$Hence,\ option\ A\ is\ the\ correct\ option$$

[Jeff@TargetTestPrep]

A chemical supply company has 60 liters of a 40% HNO3 solution. How many liters of pure undiluted HNO3 must the chemist add so that the resultant solution is a 50% solution?

A. 12
B. 15
C. 20
D. 24
E. 30

We can lert n = the amount of pure undiluted HNO3 to be added and create the equation:

(0.4 x 60 + n)/(60 + n) = 1/2

2(24 + n) = 60 + n

48 + 2n = 60 + n

n = 12

Answer: A