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Chlorine exists in two isotopic forms Cl-37 and Cl-35, but its atomic mass is 35.5. This indicates the ratio of Cl-37 and Cl-35 approximately:
(a)-1: 2
(b)-1: 1
(c)-1: 3
(d)-3: 1

Answer
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Hint: You can use this formula to get your final answer-
$\text{Average atomic mass = }\sum\limits_{i=1}^{n}{\text{(atomic mass of an isotope) x (fractional abundance)}}$
Solve it carefully the question is asking you to indicate the ratio of Cl-37 and Cl-35, not the vice versa.

Complete Solution :
Isotopes are those whose atomic number is same but the atomic masses are different like the isotopes of hydrogen, i.e., H-1, H-2, H-3. So, there are two isotopes of chlorine atom in which there are two masses 35 and 37, denoted as Cl-35 and Cl-37.

The average atomic masses are calculated by multiplying atomic masses with their respective relative abundance.
To calculate the ratio of abundances of the given isotopes, we use the equation given by average atomic mass, which is:
$\text{Average atomic mass = }\sum\limits_{i=1}^{n}{\text{(atomic mass of an isotope) x (fractional abundance)}}$
Let the fractional abundance for Cl-35 isotope be 'x'.
So, we can say fractional abundance for Cl-37 isotope will be (1 - x)

For Cl-35:
Mass of Cl-35 isotope = 35 u
Fractional abundance of Cl-35 isotope = x

For Cl-37:
Mass of Cl-37 isotope = 37 u
Fractional abundance of Cl-37 isotope = (1 - x)
The average atomic mass of chlorine = 35.5
Now by inserting all values in the above equation, we get:

$35.5=[(35x)+(37(1-x))]$
$35.5=-2x+37$
$x=0.75$
Fractional abundance for Cl-35 isotope = 0.75
Fractional abundance for Cl-37 isotope = 1 - 0.75 = 0.25

- Now just multiple these with 100 to get their percentages,
Percentage abundance for Cl-35 isotope = 75 %
Percentage abundance for Cl-37 isotope = 25 %
The ratio of Cl-37 and Cl-35 isotopes will be = 1: 3
So, the correct answer is “Option C”.

Note: Like isotopes whose atomic number is same but the atomic masses are different, we have isobars in which the different elements have different atomic numbers and have same atomic masses.