I take it we are talking about Verbal Reasoning and Quantitative Reasoning here, rather than English and Maths?
The mean of each distribution is 100 and the standard deviation is 15.
115 is at 84.13th percentile
122 is at 92.88th percentile
The mean of the two joint probability distribution is clearly 200. However the standard deviation of the joint probability distribution is determined by the covariance of the two distributions. This is unknown to us, however in the VR and Quant scores are identical, then clearly the standard deviation of the joint distribution doubles, i.e. it is 30.
A joint score of 237 in that case would be on the 89.13th percentile.
On the other hand if the covariance were zero, i.e. there was no correlation between the two scores whatsoever, then if you got 115 on one test, the most likely score on the other would be 100, and likewise, if you got say 60 on one test, then the most likely score on the other test would be 100.
Since there is no correlation between the two, the variance of the JPD in this case is equal to the sum of the respective variances, and since they both have the same variance, this would imply that we double the variance, and therefore the standard deviation of JPD is 15 * sqrt(2), i.e. 21.21.
If the mean is 200 and the sd is 21.21, then a score of 237 is on the 95.94th percentile.
Accordingly the combined percentile must be between 89.13 and 95.94.
Probably closer to 89.13 than 95.94 I might add, as I would suggest that the two are highly correlated, BUT 61 is clearly nonsense.
The only way you could get 61 is if you had a third score (non-VR?), which given the scores in the OP, and a population mean of 300 and standard deviation somewhere between sqrt(3) 15 and 3 15, you would need a score for non-VR of less than 76 (this if while CAT scores are randomly distributed among individuals, each component score of a given individual is 100% correlated with another component score for that same individual) and 70 (if there is no correlation at all).
So one or more of the numbers quoted in the OP is wrong, but most likely the overall percentile.