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NCA-GENM sample questions with answers

10 free practice questions for the NVIDIA-Certified Associate: Generative AI Multimodal exam. Try each one, then open the answer to see why the right option wins and every other option loses.

Question 1Core Machine Learning and AI Knowledge

An associate evaluates a visual question-answering pipeline on a benchmark of 600 images with five questions about each image, 3,000 questions in total. Accuracy is 71%. The associate computes a 95% confidence interval by bootstrap resampling the 3,000 question-level scores and gets a narrow interval. A reviewer notes that questions about the same image tend to be answered right or wrong together. How should the interval be computed instead?

  1. A.

    Keep resampling individual questions, since 3,000 separate scores are plenty for the bootstrap to give an accurate interval

  2. B.

    Resample whole images with all five of their questions in each bootstrap draw, then recompute accuracy per draw

  3. C.

    Replace the bootstrap with a normal-approximation interval computed from the 3,000 question-level scores

  4. D.

    Raise the number of bootstrap iterations from 10,000 to 100,000 so that the interval becomes more accurate

Show answer

Answer: B

When scores are correlated within images, bootstrap the images, not the individual questions.

  • A. More questions do not help when they are not independent; question-level resampling treats 3,000 correlated scores as 3,000 separate pieces of evidence.
  • B. Resampling the image, the unit that makes answers correlated, carries the within-image dependence into every draw, so the interval reflects it.
  • C. A normal approximation on question-level scores makes the same independence assumption, so it is just as narrow.
  • D. More iterations reduce Monte Carlo noise in the bootstrap itself but do not change what is being resampled, so the bias stays.
Question 2Core Machine Learning and AI Knowledge

An evaluation set for a support assistant contains 50% image-plus-text questions, 30% text-only and 20% image-only. Production logs show that live traffic is 10% image-plus-text, 80% text-only and 10% image-only. Pipeline A beats Pipeline B on the evaluation set overall, but B is clearly better on text-only questions. The team must estimate which pipeline will perform better in production. What analysis should it use?

  1. A.

    Compare the pipelines only on image-plus-text questions, since that slice has the most items and the tightest interval

  2. B.

    Compute each pipeline's accuracy per modality slice, then weight the slices by the production traffic mix to get expected accuracy

  3. C.

    Give each of the three modality slices an equal weight, since no single modality should dominate the comparison

  4. D.

    Use the overall evaluation-set accuracy, because it already includes every modality and so represents all users

Show answer

Answer: B

Estimate production performance by weighting per-slice scores with the production mix, not the evaluation set's mix.

  • A. Scoring one slice ignores the text-only questions that make up 80% of production, where B is clearly better.
  • B. Reweighting per-slice scores to the deployment mix answers the question asked: expected accuracy on the traffic the system will actually see.
  • C. Equal weights answer a different question, balanced performance across modalities, not expected production accuracy.
  • D. The overall figure weights slices by the evaluation set's mix, which over-represents image-plus-text five-fold compared with production.
Question 3Core Machine Learning and AI Knowledge

A dermatology research group has 12,000 skin-lesion photos from 1,500 patients, with several photos per patient, and melanoma confirmed in about 6% of patients. It wants 5-fold cross-validation in which no patient's photos appear in both the training and validation data of a fold. Every validation fold must also hold a similar share of melanoma cases, so that fold scores can be compared. Which splitter meets both requirements?

  1. A.

    GroupKFold with the patient ID passed as the groups argument, which keeps every patient's photos in one fold

  2. B.

    StratifiedGroupKFold with the patient ID passed as the groups argument and the melanoma label as y

  3. C.

    ShuffleSplit repeated five times with 20% of photos drawn at random for validation in each repeat

  4. D.

    StratifiedKFold on the melanoma label, which keeps the share of melanoma cases similar in every fold

Show answer

Answer: B

StratifiedGroupKFold is the splitter that respects patient groups and balances the rare class at the same time.

  • A. It prevents patient leakage but does not try to balance class proportions, so with a 6% positive rate some folds can end up with far fewer melanoma cases.
  • B. It keeps each patient's photos together in one fold and also approximately balances the melanoma share across folds, meeting both requirements.
  • C. It samples individual photos at random, so it neither keeps patients together nor controls the melanoma share.
  • D. It balances the class share but splits photos individually, so the same patient's photos land in both training and validation.
Question 4Core Machine Learning and AI Knowledge

A property platform predicts sale price from a 512-dimension image embedding of each listing photo plus 12 tabular features. Management asks how much the photos contribute. An associate computes permutation importance for every column on held-out data; each embedding dimension shows almost zero importance, so the draft report says photos barely matter. What analysis answers the question properly?

  1. A.

    Permute all 512 embedding columns together as one block on the held-out data and measure the drop in score

  2. B.

    Report the absolute size of each embedding dimension's learned coefficient as a measure of how much it matters

  3. C.

    Add up the 512 per-dimension permutation importances, since the total gives the combined contribution of the photos

  4. D.

    Use the tree model's impurity-based feature importances instead, since they are computed for every split in training

Show answer

Answer: A

The photo information is spread across hundreds of correlated columns, so it must be permuted as one group to measure its contribution.

  • A. Shuffling the whole block breaks every route to the photo information at once, so the score drop measures the modality's real contribution.
  • B. Coefficient magnitudes depend on scale and on correlation between dimensions, and a price model need not be linear at all.
  • C. Each dimension's importance is understated because correlated dimensions stand in for it, so the sum still understates the photos' role.
  • D. Impurity-based importances come from training data, can favour features that do not generalise and are biased toward high-cardinality features.
Question 5Core Machine Learning and AI Knowledge

Corvid's senior engineer now wants each encoder to receive its own training signal, and wants the audio branch's influence to shrink once it starts overfitting. The shipped product must remain one end-to-end network whose fused prediction is used. Per-modality validation curves are available every epoch. Which training change fits all of these constraints?

  1. A.

    Remove dropout from the audio encoder and raise its learning rate so that it catches up with the video branch before the fused classifier saturates

  2. B.

    Attach a classifier to each modality's features and to the fused features, and train on a weighted sum of the three losses with weights set from each branch's overfitting behaviour

  3. C.

    Train separate audio-only and video-only models and average their predicted probabilities at inference, removing the fused classifier from the design entirely

  4. D.

    Attach auxiliary unimodal classifiers but give every loss the same fixed weight for the whole run, since equal weighting alone keeps the modalities balanced

Show answer

Answer: B

Per-modality heads with a weighted, overfitting-aware blend of losses give each encoder its own signal and let the audio weight fall as it overfits.

  • A. Incorrect: removing regularisation and speeding up the already-overfitting branch makes the imbalance worse.
  • B. Correct: it provides per-encoder supervision, lets the audio weight fall as it overfits, and keeps one end-to-end fused network.
  • C. Incorrect: an ensemble of two separate models violates the requirement that one fused end-to-end network is shipped.
  • D. Incorrect: fixed equal weights give per-encoder signals but cannot shrink audio's influence as it overfits, which is the deciding requirement.
Question 6Core Machine Learning and AI Knowledge

A team compares two text-to-image models with human ratings of image quality on a 1 to 5 scale. Eight raters are available. The draft plan gives Model A's 400 images to raters 1 to 4 and Model B's 400 images to raters 5 to 8, because that is simpler to organise. The raters are known to differ in how harsh they are. What change most improves the validity of the comparison?

  1. A.

    Keep the split and convert each rater's scores to ranks, so that harshness no longer affects the comparison

  2. B.

    Have every rater score images from both models for the same prompts, in random order, so rater harshness affects both equally

  3. C.

    Keep the split but double the number of images per model, so the larger sample averages out the differences between raters

  4. D.

    Keep the split but drop the harshest and most lenient rater from each group so the two groups look more alike

Show answer

Answer: B

Cross raters with models on matched prompts; if raters are nested within models, rater harshness is confounded with model quality.

  • A. Ranking within each rater removes scale differences but still compares Model A's raters with Model B's raters, not the models.
  • B. Crossing raters with models, on matched prompts, means each rater's leniency applies to both models and cancels in the comparison.
  • C. More images per rater do not help, because the rater groups are fixed; the confound between rater group and model remains whatever the sample size.
  • D. Trimming raters makes the groups smaller without breaking the link between which raters scored which model.
Question 7Core Machine Learning and AI Knowledge

A retailer wants to know whether image features help predict product returns. An associate fine-tunes an image encoder on all 80,000 labelled product photos to predict the return label, extracts embeddings from it, and then runs 5-fold cross-validation of a logistic regression on those embeddings. Validation AUC is 0.93, but a model trained the same way scores 0.74 on next quarter's products. What is the main flaw?

  1. A.

    Next quarter's products are simply harder, so the drop reflects the new data rather than any flaw in the evaluation

  2. B.

    Embeddings from any pretrained encoder always leak labels, so image features can never be evaluated with cross-validation

  3. C.

    The encoder was fine-tuned on every photo's label, including validation photos, so the embeddings already encode the answers

  4. D.

    Logistic regression is too simple to use image embeddings, so it overfits the training folds and fails later

Show answer

Answer: C

Supervised feature learning is part of the model: fitting it on all labels before cross-validation leaks the validation labels.

  • A. A harder quarter could lower the score, but the evaluation design already explains an inflated 0.93 without that assumption.
  • B. An encoder that never saw these labels, used frozen, does not leak; the problem is supervised training on the validation rows.
  • C. Supervised fine-tuning on all labels lets validation labels shape the features, so each fold's validation rows were effectively seen.
  • D. A simple linear model is less prone to overfitting, not more, and it cannot explain a gap created by the features themselves.
Question 8Core Machine Learning and AI Knowledge

Garganey Retail will fine-tune an image-text model for product matching with the TAO CLIP recipe. Its GPUs fit only modest per-device batches, and the team wants a loss that does not need a normalisation over every pairwise similarity in the global batch. Both train.loss_type options, clip and siglip, would train a usable model. Which should the team choose, and why?

  1. A.

    clip, because its softmax loss scores each image-text pair independently of the rest of the batch and was designed specifically for small per-device batch sizes

  2. B.

    siglip, because its sigmoid loss scores each image-text pair independently without a batch-wide normalisation and holds up better at smaller batch sizes

  3. C.

    siglip, because its sigmoid loss uses no negative pairs at all and learns only by pulling each matched image and caption embedding closer together

  4. D.

    clip, because a softmax loss is the only contrastive formulation that can be exported and deployed after training with the TAO Toolkit

Show answer

Answer: B

The sigmoid (SigLIP) loss treats each pair as an independent binary decision, so it needs no global softmax normalisation and performs better at small batches.

  • A. Incorrect: this describes the sigmoid loss's property but attributes it to the softmax loss, which does normalise over the batch.
  • B. Correct: siglip, for the right reason: pairwise sigmoid scoring needs no global normalisation and suits smaller batches.
  • C. Incorrect: right loss, wrong reason: the sigmoid loss still uses non-matching pairs as negatives, scored independently.
  • D. Incorrect: TAO does not tie export or deployment to the softmax formulation.
Question 9Core Machine Learning and AI Knowledge

A pathology lab has two million labelled microscope tiles and ample GPU time. Its images look nothing like the natural photographs an available backbone was pretrained on. One engineer proposes random initialisation because the domains differ; another proposes a frozen backbone with a new head. Which approach is best supported by transfer-learning evidence?

  1. A.

    Initialise from the pretrained weights and fine-tune the whole network on the lab's labelled tiles

  2. B.

    Train the network from random initial weights, since pretrained features from a distant domain only hinder learning

  3. C.

    Keep the pretrained backbone frozen and train only a new classification head on the extracted tile features

  4. D.

    Fine-tune only the final classification layer first, then stop, since two million tiles cannot improve the backbone

Show answer

Answer: A

With plenty of data in a distant domain, full fine-tuning from pretrained initialisation adapts every layer while keeping the benefit of a good starting point.

  • A. Correct: abundant data allows adapting every layer, and pretrained initialisation still beats random weights even across domains.
  • B. Distant-domain features are less useful but still better than random initialisation, so starting from scratch gives up a benefit.
  • C. Freezing the backbone keeps features that fit a distant domain poorly and wastes the large labelled dataset.
  • D. With two million tiles the backbone can be improved substantially; stopping after the head wastes both data and compute.
Question 10Core Machine Learning and AI Knowledge

An e-commerce team combines an image model and a text model for product categorisation with a second-stage model that takes both models' predicted probabilities as inputs. To build the second stage's training data, an associate runs each first-stage model on the same training rows it was trained on. The stacked model looks excellent in validation but underperforms both single models in production. What is the most likely flaw?

  1. A.

    The second stage learned from in-sample predictions, which are overconfident; it should be trained on out-of-fold predictions

  2. B.

    The image and text probabilities are on different scales, so they must be standardised before the second stage

  3. C.

    The second-stage model is too simple, and a deeper neural network would learn to correct the first-stage errors

  4. D.

    Stacking can never beat the best single model, so the second stage should be removed and the best model kept

Show answer

Answer: A

A stacking model must be trained on out-of-fold predictions, or it learns from overconfident in-sample outputs.

  • A. First-stage models predict their own training rows too well, so the combiner learns to trust them too much; out-of-fold predictions mimic unseen data.
  • B. Both inputs are probabilities between 0 and 1, and rescaling would not remove the overconfidence of in-sample predictions.
  • C. A more complex combiner would fit the misleading in-sample predictions even more closely.
  • D. Stacking often beats single models when trained correctly; the problem here is how its training data was produced.

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