Free GMAT Focus Edition mock test — 64 questions, 3 sections 🌐

7 original practice questions written by Myndaq in the GMAT style. They are not taken from this mock paper or from official past papers. Try each one, then open the answer.

1. Question 1Quantitative ReasoningOne correct answer

   A stationery shop sold 300 notebooks in March. In April its notebook sales were 15 percent lower than in March, and in May its sales were 20 percent higher than in April. How many notebooks did the shop sell in May?

   1. A. 255
   2. B. 285
   3. C. 306
   4. D. 315
   5. E. 330
   Show answer and explanation

   Answer: C

   April sales are 300 × 0.85 = 255. May sales are 255 × 1.20 = 306.

   The trap is 315, which comes from treating the two changes as a single net change of -15 + 20 = +5 percent and computing 300 × 1.05. That is wrong because the 20 percent increase applies to April's smaller base, not to March's. 255 is April's figure, not May's, and 330 would require an increase of 10 percent on top of the unreduced March total.

2. Question 2Quantitative ReasoningOne correct answer

   If 2x - 3y = 13 and x + y = 9, what is the value of x² - y²?

   1. A. 45
   2. B. 54
   3. C. 63
   4. D. 72
   5. E. 81
   Show answer and explanation

   Answer: C

   From the second equation, x = 9 - y. Substituting into the first gives 2(9 - y) - 3y = 13, so 18 - 5y = 13 and y = 1. Then x = 8.

   Since x² - y² = (x + y)(x - y) = 9 × 7 = 63. Direct check: 64 - 1 = 63.

   81 is (x + y)², and 72 comes from multiplying x by y and then by 9; neither factors the difference of squares correctly.

3. Question 3Quantitative ReasoningOne correct answer

   The average (arithmetic mean) of seven consecutive even integers is 46. What is the average of the three largest of these seven integers?

   1. A. 46
   2. B. 48
   3. C. 49
   4. D. 50
   5. E. 52
   Show answer and explanation

   Answer: D

   For an odd number of consecutive even integers, the mean equals the middle term, so the fourth integer is 46 and the seven integers are 40, 42, 44, 46, 48, 50 and 52.

   The three largest are 48, 50 and 52. Their mean is again the middle value, 50; adding them gives 150, and 150 divided by 3 is 50.

   46 is the mean of all seven integers, not of the top three. 48 comes from forgetting that the integers are even and listing seven consecutive integers 43 to 49, whose three largest average 48. 49 is the mean of the four largest, 46, 48, 50 and 52. 52 is the largest integer rather than an average.

4. Question 4Verbal ReasoningOne correct answer

   Last year the council of the town of Verdale repainted every pedestrian crossing on Marchmont Street, the town's main shopping street, using high-visibility markings. In the twelve months since the repainting, the number of reported collisions between vehicles and pedestrians on Marchmont Street has fallen by roughly a third compared with the twelve months before. The council has concluded that the new markings caused the decline and has proposed repainting the crossings on every street in Verdale.

   Which of the following, if true, most seriously weakens the council's conclusion?

   1. A. The high-visibility markings cost substantially more per crossing than the conventional paint they replaced.
   2. B. For most of the past twelve months, building work forced a diversion that sent almost all through traffic away from Marchmont Street.
   3. C. Several drivers surveyed after the repainting said that the crossings were easier to see at night than before.
   4. D. Collisions between vehicles and cyclists on Marchmont Street are recorded separately and were not included in either figure.
   5. E. Marchmont Street carries more pedestrians on a typical day than any other street in Verdale.
   Show answer and explanation

   Answer: B

   The council's reasoning moves from a correlation, repainting followed by fewer collisions, to a causal claim. The most damaging objection supplies a different cause for the same decline. If a diversion removed almost all through traffic for most of the year, far fewer vehicles were available to collide with pedestrians, so the fall need not be due to the markings at all. That is choice B.

   Cost, choice A, bears on whether the town-wide plan is affordable but not on whether the markings work. Choice C reports an effect that supports the council rather than undermining it. Choice D would matter only if cyclist collisions had been counted inconsistently across the two periods, which it says they were not. Choice E explains why Marchmont Street was chosen first but says nothing about the cause of the decline.

5. Question 5Verbal ReasoningOne correct answer

   Urban populations of several songbird species sing at a higher minimum frequency than rural populations of the same species. The long-standing explanation is acoustic: low-frequency traffic noise masks the lowest notes, so birds whose songs sit higher are heard by more potential mates, and their songs are the ones young birds copy. A competing account points instead to body size. City birds tend to be slightly smaller, and a smaller vocal apparatus produces higher notes for straightforwardly physical reasons; on this view the frequency shift is a by-product of whatever makes city birds small, such as diet or earlier fledging, rather than a response to noise.

   A recent set of continuous recordings made in a park bordered by a motorway complicates both accounts. Across the course of each day, the minimum frequency of the park birds rose and fell in step with traffic volume, a change far too rapid to be attributed to body size. Yet the same recordings show that even at three in the morning, when traffic had almost ceased, the park birds never sang as low as birds recorded in nearby farmland. Neither masking nor morphology on its own accounts for both the hour-by-hour flexibility and the gap that persists when the noise is gone.

   The passage indicates that the recordings made at three in the morning are significant chiefly because they

   1. A. show that traffic noise has no measurable effect on the frequency at which urban birds sing
   2. B. indicate that masking by traffic noise cannot by itself explain the whole of the urban-rural difference in song
   3. C. establish that the birds in the park are smaller than the birds recorded in nearby farmland
   4. D. demonstrate that rural birds would raise their minimum frequency if they were exposed to motorway noise
   5. E. suggest that a bird's minimum singing frequency is fixed by its body size alone
   Show answer and explanation

   Answer: B

   The point of the three o'clock observation is that the urban-rural gap survives even when the masking noise is effectively absent. If masking were the whole story, the gap should close at that hour. So the recordings show that masking alone is not a sufficient explanation, which is choice B.

   Choice A overstates the result and contradicts the preceding sentence, in which frequency tracks traffic volume through the day. Choice C attributes to the recordings a finding about body size that the passage never says they made. Choice D concerns rural birds under conditions that were not tested. Choice E is ruled out by the rapid daily variation, which the passage says body size cannot explain.

6. Question 6Data InsightsOne correct answer

   Last week a café sold hot drinks and cold drinks only. How many cold drinks did it sell?

   (1) The café sold 180 more hot drinks than cold drinks. (2) Hot drinks accounted for 70 percent of all drinks the café sold.

   1. A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
   2. B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
   3. C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
   4. D. EACH statement ALONE is sufficient.
   5. E. Statements (1) and (2) TOGETHER are NOT sufficient.
   Show answer and explanation

   Answer: C

   Let h be hot drinks and c be cold drinks.

   Statement (1) gives h - c = 180. One equation in two unknowns leaves c undetermined: c could be 20 with h = 200, or 400 with h = 580. Not sufficient.

   Statement (2) gives h = 0.7(h + c), which simplifies to 0.3h = 0.7c, that is h/c = 7/3. This fixes the ratio but not the size: c = 30 and c = 300 both fit. Not sufficient.

   Together, h = 7c/3 and 7c/3 - c = 180, so 4c/3 = 180 and c = 135 (with h = 315). A unique value follows, so both together are sufficient and the answer is the third option.

7. Question 7Data InsightsOne or more correct answers

   A distributor operates four depots. For last quarter it recorded, for each depot, the number of orders shipped and the percentage of those orders that reached the customer on time.

   Ashby: 4,200 orders shipped, 96 percent on time. Brill: 1,500 orders shipped, 88 percent on time. Corby: 6,000 orders shipped, 91 percent on time. Dunley: 2,300 orders shipped, 99 percent on time.

   Select every statement that must be true on the basis of the figures above.

   1. A. Corby had more late orders than Brill and Dunley had between them.
   2. B. More than half of all orders shipped last quarter came from depots whose on-time rate was above 95 percent.
   3. C. Ashby had more late orders than Dunley.
   4. D. Taken together, the four depots delivered more than 93 percent of their orders on time.
   5. E. Brill had the smallest number of late orders of the four depots.
   Show answer and explanation

   Answer: A, C, D

   Late orders are shipped orders times the late rate. Ashby: 4,200 × 4 percent = 168. Brill: 1,500 × 12 percent = 180. Corby: 6,000 × 9 percent = 540. Dunley: 2,300 × 1 percent = 23.

   First option: 540 is greater than 180 + 23 = 203, so it is true. Third option: 168 is greater than 23, so it is true even though Ashby's rate is far better than Brill's. Fourth option: total shipped is 14,000 and total late is 168 + 180 + 540 + 23 = 911, so on-time orders are 13,089, about 93.5 percent, which is above 93 percent and therefore true.

   The second option is false: the depots above 95 percent are Ashby and Dunley, with 4,200 + 2,300 = 6,500 orders, which is less than half of 14,000. The fifth option is false because Dunley, with 23 late orders, had fewer than Brill's 180; a high shipped volume is not the same as a high late count.