Qwen2.5-VL-32B: Smarter and Lighter

Qwen2.5-VL-32B: Smarter and Lighter | Qwen

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Qwen2.5-VL-32B: Smarter and Lighter March 24, 2025 · 10 min · 2026 words · Qwen Team | Translations: 简体中文

QWEN CHAT GITHUB HUGGING FACE MODELSCOPE DISCORD Introduction # At the end of January this year, we launched the Qwen2.5-VL series of models, which received widespread attention and positive feedback from the community. Building on the Qwen2.5-VL series, we continued to optimize the model using reinforcement learning and open-sourced the new VL model with the beloved 32B parameter scale under the Apache 2.0 license — Qwen2.5-VL-32B-Instruct . Compared to the previously released Qwen2.5-VL series models, the features of this 32B VL model are as follows: Responses More Aligned with Human Preferences : Adjusted the output style to provide more detailed, better-formatted answers that align more closely with human preferences. Mathematical Reasoning : Significant improvement in the accuracy of solving complex mathematical problems. Fine-grained Image Understanding and Reasoning : Enhanced accuracy and detailed analysis in tasks such as image parsing, content recognition, and visual logic deduction.

Performance # Extensive benchmarking against state-of-the-art (SoTA) models of comparable scale, Qwen2.5-VL-32B-Instruct has demonstrated superiority over baselines, e.g., Mistral-Small-3.1-24B and Gemma-3-27B-IT, even surpassing the larger Qwen2-VL-72B-Instruct . Notably, it achieves significant advantages in multimodal tasks such as MMMU , MMMU-Pro , and MathVista , which focus on complex, multi-step reasoning. On MM-MT-Bench , a benchmark emphasizing subjective user experience evaluation, Qwen2.5-VL-32B-Instruct outperforms its predecessor Qwen2-VL-72B-Instruct by a substantial margin. In addition to excelling in visual capabilities, Qwen2.5-VL-32B-Instruct has also achieved top-tier performance in pure text capabilities at the same scale. Demo Cases # Fine-grained Image Understanding and Reasoning Next User I am driving a large truck on this road, and it is now 12 o’clock. Can I reach a place 110 kilometers away before 13 o’clock?

Qwen2.5-VL-32B To determine whether you can reach a place 110 kilometers away before 13:00, we need to analyze the information provided: Current Time : It is 12:00. Distance to Destination : 110 kilometers. Speed Limit for Trucks : The speed limit sign indicates that the maximum speed for trucks is 100 km/h .

Step-by-Step Analysis: # 1. Calculate the Required Travel Time # The formula to calculate travel time is: \[ \text{Time} = \frac{\text{Distance}}{\text{Speed}} \] Given: Distance = 110 km Maximum Speed = 100 km/h

Substitute the values: \[ \text{Time} = \frac{110 \text{ km}}{100 \text{ km/h}} = 1.1 \text{ hours} \] 2. Convert Time to Hours and Minutes # 1.1 hours is equivalent to: \[ 1 \text{ hour} + 0.1 \times 60 \text{ minutes} = 1 \text{ hour} + 6 \text{ minutes} = 1 \text{ hour and } 6 \text{ minutes} \] 3. Determine Arrival Time # If you start at 12:00 and it takes 1 hour and 6 minutes to travel 110 kilometers, your arrival time will be: \[ 12:00 + 1 \text{ hour and } 6 \text{ minutes} = 13:06 \] 4. Compare with Target Time # You want to arrive before 13:00. However, based on the calculations, you will arrive at 13:06, which is after 13:00. Conclusion: # Given the speed limit of 100 km/h for trucks, it is not possible to reach a place 110 kilometers away before 13:00. Thus, the final answer is: \[ \boxed{\text{No}} \]

Mathematical Reasoning Next User As shown in the figure, the straight lines AB and CD intersect at point O, OD bisects ∠AOE, ∠BOC = 50.0, then ∠EOB = ()

Qwen2.5-VL-32B Problem Analysis # We are given the following information: Straight lines $ AB $ and $ CD $ intersect at point $ O $. $ OD $ bisects $ \angle AOE $. $ \angle BOC = 50^\circ $. We need to find $ \angle EOB $.

Step-by-Step Solution # Step 1: Understand the given information and relationships # Since $ AB $ and $ CD $ are straight lines intersecting at $ O $, the angles around point $ O $ sum up to $ 360^\circ $. The vertically opposite angles are equal. Therefore: $$ \angle AOD = \angle BOC \quad \text{and} \quad \angle DOB = \angle COA. $$ Given $ \angle BOC = 50^\circ $, we have: $$ \angle AOD = 50^\circ. $$

Step 2: Use the angle bisector property # It is given that $ OD $ bisects $ \angle AOE $. This means: $$ \angle AOD = \angle DOE. $$ Since $ \angle AOD = 50^\circ $, we have: $$ \angle DOE = 50^\circ. $$

Step 3: Determine $ \angle AOE $ # The angle $ \angle AOE $ is the sum of $ \angle AOD $ and $ \angle DOE $: $$ \angle AOE = \angle AOD + \angle DOE = 50^\circ + 50^\circ = 100^\circ. $$

Step 4: Use the straight line property # Since $ AB $ is a straight line, the angles on a straight line sum up to $ 180^\circ $. Therefore: $$ \angle AOE + \angle EOB = 180^\circ. $$ Substituting $ \angle AOE = 100^\circ $, we get: $$ 100^\circ + \angle EOB = 180^\circ. $$ Solving for $ \angle EOB $: $$ \angle EOB = 180^\circ - 100^\circ = 80^\circ. $$

Final Answer # $$ \boxed{80} $$

Mathematical Reasoning Next User 侏罗纪蜘蛛网是一种非常有规律的蜘蛛网，如图，它是由无数个正方形环绕而成的，且最外边第一个正方形A_{1}B_{1}C_{1}D_{1}的面积为a_{1}=1，分别取正方形A_{1}B_{1}C_{1}D_{1}各边的三等分点A_{2}，B_{2}，C_{2}，D_{2}，作第二个正方形A_{2}B_{2}C_{2}D_{2}，且正方形A_{2}B_{2}C_{2}D_{2}的面积为a_{2}，\ldots \ldots ，依次类推，第n个正方形A_{n}B_{n}C_{n}D_{n}的面积为a_{n}. $求{a_{n}}的通项公式

Qwen2.5-VL-32B 问题分析 # 我们需要找到第 $ n $ 个正方形的面积 $ a_n $ 的通项公式。已知： 第一个正方形 $ A_1B_1C_1D_1 $ 的面积为 $ a_1 = 1 $。 每次取前一个正方形各边的三等分点，构造一个新的正方形。

我们需要确定每次构造新正方形时，其面积与前一个正方形面积之间的关系。 第一步：设第一个正方形的边长 # 设第一个正方形 $ A_1B_1C_1D_1 $ 的边长为 $ s_1 $。由于其面积为 $ a_1 = 1 $，因此： $$ s_1 = \sqrt{a_1} = \sqrt{1} = 1. $$ 第二步：构造第二个正方形 # 取 $ A_1B_1C_1D_1 $ 各边的三等分点，得到新的正方形 $ A_2B_2C_2D_2 $。我们需要计算这个新正方形的边长。 1. 确定新正方形的顶点坐标 # 假设第一个正方形 $ A_1B_1C_1D_1 $ 的顶点坐标如下： $ A_1 = (0, 1) $, $ B_1 = (1, 1) $, $ C_1 = (1, 0) $, $ D_1 = (0, 0) $.

取各边的三等分点： $ A_2 $ 是 $ A_1B_1 $ 的三等分点，坐标为 $ \left( \frac{1}{3}, 1 \right) $, $ B_2 $ 是 $ B_1C_1 $ 的三等分点，坐标为 $ \left( 1, \frac{2}{3} \right) $, $ C_2 $ 是 $ C_1D_1 $ 的三等分点，坐标为 $ \left( \frac{2}{3}, 0 \right) $, $ D_2 $ 是 $ D_1A_1 $ 的三等分点，坐标为 $ \left( 0, \frac{1}{3} \right) $.

2. 计算新正方形的边长 # 新正方形 $ A_2B_2C_2D_2 $ 的边长可以通过计算任意两点之间的距离得到。例如，计算 $ A_2 $ 和 $ B_2 $ 之间的距离： $$ A_2 = \left( \frac{1}{3}, 1 \right), \quad B_2 = \left( 1, \frac{2}{3} \right). $$ 两点间的距离公式为： $$ s_2 =...