Abstract
Spoken language models (SLMs) enable natural human-computer interaction, but their reasoning ability still lags behind that of text-based large language models, especially on spoken mathematical question answering tasks. One important reason is that SLMs reason over purely verbalized mathematical expressions, which are harder to interpret than symbolic text. However, directly transferring text-based reasoning to SLMs is nontrivial due to architectural constraints and the additional computational requirements. To address this challenge, we propose Efficient Chain-of-Modality Reasoning (ECoM Reasoning), the first framework to introduce compressed reasoning into SLMs. By compressing the textual component so that it jointly serves as speech guidance and reasoning representation, ECoM Reasoning improves reasoning accuracy while using a smaller token budget than the standard Chain-of-Modality (CoM) architecture, which generates intermediate text before speech. To train this capability, we further propose Progressive Compression, a curriculum-based strategy that gradually trains the model from full-form reasoning to compressed reasoning. Experiments on spoken mathematical question answering benchmarks show that ECoM Reasoning improves accuracy by 21% over standard CoM without explicit reasoning, and by 3% over CoM with full reasoning traces while using only 40% of the text tokens, demonstrating that it enhances SLM reasoning while remaining inference-efficient.
Examples of Token-Level Compression
Example 1
Question
a and b can do a piece of work in 4 days. With the help of c they finish the work in 3 days. c alone can do that piece of work in ?
Answer: 12 days
100%
c = 1 / 3 – 1 / 4 = 1 / 12 = > 12 days answer : d
80%
c = 1 / 3 – 1 / 4 = 1 12 12 days answer d
60%
= 1 / 3 – 1 4 12 12 days answer d
40%
1 3 1 4 12 12 days answer
20%
3 4 12 days
Example 2
Question
Bobby can deadlift 300 pounds at 13. When he is 18 he can deadlift 100 pounds more than 250% of his previous deadlift. How many pounds did he add per year?
Answer: 110
100%
His new deadlift is 300*2.5+100=850 pounds.
So he added 850-300=550 pounds to his deadlift.
This took 18-13=5 years.
So he added 550/5=110 pounds per year.
80%
new deadlift is 300*2.5+100=850 pounds added 850-300=550 pounds to deadlift took 18-13=5 years added 550/5=110 pounds per year
60%
deadlift 300*2.5+100=850 pounds added 850-300=550 pounds 18-13=5 years added 550/5=110 pounds
40%
deadlift 300*2.5+100=850 added 18-13=5 years 550/5=110 pounds
20%
300+100 550/5=110 pounds
Example 3
Question
Problem 3. Vasya solved problems for 15 days - at least one problem each day. Each day (except the first), if the weather was cloudy, he solved one more problem than the previous day, and if it was sunny, one less problem. In the first 9 days, Vasya solved 13 problems.
a) What was the weather like on the 10th day? How many problems did he solve on that day? [4 points]
b) What is the maximum number of problems Vasya could solve on the 15th day? [2 points] (B. Frenkin)
a) What was the weather like on the 10th day? How many problems did he solve on that day? [4 points]
b) What is the maximum number of problems Vasya could solve on the 15th day? [2 points] (B. Frenkin)
Answer: 7
100%
Answer. a) The 10th day was cloudy, Vasya solved 2 problems; b) 7 problems.
Solution. a) See the solution to problem 2 for 6th grade (p. 4).
b) Since on the 10th day Vasya solved 2 problems (see part a), and in each subsequent day he can solve a maximum of 1 more problem, then in 5 days he can solve a maximum of $2+5=7$ problems. This will happen if the weather was cloudy from the 11th to the 15th day.
80%
Answer. a) 10th day cloudy Vasya solved 2 problems; b) 7 problems. Solution. a) See solution to problem 2 for 6th grade (p. 4). b) Since 10th day Vasya solved 2 problems (see part a), each subsequent day can solve maximum 1 more problem, in 5 days can solve maximum $2+5=7$ problems This happen if weather cloudy from 11th to 15th day.
60%
Answer 10th day cloudy Vasya solved 2 problems 7 problems. Solution See solution problem 2 6th grade (p. 4) 10th day Vasya solved 2 problems part each subsequent day solve maximum 1 more problem 5 days solve maximum $2+5=7$ problems if weather cloudy 11th to 15th day.
40%
10th day cloudy Vasya solved 2 7 problems problem 2 6th grade. 4) 10th day Vasya solved 2 problems 1 5 days $2+5=7$ problems weather cloudy 11th to 15th
20%
10th cloudy Vasya 2 7 6th grade Vasya 5 $2+5=7 cloudy 11th 15th
Cases of Different Framework Response
Case 1: Addition Problem
Question
Jessica spent $10.22 on a cat toy, and a cage cost her $11.73. What was the total cost of Jessica's purchases?
Ground Truth: 21.95
CoM
Jessica spent $10.22 on a cat toy and a cage cost her $11.73 what was the total cost of Jessica's purchases? To find the total cost of Jessica's purchases, you need to add the cost of the cat toy and the cost of the cage. Cost of cat toy: $10.22. Cost of cage: $11.73. Total cost = $10.22 + $11.73 = $22.00. So, the total cost of Jessica's purchases is $22.00.
CoM Reasoning
Jessica spent $10.22 on a cat toy and a cage for her gecko. The cage cost her $11.73. What was the total cost of Jessica's purchases? To determine the total cost of Jessica's purchases, we need to add the cost of the cat toy and the cage together. Let's break it down step by step: 1. Identify the cost of the cat toy: $10.22. 2. Identify the cost of the cage: $11.73. Now, we add these two amounts together: 10.22 + 11.73. To make the addition easier, we can align the decimal points. Starting from the rightmost digit (the hundredths place), we add the digits in each column: In the hundredths place: 2 + 3 = 5. In the tenths place: 2 + 7 = 9. In the dollars place: 1 + 1 = 2. Putting it all together, we get: 21.95. Therefore, the total cost of Jessica's purchases is 21.95. The answer is 21.95.
ECoM Reasoning
cat toy $10.22 cage $11.73 $21.95. The answer is 21.95.
Case 2: Distance Problem
Question
If the cave is 919 feet deep and they need to travel 1307 feet. How far is the end of the cave?
Ground Truth: 2226
CoM
If the cave is 919 feet deep and they need to travel 1307 feet how far is the end of the cave? To find the distance from the end of the cave to the starting point, we need to subtract the depth of the cave from the total distance they need to travel. Distance to travel: 1307 feet. Depth of cave: 919 feet. Distance from end of cave to starting point = Distance to travel - Depth of cave = 1307 - 919 = 388 feet. So, the end of the cave is 388 feet away from the starting point.
CoM Reasoning
If the cave is 919 feet deep and they need to travel 1307 feet, how far is the end of the cave? To determine how far the end of the cave is, we need to understand that the cave is divided into two parts by the path they are traveling on. The path they are traveling on is 1307 feet long, and this path includes the depth of the cave. 1. Identify the total length of the path: 1307 feet. 2. Identify the depth of the cave: 919 feet. 3. Since the path includes both the depth of the cave and the horizontal distance to the end of the cave, we need to subtract the depth of the cave from the total length of the path to find the horizontal distance. Let's denote the horizontal distance to the end of the cave as x. According to the problem, the horizontal distance x is the remaining part of the path after accounting for the depth of the cave. We can set up the following equation: x = 1307 - 919. Now, let's perform the subtraction: x = 1307 - 919 = 388. Therefore, the distance from the start of the path to the end of the cave is 388 feet. The answer is 388.
ECoM Reasoning
cave 919 feet deep 1307 feet end 919 + 1307 2226 feet. The answer is 2226.
Case 3: Code Execution
Question
Here's a piece of code: sum starts at 0; i goes from 4 to 9; each time, i is added to sum; finally, print sum. What will be printed?
Ground Truth: 39
CoM
Here's a piece of code: sum = 0; i = 4; while (i > 0) { sum += i; i-- } Finally print sum. What will be printed? The code will print: 10. Here's a step-by-step explanation: 1. Initialize sum to 0 and i to 4. 2. Enter the while loop. 3. In the first iteration, i is 4, so sum is updated to 4 (4 + 4 = 8). 4. In the second iteration, i is 3, so sum is updated to 8 (8 + 3 = 11). 5. In the third iteration, i is 2, so sum is updated to 10 (11 + 2 = 13). 6. In the fourth iteration, i is 1, so sum is updated to 10 (10 + 1 = 11). 7. In the fifth iteration, i is 0, so the condition i > 0 is false, and the loop terminates. Therefore, the final value of sum is 10.
CoM Reasoning
Here is a piece of code: sum = 0; i goes from 4 to 9 each time i is added to sum. Finally print sum. What will be printed? Sum = 0; i goes from 4 to 9 each time i is added to sum. Finally print sum. What will be printed? Option a. The answer is 14.
ECoM Reasoning
sum 4 5 6 7 8 9 39. The answer is 39.
Case 4: Knowledge Question
Question
Who wrote the famous book "Pride and Prejudice"?
Ground Truth: Jane Austen
CoM
Who wrote the famous book "Pride and Prejudice"? The famous book "Pride and Prejudice" was written by Jane Austen, an English novelist. The novel was first published in 1813 and is considered one of the greatest works of English literature. It is a classic of the Austen family and is still widely read and studied today.
CoM Reasoning
Who wrote the famous book "Pride and Prejudice"? The famous book "Pride and Prejudice" was written by Jane Austen, a renowned English novelist. The novel is set in the late 18th century and early 19th century, primarily focusing on the social dynamics and personal relationships among the main characters. The story revolves around Elizabeth Bennet, a young woman from a modest family, and her experiences with marriage, social standing, and personal growth. The novel is celebrated for its rich character development, sharp social commentary, and its exploration of themes such as class, gender, and individuality. Jane Austen (1775-1817) was born into a family of modest means in Steventon, near Reading, England. Her family was well-connected in the literary and social circles of the time, which provided her with a unique perspective on life and literature. Austen's writing style is characterized by her meticulous attention to detail, her ability to bring characters to life with depth and nuance, and her keen observations of human nature. "Pride and Prejudice" is one of her most famous works, published in 1813. The novel was initially published in two volumes, with the first volume containing the first 34 chapters and the second volume containing the remaining 34 chapters. The novel was well-received upon its publication and has since become a classic of English literature, widely studied and enjoyed by readers around the world. In addition to "Pride and Prejudice," Jane Austen also wrote several other notable works, including "Sense and Sensibility," "Northanger Abbey," and "Emma." Her novels are celebrated for their depth, complexity, and enduring popularity, and she is often referred to as the "Queen of English Literature." In conclusion, Jane Austen is the author of the famous book "Pride and Prejudice," and her writing has had a profound impact on the world of literature and culture. The answer is Jane Austen.
ECoM Reasoning
Jane Austen. The answer is Jane Austen.
Cases of Different Compression Ratio
MultiArith
Question
A book store had 41 books in the bargain bin. If they sold 33 books, but then put 2 more in the bin, how many books would be in the bin?
Ground Truth: 10
ECoM Reasoning 100%
The number of books in the bin is 41 - 33 = 8 books. After adding 2 books, the number of books in the bin is 8 + 2 = 10 books. The answer is 10.
ECoM Reasoning 80%
Barren bin 41 - 33 = 8 books. Adding 2 books bin 8 + 2 = 10 books. The answer is 10.
ECoM Reasoning 60%
Bookstore sold 33 books 41 - 33 = 8 books put 2 books 8 + 2 = 10 books. The answer is 10.
ECoM Reasoning 40%
Bookstore sold 33 41 - 33 8 put 2 8 + 2 10. The answer is 10.
ECoM Reasoning 20%
41 33 74 2 76. The answer is 76.
ECoM Reasoning 0%
The answer is 10.
AddSub
Question
Joan had 5 dimes in her bank. She spent 2 of her dimes. How many dimes does she have now?
Ground Truth: 3
ECoM Reasoning 100%
Joan had 5 dimes and spent 2, so she has 5-2 = 3 dimes left. The answer is 3.
ECoM Reasoning 80%
John spent 2 dimes 10 dimes - 2 dimes = 8 dimes. The answer is 8.
ECoM Reasoning 60%
John spent 2 dimes 5 - 2 3 dimes. The answer is 3.
ECoM Reasoning 40%
John 5 - 2 3. The answer is 3.
ECoM Reasoning 20%
5 2 3. The answer is 3.
ECoM Reasoning 0%
The answer is 3.