{"id":42733,"date":"2023-09-07T18:08:05","date_gmt":"2023-09-07T18:08:05","guid":{"rendered":"https:\/\/mp.moonpreneur.com\/blog\/?p=42733"},"modified":"2026-01-13T11:53:35","modified_gmt":"2026-01-13T11:53:35","slug":"top15-hardest-sat-math-questions-1","status":"publish","type":"post","link":"https:\/\/mp.moonpreneur.com\/blog\/top15-hardest-sat-math-questions-1\/","title":{"rendered":"Top 15 Hardest SAT Math Questions"},"content":{"rendered":"\t\t
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Update:<\/strong>\u00a0This article was last updated on\u00a013th January 2026<\/strong>\u00a0to reflect the accuracy and up-to-date information on the page.<\/span><\/p>\t\t\t\t\t<\/div>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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Category: Calculator portion, <\/span>Multiple Choice<\/span><\/span><\/h4>

Question 10<\/span>: <\/span>The world\u2019s population has grown at an average rate of 1.9 percent per year since 1945. There were approximately 4 billion people in the world in 1975. Which of the following functions represents the world\u2019s population P, in billions of people, t years since 1975?<\/span><\/span><\/p>

A) <\/span>\"15<\/span><\/span><\/span><\/p>

B) <\/span>\"15<\/span><\/span><\/span><\/p>

C) <\/span>\"15<\/span><\/span><\/span><\/p>

D) <\/span>\"15<\/span><\/span><\/span><\/p>

Ans.<\/span> The world’s population has been growing consistently at an annual rate of 1.9% since 1945. This means it has been increasing by a constant factor of 1.019 each year since 1945.\u00a0<\/span>Choice A is the correct answer. This is because the world’s population has been consistently growing at an average annual rate of 1.9% since 1945. This means it has been increasing by a constant factor of 1.019 each year since 1945.<\/span>To estimate the world’s population in different years, we can use this constant growth factor. For example, if the world’s population was approximately 4 billion in 1975, then in 1976, it would have been about 4 multiplied by 1.019. In 1977, it would be 4 multiplied by 1.019 squared (1.019^2), and so on.<\/span><\/span><\/p>

Therefore, we can represent the world’s population, denoted as P(t), t years since 1975 with the function P(t) = 4(1.019)^t.<\/span><\/p>

Choices C and D are incorrect because they propose linear models, which do not account for the constant exponential growth observed in the world’s population.<\/span><\/p>

Choice B is also incorrect because it suggests a 90% increase in population each year, which does not align with the consistent 1.9% growth rate.<\/span><\/p>

Recommended reading:<\/strong> 20 Hardest SAT Math Questions You Must Try<\/a><\/p>

Question 11<\/span>: A rectangle was altered by increasing its length by 10 percent and decreasing its width by p percent. If these alterations decreased the area of the rectangle by 12 percent, what is the value of p?<\/span><\/div><\/div>

A) 12<\/span><\/p>

B) 15<\/span><\/p>

C) 20<\/span><\/p>

D) 22<\/span><\/p>

Ans. The correct option is C)<\/span><\/p>

To find the area of a rectangle, we use the formula: Area = Length \u00d7 Breadth.<\/span><\/p>

Let A1 = L1 \u00d7 B1, where L1 and B1 are the length and breadth of the original rectangle.<\/span><\/p>

Given that L2 = 10% + L1 = 1.1L1, and B2 = B1 – (100\/100)B1 = (100\/100 – p)B1, we have:<\/span><\/p>

A2 = 0.88A1.<\/span><\/p>

This equation can be expanded as:<\/span><\/p>

L2 \u00d7 B2 = 0.88(L1 \u00d7 B1),<\/span><\/p>

1.1L1 \u00d7 ((100\/100 – p)B1) = 0.88(L1 \u00d7 B1).<\/span><\/p>

Now, simplify further:<\/span><\/p>

1.1 \u00d7 (100B1 – pB1) = 0.88(L1 \u00d7 B1),<\/span><\/p>

110B1 – 1.1pB1 = 0.88(L1 \u00d7 B1).<\/span><\/p>

Rearrange the terms:<\/span><\/p>

110B1 – 1.1pB1 = 0.88(L1 \u00d7 B1),<\/span><\/p>

22B1 = 1.1p.<\/span><\/p>

Now, solve for p:<\/span><\/p>

22B1 = 1.1p,<\/span><\/p>

p = 20.<\/span><\/p>

So, the value of p is indeed 20.<\/span><\/p>

Choice A is not the correct answer and could be due to a confusion between the 12 percent decrease in area and the percent decrease in width.<\/span><\/p>

Choice B is also incorrect because reducing the width by 15 percent leads to a 6.5 percent decrease in area, not the 12 percent decrease observed.<\/span><\/p>

Choice D is not the right answer and might be a result of adding the percentages provided in the question (10 + 12).<\/span><\/p>

Recommended reading: <\/strong>10 World\u2019s Hardest Math Problems With Solutions and Examples That Will Blow Your Mind<\/a><\/p>

Question 12<\/span>: A square field measures 10 meters by 10 meters. Ten students each mark off a randomly selected region of the field; each region is square and has side lengths of 1 meter, and no two regions overlap. The students count the earthworms contained in the soil to a depth of 5 centimeters beneath the ground\u2019s surface in each region. The results are shown in the table below.<\/span><\/span><\/p>

\"Hardest<\/a><\/span><\/p>

Which of the following is a reasonable approximation of the number of earthworms to a depth of 5 centimeters beneath the ground\u2019s surface in the entire field?<\/span><\/p>

A) 150<\/span><\/p>

B) 1,500<\/span><\/p>

C) 15,000<\/span><\/p>

D) 150,000<\/span><\/p>

Ans. <\/span>We have a square field with a side length of 10 meters. Within this field, there is a smaller square region with a side length of 1 meter.<\/span><\/span><\/p>

To estimate the total number of earthworms in the entire field, we consider that there are 100 such smaller square regions (10×10 grid) within the larger square field.<\/span><\/p>

Since we have the analysis data available for 10 of these smaller regions, we can use this information to approximate the total number of earthworms in the entire field.<\/span><\/p>

To do this, we multiply the total number found in these ten regions by 10 to estimate the total number of earthworms in the field.<\/span><\/p>

We then sum up the numbers obtained from these ten regions:<\/span>
<\/span>107 + 147 + 146 + 135 + 149 + 141 + 150 + 154 + 176 + 166 = 1471<\/span><\/span><\/p>

Finally, we multiply the sum by 10 to get an approximation of the total number of earthworms:<\/span>
<\/span>1471 \u00d7 10 = 14710, which can be approximated to about 15000.<\/span><\/span><\/p>

Recommended reading: <\/strong>How Many Questions Are on the SAT? 2025 Update<\/a><\/p>

Question 13<\/span>: Of the following four types of savings account plans, which option would yield exponential growth of the money in the account?<\/span><\/span><\/p>

A) Each successive year, 2% of the initial savings is added to the value of the account.<\/span><\/p>

B) Each successive year, 1.5% of the initial savings and $100 is added to the value of the account.<\/span><\/p>

C) Each successive year, 1% of the current value is added to the value of the account.<\/span><\/p>

D) Each successive year, $100 is added to the value of the account.<\/span><\/p>

Ans. Choice C is the correct answer. This is because it describes exponential growth.<\/span><\/p>

Exponential growth is characterized by an increase of a quantity at a relative rate, meaning it increases by the same factor over equal increments of time. In this case, the value of the account increases by 1% each year, which means it is multiplied by the same factor, 1.01, each year. This is a clear indication of exponential growth.<\/span><\/p>

On the other hand: <\/span>Choices A and B are incorrect because they describe linear growth, where the rate depends only on the initial value, resulting in the value increasing by the same amount each year. <\/span>Choice D is also incorrect as it represents linear growth, where the increase is constant each year.<\/span><\/span><\/p>

Recommended reading: <\/strong>An overview of the SAT Math Test Sections<\/a><\/p>

Question 14<\/span>: The 22 students in a health class conducted an experiment in which they each recorded their pulse rates, in beats per minute, before and after completing a light exercise routine. The dot plots below display the results.<\/span><\/span><\/p>

\"Hardest<\/a><\/span><\/p>

Let s1 and r1 be the standard deviation and range, respectively, of the data before exercise, and let s2 and r2 be the standard deviation and range, respectively, of the data after exercise. Which of the following is true?<\/span><\/p>

  1. s1 = s2 and r1 = r2<\/span><\/p><\/li>

  2. s1 < s2 and r1 < r2<\/span><\/p><\/li>

  3. s1 > s2 and r1 > r2<\/span><\/p><\/li>

  4. s1 \u2260 s2 and r1 = r2<\/span><\/p><\/li><\/ol>

    Ans. Choice D is the correct answer. This is because the two data sets have the same range, which is 32.<\/span><\/p>

    Here’s an explanation of why Choice D is correct:<\/span><\/p>