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PSAT Club

PSAT Club. Math – Grid-Ins. General Hints. Here are some general hints for answering Student-Produced Response questions, also called Grid-Ins. Since answer choices aren't given, a calculator may be helpful in avoiding careless mistakes on these questions.

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PSAT Club

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  1. PSAT Club Math – Grid-Ins

  2. General Hints • Here are some general hints for answering Student-Produced Response questions, also called Grid-Ins. • Since answer choices aren't given, a calculator may be helpful in avoiding careless mistakes on these questions. • It's suggested that you write your answer in the boxes above the grid to avoid errors in gridding. • The grid can hold only four places and can accommodate only positive numbers and zero. • Do not worry about which column to begin gridding the answer. As long as the answer is gridded completely, you will receive credit. • Unless a problem indicates otherwise, an answer can be entered on the grid either as a decimal or as a fraction. • You don't have to reduce fractions like 3/24 to their lowest terms. • Convert all mixed numbers to improper fractions before gridding the answer. • If the answer is a repeating decimal, you must grid the most accurate value the grid will accommodate. • Some questions may have more than one right answer. • You don't lose any points for a wrong answer. • Know the gridding rules before taking the test.

  3. Directions • Directions: • Grid-ins (student-produced response questions) require you to solve the problem and enter your answer by marking the ovals in the special grid.

  4. Practice Question 1 • If x is a positive integer, what is one possible value of the units digit of 1032x after it has been multiplied out?

  5. Answer 1 • If x is a positive integer, what is one possible value of the units digit of 1032x after it has been multiplied out? • Explanation:For an integer, the units digit is the digit furthest to the right. It is sometimes referred to as the ones digit. For example, in a three-digit integer such as 125, the digit 5 is the units digit. In this problem, 1032x can be rewritten as (1032)x. The number 1032 has a units digit of 9. When a number with a units digit of 9 is raised to the x power, the resulting number will have a units digit of 1 (since 92 = 81) or a units digit of 9 (since 93 = 729). For all positive integer values of x, the units digit of 1032)x will be either 1 or 9.

  6. Practice Question 2 • If XYZ above is equilateral, what is the value of r + s + t + u ?

  7. Answer 2 • In the figure above, AD is a diameter of the circle with center O and AO = 5. What is the length of arc BCD ? • (A) • (B) • (C) • (D)CORRECT ANSWER • (E) • Explanation: • To solve this problem, it is helpful to draw segment OB in the figure. Since OB and OD are both radii of the circle, they both equal 5. Therefore, the angles opposite these congruent sides of BOD are congruent and OBD = 36°. The third angle of the triangle, BOD, equals 180°- 36°- 36° = 108°. Arc BCD is a fraction of the circumference of the circle and more specifically equals , which equals The correct answer is choice (D).

  8. Answer 2 • If XYZ above is equilateral, what is the value of r + s + t + u ?

  9. Practice Question 2 • If XYZ above is equilateral, what is the value of r + s + t + u ? • Correct Answer: 300 • Explanation:Since XYZ is equilateral, r = u = 60. The sum of s + t equals 180 since the two angles form a straight line. Therefore, r + s + t + u = 60 + 60 + 180 = 300.

  10. Practice Question 3 • How many different three-digit numbers between 100 and 1,000 have 5 as the tens digit?

  11. Answer 3 • How many different three-digit numbers between 100 and 1,000 have 5 as the tens digit? • Correct Answer: 90 • Explanation:In the 100's, the numbers with 5 as the tens digit are 150, 151, 152, ..., 159 (ten numbers). Also, there will be 10 such numbers in the 200's, in the 300's, etc. Therefore, there are 10 x 9 = 90 three-digit numbers between 100 and 1,000 that have 5 as the tens digit.

  12. Practice Question 4 • The measures of the lengths of the three sides of a triangle are prime numbers. If two of the sides are 5 and 23, what is one possible value for the length of the third side?

  13. Answer 4 • The measures of the lengths of the three sides of a triangle are prime numbers. If two of the sides are 5 and 23, what is one possible value for the length of the third side? • Correct Answer: 19 or 23 • Explanation:Since two of the sides of the triangle are 5 and 23, let y  represent the third side. By the triangle inequality • 5 + y  > 23  or  y  > 18 • 5 + 23 > y   or  y  < 28 • 23 + y  > 5  or  y  > -18 • The lengths of the three sides of the triangle are prime numbers so that y  must be a prime number between 18 and 28. Thus, the two possible values for the length of the third side of the triangle are 19 and 23. Either answer may be gridded as the correct answer to the problem.

  14. Practice Question 5 • The sum of the digits is 6. • Each digit is different. • The number is odd. • What is the greatest 4-digit number that has all of the characteristics listed above?

  15. Answer 5 • The sum of the digits is 6. • Each digit is different. • The number is odd. • What is the greatest 4-digit number that has all of the characteristics listed above? • Correct Answer: 3,201 • Explanation:Since the sum of the digits of the four-digit number is 6, none of the digits can be greater than 6. The greatest four-digit number whose digits sum to 6 is 6,000. However, each digit must be different and the number must be odd. The greatest four-digit number having the given characteristics will have the largest digit in the thousands place. To maximize the number in the thousands place, let the units digit be 1. The thousands place cannot be 5 since 5,001 does not have four different digits. The thousands place cannot be 4 since 4,101 still does not have four different digits. If the thousands place is 3, then the number could be 3,201 and this is the greatest four-digit number that satisfies all three given conditions.

  16. Practice Question 6 • The sum of r and p is equal to twice s, and p is 36 less than twice the sum of r and s. What is the value of r ?

  17. Answer 6 • The sum of r and p is equal to twice s, and p is 36 less than twice the sum of r and s. What is the value of r ? • Correct Answer: 12 • Explanation:The phrase "the sum of r and p is equal to twice s" can be written as r + p = 2s. The phrase "p is 36 less than twice the sum of r and s" can be written as p = 2(r + s) - 36. From the first equation, p = 2s - r. Substituting this expression for p into the second equation yields • 2s - r = 2(r + s) - 362s - r = 2r + 2s - 36-r =2r - 36-3r =-36r =12 The value of r is 12.

  18. Practice Question 7 • RESULTS OF AN ELECTION • Candidate Number of Votes • X 7,400 • Y 2,375 • Z 5,250 • The results for three candidates in an election are shown in the table above. If each voter voted for exactly one candidate, what is the fewest number of voters who would have had to vote differently in order for Candidate Z to have received more votes than Candidate X?

  19. Answer 7

  20. Practice Question 7 • RESULTS OF AN ELECTION • Candidate Number of Votes • X 7,400 • Y 2,375 • Z 5,250 • The results for three candidates in an election are shown in the table above. If each voter voted for exactly one candidate, what is the fewest number of voters who would have had to vote differently in order for Candidate Z to have received more votes than Candidate X?

  21. Practice Question 7 • RESULTS OF AN ELECTION • Candidate Number of Votes • X 7,400 • Y 2,375 • Z 5,250 • The results for three candidates in an election are shown in the table above. If each voter voted for exactly one candidate, what is the fewest number of voters who would have had to vote differently in order for Candidate Z to have received more votes than Candidate X? • Correct Answer: 1076 • Explanation:For this question, let k be the number of voters who changed their vote. Since you want to make k as small as possible, the k voters should come from those who voted for Candidate X. To determine an answer to the problem, you would need to solve the inequality 5,250 + k > 7,400 - k. Solving this inequality yields 2k > 2150 or k > 1075. Therefore, 1,076 voters who had voted for Candidate X would have to change their vote and vote for Candidate Z in order for Candidate Z to receive more votes than Candidate X. The correct answer to this question is 1076.

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