Word problems based on APs

Correct answer: C. 14

This forms an arithmetic progression with first term 12 and common difference 5. Hence, \(a_n=12+(n-1)\times5\). Putting \(a_n=77\), we get \(12+5(n-1)=77\), so \(5(n-1)=65\), \(n-1=13\), and \(n=14\). Therefore, 77 instruments will be placed on the 14th day. On the 13th day, there would be only 72 instruments. Exam tip: in AP word problems, identify the first term and common difference, then equate the target value to \(a_n\).

Correct answer: D. (10)

The governing concept is the nth-term formula for an arithmetic progression. Hourly production forms the sequence 150, 140, 130, and so on, so the first term is a = 150 and the common difference is d = -10 because production decreases by 10 each hour. For the nth hour, a_n = a + (n - 1)d. Setting the production equal to 60 gives 60 = 150 + (n - 1)(-10). Thus 60 = 150 - 10n + 10 = 160 - 10n, so 10n = 100 and n = 10. Therefore option D is correct. Options A, B and C represent earlier hours and give 90, 80 and 70 packets respectively, not 60.

Correct answer: C. 28

The correct answer is 28 seats, option C. Let the number of seats in the first row be a. The rows form an AP with n = 16, common difference d = 4, and total S_16 = 928. Apply S_n = n/2[2a + (n − 1)d]: 928 = 16/2[2a + 15 × 4] = 8(2a + 60). Dividing by 8 gives 116 = 2a + 60, so 2a = 56 and a = 28. Therefore the first row had 28 seats. Option A, 24, option B, 26, and option D, 30, would give totals different from 928. Check the result: the last row is 28 + 15 × 4 = 88 seats, and the average row size is (28 + 88)/2 = 58; 16 × 58 = 928. This confirms the answer and illustrates why the AP sum formula is suitable.

Correct answer: C. 10

This is an arithmetic-progression word problem involving the nth term. The first-floor count is the first term, a = 150, and the number of windows decreases by 10 on every successive floor, so the common difference is d = −10. For the nth floor, use aₙ = a + (n − 1)d. Setting the required count equal to 60 gives 60 = 150 + (n − 1)(−10). Thus −90 = −10(n − 1), so n − 1 = 9 and n = 10. Therefore, option C is correct. The sequence is 150, 140, 130, 120, 110, 100, 90, 80, 70, 60. Option A would give 80 windows, option B 70, and option D 50, so none of them satisfies the condition.

Correct answer: B. 9800 रुपये

The governing concept is the sum of the terms of an arithmetic progression. The daily incomes form 700, 850, 1000, and so on, so the first term is a = 700, the common difference is d = 150, and the number of terms is n = 8. Use Sₙ = n/2[2a + (n − 1)d]. Therefore, S₈ = 8/2[2(700) + 7(150)] = 4[1400 + 1050] = 4 × 2450 = 9800 rupees. Hence option B is correct. The question asks for the total of all eight days, not merely the income on day eight. The eighth-day income would be 700 + 7(150) = 1750 rupees, which is only one term. The other choices arise from incomplete or incorrect addition.

Correct answer: C. 10

This problem uses the nth-term formula for a decreasing arithmetic progression. The first term is a = 180 because the first floor has 180 windows. Since every higher floor has 15 fewer windows, the common difference is d = −15. For the nth floor, aₙ = a + (n − 1)d. Put aₙ = 45: 45 = 180 + (n − 1)(−15). Hence −135 = −15(n − 1), so n − 1 = 9 and n = 10. Thus option C is correct. A direct check gives 180 on floor 1, 165 on floor 2, and after nine decreases, 180 − 9×15 = 45 on floor 10. Floor 8 has 75, floor 9 has 60, and floor 11 has 30 windows, so the distractors do not meet the requirement.

Correct answer: C. 2184

This is an arithmetic-progression word problem because the number of seats increases by the same amount, 7, from one row to the next. The first term is a = 34, the common difference is d = 7, and the number of rows is n = 21. Since the question asks for the total, use S_n = n/2[2a + (n − 1)d]. Substitution gives S_21 = 21/2[2(34) + 20(7)] = 21/2[68 + 140] = 21/2 × 208 = 21 × 104 = 2184. Therefore, option C is correct. The distractors can result from using the nth-term formula instead of the sum formula, omitting a row, or making an arithmetic error. The constant increase confirms that AP methods are appropriate.

Correct answer: B. 3

The row sizes form an arithmetic progression because each row increases by the same amount. The first term is a = 25, the number of rows is n = 22, and the total is S_22 = 1243. Apply S_n = n/2[2a + (n − 1)d]: 1243 = 22/2[2(25) + 21d] = 11(50 + 21d). Dividing by 11 gives 113 = 50 + 21d. Hence 21d = 63 and d = 3. Therefore, option B is correct. This value can be checked directly: the last row is 25 + 21(3) = 88, and the average row size is (25 + 88)/2 = 56.5; multiplying by 22 gives 1243. The other options fail the given total.

Correct answer: C. 7

The governing concept is the nth-term formula of an arithmetic progression. The first row gives a_1 = 33 and the 18th row gives a_18 = 152. For an AP, a_n = a + (n − 1)d, where d is the equal increase. Substituting the known values gives 152 = 33 + (18 − 1)d = 33 + 17d. Therefore, 119 = 17d and d = 7. Hence option C is correct. A quick check confirms it: starting at 33 and adding 7 seventeen times gives 33 + 119 = 152. If the increase were 5, 6, or 8, the 18th-row counts would be 118, 135, or 169, respectively, so those distractors cannot satisfy the data.

Correct answer: A. 1530

The successive gaps are 18, 27, 36, and so on, so they form an arithmetic progression. The first term is a = 18, the common difference is d = 9, and the number of gaps is n = 17. Because the question asks for total distance, not just the last gap, use S_n = n/2[2a + (n − 1)d]. Thus S_17 = 17/2[2(18) + 16(9)] = 17/2[36 + 144] = 17/2 × 180 = 17 × 90 = 1530 metres. Therefore, option A is correct. The 17th gap alone would be 18 + 16×9 = 162 metres, so confusing a_n with S_n could lead to an incorrect distractor. The other values also do not satisfy the required sum.

Correct answer: D. 18

The seat counts form an arithmetic progression with first term a = 36 and common difference d = 6. If there are n rows, the total is S_n = n/2[2a + (n − 1)d]. Substituting the data gives 1566 = n/2[72 + 6(n − 1)] = n/2(6n + 66) = 3n(n + 11). Therefore, n(n + 11) = 522. Testing the positive whole-number choices, n = 18 gives 18(18 + 11) = 18 × 29 = 522, so the number of rows is 18. Option D is correct. Direct verification gives the last row as 36 + 17×6 = 138; the average of the first and last rows is 87, and 87×18 = 1566. The other choices do not produce the stated total.

Correct answer: C. 4

The governing concept is the sum of the terms of an arithmetic progression. The daily numbers of instruments form an AP with first term a = 16, number of days n = 25, total S_n = 1600, and common difference d. Use S_n = n/2[2a + (n − 1)d]. Thus 1600 = 25/2[2(16) + 24d] = 25/2(32 + 24d). Multiplying by 2 and dividing by 25 gives 128 = 32 + 24d. Hence 24d = 96 and d = 4. Therefore, the number placed increased by four instruments each day, so option C is correct. The other choices would produce different totals when substituted into the same sum formula. This is a sum-based AP problem, not merely an nth-term question.

Correct answer: C. 10

The numbers of windows form an arithmetic progression with first term a = 220 and common difference d = −20, because the number decreases by 20 on each successive floor. The nth-term formula is aₙ = a + (n − 1)d. Substituting the required number 40 gives 40 = 220 + (n − 1)(−20). Hence, −180 = −20(n − 1), so n − 1 = 9 and n = 10. Therefore, the tenth floor has 40 windows, making option C correct. Checking the sequence confirms this: floor 1 has 220 windows, floor 2 has 200, and after nine decreases of 20, floor 10 has 40. The other choices correspond to 60, 80 or 20 windows.

Correct answer: C. 12

The governing concept is the nth term of an arithmetic progression with a negative common difference. The first floor gives a = 260, and because each successive floor has 18 fewer windows, d = −18. For the nth floor, a_n = a + (n − 1)d. Setting the term equal to 62 gives 62 = 260 − 18(n − 1). Hence −198 = −18(n − 1), so n − 1 = 11 and n = 12. A direct check confirms this: from the first to the twelfth floor there are 11 decreases, and 260 − 11 × 18 = 260 − 198 = 62. Thus option C is correct. The other answers result from counting 12 decreases instead of 11 or mishandling the negative difference.