Word problems based on APs

Correct answer: D. (480000)

The profits form the AP (60000,68000,76000,\ldots) and (S_6=480000). Exam tip: treat yearly profits as separate terms.

Correct answer: C. (870)

The tile numbers are (9,16,23,\ldots) and (S_{15}=870). Exam tip: find the total of a pattern using the sum formula.

Correct answer: C. (1620)

The sales form the AP (120,135,150,\ldots) and (S_9=1620). Exam tip: treat daily sales as AP terms.

Correct answer: B. (50)

Using (S_{12}=4500) and (a=100) gives (d=50). Exam tip: find the common difference from total and first term.

Correct answer: C. 14

The amount filled per minute forms an arithmetic progression with first term 20 and common difference 9. Therefore, the amount filled in the nth minute is \(a_n=20+(n-1)\times 9\). Using \(20+(n-1)\times 9=137\), we get \((n-1)\times 9=117\), so \(n-1=13\) and \(n=14\). Hence, 137 litres are filled in the 14th minute. In the 13th minute, the amount would be only 128 litres. Exam tip: equate the target amount to the nth term \(a_n\) of the AP and solve for n.

Correct answer: B. (1650)

The patient numbers are (35,40,45,\ldots) and (S_{20}=1650). Exam tip: use the sum formula for total patients.

Correct answer: C. (810)

The sack numbers are (45,53,61,\ldots) and (S_{10}=810). Exam tip: treat each row as one term of the AP.

Correct answer: C. (860)

The video numbers are (5,9,13,\ldots) and (S_{20}=860). Exam tip: take the number of weeks as the number of terms.

Correct answer: C. (456)

The lane numbers are (6,9,12,\ldots) and (S_{16}=456). Exam tip: treat a sequence with equal increase as an AP.

Correct answer: C. (80)

The number of drawings is (14,20,26,\ldots) and (a_{12}=80). Exam tip: work done on the (n)th day is the (n)th term.

Correct answer: C. (666)

The marks form the AP (3,7,11,\ldots) and (S_{18}=666). Exam tip: take the number of questions as (n) and find the sum.

Correct answer: C. (864)

The correct answer is C. The gates form an arithmetic progression: first term \\(a=48\\), common difference \\(d=12\\), and number of terms \\(n=9\\). The total is \\(S_n=\frac{n}{2}[2a+(n-1)d]\\). Substituting gives \\(S_9=\frac{9}{2}[2(48)+8(12)] =\frac{9}{2}(96+96)=\frac{9}{2}\times192=864\\). Therefore option C is correct. Option A, 816, is too small and results from an incorrect total or missing increase. Option B, 840, is also not the sum of the nine terms. Option D, 888, is too large and usually comes from adding an extra increase. The terms are 48, 60, 72, 84, 96, 108, 120, 132 and 144; their sum confirms 864. Memory cue: identify first term, difference and number of terms before using the AP sum formula.

Correct answer: C. (12)

The instrument numbers are (20,25,30,\ldots) and (S_n=570) gives (n=12). Exam tip: when a total target is given, find the number of terms.

Correct answer: C. (690)

The distances form the AP (18,22,26,\ldots) and (S_{15}=690). Exam tip: treat each hour's distance as a term.

Correct answer: B. (1500)

The weekly sales are (70,80,90,\ldots) and (S_{12}=1500). Exam tip: use the sum formula to add sales across weeks.

Correct answer: C. (9500)

The direct answer is C: ₹9,500. The daily incomes form an arithmetic progression because the first income is ₹500 and the same amount, ₹100, is added each day. Thus, the first term is \\(a=500\\), the common difference is \\(d=100\\), and the number of terms is \\(n=10\\). The tenth-day income is \\(a_{10}=a+(n-1)d=500+9(100)=1400\\). The total can be found by \\(S_n=\\frac{n}{2}[2a+(n-1)d]\\): \\(S_{10}=\\frac{10}{2}[2(500)+9(100)]=5(1900)=9500\\). Option A, ₹9,000, is wrong because it does not add the ten correct terms. Option B, ₹9,250, is wrong because the arithmetic calculation is incomplete or inaccurate. Option C, ₹9,500, is correct. Option D, ₹9,750, is also too high and does not follow the given first term and increase. A quick check is to list the terms: 500, 600, 700, 800, 900, 1000, 1100, 1200, 1300 and 1400; their sum is ₹9,500. Memory cue: identify first term, common difference and number of terms before using the AP sum formula.

Correct answer: C. (871)

The page numbers are (25,32,39,\ldots) and (S_{13}=871). Exam tip: use the sum formula for total printing.

Correct answer: C. (1200)

The cloth sales form (80,100,120,\ldots) and (S_8=1200). Exam tip: use the sum formula for sales across all days.

Correct answer: C. (1020)

The member numbers are (12,18,24,\ldots) and (S_{17}=1020). Exam tip: add all days for total members.

Correct answer: B. (4)

Using (S_{15}=720) and (a=20) gives (d=4). Exam tip: find the common difference from total and first term.

Correct answer: C. (665)

The plate numbers are (15,20,25,\ldots) and (S_{14}=665). Exam tip: treat each table as one term of the AP.

Correct answer: C. (730)

The doses form the AP (8,11,14,\ldots) and (S_{20}=730). Exam tip: add the terms for the total dose.

Correct answer: C. (1225)

The water amounts are (100,125,150,\ldots) and (S_7=1225). Exam tip: treat hourly filled water as AP terms.

Correct answer: C. 15

Here, the first term is 90 and the common difference is 10. Therefore, the number of newspapers sold on the nth day is
\(a_n=90+(n-1)\times10\). Putting
\(90+(n-1)\times10=230\) gives
\(n-1=14\), so
\(n=15\). Day 14 is a close distractor, but only 220 newspapers would be sold that day. Exam tip: equate the target value to the nth-term formula of the AP and solve for n.

Correct answer: C. (22)

Putting (S_n=1859) in the sum formula gives (n=22). Exam tip: when total seats are given, form an equation for (n).