In a machine, the load is (980) units at the first stage and decreases by (45) units at each next stage. At which stage will the load be (305) units?
Answer and explanation
Correct answer: (16)th
The loads form an arithmetic progression because the same amount, 45 units, is subtracted at every successive stage. Here the first term is a = 980 and the common difference is d = −45. For the nth term, use a_n = a + (n − 1)d. Substituting the required load gives 305 = 980 + (n − 1)(−45). Thus 675 = 45(n − 1), so n − 1 = 15 and n = 16. Therefore, the load is 305 units at the 16th stage, making option C correct. Options A, B and D result from an incorrect count of the number of decreases or from using the wrong sign for the common difference.
Frequently asked questions
What is the correct answer to this question?
(16)th
Why is this the correct answer?
The loads form an arithmetic progression because the same amount, 45 units, is subtracted at every successive stage. Here the first term is a = 980 and the common difference is d = −45. For the nth term, use a_n = a + (n − 1)d. Substituting the required load gives 305 = 980 + (n − 1)(−45). Thus 675 = 45(n − 1), so n − 1 = 15 and n = 16. Therefore, the load is 305 units at the 16th stage, making option C correct. Options A, B and D result from an incorrect count of the number of decreases or from using the wrong sign for the common difference.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Arithmetic Progressions (AP). Topic: Word problems based on APs.
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