12 ...4)⇒ e + f = 15 ...5)⇒ d + e = 20 ...6)By USING (1) + (2) + (3) – (4) – (5) – (6)⇒ a + b + c + d + f + g = 60 + 50 + 48 – 12 – 15 - 20 = 111Total number of students in school = a + b + c + d + f + g + e = 111 + e.Maximum number of students will depend on the maximum value of e.From the EQUATIONS (4), (5) and (6) we have the following three cases:Case 1: Let b = 0⇒ e = 12.By substituting the value of e in (5), we get⇒ f = 15 – e = 3By substituting the value of e in (5), we get⇒ d = 20 – e = 8Case 2: Let f = 0⇒ e = 15.By substituting the value of e in (4), we get⇒ b = 12 – e = -3We got b as negative, so we need to discard this case ∵ b ≥ 0.Case 3: Let d = 0⇒ e = 20.By substituting the value of e in (4), we get⇒ b = 12 – e = -8We got b as negative, so we need to discard this case ∵ b ≥ 0.So the maximum possible value of e is from case 1 i.e. e = 12.∴ Maximum possible number of students in school = 111 + e = 123.

"> 12 ...4)⇒ e + f = 15 ...5)⇒ d + e = 20 ...6)By USING (1) + (2) + (3) – (4) – (5) – (6)⇒ a + b + c + d + f + g = 60 + 50 + 48 – 12 – 15 - 20 = 111Total number of students in school = a + b + c + d + f + g + e = 111 + e.Maximum number of students will depend on the maximum value of e.From the EQUATIONS (4), (5) and (6) we have the following three cases:Case 1: Let b = 0⇒ e = 12.By substituting the value of e in (5), we get⇒ f = 15 – e = 3By substituting the value of e in (5), we get⇒ d = 20 – e = 8Case 2: Let f = 0⇒ e = 15.By substituting the value of e in (4), we get⇒ b = 12 – e = -3We got b as negative, so we need to discard this case ∵ b ≥ 0.Case 3: Let d = 0⇒ e = 20.By substituting the value of e in (4), we get⇒ b = 12 – e = -8We got b as negative, so we need to discard this case ∵ b ≥ 0.So the maximum possible value of e is from case 1 i.e. e = 12.∴ Maximum possible number of students in school = 111 + e = 123.

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What can be the maximum number of students in the school?

Current Affairs General Awareness in Current Affairs 9 months ago

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Calculation:⇒ a + b + d + e = 60 ...1)⇒ d + e + f + g = 50 ...2)⇒ b + c + e + f = 48 ...3)⇒ b + e = 12 ...4)⇒ e + f = 15 ...5)⇒ d + e = 20 ...6)By USING (1) + (2) + (3) – (4) – (5) – (6)⇒ a + b + c + d + f + g = 60 + 50 + 48 – 12 – 15 - 20 = 111Total number of students in school = a + b + c + d + f + g + e = 111 + e.Maximum number of students will depend on the maximum value of e.From the EQUATIONS (4), (5) and (6) we have the following three cases:Case 1: Let b = 0⇒ e = 12.By substituting the value of e in (5), we get⇒ f = 15 – e = 3By substituting the value of e in (5), we get⇒ d = 20 – e = 8Case 2: Let f = 0⇒ e = 15.By substituting the value of e in (4), we get⇒ b = 12 – e = -3We got b as negative, so we need to discard this case ∵ b ≥ 0.Case 3: Let d = 0⇒ e = 20.By substituting the value of e in (4), we get⇒ b = 12 – e = -8We got b as negative, so we need to discard this case ∵ b ≥ 0.So the maximum possible value of e is from case 1 i.e. e = 12.∴ Maximum possible number of students in school = 111 + e = 123.

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Posted on 18 Nov 2024, this text provides information on Current Affairs related to General Awareness in Current Affairs. Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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