Identify the correct mathematical statements:
A. The expression 1/(5×6) + 1/(6×7) + 1/(7×8) + …… + 1/(24×25) is equal to 4/25.
B. Of the three numbers, second is twice the first and is also thrice of the third. If the average of three numbers is 55, then largest number is 90.
C. If (a + b):(b + c):(c + a) = 6:7:8 and a + b + c = 14, then value of c is 8.
D. (0.2×0.2 + 0.02×0.02 − 0.4×0.02) ÷ 0.18 is equal to 0.09.
Choose the correct answer from the options given below:सही गणितीय कथनों की पहचान कीजिए:
A. व्यंजक 1/(5×6) + 1/(6×7) + 1/(7×8) + …… + 1/(24×25) का मान 4/25 है।
B. तीन संख्याओं में से दूसरी संख्या पहली से दोगुनी है और तीसरी से तिगुनी भी है। यदि तीनों संख्याओं का औसत 55 है, तो सबसे बड़ी संख्या 90 है।
C. यदि (a + b):(b + c):(c + a) = 6:7:8 तथा a + b + c = 14, तो c का मान 8 है।
D. (0.2×0.2 + 0.02×0.02 − 0.4×0.02) ÷ 0.18 का मान 0.09 है।
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Correct answer: D. A and B onlyकेवल A और B
Explanation
A: the series telescopes, 1/5 − 1/6 + 1/6 − 1/7 + … + 1/24 − 1/25 = 1/5 − 1/25 = 4/25, so A is true. B: let the second = 6k, then first = 3k and third = 2k; sum = 11k = 3 × 55 = 165, k = 15, so the numbers are 45, 90, 30 and the largest is 90 — true. C: the three ratios add to 2(a+b+c) = 28 over 21 parts, so 1 part = 4/3 and a+b = 8, giving c = 14 − 8 = 6, not 8 — false. D: (0.04 + 0.0004 − 0.008) ÷ 0.18 = 0.0324 ÷ 0.18 = 0.18, not 0.09 — false. Hence A and B only.A: श्रेणी दूरबीनी (telescoping) है, 1/5 − 1/6 + 1/6 − 1/7 + … + 1/24 − 1/25 = 1/5 − 1/25 = 4/25, अतः A सही है। B: मान लीजिए दूसरी संख्या = 6k, तब पहली = 3k और तीसरी = 2k; योग = 11k = 3 × 55 = 165, k = 15, अतः संख्याएँ 45, 90, 30 हैं और सबसे बड़ी 90 है — सही। C: तीनों अनुपातों का योग 2(a+b+c) = 28 है जो 21 भागों में है, अतः 1 भाग = 4/3 और a+b = 8, जिससे c = 14 − 8 = 6 होता है, 8 नहीं — गलत। D: (0.04 + 0.0004 − 0.008) ÷ 0.18 = 0.0324 ÷ 0.18 = 0.18 है, 0.09 नहीं — गलत। अतः उत्तर केवल A और B है।More Mathematical Reasoning and Aptitude questions
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