Miscellaneous Series — JEE Main Mathematics MCQs with Solutions
Free JEE Main Mathematics Miscellaneous Series MCQs with step-by-step solutions (20 questions). Part of Sequences and Series. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Miscellaneous Series · easy · numerical
The recurring decimal $0.\overline{3}$ (i.e. $0.333\dots$) expressed as a fraction is:
A. $\dfrac13$ ✓ Correct
B. $\dfrac{33}{100}$
C. $\dfrac{3}{10}$
D. $\dfrac{1}{30}$
Solution: As an infinite G.P. $\dfrac{3}{10} + \dfrac{3}{100} + \cdots = \dfrac{3/10}{1 - 1/10} = \dfrac13$.
Q2 — Miscellaneous Series · easy · numerical
The value of $0.\overline{6}$ as a fraction is:
A. $\dfrac35$
B. $\dfrac67$
C. $\dfrac23$ ✓ Correct
D. $\dfrac{6}{11}$
Solution: $\dfrac{6/10}{1 - 1/10} = \dfrac{6}{9} = \dfrac23$.
Q3 — Miscellaneous Series · easy · numerical
The value of $0.\overline{7}$ as a fraction is:
A. $\dfrac{70}{99}$
B. $\dfrac{7}{11}$
C. $\dfrac79$ ✓ Correct
D. $\dfrac{7}{10}$
Solution: A single repeating digit over $9$: $0.\overline{7} = \dfrac79$.
Q4 — Miscellaneous Series · medium · numerical
The recurring decimal $0.\overline{27}$ (i.e. $0.2727\dots$) equals:
A. $\dfrac{3}{11}$ ✓ Correct
B. $\dfrac{27}{90}$
C. $\dfrac{27}{100}$
D. $\dfrac{2}{7}$
Solution: $0.\overline{27} = \dfrac{27}{99} = \dfrac{3}{11}$.
Q5 — Miscellaneous Series · medium · numerical
The value of $0.\overline{18}$ as a fraction is:
A. $\dfrac{1}{6}$
B. $\dfrac{9}{50}$
C. $\dfrac{18}{100}$
D. $\dfrac{2}{11}$ ✓ Correct
Solution: $0.\overline{18} = \dfrac{18}{99} = \dfrac{2}{11}$.
Q6 — Miscellaneous Series · hard · numerical
The mixed recurring decimal $0.4\overline{5}$ (i.e. $0.4555\dots$) equals:
A. $\dfrac{45}{99}$
B. $\dfrac{5}{11}$
C. $\dfrac{9}{20}$
D. $\dfrac{41}{90}$ ✓ Correct
Solution: $0.4\overline{5} = \dfrac{45 - 4}{90} = \dfrac{41}{90}$.
Q7 — Miscellaneous Series · hard · numerical
The mixed recurring decimal $0.2\overline{3}$ (i.e. $0.2333\dots$) equals:
A. $\dfrac{21}{100}$
B. $\dfrac{23}{99}$
C. $\dfrac{7}{30}$ ✓ Correct
D. $\dfrac{1}{4}$
Solution: $0.2\overline{3} = \dfrac{23 - 2}{90} = \dfrac{21}{90} = \dfrac{7}{30}$.
Q8 — Miscellaneous Series · medium · numerical
The value of $1.\overline{6}$ as an improper fraction is:
A. $\dfrac{16}{9}$
B. $\dfrac{11}{6}$
C. $\dfrac53$ ✓ Correct
D. $\dfrac{8}{5}$
Solution: $1.\overline{6} = 1 + \dfrac69 = 1 + \dfrac23 = \dfrac53$.
Q9 — Miscellaneous Series · hard · numerical
The recurring decimal $0.\overline{123}$ (i.e. $0.123123\dots$) equals:
A. $\dfrac{41}{333}$ ✓ Correct
B. $\dfrac{123}{990}$
C. $\dfrac{1}{8}$
D. $\dfrac{123}{1000}$
Solution: $0.\overline{123} = \dfrac{123}{999} = \dfrac{41}{333}$.
Q10 — Miscellaneous Series · medium · numerical
The infinite G.P. $0.5 + 0.05 + 0.005 + \cdots$ sums to:
A. $\dfrac12$
B. $\dfrac59$ ✓ Correct
C. $\dfrac{5}{11}$
D. $0.55$
Solution: $\dfrac{0.5}{1 - 0.1} = \dfrac{0.5}{0.9} = \dfrac59$.
Q11 — Miscellaneous Series · medium · numerical
The value of $0.\overline{05}$ (i.e. $0.050505\dots$) is:
A. $\dfrac{1}{18}$
B. $\dfrac{5}{100}$
C. $\dfrac{1}{20}$
D. $\dfrac{5}{99}$ ✓ Correct
Solution: Two repeating digits over $99$: $\dfrac{5}{99}$.
Q12 — Miscellaneous Series · medium · numerical
The value of $2.\overline{3}$ as a fraction is:
A. $\dfrac{23}{9}$
B. $\dfrac73$ ✓ Correct
C. $\dfrac{21}{9}$
D. $\dfrac{20}{9}$
Solution: $2.\overline{3} = 2 + \dfrac39 = 2 + \dfrac13 = \dfrac73$.
Q13 — Miscellaneous Series · medium · numerical
The sum $0.7 + 0.07 + 0.007 + \cdots$ (to infinity) is:
A. $0.77$
B. $\dfrac79$ ✓ Correct
C. $\dfrac{7}{11}$
D. $\dfrac{7}{10}$
Solution: $\dfrac{0.7}{1 - 0.1} = \dfrac{0.7}{0.9} = \dfrac79$.
Q14 — Miscellaneous Series · hard · numerical
The mixed recurring decimal $0.1\overline{6}$ (i.e. $0.1666\dots$) equals:
A. $\dfrac{1}{60}$
B. $\dfrac16$ ✓ Correct
C. $\dfrac{5}{33}$
D. $\dfrac{16}{99}$
Solution: $0.1\overline{6} = \dfrac{16 - 1}{90} = \dfrac{15}{90} = \dfrac16$.
Q15 — Miscellaneous Series · medium · numerical
The value of $0.\overline{36}$ as a fraction in lowest terms is:
A. $\dfrac{4}{11}$ ✓ Correct
B. $\dfrac{36}{100}$
C. $\dfrac{9}{25}$
D. $\dfrac{18}{55}$
Solution: $0.\overline{36} = \dfrac{36}{99} = \dfrac{4}{11}$.
Q16 — Miscellaneous Series · medium · numerical
The infinite series $1 - \dfrac12 + \dfrac14 - \dfrac18 + \cdots$ sums to:
A. $2$
B. $\dfrac34$
C. $\dfrac12$
D. $\dfrac23$ ✓ Correct
Solution: G.P. with $a = 1, r = -\tfrac12$: $\dfrac{1}{1 + \tfrac12} = \dfrac23$.
Q17 — Miscellaneous Series · hard · numerical
A ball dropped from a height of $16$ m rebounds to half its previous height each time. The total distance it travels before coming to rest is:
A. $96$ m
B. $48$ m ✓ Correct
C. $32$ m
D. $64$ m
Solution: Total $= 16 + 2(8 + 4 + 2 + \cdots) = 16 + 2 \cdot \dfrac{8}{1 - \tfrac12} = 16 + 32 = 48$ m.
Q18 — Miscellaneous Series · medium · theory
The statement $0.\overline{9} = 1$ is:
A. false; it is slightly less than $1$
B. true only approximately
C. undefined
D. true (the infinite G.P. sums exactly to $1$) ✓ Correct
Solution: $0.\overline{9} = \dfrac{9/10}{1 - 1/10} = 1$ exactly.
Q19 — Miscellaneous Series · hard · numerical
The telescoping trigonometric-type sum $\sum_{r=1}^{\infty} \tan^{-1}\!\left(\dfrac{1}{1 + r + r^2}\right)$ equals:
A. $\dfrac{\pi}{3}$
B. $\dfrac{\pi}{2}$
C. $\pi$
D. $\dfrac{\pi}{4}$ ✓ Correct
Solution: Since $\tan^{-1}\dfrac{1}{1+r+r^2} = \tan^{-1}(r+1) - \tan^{-1}(r)$, the sum telescopes to $\dfrac{\pi}{2} - \dfrac{\pi}{4} = \dfrac{\pi}{4}$.
Q20 — Miscellaneous Series · hard · numerical
The value of $0.\overline{1} + 0.\overline{2}$ (i.e. $0.111\dots + 0.222\dots$) as a fraction is:
A. $\dfrac13$ ✓ Correct
B. $\dfrac{1}{9}$
C. $\dfrac{2}{9}$
D. $\dfrac{3}{10}$
Solution: $0.\overline{1} + 0.\overline{2} = \dfrac19 + \dfrac29 = \dfrac39 = \dfrac13$.