Write four more rational numbers in each of the following patterns:
(i) \(\frac{-3}{5},\frac{-6}{10},\frac{-9}{15},\frac{-12}{20}\)...
(ii) \(\frac{-1}{4},\frac{-2}{8},\frac{-3}{12},\)...
(iii) \(\frac{-1}{6},\frac{2}{-12},\frac{3}{-18},\frac{4}{-24}\),....
(iv)\(\frac{-2}{3},\frac{2}{-3},\frac{4}{-6},\frac{6}{-9}\),...
(i) \(\frac{-3}{5},\frac{-6}{10},\frac{-9}{15},\frac{-12}{20}\)...
(ii) \(\frac{-1}{4},\frac{-2}{8},\frac{-3}{12},\)...
(iii) \(\frac{-1}{6},\frac{2}{-12},\frac{3}{-18},\frac{4}{-24}\),....
(iv)\(\frac{-2}{3},\frac{2}{-3},\frac{4}{-6},\frac{6}{-9}\),...
Solution and Explanation
(i)\(\frac{-3}{5},\frac{-6}{10},\frac{-9}{15},\frac{-12}{20}\),...
= \(\frac{-3}{5},\frac{-3\times2}{5\times2},\frac{-3\times3}{5\times3},\frac{-3\times4}{5\times4}\),...
It can be observed that the numerator is a multiple of 3 while the denominator is a multiple of 5 and as we increase them further, these multiples are increasing. Therefore, the next four rational numbers in this pattern are
= \(\frac{-3\times5}{5\times5},\frac{-3\times6}{5\times6},\frac{-3\times7}{5\times7},\frac{-3\times8}{5\times8}\)...
= \(\frac{-15}{25},\frac{-18}{30},\frac{-21}{35},\frac{-24}{40}\)...
(ii) \(\frac{-1}{4},\frac{-2}{8},\frac{-3}{12},\)...
= \(\frac{-1}{4},\frac{-1\times2}{4\times2},\frac{-1\times3}{4\times3}.\)..
The next four rational numbers in this pattern are.
=\(\frac{-1\times4}{4\times4},\frac{-1\times5}{4\times5},\frac{-1\times6}{4\times6},\frac{-1\times7}{4\times7}.\)...
\(= \frac{-4}{16},\frac{-5}{20},\frac{-6}{24},\frac{-7}{28}...\)
(iii)\(\frac{-1}{6},\frac{2}{-12},\frac{3}{-18},\frac{4}{-24},....\)
\(= \frac{-1}{6},\frac{1\times2}{-6\times2},\frac{1\times3}{-6\times3},\frac{1\times4}{-6\times4}...\)
The next four rational numbers in this pattern are
\(\frac{1\times5}{-6\times5},\frac{1\times6}{-6\times6},\frac{1\times7}{-6\times7},\frac{1\times8}{-6\times8}....\)
\(\frac{5}{-30},\frac{6}{-36},\frac{7}{-42},\frac{8}{-48}...\)
(iv) \(\frac{-2}{3},\frac{2}{-3},\frac{4}{-6},\frac{6}{-9},...\)
\(\frac{-2}{3},\frac{2}{-3},\frac{2\times2}{-3\times2},\frac{2\times3}{-3\times3}...\)
The next four rational numbers in this pattern are
\(\frac{2\times4}{-3\times4},\frac{2\times5}{-3\times5},\frac{2\times6}{-3\times6},\frac{2\times7}{-3\times7}...\)
\(\frac{8}{-12},\frac{10}{-15},\frac{12}{-18},\frac{14}{-21}....\)
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(i)\(\frac{7}{24}\)-\(\frac{17}{36}\)
(ii) \(\frac{5}{63}\)-(\(-\frac{6}{21}\))
(iii) -\(\frac{6}{13}\)-(-\(\frac{7}{15}\))
(iv) -\(\frac{3}{8}\)-\(\frac{7}{11}\)
(v)\(-\frac{21}{9}\)-6- CBSE Class VII
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(ii) \(\frac{5}{3}\)+\(\frac{3}{5}\)
(iii) -\(\frac{9}{10}\)+\(\frac{22}{15}\)
(iv) \(\frac{-3}{-11}\)+\(\frac{5}{9}\)
(v)\(-\frac{8}{19}\)+\(\frac{(-2)}{57}\)
(vi) \(-\frac{2}{3}\)+0
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- Some pictures (a) to (c) are given below. Tell which of them show:
\((a)\)
\((b)\)
\((c)\)
(i) 3 × \(\frac{1}{5}\) =\(\frac{3}{5}\)
(ii) 2 × \(\frac{1}{3}\) =\(\frac{2}{3}\)
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(ii) Determine the integer whose product with (–1) is
(a) –22
(b) 37
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- In a class test (+3) marks are given for every correct answer and (–2) marks are given for every incorrect answer and no marks for not attempting any question. (i) Radhika scored 20 marks. If she has got 12 correct answers, how many questions has she attempted incorrectly? (ii) Mohini scores –5 marks in this test, though she has got 7 correct answers. How many questions has she attempted incorrectly?
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(c) (–21) × (–30)
(d) (–316) × (–1)
(e) (–15) × 0 × (–18)
(f) (–12) × (–11) × (10)
(g) 9 × (–3) × (– 6)
(h) (–18) × (–5) × (– 4)
(i) (–1) × (–2) × (–3) × 4
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