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Question
Match each of the entries in Column I with the appropriate entries in Column II.
Column I | Column II |
(i) x + 5 = 9 | (A) `- 5/3` |
(ii) x – 7 = 4 | (B) `5/3` |
(iii) `x/12` = – 5 | (C) 4 |
(iv) 5x = 30 | (D) 6 |
(v) The value of y which satisfies 3y = 5 | (E) 11 |
(vi) If p = 2, then the value of `1/3 (1 - 3p)` | (F) – 60 |
(G) 3 |
Solution
Column I | Column II |
(i) x + 5 = 9 | (C) 4 |
(ii) x – 7 = 4 | (E) 11 |
(iii) `x/12` = –5 | (F) –60 |
(iv) 5x = 30 | (D) 6 |
(v) The value of y which satisfies 3y = 5 | (B) `5/3` |
(vi) If p = 2, then the value of `1/3 (1 - 3p)` | (A) `- 5/3` |
Explanation:
(i) Given equation is x + 5 = 9
⇒ x = 9 – 5 .....[Transposing 5 to RHS]
⇒ x = 4
(ii) Given equation is x – 7 = 4
⇒ x = 4 + 7 .....[Transposing (–7) to RHS]
⇒ x = 11
(iii) Given equation is `x/12` = –5
⇒ `12 xx x/12 = - 5 xx 12` ......[Multiplying both sides by 12]
⇒ x = – 60
(iv) Given equation is 5x = 30
⇒ `(5x)/5 = 30/5` .....[Dividing both sides by 5]
⇒ x = 6
(v) Given equation is 3y = 5
⇒ `(3y)/3 = 5/3` .....[Dividing both sides by 3]
⇒ y = `5/3`
(vi) Given, p = 2
Put the value of p in the equation = `1/3 xx (1 - 3p)`, we get
= `1/3 (1 - 3 xx 2)`
= `1/3 xx (1 - 6)`
= `1/3 xx (-5)`
= `- 5/3`
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