Match the List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| (For given Euclidean Algorithm, \(a=bq+r\), where \(a\) & \(b\) are real) | Values of \(q\) and \(r\) | ||
| A. | \(a=-112,\ b=-7\) | I. | \(q=-13,\ r=1\) |
| B. | \(a=118,\ b=-9\) | II. | \(q=14,\ r=3\) |
| C. | \(a=-109,\ b=6\) | III. | \(q=-19,\ r=5\) |
| D. | \(a=115,\ b=8\) | IV. | \(q=16,\ r=0\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → A-IV, B-I, C-III, D-II
In Euclid’s division, \(a=bq+r\), the remainder must satisfy \(0\le r\lt|b|\): it is never negative and always smaller than the divisor. (The values here are integers, as Euclid’s division needs.) So \(q\) is chosen to make \(bq\) the largest multiple of \(b\) that does not exceed \(a\), and then \(r=a-bq\).
(A) a = −112, b = −7 — this matches (IV) q = 16, r = 0.
\(-7\) goes into \(-112\) 16 times with nothing left over, so \(-112=(-7)(16)+0\)
(B) a = 118, b = −9 — this matches (I) q = −13, r = 1.
\((-9)\times(-13)=117\), the largest multiple of 9 not above 118
The remainder is \(118-117=1\): \(118=(-9)(-13)+1\)
(C) a = −109, b = 6 — this matches (III) q = −19, r = 5.
\(6\times(-19)=-114\); taking \(q=-18\) would give \(-108\), which is above \(-109\) and would leave a negative remainder
The remainder is \(-109-(-114)=5\): \(-109=6(-19)+5\)
(D) a = 115, b = 8 — this matches (II) q = 14, r = 3.
\(8\times14=112\), the largest multiple of 8 not above 115
The remainder is \(115-112=3\): \(115=8(14)+3\)