Find the smallest no, which leaves remainders 8 & 12 when divided by 28 & 32 respectively.

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200

204

208

210

answer is B.

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Detailed Solution

28 = 2 × 2 × 7
32 = 2 × 2 × 2 × 2 ×2
LCM = 224
Smallest no. which when divided by 28 and 32 leaves remainder 8 and 12 is 224 – (8 +12)
= 204
Exam type - CBSE
Subject- Mathematics
Chapter - Understanding Quadrilaterals
Topic- Kinds of Quadrilaterals
Subtopic- Kite

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