Find the smallest number which when divided by $20, 25, 35$ and $40$ leaves remainders $14, 19, 29$ and $34$ respectively.
Step-by-Step Solution
Key Concept: Notice $20-14=6, 25-19=6, 35-29=6, 40-34=6$. Required number is $\text{LCM}(20, 25, 35, 40) - 6$.
Common difference $= 6$. Prime factorisations:
$20 = 2^2 \times 5$, $25 = 5^2$, $35 = 5 \times 7$, $40 = 2^3 \times 5$. [1.0 Mark]
$\text{LCM}(20, 25, 35, 40) = 2^3 \times 5^2 \times 7 = 8 \times 25 \times 7 = 1400$. [1.0 Mark]
Required number $= 1400 - 6 = 1394$. [1.0 Mark]
---
🎯 Official CBSE Marking Scheme:
Common difference calculation $= 6$: 1.0 Mark
Evaluating $\text{LCM} = 1400$: 1.0 Mark
Subtracting 6 to get 1394: 1.0 Mark
Correct Answer: