Let the four numbers be \( a,\; b,\; c,\; d \).
Average of first three numbers = 16
\[ \frac{a + b + c}{3} = 16 \]
\[ a + b + c = 48 \]
Average of last three numbers = 15
\[ \frac{b + c + d}{3} = 15 \]
\[ b + c + d = 45 \]
Given \( d = 20 \)
\[ b + c + 20 = 45 \]
\[ b + c = 25 \]
From first equation:
\[ a + 25 = 48 \]
\[ a = 23 \]
Correct Answer: Option D
চারটি সংখ্যা ধরা যাক \( a,\; b,\; c,\; d \)
প্রথম তিনটির গড় = 16
\[ \frac{a + b + c}{3} = 16 \]
\[ a + b + c = 48 \]
শেষ তিনটির গড় = 15
\[ \frac{b + c + d}{3} = 15 \]
\[ b + c + d = 45 \]
প্রদত্ত \( d = 20 \)
\[ b + c + 20 = 45 \]
\[ b + c = 25 \]
প্রথম সমীকরণ থেকে:
\[ a + 25 = 48 \]
\[ a = 23 \]
সঠিক উত্তর: Option D
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