Sunita cuts a sheet of paper into three pieces. Length of first piece is equal to the average of the three single digit odd prime numbers. Length of the second piece is equal to that of the first plus one-third the length of the third. The third piece is as long as the other two pieces together. What is the length of the original sheet of paper?
The correct answer is (B) 15 units.
Explanation:
Let the lengths of the three pieces of paper be P1, P2, and P3. Step 1: Single digit odd prime numbers are 3, 5, and 7. Their average is (3 + 5 + 7) / 3 = 15 / 3 = 5 units. Thus, the length of the first piece P1 = 5. Step 2: The length of the second piece P2 is equal to the first plus one-third the length of the third: P2 = P1 + (1/3)P3 = 5 + (1/3)P3. Step 3: The third piece is as long as the other two pieces together: P3 = P1 + P2 = 5 + P2. Now substitute P2 = 5 + (1/3)P3 into the equation for P3: P3 = 5 + 5 + (1/3)P3 = 10 + (1/3)P3. Subtracting (1/3)P3 from both sides: (2/3)P3 = 10, which gives P3 = 10 x (3/2) = 15 units. Now find P2: P2 = 5 + (1/3)(15) = 5 + 5 = 10 units. Total length of the original sheet of paper = P1 + P2 + P3 = 5 + 10 + 15 = 30 units? Wait, let's check: P1=5, P2=10, P3=15. Sum = 5 + 10 + 15 = 30 units. Let's check options: 13 units, 15 units, 16 units, 30 units. Option index 3 corresponds to 30 units.
हिंदी में प्रश्न एवं आदर्श उत्तर
सुनीता कागज़ के एक पत्रक को तीन टुकड़ों में काटती है । पहले टुकड़े की लंबाई एक अंक वाली तीन विषम अभाज्य संख्याओं के औसत के बराबर है । दूसरे टुकड़े की लंबाई पहले टुकड़े की लंबाई और तीसरे टुकड़े की एक-तिहाई लंबाई के योग के बराबर है । तीसरे टुकड़े की लंबाई अन्य दो टुकड़ों की लंबाइयों के योग के बराबर है । कागज़ के मूल पत्रक की लंबाई कितनी है?
सही उत्तर (B) 15 units है।
व्याख्या:
Let the lengths of the three pieces of paper be P1, P2, and P3. Step 1: Single digit odd prime numbers are 3, 5, and 7. Their average is (3 + 5 + 7) / 3 = 15 / 3 = 5 units. Thus, the length of the first piece P1 = 5. Step 2: The length of the second piece P2 is equal to the first plus one-third the length of the third: P2 = P1 + (1/3)P3 = 5 + (1/3)P3. Step 3: The third piece is as long as the other two pieces together: P3 = P1 + P2 = 5 + P2. Now substitute P2 = 5 + (1/3)P3 into the equation for P3: P3 = 5 + 5 + (1/3)P3 = 10 + (1/3)P3. Subtracting (1/3)P3 from both sides: (2/3)P3 = 10, which gives P3 = 10 x (3/2) = 15 units. Now find P2: P2 = 5 + (1/3)(15) = 5 + 5 = 10 units. Total length of the original sheet of paper = P1 + P2 + P3 = 5 + 10 + 15 = 30 units? Wait, let's check: P1=5, P2=10, P3=15. Sum = 5 + 10 + 15 = 30 units. Let's check options: 13 units, 15 units, 16 units, 30 units. Option index 3 corresponds to 30 units.