प्रत्येक 2 gm, 5 gm, 10 gm, 25 gm, 50 gm भार वाले वृहद संख्या में चांदी के सिक्के हैं । निम्नलिखित कथनों पर विचार कीजिए :
1. 78 gm सिक्कों को खरीदने के लिए कम-से-कम 7 सिक्के खरीदना आवश्यक है ।
2. इन सिक्कों का उपयोग कर 78 gm भार वजन करने के लिए 7 से कम सिक्के उपयोग में लाए जा सकते हैं ।
उपर्युक्त कथनों में से कौन-सा/से सही है/हैं ?
सही उत्तर (A) 1 only है।
व्याख्या:
We have silver coins weighing 2 gm, 5 gm, 10 gm, 25 gm, and 50 gm (in large numbers). Let us evaluate Statement-1: 'To buy 78 gm of coins one must buy at least 7 coins.' If we want to buy 78 gm using the minimum number of coins, we should use the largest possible denominations: 50 gm (1 coin) = 50 gm (remaining 28 gm). Next largest: 25 gm (1 coin) = 25 gm (remaining 3 gm). Next largest: 10 gm (0), 5 gm (0). Remaining 3 gm can be made using one 2 gm coin and one 1 gm? Wait, coin denominations are 2, 5, 10, 25, 50. Remaining 3 gm can be made using 2 gm (1 coin) and 2 gm (1 coin)? Wait, 2 + 2 = 4 > 3. To make 3 gm using available coins (2 gm minimum), wait! Can we make 3 gm? 2 gm + 2 gm = 4 gm (not 3 gm). Wait, what coin denominations are available? 2 gm, 5 gm, 10 gm, 25 gm, 50 gm. There is NO 1 gm coin! So to make 78 gm by buying coins: Let's check combinations: 50 + 25 = 75 (2 coins). Remaining 3 gm cannot be made with 2 gm coins alone (since 2+2=4). Wait, 50 + 10 + 10 + 5 + 2 + 1? No 1 gm coin. Let's test combinations summing to 78: 50 + 10 + 10 + 5 + 2 + 2 = 79. What about 50 + 10 + 10 + 5 + 2 = 77 (not 78). What about 50 + 10 + 5 + 5 + 5 + 2 + 2 = 79. Let's check: 25 + 25 + 25 + ? No. What about 50 + 10 + 10 + 5 + 2 + ... wait, let's find a combination of coins that sum to 78: 50 + 10 + 10 + 5 + 2 + ? We need 78. Coins available: 2, 5, 10, 25, 50. Let's test 7 coins: 50 + 10 + 10 + 5 + 2 + 1? No 1 gm. What about 25 + 25 + 10 + 10 + 5 + 2 + 1? No. Wait, let's check Statement-2: 'To weigh 78 gm using these coins one can use less than 7 coins.' In weighing (using a physical balance), we can put weights on both pans (subtraction is allowed!). To weigh 78 gm: Put 100 gm on one pan and (50 + 25 + 5) = 80? No. Put 78 on one side: 50 + 25 + 5 = 80 on one side, and 2 on the other side! 80 - 2 = 78 gm! How many coins used here? 50 (1), 25 (1), 5 (1), 2 (1) = 4 coins! 4 coins is less than 7 coins. Thus Statement-2 is correct (we can weigh 78 gm using 4 coins: 50, 25, 5 on one pan, and 2 on the other pan, since 50 + 25 + 5 - 2 = 78 gm). What about Statement-1? Buying 78 gm of coins requires 7 coins because without weighing (only addition), to make 78: 50 + 25 + 2 + ... wait, 50 + 25 = 75, need 3, but smallest coin is 2, so 50 + 25 + 2 + 2 = 79 (too much), 50 + 10 + 10 + 5 + 2 + 1? No 1. Let's check 25 + 25 + 25 + 2 + 1? No. To buy 78 gm strictly by addition requires at least 7 coins (e.g., 25 + 25 + 10 + 10 + 5 + 2 + 2 = 79? No, 25+25+10+10+5+2+2 = 79; 25+25+10+5+5+5+2 = 77; 50 + 10 + 5 + 5 + 5 + 2 + 2 = 79; actually 7 or more coins are needed). Thus both Statement 1 and Statement 2 are correct. Therefore, option (c) is correct.
In English (Question & Model Answer)
There are large number of silver coins weighing 2 gm, 5 gm, 10 gm, 25 gm, 50 gm each. Consider the following statements:
1. To buy 78 gm of coins one must buy at least 7 coins.
2. To weigh 78 gm using these coins one can use less than 7 coins.
Which of the statements given above is/are correct?
The correct answer is (A) 1 only.
Explanation:
We have silver coins weighing 2 gm, 5 gm, 10 gm, 25 gm, and 50 gm (in large numbers). Let us evaluate Statement-1: 'To buy 78 gm of coins one must buy at least 7 coins.' If we want to buy 78 gm using the minimum number of coins, we should use the largest possible denominations: 50 gm (1 coin) = 50 gm (remaining 28 gm). Next largest: 25 gm (1 coin) = 25 gm (remaining 3 gm). Next largest: 10 gm (0), 5 gm (0). Remaining 3 gm can be made using one 2 gm coin and one 1 gm? Wait, coin denominations are 2, 5, 10, 25, 50. Remaining 3 gm can be made using 2 gm (1 coin) and 2 gm (1 coin)? Wait, 2 + 2 = 4 > 3. To make 3 gm using available coins (2 gm minimum), wait! Can we make 3 gm? 2 gm + 2 gm = 4 gm (not 3 gm). Wait, what coin denominations are available? 2 gm, 5 gm, 10 gm, 25 gm, 50 gm. There is NO 1 gm coin! So to make 78 gm by buying coins: Let's check combinations: 50 + 25 = 75 (2 coins). Remaining 3 gm cannot be made with 2 gm coins alone (since 2+2=4). Wait, 50 + 10 + 10 + 5 + 2 + 1? No 1 gm coin. Let's test combinations summing to 78: 50 + 10 + 10 + 5 + 2 + 2 = 79. What about 50 + 10 + 10 + 5 + 2 = 77 (not 78). What about 50 + 10 + 5 + 5 + 5 + 2 + 2 = 79. Let's check: 25 + 25 + 25 + ? No. What about 50 + 10 + 10 + 5 + 2 + ... wait, let's find a combination of coins that sum to 78: 50 + 10 + 10 + 5 + 2 + ? We need 78. Coins available: 2, 5, 10, 25, 50. Let's test 7 coins: 50 + 10 + 10 + 5 + 2 + 1? No 1 gm. What about 25 + 25 + 10 + 10 + 5 + 2 + 1? No. Wait, let's check Statement-2: 'To weigh 78 gm using these coins one can use less than 7 coins.' In weighing (using a physical balance), we can put weights on both pans (subtraction is allowed!). To weigh 78 gm: Put 100 gm on one pan and (50 + 25 + 5) = 80? No. Put 78 on one side: 50 + 25 + 5 = 80 on one side, and 2 on the other side! 80 - 2 = 78 gm! How many coins used here? 50 (1), 25 (1), 5 (1), 2 (1) = 4 coins! 4 coins is less than 7 coins. Thus Statement-2 is correct (we can weigh 78 gm using 4 coins: 50, 25, 5 on one pan, and 2 on the other pan, since 50 + 25 + 5 - 2 = 78 gm). What about Statement-1? Buying 78 gm of coins requires 7 coins because without weighing (only addition), to make 78: 50 + 25 + 2 + ... wait, 50 + 25 = 75, need 3, but smallest coin is 2, so 50 + 25 + 2 + 2 = 79 (too much), 50 + 10 + 10 + 5 + 2 + 1? No 1. Let's check 25 + 25 + 25 + 2 + 1? No. To buy 78 gm strictly by addition requires at least 7 coins (e.g., 25 + 25 + 10 + 10 + 5 + 2 + 2 = 79? No, 25+25+10+10+5+2+2 = 79; 25+25+10+5+5+5+2 = 77; 50 + 10 + 5 + 5 + 5 + 2 + 2 = 79; actually 7 or more coins are needed). Thus both Statement 1 and Statement 2 are correct. Therefore, option (c) is correct.