NIMCET 2015 — Reasoning PYQ

1. For 78 paise:

• $1$ coin of $50$ paise (Remaining: $28$)

• $1$ coin of $25$ paise (Remaining: $3$)

• Since we only have $2$ and $5$ paise coins left, we cannot use a $5$. We must use $2$ paise coins. However, $3$ is not divisible by $2$.

• Adjustment: Use $1\times 50$, $0\times 25$, $2\times 10$, $1\times 5$, $1\times 2$ (Total $67$ - not $78$).

• Best Combination for 78:

  • $1\times 50$ paise

  • $1\times 10$ paise

  • $1\times 10$ paise

  • $1\times 4$ (No, use $2+2+2+2$ is too many)

  • Let's try: $50+25+2+1$ (No $1$ available).

  • We must reach $78$ using $\{2,5,10,25,50\}$.

  • $50+10+10+2+2+2+2=78$ ($7$ coins)

  • $50+25+?$: Not possible.

  • $50+10+5+5+2+2+2+2$ (Too many)

  • $25+25+25+2+1$ (No)

  • Correct smallest for 78: $50+10+10+2+2+2+2=7$ coins.

2. For 69 paise:

• $1\times 50$ paise (Remaining: $19$)

• $1\times 10$ paise (Remaining: $9$)

• $1\times 5$ paise (Remaining: $4$)

• $2\times 2$ paise (Remaining: $0$)

• Total for 69: $1+1+1+2=5$ coins.

3. For Rs. 1.01 (101 paise):

• $2\times 50$ paise (Remaining: $1$) - Not possible as we have no $1$ paisa coin.

• $1\times 50$ paise $+2\times 25$ paise (Remaining: $1$) - Not possible.

• We need a combination that ends in $1$. Since all coins except $5$ and $25$ are even, and we need an odd sum, we need to use $5$ or $25$.

• $1\times 50+1\times 25+1\times 10+1\times 10+3\times 2=101$

• Coins: $1(50)+1(25)+2(10)+3(2)=7$ coins.

Total smallest number of coins:

Conclusion:

The smallest number of coins required is $19$.

Correct Option: (a) 19