Question

Ten trucks numbered 1 to 10, are carrying packets of sugar. Each packet weighs either $999gms$ or $1000gms$ and each truck carries only the packets of equal weights. The combined weight of 1 packet selected from the first truck, 2 packets from the second, 4 packet from the third, and so on and $2^9$ packets from the tenth truck is $1022870gms$. The trucks that have the lighter bags are:

• A. $1,3,5$
• B. $2,4,5$
• C. $1,9$
• D. $2,8$

Answer 1

The correct option is D $2,8$

Let $n_1$ be the packets with $999grams$

and $n_2$ be the packets with $1000grams$

1. Truck = $1$
2. Truck = $2$
3. Truck = $4$
4. Truck = $8$
5. Truck = $16$
6. Truck = $32$
7. Truck = $64$
8. Truck = $128$
9. Truck = $256$
10. Truck = $512$

$n_1\left(999\right)+n_2\left(1000\right)=1022870$

$n_1$ should be a multiple of $10$ because the total contain $0$ at unit place.

Also, for $n_2$ to be a natural number, the last 3 digits of total must be equal to the last 3 digits of total weight of $999gms$ packets.

Options for $n_1$ can be $(2+8),(2+128),(4+16),(4+256),(8+32),(8+512),(16+64),(32+128),(64+256),(128+512)$

$(2+128)\times 999=129870$

Hence Truck 2 and 8 contain the $999gms$ packets.