You have five coins, apparently alike, but actually of different weights. You also have a two arm scale.

Can you manage to sort the coins in ascending order, using the scale only seven times?

Bonus question: can it be done in fewer weighings?

Choose two weights for group 1 and choose another two weights for group 2. Compare the two group 1 weights against each other and compare the two group 2 weights against each other. Compare the heavier of group 1 against group 2.

After these three weighings, relabel the weights as follows: Label the weights such that A>B, A>C>D, and E is unknown. A and B are from one of the groups, C and D are from the other and E is the weight in neither group.

These first three weighings reduced the possible orderings from 120 to 15, and the following weighings continue to divide the possibilities in half. The next two weighings find the position of E in ACD, and the final two weighings find the position of B in the last four positions.

Weigh E against C:
Case 1: E>C
Weigh A against E:
Case 1.1 E>A
Weigh B against C
Case 1.1.1 B>C
The order is EABCD
Case 1.1.2 C>B
Weigh B against D
If B>D then the order is EACBD
If D>B then the order is EACDB
Case 1.2 A>E
Weigh B against C
Case 1.2.1 B>C
Weigh B against E
If B>E then the order is ABECD
If E>B then the order is AEBCD
Case 1.2.2 C>B
Weigh B against D
If B>D then the order is AECBD
If D>B then the order is AECDB
Case 2: C>E
Weigh E against D
Case 2.1 E>D
Weigh B against E
Case 2.1.1 B>E
Weigh B against C
If B>C then the order is ABCED
If C>B then the order is ACBED
Case 2.1.2 E>B
Weigh B against D
If B>D then the order is ACEBD
If D>B then the order is ACEDB
Case 2.2 D>E
Weigh B against D
Case 2.2.1 B>D
Weigh B against C
If B>C then the order is ABCDE
If C>B then the order is ACBDE
Case 2.2.2 D>B
Weigh B against E
If B>E then the order is ACDBE
If E>B then the order is ACDEB