Calculate the degree of unsaturation: the answer is 1; it can have a carbon-carbon double bond, a carbonyl, or a ring.
Since the degree of unsaturation indicates that the compound could have a carbonyl, let's look for that first. There is a band at 1726, suggesting a saturated aldehyde (1740-1720). Whenever you suspect a compound is an aldehyde, look for bands in the region 2830-2695. They often appear as shoulder-peaks, just to the right of the C-H alkane stretch peaks. In the spectrum below, note the two bands in this region, suggesting that the compound is indeed an aldehyde.
Since the IR spectrum indicates an aldehyde, look for this functionality in the NMR spectrum. The aldehydic proton appears in the NMR from 9-10, usually as a small singlet.
The spectrum above shows a small singlet corresponding to one proton at 9.2 ppm, confirming that the compound is an aldehyde. Protons on the carbon adjacent to the aldehyde carbonyl will show up at 2-2.7 ppm; this is the triplet peak of 2 protons at 2.4 ppm on the above spectrum. Thus, so far we know that there is an aldehyde group next to a methylene group which is next to a carbon that has two hydrogens:
This accounts for 3 of the 5 carbons in the molecule. The un-colored hydrogens in the above structure could correspond to the peak of 2 hydrogens centered at 1.6 ppm; this peak is a pentet indicating that these protons are adjacent to carbons with a total of 4 hydrogens. The peak centered at 1.35 ppm has two hydrogens and is a sextet, indicating it is next to carbons that have a total of 5 hydrogens. Finally, the peak at 0.9 ppm has 3 hydrogens and is a triplet, indicating it is a methyl group adjacent to a carbon that has 2 hydrogens. Therefore, it looks like the molecule is a straight-chain of 5 carbons with the aldehyde group at one end:
Note that the closer a group is to the carbonyl function, the further downfield it is shifted. Here is how the NMR correlates to the structure: