When neopentyl alcohol is heated with an acid, it slowly converted into an 85:15 mixture of alkenes A and B, respectively. What are these alkenes?

Neopentyl alcohol is $$HO{-}CH_2{-}C(CH_3)_3$$ (IUPAC : 2,2-dimethyl-1-propanol). On heating with a protonic acid (conc. $$H_2SO_4$$, $$H_3PO_4$$ etc.) it undergoes acid-catalysed dehydration by the $$E_1$$ mechanism.

Step 1 : Protonation and loss of water
$$HO{-}CH_2{-}C(CH_3)_3 + H^+ \;\longrightarrow\; HOH^+-CH_2{-}C(CH_3)_3$$
$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\longrightarrow\; CH_2^{+}{-}C(CH_3)_3 + H_2O$$
The leaving of water gives a highly unstable primary carbocation.

Step 2 : 1,2-Methyl shift
A neighbouring methyl group migrates with its bonding pair (1,2-alkyl shift) to stabilise the cation:

$$CH_2^{+}{-}C(CH_3)_3 \;\xrightarrow[\text{methyl shift}]{}\; (CH_3)_2C^{+}{-}CH(CH_3)$$
The rearranged carbocation is tertiary, hence far more stable.

Step 3 : Elimination ($$E_1$$) from the tertiary carbocation
The positive carbon is adjacent to three kinds of $$\beta$$-hydrogen:

Case A - Removal of the single hydrogen on the $$CH(CH_3)$$ group
$$\;\;(CH_3)_2C^{+}{-}CH(CH_3) \; \xrightarrow[-H^+]{E_1}\; (CH_3)_2C=CHCH_3$$
This gives the tri-substituted alkene 2-methyl-2-butene.

Case B - Removal of a hydrogen from either of the two equivalent methyl groups attached directly to the carbocation centre
$$\;\;(CH_3)_2C^{+}{-}CH(CH_3) \; \xrightarrow[-H^+]{E_1}\; CH_2=C(CH_3)CH_2CH_3$$
This furnishes the di-substituted alkene 2-methyl-1-butene.

Step 4 : Product distribution
According to Zaitsev’s rule, the more substituted alkene (2-methyl-2-butene) is formed preferentially. Experimentally the ratio is about 85 % (A) : 15 % (B).

Hence
A (85 %) = 2-methyl-2-butene $$[(CH_3)_2C=CHCH_3]$$
B (15 %) = 2-methyl-1-butene $$[CH_2=C(CH_3)CH_2CH_3]$$

Option B which is: 2-methyl-2-butene and 2-methyl-1-butene.