I have a solution to the same problem here, however, with a slight change in the given such as: 2 times; 68 Joules. When examining this solution, I can assure you that you can now answer the problem on your own.
Using the conservation of momentum,

0 = MaVa + MbVb

where

Ma = mass of particle A

Va = final velocity of particle A

Mb = mass of particle B

Vb = final velocity of particle B

Since Ma = 4(Mb), then the above formula becomes

0 = 4(Mb)(Va) + Mb(Vb)

Va = – (Vb/4)

(KE)a = kinetic energy of particle A = (1/2)(Ma)(Va)^2

(KE)b = kinetic energy of particle B = (1/2)(Mb)(Vb)^2

Taking the ratio of the two kinetic energies,

(KE)b/(KE)a = [(1/2)(Mb)(Vb)^2]/[(1/2)(Ma)(Va)^2]

and since

Ma = 4(Mb)

and

Va = -Vb/4

then the above simplifies to

(KE)b/(KE)a = [(Mb)(Vb)^2]/[(4Mb)(Vb/4)^2]

(KE)b/(KE)a = 1/(4*1/16) = 4

Therefore,

(KE)b = 4(KE)a — call this Equation A

Using the law of conservation of energy,

58 = (KE)a + (KE)b

and since (KE)b = 4(KE)a (from Equation A), the above simplifies to

58 = (KE)a + 4(KE)a

5(KE)a = 58

(KE)a = 11.6 joules

Hence,

(KE)b = 4(11.6) = 46.4 joules

ANSWER:

Kinetic energy of particle A = 11.6 joules

Kinetic energy of particle B = 46.4 joules

Hope this helps.