Q.34 In 2005, Peter’s age was exactly four times that of his son, Quentin. In 2021, Peter will be
exactly twice Quentin’s age.
What is the difference between their ages?
A 16
B 20
C 24
D 28
E 36
Hi! so for this we can do simultaneous equations.
in 2005 → P = 4Q
in 2021, they both age by 16 years → P+16 = 2(Q+16)
the equations will be
P-4Q=0
P-2Q=16
solve by cancelling P out to find Q.(simultaneous equation method) then substitute Q in the first equation to get P, then you can subtract their ages to find the difference. Hope this helps!
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24
Explanation:
In 2005, let Quentin’s age=q & Peter =4q
In 2021 (i.e 16 years later) ==> (4q+16)=2(q+16)
On solving, we get q=8 & Peter’s age is 8x4=32
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Thank you both ! it makes sense now 
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