TheCATExam
Sign in

CAT 2018 Slot 1 — QA Question 26

Basics (Functions)Easy

If f(x + 2) = f(x) + f(x + 1) for all positive integers x, and f(11) = 91, f(15) = 617, then f(10) equals

Answer & solution

Correct answer: 54

Answer: 54

Solution

Easy

The recurrence f(x+2)=f(x)+f(x+1)f(x+2)=f(x)+f(x+1) lets us climb from low terms up to f(15)f(15). Express f(15)f(15) in terms of the two given anchors f(11)f(11) and f(10)f(10), then solve the resulting linear equation.

1

Expand f(15)f(15) downward using the recurrence. Repeatedly replace the highest term until everything is in f(12)f(12), f(11)f(11), f(10)f(10).

f(15)=f(14)+f(13)⇒ f(15)=(f(13)+f(12))+f(13)=2f(13)+f(12)⇒ f(15)=2(f(12)+f(11))+f(12)=3f(12)+2f(11)⇒ f(15)=3(f(11)+f(10))+2f(11)=5f(11)+3f(10)\begin{aligned} &f(15)=f(14)+f(13)\\ &\Rightarrow\ f(15)=\big(f(13)+f(12)\big)+f(13)=2f(13)+f(12)\\ &\Rightarrow\ f(15)=2\big(f(12)+f(11)\big)+f(12)=3f(12)+2f(11)\\ &\Rightarrow\ f(15)=3\big(f(11)+f(10)\big)+2f(11)=5f(11)+3f(10) \end{aligned}
2

Plug in the given values f(15)=617f(15)=617 and f(11)=91f(11)=91.

617=5(91)+3f(10)(from step 1)⇒ 617=455+3f(10)⇒ 3f(10)=162⇒ f(10)=54\begin{aligned} &617 = 5(91)+3f(10) \quad\text{(from step 1)}\\ &\Rightarrow\ 617 = 455 + 3f(10)\\ &\Rightarrow\ 3f(10) = 162\\ &\Rightarrow\ f(10) = 54 \end{aligned}
f(10)=54f(10) = 54

Related Basics (Functions) questions

See all Functions & Graphs questions →
CAT 2018 Slot 1 QA Q26: If f(x + 2) = f(x) + f(x + 1) for all positive integers x, and f(11) = 91, f(15) = 617, then f(10) equals — Solution | TheCATExam