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CAT 2020 Slot 3 — QA Question 8

Number TheoryEasy

Let m and n be natural numbers such that n is even and 0.2 < m20,nm,n11 < 0..5. Then m – 2n equals

Answer & solution

Correct answer: 1

  • A

    4

  • B

    3

  • C

    2

  • 1

Solution

Medium

Each of the three fractions m20,nm,n11\dfrac{m}{20},\dfrac{n}{m},\dfrac{n}{11} lies strictly between 0.20.2 and 0.50.5. Convert the simple bounds to integer ranges, use "nn even" to fix nn, then the middle fraction to fix mm.

1

Bound mm from m20\dfrac{m}{20}.

\begin{aligned} &0.2<\frac{m}{20}<0.5\\ &\Rightarrow\ 4
2

Bound nn from n11\dfrac{n}{11}, then use "nn even".

\begin{aligned} &0.2<\frac{n}{11}<0.5\\ &\Rightarrow\ 2.2
3

Use the middle fraction nm\dfrac{n}{m} to fix mm. With n=4n=4:

\begin{aligned} &0.2<\frac{n}{m}<0.5\\ &\Rightarrow\ 0.2<\frac{4}{m}<0.5\\ &\Rightarrow\ 8Combined with step 1 (m≤9m\le9), the only value is m=9m=9.

4

Compute m−2nm-2n.

m−2n=9−2×4⇒ m−2n=1\begin{aligned} &m-2n=9-2\times4\\ &\Rightarrow\ m-2n=1 \end{aligned}
m−2n=1m-2n=1

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CAT 2020 Slot 3 QA Q8: Let m and n be natural numbers such that n is even and 0.2 < m 20 , n m , n 11 < 0..5. Then m &ndash; 2n equal — Solution | TheCATExam