CMAT Simplification Questions

(A): $$\sqrt{3} - \sqrt{2}$$ = $$\dfrac{1}{\sqrt{\ 3}+\sqrt{\ 2}}$$

(B): $$\sqrt{4} - \sqrt{3}$$ = $$\dfrac{1}{\sqrt{\ 4}+\sqrt{\ 3}}$$

(C): $$\sqrt{5} - \sqrt{4}$$ = $$\dfrac{1}{\sqrt{5}+\sqrt{\ 4}}$$

(D): $$\sqrt{2} - 1$$ = $$\dfrac{1}{\sqrt{2}+\sqrt{\ 1}}$$

Now, $$\left(\sqrt{\ 5}+\sqrt{\ 4}\right)>\left(\sqrt{\ 4}+\sqrt{\ 3}\right)>\left(\sqrt{\ 3}+\sqrt{\ 2}\right)>\left(\sqrt{\ 2}+1\right)$$

So, $$\dfrac{1}{\sqrt{\ 5}+\sqrt{\ 4}}<\dfrac{1}{\sqrt{\ 4}+\sqrt{\ 3}}<\dfrac{1}{\sqrt{\ 3}+\sqrt{\ 2}}<\dfrac{1}{\sqrt{\ 2}+1}$$

So, $$D>A>B>C$$

A: $$\sqrt{401} -\sqrt{399}$$ = $$\dfrac{2}{\sqrt{\ 401}+\sqrt{\ 399}}$$

B: $$\sqrt{301} -\sqrt{299}$$ = $$\dfrac{2}{\sqrt{\ 301}+\sqrt{\ 299}}$$

C: $$\sqrt{101} -\sqrt{99}$$ = $$\dfrac{2}{\sqrt{\ 101}+\sqrt{\ 99}}$$

D: $$\sqrt{201} -\sqrt{199}$$ = $$\dfrac{2}{\sqrt{\ 201}+\sqrt{\ 199}}$$

Now, $$\sqrt{\ 401}+\sqrt{\ 399}>\sqrt{\ 301}+\sqrt{\ 299}>\sqrt{\ 201}+\sqrt{\ 199}>\sqrt{\ 101}+\sqrt{\ 99}$$

or, $$\dfrac{2}{\sqrt{\ 401}+\sqrt{\ 399}}<\dfrac{2}{\sqrt{\ 301}+\sqrt{\ 299}}<\dfrac{2}{\sqrt{\ 201}+\sqrt{\ 199}}<\dfrac{2}{\sqrt{\ 101}+\sqrt{\ 99}}$$

So, $$A<B<D<C$$

or, $$C>D>B>A$$

So, option D is the correct answer.

Evaluating A:

$$\dfrac{5}{9}\times\ \left(225.4-45.4\right)=\dfrac{5}{9}\times\ 180=100=\left(10\right)^2$$

So, A= 10

Evaluating B:

$$\sqrt{1024}\times40+(20)^2+0.5\%$$ of $$9600+469=(?)^{3}$$

=$$32\ \times\ 40+400+\dfrac{0.5}{100}\times\ 9600+469=32\times\ 40+400+48+469=2197=13^3$$

So, B=13

Evaluating C:

$$\dfrac{8^2}{16^2}\times\ 32^3\times\ \sqrt{\ 256}=\dfrac{1}{2^2}\times\ \left(2^5\right)^3\times\ 2^4=2^{17}$$

So, C=17

Evaluating D:

$$18\dfrac{1}{3}+9\dfrac{2}{3}-10\dfrac{1}{3}=8+9\dfrac{2}{3}=1\dfrac{2}{3}+8+8=1\dfrac{2}{3}+16$$

So, D = 16

So, the correct order of arrangement is $$C>D>B>A$$

A. $$\sqrt{\dfrac{0.81\times\ 0.484}{0.064\times6.25}}=\sqrt{\dfrac{0.81\times\ 4.84}{0.64\times6.25}}=\sqrt{\dfrac{\left(0.9\right)^2\times\ \left(2.2\right)^2}{\left(0.8\right)^2\times\left(2.5\right)^2}}=\dfrac{0.9\times2.2}{0.8\times2.5}=0.99$$

B. $$\sqrt{\dfrac{0.204\times\ 42}{0.07\times3.4}}=\sqrt{\dfrac{20.4\times42}{3.4\times7}}=\sqrt{6\times6}=6$$

C. $$\sqrt{\dfrac{0.081\times0.324\times\ 4.624}{1.5625\times0.0289\times\ 72.9\times64}}=\sqrt{\dfrac{\left(0.9\right)^2\times\left(0.18\right)^2\times\left(6.8\right)^2}{\left(1.25\right)^2\times\left(0.17\right)^2\times\left(27\right)^2\times8^2}}=\dfrac{0.9\times0.18\times6.8}{1.25\times0.17\times27\times8}=0.024$$

D. $$\sqrt{\dfrac{9.5\times0.085}{0.0017\times0.19}}=\sqrt{50\times50}=50$$