AP ICET 2018 Question Paper Shift-1 (2nd May)

$$\sqrt{21 - 4\sqrt{5} + 8\sqrt{3} - 4\sqrt{15}} =$$ =  $$\sqrt{2^2+\left(-\sqrt{\ 5}\right)^2+\left(2\sqrt{\ 3}\right)^2+2\cdot2\cdot-\sqrt{5}+2\cdot\left(-\sqrt{\ 5}\right)\left(2\sqrt{3}\right)+2\cdot\left(2\sqrt{3}\right)\cdot2}=$$

Which is in the form $$a^2+b^2+c^2+2ab+2bc+2ca\ =\ \left(a+b+c\right)^2$$

hence $$\sqrt{21 - 4\sqrt{5} + 8\sqrt{3} - 4\sqrt{15}} =$$ = $$2-\sqrt{\ 5}+2\sqrt{\ 3}$$

The number will be of the form (75k+28)

75k+28 = 75k+25+3 = 25(3k+1)+3 = 25p+3

$$\therefore\ $$The remainder when (75k+28) is divided by 25 = 3

Hence, the correct answer is Option C

The integers between 200 and 400 divisible by 7 are 203,210...... 399

$$\therefore\ $$Required Sum = 203 + 210 +....+399

$$=7[29+30+.....+57]$$

$$=7[\frac{29}{2}\left(29+57\right)]$$

$$=7[\frac{29}{2}\left(86\right)]$$

$$=7\times29\times43$$

$$=8729$$

Hence, the correct answer is Option D

LCM of 27, 36, 45 = 540

A, B, C will again meet at P after every 540 seconds.

$$\therefore\ $$Number of rounds made my C in 540 seconds = $$\frac{540}{45}$$ = 12

Hence, the correct answer is Option A

53 is not divisible by 23, hence option A is eliminated

79 is not divisible by 23, hence option C is eliminated

84 is not divisible by 23, hence option D is eliminated

Let a = $$0.\overline{12}$$ ........(1)

$$\Rightarrow$$ 100a = $$12.\overline{12}$$ ........(2)

Solving equation(2) - equation(1)

99a = 12

$$\Rightarrow$$ a = $$\frac{12}{99}$$

$$\Rightarrow$$ a = $$\frac{4}{33}$$

$$\Rightarrow$$ $$0.\overline{12}$$ = $$\frac{4}{33}$$

Hence, the correct answer is Option D

Given, $$7p > 5$$

$$\Rightarrow$$  $$p>\frac{5}{7}$$

$$\Rightarrow$$  $$p>0.714$$

The smallest prime number greater than 0.714 is 2

$$\Rightarrow$$ $$p=2$$

$$7q < 497$$

$$\Rightarrow$$  $$q<71$$

The largest prime number less than 71 is 67

$$\Rightarrow$$  $$q=67$$

$$\therefore\ $$ $$\frac{p}{q}=\frac{2}{67}$$

$$\frac{3}{5}$$ = 0.6

$$\frac{6}{7}$$ = 0.8

$$\frac{5}{8}$$ = 0.62

$$\frac{7}{9}$$ = 0.77

$$\therefore\ $$The greatest among $$\frac{3}{5}, \frac{6}{7}, \frac{5}{8}, \frac{7}{9}$$ is $$\frac{6}{7}$$

Hence, the correct answer is Option B

a = $$\sqrt[\ 3]{4}\ =\ \sqrt[\ 12]{256}$$

b = $$\sqrt[\ 4]{5}\ =\ \sqrt[\ 12]{125}$$

c = $$\sqrt[\ 4]{7}\ =\ \sqrt[\ 12]{343}$$

d=$$\sqrt[\ 3]{9}\ =\ \sqrt[\ 12]{6561}$$

hence b<a<c<d

Let numerator = x

and denominator = y

New numerator = 1.2x

New denominator = 0.9y

Given,

1.2x/0/9y = 16/21

=> x/y = 4/7