TS ICET 2019 Question Paper Shift-2 (23rd May)

$$\sqrt{18 + x \sqrt{2}} = \sqrt{12} + \sqrt{6} \Rightarrow x =$$

Take square both side

$$({\sqrt{18 + x \sqrt{2}}})^2 = ({\sqrt{12} + \sqrt{6}})^2 \Rightarrow x =$$

$${18 + x \sqrt{2}} = 12 + 6 + 2 * \sqrt{12} * \sqrt{6} \Rightarrow x =$$

$${18 + x \sqrt{2}} = 18 + 2 * \sqrt{6} * \sqrt{2} * \sqrt{6} \Rightarrow x =$$

18 will be eliminated from both the sides.

$${x \sqrt{2}} = 2 * \sqrt{6} * \sqrt{2} * \sqrt{6}$$

x =2 * 6

x = 12

Hence, option C is correct.

Let the number be N.

As the number leaves 48 as remainder when divided by 783, then

N = 783 x k + 48 (Here, k is an integer)

N = 29 x 27 x k + 29 + 19

N = 29 (27k + 1) + 19

Let 27k + 1 = m

then N = 29m + 19

So, remainder = 19

Hence, option C is correct.

Factors of 1248 = 2 x 2 x 2 x 2 x 2 x 3 x 13

It can be written as 

1248 = $$2^{5}$$ x $$3^{1}$$ x $$13^{1}$$

Factors are written in the form: $$2^{p}$$ x $$3^{q}$$ x $$13^{r}$$

Total number of divisors = (p+1) x (q+1) x (r+1)

= (5+1) x (1+1) x (1+1)

= 6 x 2 x 2 = 24

1 and the number itself are excluded, then

24 - 2 = 22

So, the number of divisors will be 22.

Hence, option C is correct.

Take L.C.M. of 7, 11, 13

L.C.M. (7, 11, 13) = 1001

3000 = 1001 x m + k (where k is remainder)

Put value of m = 2

3000 = 1001 x 2 + k

3000 = 2002 + k

k = 998

So, the least number which when subtracted from 3000 is exactly divisible by 7, 11, 13.

Hence, option D is correct.

Since 10 is remainder in each case, subtract 10 from every number.

1354 - 10 = 1344

1866 - 10 = 1856

2762 - 10 = 2752

HCF (1344, 1856, 2752) = 64

So, 64 will be the greatest number.

Hence option C is correct.

Let  x = $$3.7\overline{84}$$ 

We can write it as 

x = 3.7848484...

Multiply the equation with 10 on both the sides

10x = 37.8484... equation 1

Multiply equ 1 with 100 on both the sides

1000x = 3784.8484... equation 2

Subtract equation 1 from equation 2

1000x - 10x = 3784.8484... - 37.8484...

990x = 3747

x = $$\frac{3747}{990}$$

where $$\frac{3747}{990}$$ is in $$\frac{p}{q}$$ form

For $$GCD(p, q) = 1$$

$$\frac{3747}{990}$$ = $$\frac{1249}{330}$$

p + q = 1249 + 330

         = 1579

Hence, option B is correct answer.

3+$$\frac{1}{4+\frac{1}{5+\frac{1}{6}}}$$

3+$$\frac{1}{4+\frac{1}{\frac{31}{6}}}$$

3+$$\frac{1}{4+\frac{6}{31}}$$

3+$$\frac{1}{\frac{124+6}{31}}$$

3+$$\frac{31}{130}$$

$$\frac{421}{130}$$

Hence, option A is correct.

$$\sqrt{2}$$, $$\sqrt[3]{3}$$, $$\sqrt[5]{5}$$, $$\sqrt[7]{7}$$

$${2^\frac{1}{2}}$$, $${3^\frac{1}{3}}$$, $${5^\frac{1}{5}}$$, $${7^\frac{1}{7}}$$

Now, take L.C.M. of denominators of power

L.C.M. of (2, 3, 5, 7) = 210

$${2^\frac{1*105}{2*105}}$$, $${3^\frac{1*70}{3*70}}$$, $${5^\frac{1*42}{5*42}}$$, $${7^\frac{1*30}{7*30}}$$

$${2^\frac{105}{210}}$$, $${3^\frac{70}{210}}$$, $${5^\frac{42}{210}}$$, $${7^\frac{30}{210}}$$

$$\sqrt[210]{{2}^{105}}$$, $$\sqrt[210]{{3}^{70}}$$, $$\sqrt[210]{{5}^{42}}$$, $$\sqrt[210]{{7}^{30}}$$

The largest power is of 2. So, $$\sqrt{2}$$ will be the largest number.

Hence, option A is correct. 

As, the denominators are 4, 6, 8, 9 in each option

take L.C.M. of 4, 6, 8, 9 = 72

Now, make all the denominators equal to 72.

Since 1 is common in each question, we can ignore it,

Option A: $$1\frac{1*18}{4*18}>1\frac{5*12}{6*12}>1\frac{7*9}{8*9}>1\frac{8*8}{9*8}$$

               $$1\frac{18}{72}>1\frac{60}{72}>1\frac{63}{72}>1\frac{64}{72}$$ 

               (condition does not satisfy)

Option B: $$1\frac{1*18}{4*18}>1\frac{7*9}{8*9}>1\frac{8*8}{9*8}>1\frac{5*12}{6*12}$$

              $$1\frac{18}{72}>1\frac{63}{72}>1\frac{64}{72}>1\frac{60}{72}$$ 

               (condition does not satisfy)

Option C: $$1\frac{8*8}{9*8}>1\frac{7*9}{8*9}>1\frac{5*12}{6*12}>1\frac{1*18}{4*18}$$

              $$1\frac{64}{72}>1\frac{63}{72}>1\frac{60}{72}>1\frac{18}{72}$$ 

               (condition satisfies)

Option D: $$1\frac{8*8}{9*8}>1\frac{5*12}{6*12}>1\frac{1*18}{4*18}>1\frac{7*9}{8*9}$$

              $$1\frac{64}{72}>1\frac{60}{72}>1\frac{4}{72}>1\frac{63}{72}$$ 

              (condition does not satisfy)

Hence, option C is correct.

Given: Amount of mixture = 40L

Water content in the mixture = 10% of 40 = 40 x $$\frac{10}{100}$$ = 4L

So, milk content in the mixture will be = 40 - 4 = 36L

Let the amount of water need to be added to make the water content to 20% be xL.

Then, $$\frac{4+x}{40+x}$$ x 100 = 20

          x = 5L

Hence, option C is correct.