AP ICET 2013 Question Paper

$$\left(\frac{\sqrt{5} + \sqrt{7}}{\sqrt{5} - \sqrt{7}} + \frac{\sqrt{5} - \sqrt{7}}{\sqrt{5} + \sqrt{7}}\right)^2$$ = {$$\frac{\left(\left(\sqrt{\ 5}+\sqrt{\ 7}\right)^2-\ \left(\sqrt{\ 5}-\sqrt{\ 7}\right)^2\right)}{\left(\sqrt{\ 5}^{ }+\sqrt{\ }7\right)\left(\sqrt{\ 5}^{ }-\sqrt{\ }7\right)}$$}^2

= $$\left(\frac{\left(\left(5\ +7\ +2.\sqrt{\ 5}.\sqrt{\ 7}\right)+\left(5\ +7\ -2.\sqrt{\ 5}.\sqrt{\ 7}\right)\right)}{5-7}\right)^2$$ = $$\left(\frac{\left(12+12\right)}{-2}\right)^2=\ \left(-12\right)^2\ =144\ Answer$$

Solution'

Concept used: Divisibility rule of 11:The difference of sum of digits at even and odd places should be 0 or multiple of 11

Lets take each no. one by one

So the required no. is 1873410 Answer

Principle used to solve :Rem (A x B) = Rem. (A) x Rem. (B)

A/q

When divided by 11

Rem. (73 x 79x 81) = Rem. (73) x Rem.(79) x Rem.(81) 

Rem. (73) = 7                    

Rem. (79) = 2

Rem. (81) = 4

So Rem. (73 x 79x 81) = Rem (7 x 2 x 4) = Rem (56) = 1

Hence Remainder =1 Answer 

Solution

18 = 6.3

24 = 6.4

36 = 6.6

LCM = 6.4.3 = 72

So the no. should be divisible by 72. Out of the given options, the Highest such no. before 9999 is 9936. 

Solution

Concept Used: Product of two numbers = (L.C.M of numbers) x (G.C.D of numbers)

Here

$$a^2 . b^2 = 32400$$

Hence a.b = $$\sqrt{\ 32400}$$ = 180 =Product of Numbers 

Given L.C.M. = 60

So axb = (l.c.m) x( g.c.d)

180 = 60 x g.c.d

g.c.d = 3 Answer

Solution

$$\mid x - 6 \mid = 5$$    |x-6|=5

if x-6 > 0 ; x>6

so the mod function will become, x-6 = 5

x=11......x max.

If x-6 < 0 ; x<6

so the mod function will become, -(x-6) = 5

6 -x = 5

x=6-5 =1  .....x min

Similarly for y

$$\mid 3y - 12 \mid = 6$$    |3y-12|= 6

If 3y-12 < 0, y < 4

so the mod function will become, - (3y-12) = 6

12 -3y = 6 ; y =2 ........y min.

If 3y-12 > 0, y > 4

so the mod function will become, (3y-12) = 6

3y = 12+6 = 18

y = 6 .......y max

For,maximum value of $$\frac{x}{y} =$$ Numerator (x) should be maximum and denominator (y) should be minimum.

$$\frac{x\ \left(\max\right)}{y\left(\min\right)}\ =\ \frac{11}{2\ }$$ Answer

Solution:

Let the rational No. be $$\frac{A}{B}$$

A/Q

$$\frac{A}{B+3}$$ = $$\frac{1}{3}$$  =>3A = B+3   

3A = B+3 => A=$$\frac{1}{3}$$( B+3)   ....(i)

$$\frac{A +4}{B}$$ = $$\frac{3}{4}$$ => 4A +16 = 3B ....(ii)

4A +16 = 3B

$$\frac{4}{3}$$( B+3) + 16 = 3B

$$\frac{4B}{3} $$ + 4 + 16 = 3B

20 = 3B - $$\frac{4B}{3} $$

20 = $$\frac{5B}{3} $$

B=12

Putting in Equation (i)

A = 5

Hence required rational No. is $$\frac{5}{12}$$   Answer

Solution

$$\frac{1}{2}, \frac{5}{6}, \frac{7}{9}, \frac{4}{5}$$ 

LCM of denominators = LCM(2,6,9,5) = 90

Converting all the fractions to common base

$$\frac{1}{2}$$ = $$\frac{1 . 45}{2. 45}$$ = $$\frac{45}{90}$$

$$\frac{5}{6}$$ = $$\frac{5 . 15}{6 .15}$$ = $$\frac{75}{90}$$

$$\frac{7}{9}$$ = $$\frac{7 . 10}{9 . 10}$$ = $$\frac{70}{90}$$

$$\frac{4}{5}$$ = $$\frac{4 . 18}{5 . 18}$$ = $$\frac{72}{90}$$

As denominators are same, so fraction with biggest numerator is biggest and with smallest numarator is smallest.

Biggest fraction = $$\frac{75}{90}$$

Smallest fraction = $$\frac{45}{90}$$

Difference = $$\frac{75}{90}$$ - $$\frac{45}{90}$$ = $$\frac{30}{90}$$ = $$\frac{1}{3}$$ Answer

Solution:

$$\frac{13}{14}, \frac{11}{12}, \frac{15}{16}, \frac{8}{11}$$

We can proceed with approximation;

$$\frac{13}{14}$$ = 0.92..

$$\frac{11}{12}$$ = 0.91...

$$\frac{15}{16}$$ = 0.93...

$$\frac{8}{11}$$ = 0.72....

So , Ascending Order will be 0.72,0.91,0.92,0.93

i.e.  $$\frac{8}{11}$$ , $$\frac{11}{12}$$ , $$\frac{13}{14}$$ ,  $$\frac{15}{16}$$

Solution:

A/q 

7% of 5% of 8% of x is 105

$$\frac{7}{100}$$ x $$\frac{5}{100}$$ x $$\frac{8}{100} x$$ = 105

7.5.8.$$x\ $$ = 105000 x 1000

280$$x\ $$ = 105000000

$$x\ $$ = 375000 Answer