**Directions(1-5):** Find the relation between x and y and choose a correct option.

- I. 3√x– 18/ √x = √x

II. 2√y + (y−36)/ √y = –√yx>=yx=y or relation cannot be established.y>=xx>yy>xOption B

3√x– 18/ √x = √x

=>3x–18 = x

=>2x = 18

=>x = 9

2√y +( y−36)/ √y = –√y

=>2y+y–36 = –y

=>4y = 36

=>y = 9

x=y - I. 6 y^2 + 1/2 = 7/2 y

II. 12 x^2 + 2 = 10 xy>=xx>yx>=yy>xx=y or relation cannot be established.Option C

6y^2 + 1/2 = 7/2 y

12y^2 – 7 y + 1 = 0

12y^2 – 44y – 3y + 1 = 0

4y (3y – 1) – 1 (3y –1) = 0

(3y – 1) (4y – 1) = 0

y = 1/3, ¼

12x^2 – 10x + 2 = 0

6x^2 – 5x + 1 = 0

3x (2x –1) – 1 (2x –1)= 0

x = 1/3, ½

x>=y - I. x ^4 – 1000 = 296

II. y^2 + 330 = 346x=y or relation cannot be established.x>=yx>yy>xy>=xOption A

x = + 6,-6

y = + 6, -6

No relation - I. 21x^2 – 26x + 5 = 0

II. y = √0.1024y>xx>yy>=xx=y or relation cannot be established.x>=yOption D

x = 0.23, 1

y = + 0.32

No relation - I. (6^ 3+9^2)/ 11 = x^3

II. 15y^3= (36×18) + 12y^3x>=yy>xx=y or relation cannot be established.y>=xx>yOption B

(6^3+9^2)/ 11 = x^3

=>216+81 11 = x ^3

=>297 11 = x ^3

=>x ^3 = 27

x = +3

15y^3= (36×18) + 12y^3

=>15y^3 –12y^3 = (36×18)

=>3y^3 = (36×18)

=>3y^3 = 648

=>y ^3 = 216

y = +6

y>x - How many 3-letter words with or without meaning, can be formed out of the letters of the word, ‘LOGARITHMS’, if repetition of letters is not allowed?
700660720656685Option C

The word ‘LOGARITHMS’ has 10 different letters.

Required answer = 10P3

=10×9×8=720

- Compound Interest for two years at 20% per annum would fetch a sum of money Rs. 482, if the interest was payable half- yearly then it was payable annually. What is the sum?
3500022800330002000025200Option D

Let sum=Rs.x

C.I. when compounded half yearly = [x(1+10/100)^4−x] = 4641/10000

C.I. when compounded annually = [x(20/100)^2−x] = 11/25

4641/10000 x−11/25x = 482

=> x=20000 - A and B can do a piece of work in 45 and 40 days respectively.They began work together but A leaves after X days and B finished the rest of the work in X+14 days. After how many days did A leave?
1179105Option C

Total work in one day = (1/45+1/40) = 17/360

In X days, 17X/360

Remaining work = (1-17x/360)

Now, B can do 1 pt. work in (x+14)/(1-17x/360) days.

(x+14)/(1-17x/360) = 40

X = 9 days. - A train travelling at a speed of 75 mph enters a tunnel 3(1/2) miles long. The train is ¼ mile long. How long does it take for the train to pass through the tunnel from the moment the front enters to the moment the rear emerges?
53742Option B

Total distance covered = (7/2+1/4) miles

= 154 miles

Time taken = (154*75) hrs = 120 hrs.

= (1/20*60) min.= 3 min - In a city of an adult population, 45% of men and 25% of women are married. If no man marries more than one woman, and vice versa, What is the percentage of total population of adults who are married ?
20.25%.32.14%.28.15%.25.23%.30.12%.Option B

Let no. of men = x & total population be 100.

Total women = (100-x)

Married man = 45x /100 —————(1)

& married women = 25/100 (100-x) ——— (2)

From (1) & (2) x = 500/14

So, Married man = 225 /14

Married woman = 225/14

Total married population = men + women = 450/14

Required % = (450/14)/100 *100

= 450/14% = 32.14%.

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