TISSNET 2021

Since it is given that 2 is one of the roots of the equation,

Putting x=2 in the above equation

2(4) + 2k + 4=0

k= -6

Thus, the above equation will be in the form $$2x^2 - 6x + 4 = 0$$

Thus, the correct option is C.

Increase in length of each side = 25%

Apply successive percentage formula

we get increase in area = 25 + 25 + (25*25)/100 = 50+6.25 = 56.25

The possible outcomes will be

Aman, Danish, Anaya, Nero

Anaya, Danish, Aman, Nero

Nero, Aman, Danish, Anaya

Nero, Anaya, Danish, Aman

A rhombus has all sides of equal length.

A kite has two adjacent sides which are equal and the other two adjacent sides are also equal but the opposite sides are not equal.

The diagonals of both a rhombus and a kite bisect each other.

Both the rhombus and kite have two acute angles and two obtuse angles.

LCM(18, 20, 30)= 180

Let total work be 180 units

Abhay can complete the work in 18 days which means in one day he does 10 units of work.

Balram can complete the work in 20 days which means in one day he does 9 units of work.

Chandan can complete the work in 30 days which means in one day he does 6 units of work.

Let the remaining work done by A be x days

Therefore, 180 = 2(9+6) + 10x

10x = 150

x= 15 days

Two shots were fired at an interval of 6 minutes, but the man heard both shots at an interval of 5 minutes 52 seconds which means that between the hearing of the first shot and the second shot, he covered a distance equal to 330*8 m.

Time taken by man to cover the distance = 352 seconds

Speed of man = (330*8)/352 = 7.5 m/s

Speed in Km/hr = 7.5 * (18/5) = 27 Km/hr

Let the positive no. be N

N = kq + 21

N' = 2N = 2kq + 42

Now N' leaves a remainder of 13 when divided by q

So, N' = pq + 13

Therefore Remainder when 42 is divided by q should be 13 which is only possible when q=29.

Since the number is more than double, the total amount will be at least twice the number.

If the number is x, the amount will be more than 2x.

Thus, $$2x\ge\ x\left(1+\frac{1}{10}\right)^{^r}$$

$$\left(1+\frac{1}{10}\right)^{^{r\ }}\ge\ 2$$

For r = 8, the value of $$\left(1.1\right)^r=2.143$$

Thus, the value of r will be 8.

Thus, the correct option is B.

Alternative:

No. of years in which a sum of money invested at r% compounded annually amounts to double is 72/r years

So, no. of years in which the amount will be doubled = 72/10 = 7.2 years

So, the least no. of years for which the amount will be more than doubled will be 8 years.

Let the numbers be x and y
It is given,
xy = 45
x - y = 4
squaring on both sides, we get
$$x^2+y^2-2xy\ =\ 16$$
$$x^2+y^2$$ = 16 + 2(45) = 106

Answer is option A.

Simple Interest biennially = (1/2)*(P*r*2T)/100 = PrT/100

Simple Interest quarterly = (8)*(P*r*T/4)/100 = 2PrT/100

So, Ratio = 1:2