SRCC GBO 2022

The total sum of heights of 30 students was 30 $$\times$$ 165, which is 4950 cm. 

Now, 5 students with an average height of 172 cm joined the class, which makes the total height as 860+4950= 5810 cm.

The average height of the 35 students now will be $$\ \frac{\ 5810}{35}$$ = 166 cm. 

In any star-shaped figure, the sum of vertex angles is 180$$\times\ $$(n-4), where n is the number of vertices. 

Hence, in this case, it has to be 180$$\times\ $$(5), which is 900$$^\circ$$

The multiplying factor for a decrease of 25% is 0.75

Hence, the value of the ATM machine after two years shall be 55,000$$\times0.75\times\ 0.75$$, which is 30,937.5

In this question, we shall find the HCF of the three numbers 780, 690, and 950, which shall match with the options. 

It is clear that 10 shall be the answer, which can used to measure all the three different volumes. 

When f(x) is shifted 2 units to left, it becomes f(x-2). Now, 12 units downwards would mean f(x-2) - 12

Hence, the new equations h(x) is $$\left(x-2\right)^2-5\left(x-2\right)\ +\ 6\ -12$$, which is $$x^2 - x - 12$$

For $$f(x) = \log_{5}(x-4)-4, x > 0$$ to be valid, (x-4) has to be greater than 0. Hence, x > 4

Now, the VA is a line that guides the graph (above which the graph is valid and below which it becomes invalid). Hence, it has to be the line x=4, as x>4 for the graph to be valid.

The sum of two co-interior angles in the same side shall add up to 180$$^\circ$$.

Hence, $$\frac{3x}{4}$$+$$\left(\frac{5x}{4} - 20 \right)$$ = 180

Or 8x = 800, making x = 100

Hence, $$\frac{3x}{4}$$ = 75 and $$\left(\frac{5x}{4} - 20 \right)$$ = 105

Area of ellipse is: $$\pi\ ab$$, where a and b are the half of the length of major and minor axis. 

Substituting a with 6 and b with 1.4 ($$\pi$$ being 3.14), we get the area as 26.4 approximately.  

$$(x^2 - x - 6)$$ is a factor of the fourth degree equation. Hence, the roots of $$(x^2 - x - 6)$$ (3 and -2) will satisfy the fourth degree equation.

Putting x=3 in $$x^4 + ax^3 - x^2 - bx - 12 = 0$$, we get 81+27a-9-3b-12=0, or 27a-3b= -60..............(1)

Putting x= -2 in the equation, we get 16-8a-4+2b-12=0, or 8a-2b=0............ (2)

Solving the two, we get a as -4 and b as -16

Let the speed of the train be x km/h.

The resultant speed of the train with respect to the first person shall be (x-4) km/h, and that for the second person is (x-6) km/h.

In 4.2 seconds, the train overtakes the first person. Hence, the distance between them is $$\frac{\ 4.2}{3600}\times\ \left(x-4\right)$$

In 4.8 seconds, the train overtakes the first person. Hence, the distance between them is $$\frac{\ 4.8}{3600}\times\ \left(x-6\right)$$

These two values are equal. From this, we get x= 20 km/h.