SRCC GBO 2024

$$\frac{x^\frac{1}{2} + x^{-\frac{1}{2}}}{1-x}+\frac{1-x^{-\frac{1}{2}}}{1-\sqrt{x}}$$

$$\frac{\left(\sqrt{\ x}+\frac{1}{\sqrt{\ x}}\right)}{1-x}+\frac{\left(1-\frac{1}{\sqrt{\ x}}\right)}{1-\sqrt{\ x}}$$

$$\frac{1}{\sqrt{\ x}}\cdot\frac{\left(x+1\right)}{1-x}+\left(\frac{1}{\sqrt{\ x}}\cdot\frac{\left(\sqrt{\ x}-1\right)}{1-\sqrt{\ x}}\right)$$

$$\left(\frac{1}{\sqrt{\ x}}\cdot\frac{x+1}{1-x}\right)-\frac{1}{\sqrt[\ ]{x}}$$

$$\frac{1}{\sqrt{\ x}}\cdot\left(\frac{\left(x+1\right)}{1-x}-1\right)$$

$$\frac{1}{\sqrt{\ x}}\cdot\frac{\left(x+1-1+x\right)}{1-x}$$

$$\frac{1}{\sqrt{\ x}}\cdot\frac{2x}{1-x}$$

$$2\ \frac{\sqrt{\ x}}{1-x}$$

Substitute 7 in place of x.

$$2\ \frac{\sqrt{\ 7}}{1-7}$$

That will be $$-\frac{\sqrt{\ 7}}{3}$$

Let us find out how many subsets of B are subsets of A. For this, we must find A and B's intersection, {6,12}. The number of subsets of this set is $$2^2=4$$. Now, find the total subsets of B.

$$2^n\ where\ n\ is\ no.\ of\ elements$$

$$2^4=16$$

The answer is to subtract the four subsets that are in common with A and B. 16 - 4 =12.

I. $$(4523)_{6} = (1059)_{10}$$

$$\left(4523\right)_6=4\cdot6^3+5\cdot6^2+2\cdot6^1+3\cdot6^0=1059$$

So, this is correct.
II. $$(4523)_{6} = (1069)_{10}$$

$$\left(4523\right)_6=4\cdot6^3+5\cdot6^2+2\cdot6^1+3\cdot6^0=1059$$

So, this is incorrect.
III. $$(0.203)_{5} = (0.424)_{10}$$

$$\left(0.203\right)_5$$ = $$2\cdot\frac{1}{5}+0\cdot\frac{1}{25}+3\cdot\frac{1}{125}=0.424$$

That will be $$\left(0.424\right)_{10}$$

So, this is correct.

IV. $$(0.203)_{5} = (0.406)_{10}$$

$$\left(0.203\right)_5$$ = $$2\cdot\frac{1}{5}+0\cdot\frac{1}{25}+3\cdot\frac{1}{125}=0.424$$

So, this is incorrect.

Here, we have to find the area of the  quadrilateral EQFS.

Lets draw a line from S to Q.

From the image we can say that the area of  SFQE = area of $$\triangle\ $$SFQ + area of $$\triangle\ $$SEQ

area of $$\triangle\ $$SFQ= $$\frac{1}{2}\cdot b\cdot h$$

We know that SR and RQ are perpendicular(given in question). So, height of $$\triangle\ $$SFQ is 11.

$$=\frac{1}{2}n\cdot11$$=$$\frac{11n}{2}$$.

area of $$\triangle\ $$SEQ = $$\frac{1}{2}\cdot b\cdot h$$

We know that SP and PQ are perpendicular(given in question). So, height of $$\triangle\ $$SEQ is 7.

$$=\frac{1}{2}n\cdot7$$=$$\frac{7n}{2}$$.

$$\frac{11n}{2}+\frac{7n}{2}=225\ \ \Rightarrow\ 9n=225\ \Rightarrow n=25$$

Formulae for the compound interest is $$p\cdot\left(1+\frac{r}{100}\right)^n$$

Here n=3, p=5500. The amount he receives after three is equal to $$5500\cdot\left(1+\frac{10}{100}\right)^3$$

It is given that he receives an interest of 33.1% after three years. So, the total sum would be 133.1% of the initial investment.

$$5500\cdot\left(1+\frac{10}{100}\right)^3=5500\cdot\frac{133.1}{100}$$

$$\left(1+\frac{r}{100}\right)^3=1.331$$

$$\left(1+\frac{r}{100}\right)^3=\left(1.1\right)^3$$

From this, r=10.

As per the question, x and y sum up to 20. Assume that there are 72 units of work. Kamla does three units of work a day, and Nirmala does four. We got another equation as well: 3x+4y=72. Solving both gives us x=8, y=12. 

Let's verify each option:

$$\frac{4x}{y}+1\ is\ even\ $$

$$\frac{4x}{y}+1\ $$ = $$\frac{4*8}{12}+1\ $$ $$=\frac{8}{3}+1$$

2X + Y = 22 $$\Rightarrow 2*8+12=28$$

X - Y is an odd integer. 8-12=-4 it is even.

$$x^{2} + 1$$ is a multiple of 13

$$8^{2} + 1=65$$ is a multiple of 13.

So, only option D is correct.

$$\log_42$$ is 1/2. So the denominator will be $$x-12$$. In numrator if x goes below 10 or above 15 the value under root goes negative. X can't be 12 as the denominator goes to 0. So the intervel is $$\left[10,12\right)\ \ U\ \left(12,15\right]$$

The formulae for the sum of first n terms in AP is $$\frac{n}{2}\cdot\left(2\cdot a+\left(n-1\right)\cdot d\right)$$.

$$\frac{12504}{25}=\frac{24}{2}\cdot\left(2\cdot a+23\cdot d\right)$$

$$\frac{1042}{25}=\left(2\cdot a+23\cdot d\right)$$

Lets generate other equation, given the sum of next 24 terms is $$\frac{17112}{25}$$.

So the first 48 terms is $$\frac{29616}{25}=\frac{48}{2}\cdot\left(2a+\left(47\cdot d\right)\right)$$

Solving both we get a=429/25 d=8/25

Find out the LCM of the ropes' length. The LCM of 450, 990,1620 is 89100 cm. So, the side length can be any multiple of 891m, given that the area is less than 10,00,000. So, it has to be 891.

Given the ratio and two points on the line, we can find out the point that divides the line in that ratio using internal section formulae $$\left(\frac{\left(mx_1+nx_2\right)}{m+n},\frac{\left(my_1+ny_2\right)}{m+n}\right)$$.

The divider of line MN(point A) is as follows.

$$\left(\frac{\left(3\cdot1+2\cdot-1\right)}{3+2},\frac{\left(3\cdot1+2\cdot3\right)}{3+2}\right)$$

A$$\left(\frac{1}{5},\frac{9}{5}\right)$$

Similarly, for line ST (point B)

$$\left(\frac{\left(3\cdot2+2\cdot0\right)}{3+2},\frac{\left(3\cdot7+2\cdot-4\right)}{3+2}\right)$$

B$$\left(\frac{6}{5},\frac{13}{5}\right)$$

Also, do the same for the other line; we get points A and B. Now, find the line equation using the formula y=mx+c.

m=$$\frac{y_2-y_1}{x_2-x_1}\ =\ \frac{\left(\frac{13}{5}-\frac{9}{5}\right)}{\frac{6}{5}-\frac{1}{5}}\ =\ \frac{4}{5}$$

The euation becoms $$y=\frac{4}{5}\cdot x+c$$

Substotue one of those two points to get c value.$$\frac{13}{5}=\frac{4}{5}\cdot\frac{6}{5}+c=>c=\frac{41}{25}$$

The equation will be $$y=\frac{4}{5}\cdot x+\frac{41}{25}=>20x-25y+41=0$$