SRCC GBO 2024

Check with each of the options,

$$7^{141} + 1$$ can be written as $$\left(7^{47}\right)^3+1^3$$

$$a^3+b^3=\left(a+b\right)\left(a^2+b^2-ab\right)$$

So, $$7^{141} + 1$$ is $$\left(7^{47}+1\right)\left(7^{94}+1-7^{47}\right)$$

Hence, option B is the correct answer.

Assume that there are x horses and y sheep. They both add up to 100%. If 50% of horses were sheep, there would have been 50% more sheep than the number of horses. So, considering horses count as h/2 and sheep count as h/2+s, we can frame the equation (h/2+h/4) = s+h/2. Solving this, we get s=(h/4). Since both add to 100%, we get horses as 80%.

Factorize 1080 - $$2^3\cdot3^3\cdot5$$. Here, HCF has to be between 3 and 12; the only possible number would be 6. The remaining thing that exists is only 5. So there are two ways: one, 5, belong to x, and two, five, belong to y.

Given a chord of length r, and the lines joining the centre and the ends of the chord are also radii, these lines form an equilateral triangle.

The area of the minor part will be the area of sector OAB minus the $$\triangle\ $$le OAB will be the required answer.

The area of sector = $$\frac{\theta}{360^o}\cdot\pi\ r^2$$

Where $$\theta\ $$=$$60^o$$

Area of sector = $$\frac{60^o}{360^o}\cdot\pi\ \cdot42^2$$ = $$\frac{\pi}{6}\ \cdot42^2$$

Area of triangle OAB = $$\frac{1}{2}\cdot b\cdot h\ =\ \frac{1}{2}\cdot42\cdot\frac{42\sqrt{\ 3}}{2}\ =\ 42^2\cdot\frac{\sqrt{\ 3}}{4}$$

Area of shaded region = Area of sector OAB - area of $$\triangle\ $$le OAB = $$\frac{\pi}{6}\ \cdot42^2 - \ 42^2\cdot\frac{\sqrt{\ 3}}{4}$$ = $$42^{2} (\frac{\pi}{6} - \frac{\sqrt{3}}{4})$$

The domain of the log is (0, $$\infty\ $$). So the x could go from (-1,$$\infty\ $$). In general, the value of a log goes negative when the argument of a log is between 0 and 1 when the base is 10. In this question, the argument stays in that range when x is between -1 and 0. Hence, the graph will pass through the third quadrant when x is in that range. And when the x is positive, the log value will be positive. Hence, the graph passes through the first quadrant.

In this, there are two kinds of separations. Men and women, Indian and Non-Indian. It is given that 2/3 of all are Indians. From this, we deduce that 1/3 are Non-Indians. Of these 1/3 Non-Indians, 3/8 are women. Now we can derive the Non-Indian man, it is (1/3)*(1-(3/8))=5/24. 5/24 of all are Non-Indian men. It is given that 1/3 of all are women, so 2/3 are men. Of 2/3 men, 5/24 are Non-Indians, so Indian men are 11/24. The probability of a picked man being Non-Indian is (5/24)/(2/3). That is 5/16.

The longest chord means diameter. The given lines make a right-angled triangle. The circumcircle will touch all three vertices of the triangle. Since there is a right angle at a vertex, the side opposite to that will be the diameter. The line is 4x + 3y = 24. This line touches the x-axis at(0, 8) and the y-axis at (6,0). The distance between the points is the diameter.

The distance between the two points is $$\sqrt{\ 8^2+6^2}=10$$

I) $$ab(a + b) + bc(b + c) + ca(c +a) \leq 6abc $$

$$\frac{ab(a+b)}{abc}+\frac{bc(b+c)}{abc}+\frac{ca(c+a)}{abc}\le6$$

$$\frac{(a+b)}{c}+\frac{(b+c)}{a}+\frac{(c+a)}{b}\leq6$$

We shall rearrange the terms

$$\frac{a}{c}+\frac{c}{a}+\frac{b}{c}+\frac{c}{b}+\frac{a}{b}+\frac{b}{a}\le6$$

Here, we have six terms: three fractions and their reciprocals.

$$t+\frac{1}{t}\ge2$$ this is always true.

So three such equations add to $$\ge6$$

So, it is only sometimes valid.

II. $$\frac{a^{2} + b^{2} + c^{2}}{abc} \leq \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$$

$$\frac{a^2+b^2+c^2}{abc}\le\frac{a\cdot b+b\cdot c+c\cdot a}{abc}$$

$$a^2+b^2+c^2\le a\cdot b+b\cdot c+c\cdot a$$

Consider (a,b,c) as (1,2,3)

Substituting them in the above equation,

$$1^2+2^2+3^2\le1\cdot2+2\cdot3+3\cdot1$$

$$14\le11$$ is not true, so this is also not true.

Hence, option D is correct.

Let the speed of car A be A and car B be B. Given that the speed of A is 60kmph. Let us assume that they both meet at Y after a time of t. A travels a distance of 400 in 6hr 40m. This distance is covered by B in t time.

400= $$B\cdot t$$. --->1

Again, the distance travelled by A in t time will be covered by B in 3hr 45m, which can be written as$$\ A\cdot t=B\cdot\ \frac{15}{4}$$ -->2

Substitute B =$$400/t$$ from eq 1. We know that A is 60.

$$60\cdot t=\left(\frac{400}{t}\right)\cdot\frac{15}{4}$$

$$t^2=25$$

t=5

The question is asked for atleast two, but the answer is provided for three.

First, arrange two out of four girls in the last two positions in 4*3 ways. Now consider three boys as a unit and two unarranged girls as two units. Arranging these three units can be done in 3! ways. Again, those three boys can be arranged in 3! ways.

So, the answer is 432.