SRCC GBO 2022

Since $$\left(x^2 + \frac{1}{x^2}\right) = 8.25$$, we can rewrite the same as $$\left(x-\ \frac{\ 1}{x}\right)^2\ +\ 2x\times\ \ \frac{\ 1}{x}$$ = 8.25

This gives us $$\left(x-\ \frac{\ 1}{x}\right)^2$$ = 6.25

Or, $$\left(x-\ \frac{\ 1}{x}\right)$$ = 2.5

Now, $$\left(x\right)^3-\ \left(\frac{\ 1}{x}\right)^3$$ = 

On substituting the values, we get the answer as 30.625

Since x is inversely proportional to  $$(2y^2 - 5)$$, we get the equation as x= $$\frac{k\ }{2y^2-5}$$ where k is the proportionality constant.

Now, x= 25 when y=15. Putting this in the above equation, we get the value of k as 11,125.

When y= 10, x = $$\frac{11125\ }{200-5}$$, which is approximately equal to 57.

g(x) = $$x^4$$

Now, fog(x) is $$x^4$$ + 7, which is Option C.

The summation of 10 steps, 11 steps, and 20 steps is 41 steps. 

The section formula to find the point dividing the line is $$\ \frac{\ na+mb}{n+m}$$ where the ratio is n:m and the coordinates (x or y axis) are a and b.

Applying the same for x-coordinate:  $$\ \frac{\ 44}{8}$$, which is equal 5.5

For y-coordinate: $$\ \frac{\ 56}{8}$$ which is 7.

Hence, the point is (5.5 , 7)

Since the different combinations (both selection and placement) would produce different signals, we need to consider each case.

Case 1: When only 1 light is used

Here, we need to select and arrange 1 light out of seven. It can be done in 7 ways. However, two produce signals, we need at least two lights. Hence, we shall discard this case. 

Case 2: When 2 lights are used

This can be done by selecting two lights out of seven and then arranging them, which is $$7c_2\times\ 2!$$, which is 49 ways

Case 3: When 3 lights are used

This can be done by selecting three lights out of seven and then arranging them, which is $$7c_3\times\ 3!$$, which is 210 ways

Case 4: When 4 lights are used

This can be done by selecting four lights out of seven and then arranging them, which is $$7c_4\times\ 4!$$, which is 840 ways

Case 5: When 5 lights are used

This can be done by selecting five lights out of seven and then arranging them, which is $$7c_5\times\ 5!$$, which is 2520 ways

Case 6: When 6 lights are used

This can be done by selecting 6 lights out of seven and then arranging them, which is $$7c_6\times\ 6!$$, which is 5040 ways

Case 7: When 7 lights are used

This can be done by selecting 7 lights out of seven and then arranging them, which is $$7c_7\times\ 7!$$, which is 5040 ways

Adding Case 2 till Case 7, we get the answer as 13,699

Let the ratio in which the two varieties are mixed be a:b

Hence, $$\ \frac{40a+50b}{a+b}$$ = 43

$$\ 40a+50b=43a+43b$$

This gives us 7b = 3a or a:b = 7:3

The option has an error; this happened in the actual exam paper, so we can't change it.

Since the reduction is ₹250, which is 5% of the total bill amount.

Hence, the total due amount is x, whose 5% is ₹250.

This gives us: 5% of x is ₹250, which makes x as 5% is ₹5,000

In 8 litres of A, the quantity of petrol (40% of the total quantity) is 3.2 litres and that of kerosene is 4.8 litres. 

In 12 litres of B, the quantity of petrol (60% of the total quantity) is 7.2 litres and that of kersone 4.8 litres. 

In the final mixture, the quantity of petrol becomes 10.4 litres and that of kerosene becomes 9.6 litres, making the ratio as 13:12.