SRCC GBO 2022 Question Paper

$$6^{15} + 6^{16} + 6^{17} + 6^{18}$$

Let's take $$6^{15}$$ common from the whole.

We get $$6^{15}$$(1+$$6^{1}$$+$$6^{2}$$+$$6^{3}$$).

$$6^{15}$$ can be written as $$2^{15}$$*$$3^{15}$$.

24 can be written as $$2^{3}*3$$.

So, $$6^{15}$$ is divisible by 24. Hence, the remainder is zero.

The volume of a cube is $$s^3$$, where s is the edge measure. 

Hence, the volume of the three cubes is: 125 cub.cm, 343 cub.cm, and 729 cub.cm

When melted, the total volume is 1,197 cub.cm.

To find the cube's edge with the volume of 1,197 cub.cm, we shall find the cube root of this number, which happens to be 10.62 cm (approximately).

Since $$\beta\ =\ 5\alpha\ \ $$, we can say that the product of roots ($$5\alpha\ ^2\ \ $$) = $$\ \frac{\ c}{a}$$ ............ (1)

and sum of roots ($$6\alpha\ $$) = $$\ \frac{\ -b}{a}$$ ................. (2)

Hence, $$\alpha\ $$ = $$\ \frac{\ -b}{6a}$$    ............. (3)

Substituting (3) in (1), we get $$5\times\ \ \frac{\ b^2}{a^2}$$ = $$\ \frac{\ c}{a}$$

or, $$5b^2 = 36 ac$$

Probability of at least 3 bottles being safe = Probability of at least 3 bottles being safe + Probability of all 4 bottles being safe

To find Probability of all 4 bottles being safe, we also need to multiply the result with 4, since any of the four bottles can be damaged.

Also, probability of a bottle being damaged is $$\ \frac{\ 9}{10}$$

Hence, Probability of at least 3 bottles being safe = $$\ \ \left(\frac{\ 8}{9}\right)^4$$ + 4$$\ \times\ $$ $$\ \left(\frac{\ 8}{9}\right)^3\times\ \left(\ \frac{\ 1}{9}\right)$$

On solving this, we get the answer as $$\frac{\ 2048}{2187}$$

The common difference of $$[p_n]$$ is 2, while that of $$[qn]$$ is 3. 

The last term for $$[p_n]$$ will be 4 + 49$$\times\ $$2, which is 102

Similarly, the last term for $$[qn]$$ will be 2 + 49$$\times\ $$3, which is 149

Now, the first common term for these two series is 8. Further, the common difference for the series of common terms shall be the LCM (2,3), which is 6.

Hence, 8 + 6K < 102

Or, K < 15.666.... 

This means K can take the maximum value of 15 and minimum of 0. Hence, 16 terms are common to the two series. 

In this question, we need to assume two cases for the sign of the denominator because of the inequality.

Case 1: 2x-1 > 0, or x> $$\frac{\ 1}{2}$$

This makes x+1 > 2x-1, or x<2

Hence, the range is $$\frac{\ 1}{2}$$ < x < 2

Case 2: 2x-1 < 0, or x < $$\frac{\ 1}{2}$$

This makes x+1 < 2x-1, or x > $$\frac{\ 1}{2}$$

This case is inconsistent in nature. Hence, the answer is $$\frac{\ 1}{2}$$ < x < 2

$$\angle A$$+$$\angle B$$+$$\angle C$$ = 180

Or, $$\angle B$$+$$\angle C$$ = 126

Now, the angle bisectors divide $$\angle B$$ and $$\angle C$$ into half, hence their sum is 63

Now, in $$\triangle BOC$$, we 63+$$\angle BOC$$ = 180, or $$\angle BOC$$ = $$117^\circ$$

=> Avinash invested ₹6,400 in three business at 3%, 5% and 7% per year with simple interest.

Simple interest formula is $$SI\ =\ \frac{PRT}{100}$$

Where P is principal, R is rate of interest and T is time duration in years

Here, Let us assume the amount invested in three businesses is P1, P2 and P3.

Therefore, the interest after 1 year because of simple interest will be 3P1/100 , 5P2/100 and 7P3/100

These are equal, Hence

=> 3P1 = 5P2 = 7P3

It is also known that total investment done in three business is equal to 6400

P1 + P2 + P3 = 6400

Substituting values of P2 and P3 in terms of P1

P1+3P1/5 + 3P1/7 = 6400

We need to calculate the value of P1

(35+21+15)P1/35 = 6400

=> 71P1/35 =6400

=> P1 is approximately 3200 (Option B is correct answer)

To divide $$(21^{13} + 29^{13})$$ by 25, we shall divide 21 and 29 individually by 25. 

 By doing so, we get the remainder as $$(4^{13} + (-4)^{13})$$, which is 0. 

First, we shall consider the series as a summation if two infinite series: 

S1: $$\ \ \frac{\ 1}{5}\ +\ \ \frac{\ 1}{25}+\ ......$$

S2: $$\ \ \frac{\ 3}{25}\ +\ \ \frac{\ 3}{625}+\ ......$$

Taking S1: The sum of series shall be $$\ \ \frac{\ a}{1-r}\ $$, hence $$\ \ \frac{\ \ \frac{\ 1}{5}}{1-\ \frac{\ 1}{25}}\ $$, which is $$\ \frac{\ 25}{120}$$

Taking S2: The sum of series shall be $$\ \ \frac{\ a}{1-r}\ $$, hence $$\ \frac{\ \ \frac{\ 3}{25}}{1-\ \frac{\ 1}{25}}$$, which is $$\ \frac{\ 1}{8}$$

Adding S1 and S2, we get the answer as $$\ \frac{\ 1}{3}$$