SRCC GBO Sample Paper 2026

Let the first term of the series be a, and the common difference of the series be d. It is given that a = d and the 29th term of the series is given as 609.

$$a_{29}\ =\ a\ +\ \left(29\ -1\right)d\ =\ a\ +\ 28a\ =\ 29a\ =\ 609$$

$$a\ =\ 21\ =\ d$$

The sum of 18 terms can be calculated as,

$$S_{18}\ =\ \dfrac{n}{2}\left(2a\ +\ \left(n\ -\ 1\right)d\right)\ =\ \dfrac{18}{2}\left(2\times\ 21\ +\ \left(18\ -\ 1\right)21\right)\ =\ 9\left(19\ \times\ 21\right)\ =\ 3591\ $$

The correct answer is option C.

Prime factorization of 55440.

$$55440=2^4\times3^2\times5\times7\times11$$

We know that the total factors will be equal to (4+1)*(2+1)*(1+1)*(1+1)*(1+1) = 120

And, the total number of even factors is equal to (3+1)*(2+1)*(1+1)*(1+1)*(1+1) = 96

To find the sum of all factors, we will put all the possible powers of any prime number and add them together and multiply with other primes.

$$2^4\times3^2\times5\times7\times11$$

Sum = $$\left(2^0+2^1+2^2+2^3+2^4\right)\times\left(3^0+3^1+3^2\right)\times\left(5^0+5^1\right)\times\left(7^0+7^1\right)\times\left(11^0+11^1\right)$$

=> 31*13*6*8*12

=> 232128

I) $$(260)_7+\left(740\right)_8=\left(514\right)_{11}$$

$$(260)_7 = 2*7^2+6*7 = 140$$

$$\left(740\right)_8 = 7*8^2+4*8 = 480$$

140+480 =620

$$(620)_{10}=(514)_{11}$$

II)$$\left(824\right)_9+\left(101101\right)_2=\left(710\right)_{10}$$

$$\left(824\right)_9 = 8*9^2+2*9+4 = 670$$

$$(101101)_2=(1\times2^5)+(0\times2^4)+(1\times2^3)+(1\times2^2)+(0\times2^1)+(1\times2^0)\ =\ (45)_{10}$$

670+45 = $$(715)_{10}$$

III) $$\left(0.353\right)_6=\left(0.505\right)_{10}$$

$$3\cdot\frac{1}{6}+5\cdot\frac{1}{6\cdot6}+3\cdot\frac{1}{6\cdot6\cdot6}=\frac{1}{2}+\frac{5}{36}+\frac{1}{72}$$

This is clearly greater than 0.55. So, this is incorrect.

IV) $$\left(0.11111\right)_2=\left(0.96875\right)_{10}$$

$$\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}=0.5+0.25+0.125+0.0625+0.03125=0.96875$$

So, this is a correct statement.

I and IV are the only correct statements.

The lines given are,

3x - 4y - 5 = 0   ----(1)

6x - 8y + 10 = 0  -----(2)

(1) * 2, we get,

6x - 8y - 10 = 0 ----(1)

We can see that (1) and (2) are parallel to each other as the slopes of the lines are equal.

The distance can be calculated using the formula,

$$dis\tan ce\ =\ \dfrac{\left|c_1\ -\ c_2\right|}{\sqrt{\ a^2\ +\ b^2}}$$

$$dis\tan ce\ =\ \dfrac{\left|10\ -\left(-10\right)\right|}{\sqrt{\ 6^2\ +\ \left(-8\right)^2}}$$

$$dis\tan ce\ =\ \dfrac{20}{\sqrt{\ 36\ +\ 64}}\ =\ \dfrac{20}{10}\ =\ 2$$

The correct answer is option B.

As shown in the figure above, we must calculate the length of BD. We are given the length of AC as 16 and AB as 12. The length BC can be calculated as,

$$AC^2\ +\ AB^2\ =\ BC^2$$

$$BC\ =\ \sqrt{\ AC^2\ +\ AB^2}$$

$$BC\ =\ \sqrt{\ 12^2\ +\ 16^{\ 2}}\ =\ \sqrt{\ 144+256}\ =\ 20$$

The length BD = BC + CD = 20 + 12 = 32 cm

The correct answer is option D.

Let the value of AB be x.

Using the Pythagoras theorem, we get,

$$CD^2\ =\ x^2\ +\ 9^2$$

$$CD\ =\ \sqrt{x^2\ +\ 9^2\ }$$

The perimeter is given as,

x + 5 + $$\sqrt{x^2\ +\ 9^2\ }$$ + 14 = 46

$$\sqrt{x^2\ +\ 9^2\ }\ =\ 46\ -\ 19\ -\ x$$

$$\sqrt{x^2\ +\ 81\ }\ =\ 27\ -\ x$$

Squaring on both sides, we get,

$$x^2\ +\ 81\ \ =\ 729\ +\ x^2\ -\ 54x$$

$$54x=\ 729\ -\ 81\ =\ 648$$

$$x\ =\ 12$$ cm

$$CD\ =\ \sqrt{\ 144\ +\ 81}\ =\ 15$$ cm

Area of the quadrilateral = area if rectangle  +  area of triangle

area of rectangle = 12 * 5 = 60 sq. cm

area if triangle  =  $$\dfrac{1}{2}\ \times\ 9\ \times\ 12$$  = 54 sq. cm

Area of quadrilateral = 60 + 54 = 114 sq. cm

The correct answer is option B.

The number of diagonals formula is given as $$\dfrac{n\left(n\ -\ 3\right)}{2}$$. The number of diagonals is given as 20 in the questions. The number of sides can be calculated as.

$$\dfrac{n\left(n\ -\ 3\right)}{2}\ =\ 20$$

$$n\left(n\ -\ 3\right)\ =\ 40$$

$$n^2\ -3n\ -\ 40\ =\ 0$$

$$n^2\ -\ 8n+\ 5n\ -\ 40\ =\ 0$$

$$\left(n\ +\ 5\right)\left(n\ -\ 8\right)\ =\ 0$$

$$n\ =\ 8\ or\ n\ =\ -5$$

Since n cannot be negative, the value of n is 8.

The exterior angle can be calculated as $$\dfrac{360}{n}\ =\ \dfrac{360}{8}\ =\ 45^{\circ\ }$$

The correct answer is option B.

It is given that 65% of the total are boys. So, the probability of picking a boy is 0.65.

We are given that if a boy is picked, the probability of the boy being a class 10 student is 0.3.

Using Bayes theorem $$P\left(10\left|B\right|\right)=\frac{P\left(10\ \bigcap\ B\right)}{P\left(B\right)}$$

$$0.3=\frac{P\left(10\ \bigcap\ B\right)}{0.65}$$

$$P\left(10\ \bigcap\ B\right)=0.3\times0.65\ =\ 0.195$$

$$P\left(9\ \bigcap\ B\right)=0.65-0.195\ =\ 0.455$$

We are told that class 9 and 10 students are in 3:2. So the probability of picking class 9 students is 0.6.

We know that $$P\left(9\ \bigcap\ B\right) =\ 0.455$$. From this $$P\left(9\ \bigcap\ G\right)=0.6-0.455\ =\ 0.145$$

The sum of numbers obtained should not be more than 14.

We will calculate all the possibilities to get a sum of 15,16,17 and 18 to find the probability of getting the sum of more than 14, and subtract it from 1 to get the probability that the sum of numbers obtained on the three unbiased
dice is not more than 14.

When three dice are rolled, the maximum sum that can be obtained is 18.
So, in the given question, the sum can be 15 or 16 or 17 or 18
For the sum to be 15, there are possibilities:
1. (6,6,3) - total cases = $$\dfrac{3!}{2!}$$ = 3
2. (6,5,4) - total cases = 3! = 6
3. (5,5,5) - total cases = 1

For the sum to be 16, there can be two possibilities:
1. (6 , 6 , 4) - total cases = $$\dfrac{3!}{2!}$$ = 3
2. (6, 5, 5) - total cases = $$\dfrac{3!}{2!}$$ = 3

For the sum to be 17, there can be only one possibility:
1. (6, 6, 5) - total cases = $$\dfrac{3!}{2!}$$ = 3
For the sum to be 18, there is only one case, which is (6, 6, 6)

So, the no. of favourable outcomes such that sum is greater than 14 = (10+ 3 + 3 + 3 + 1) = 20
Total no. of outcomes = 6 * 6 * 6 = 216

Required probability =1-  $$\dfrac{20}{216}$$ = $$\dfrac{196}{216}$$ = $$\dfrac{49}{54}$$

Hence, option A is the correct answer.

The equation given is,

$$8^x\ =\ 3^{x^2}$$

Applying log on both sides, we get,

$$\log8^x\ =\ \log3^{x^2}$$

$$x\log8\ =\ x^2\log3$$

$$\ x^2\log3\ -\ x\log8\ =\ 0$$

$$\ x\left(x\log3\ -\ \log8\right)\ =\ 0$$

So, the solutions of the above equation are x = 0 and xlog3 = log8, which can also be written as $$\ x\ =\ \dfrac{\log8}{\log3}\ =\ \log_38$$

The sum of all possible values is $$0\ +\ \log_38\ =\ \log_38$$ which can also be written as $$\log_32\ +\ \log_34$$.

The correct answer is option B.