SRCC GBO 2024 Question Paper

Join the points O and P, the slope of this line is 1/2/ Join the points O and Q, the slope of this line is -2. The product of slopes of two perpendicular lines will be -1. These two lines are perpendicular. The point R lies on OP and S on OQ. Since these two lines are perpendicular, the line joining R and S will be the hypotenuse. So it is $$\sqrt{\ 30^2+40^2}=50$$.

Option B and D are both correct. The sum of interior and exterior angles is 180. Since the ratio is 9:2, let the exterior angle be 9x, and the interior angle be 2x.

$$11x=180$$

$$x=\frac{180}{11}$$

The exterior angle of a regular polygon is 360/n. The derived exterior angle from the x value is 360/11.

So, the n value is 11.

So option B, D are correct.

In this question, we need to check with each option.

A) If 1/6 part of the mixture is replaced with water.

The initial part of the water is 3/8, and that of milk is 5/8. When 1/6 of the mixture is replaced with water, the milk taken out will be added as water. $$\frac{1}{6}\cdot\frac{5}{8}=\frac{5}{48}$$ is the amount of milk that is taken out, and that will also be the amount of water added.

So, the new ratio will be $$\frac{x}{y}=\frac{\left(\frac{3}{8}+\frac{5}{48}\right)}{\frac{5}{8}-\frac{5}{48}}=\frac{23}{25}$$. But in the option it is given that x>y. So, it is the incorrect option.

B) If $$\frac{1}{5}$$, part of the mixture is replaced with water.

The initial part of the water is 3/8, and that of milk is 5/8. When 1/5 of the mixture is replaced with water, the milk taken out will be added as water. $$\frac{1}{5}\cdot\frac{5}{8}=\frac{1}{8}$$ is the amount of milk that is taken out, and that will also be the amount of water added.

So, the new ratio will be $$\frac{x}{y}=\frac{\left(\frac{3}{8}+\frac{1}{8}\right)}{\frac{5}{8}-\frac{1}{8}}=\frac{\left(\frac{4}{8}\right)}{\left(\frac{4}{8}\right)}=1:1$$. In the option it is given that x=y. So, it is a correct option.

C) If $$\frac{1}{7}$$, part of the mixture is replaced with water.

The initial part of the water is 3/8, and that of milk is 5/8. When 1/7 of the mixture is replaced with water, the milk taken out will be added as water. $$\frac{1}{7}\cdot\frac{5}{8}=\frac{5}{56}$$ is the amount of milk that is taken out, and that will also be the amount of water added.

So, the new ratio will be $$\frac{x}{y}=\frac{\left(\frac{3}{8}+\frac{5}{56}\right)}{\frac{5}{8}-\frac{5}{56}}=\frac{26}{30}$$. In the option it is given that xy. So, it is a correct option.

D) If $$\frac{1}{4}$$, part of the mixture is replaced with water.

The initial part of the water is 3/8, and that of milk is 5/8. When 1/4 of the mixture is replaced with water, the milk taken out will be added as water. $$\frac{1}{4}\cdot\frac{5}{8}=\frac{5}{32}$$ is the amount of milk that is taken out, and that will also be the amount of water added.
So, the new ratio will be $$\frac{x}{y}=\frac{\left(\frac{3}{8}+\frac{5}{32}\right)}{\frac{5}{8}-\frac{5}{32}}=\frac{17}{15}$$. In the option it is given that x>y. So, it is a correct option.

$$\log_{10}50 + \left(\frac{\log_{0.5}5}{1+\log_{2}5}\right)$$

$$\log_{0.5}5\ $$ can be written as $$\frac{\log_25}{\log_20.5}$$

It becomes -$$\log_25$$ as $$\log_20.5\ $$ is -1.

$$1+\log_25\ $$ will be $$\log_210\ $$ since 1 can be written as $$\log_22$$.

After these calculations $$ \left(\frac{\log_{0.5}5}{1+\log_{2}5}\right)$$ will trun into -$$\frac{\log_25}{\log_210}$$ which is -$$\log_{10}5$$

Now we have $$\log_{10}50 - \log_{10}5$$ which is nothing but $$\log_{10}10$$. So the answer is 1.

I) When a number has to be divisible by 52, basically it has to be divisible by 4 and 13.

Taking $$2^{10\ }$$common from the expression $$2^{10\ }*\left(2^{10}+3^{10}\right)$$. $$2^{10\ }$$ is divisible by 4 but not by 13.

So lets check if $$2^{10\ }+3^{10}$$ is divisible by 13.

Remainder of$$2^{10\ }$$when divided by 13.

$$2^{10\ }=2^5\cdot2^5$$

$$2^5\%13=32\%13=6$$

$$2^{10\ }=2^5\cdot2^5 = \left(13k+6\right)\cdot\left(13k+6\right)=13K+36=13K+10$$

So the remainder is 10.

Remainder of $$3^{10\ }$$when divided by 13.

$$3^{10\ }=3^5\cdot3^5$$

$$3^5\%13=243\%13=9$$

$$3^{10\ }=3^5\cdot3^5 = \left(13k+9\right)\cdot\left(13k+9\right) = 13K+81 = 13K+3$$

The remainder of $$2^{10 }+3^{10}$$when divided by 13 = 10+3 = 13, so it is divisible by 13.

II. $$7^{15} + 64^{5}$$ is divisible by 11.

Check for $$7^3$$ %11

343%11=2

$$7^{15}=\left(7^3\right)^5=\left(11k+2\right)^5=11K+32=11K+10$$

This gives the remainder 10.

Remainder of $$64^5$$ %11

This is same as$$\left(2^6\right)^5\%11=2^{30}\%11$$

$$2^{10}\%11=1024\%11=1$$

$$2^{30}\%11=\left(2^{10}\right)^3\%11=1^3\%11$$

This gives the remainder 1.

$$\left(7^{15}+64^5\right)\%11=\left(10+1\right)\%11=0$$

So, it is divisible by 11.

III.$$2^{20} - 49^{10}$$ is divisible by 9.

Re-write it as $$\left(2^{10}\right)^2-\left(7^{10}\right)^2$$

It is in the form of $$a^2-b^2$$=$$(a+b)\cdot(a-b)$$.

$$\left(2^{10}+7^{10}\right)\cdot\left(2^{10}-7^{10}\right)$$

Apply the same for $$\left(2^{10}-7^{10}\right)$$

$$\left(2^5+7^5\right)\cdot\left(2^5-7^5\right)\cdot\left(2^{10}+7^{10}\right)$$

Now, check each of the terms to see if they are divisible by 9. If atleast one is divisible, then the whole term is divisible by 9.

Check $$\left(2^5+7^5\right)$$

$$\left(32+343\cdot49\right)\%9$$

$$\left(9k_1+5+\left(9k_2+1\right)\cdot\left(9k_3+4\right)\right)\%9$$

where $$k_1,\ k_2,\ k_3\ are\ constant$$

$$\left(9k_1+5+\left(9k_4+4\right)\right)\%9$$

$$\left(9K+5+4\right)\%9$$

$$\left(9K+9\right)\%9$$

Since this term is divisible by 9, statement 3 is true.

IV. $$3^{15} - 8^{5}$$ is divisible by 5.

$$3^{15} - 8^{5}$$ = $$3^{15}-2^{15}$$

Rewriting it as $$\left(3^5\right)^3-\left(2^5\right)^3$$

This is in the form $$a^3-b^3=\left(a-b\right)\cdot\left(a^2+b^2+a\cdot b\right)$$

$$\left(3^5-2^5\right)\cdot\left(3^{10}+2^{10}+6^5\right)$$

$$\left(243-32\right)\cdot\left(243\cdot243+32\cdot32+6^5\right)$$

$$\left(211\right)\cdot\left(243\cdot243+32\cdot32+6^5\right)$$

When divided by 5, the remainder will depend on the unit digit. If it is below 5, that will be the remainder. If it is above 5, subtract the number from 5, the result will be the remainder.

$$\left(1\right)\cdot\left(3\cdot3+2\cdot2+1\right)=\left(1\right)\cdot\left(9+4+1\right)$$

Any power of six will have the unit digit as 6; hence, the remainder will be 1.

14%5=4.

So, the number will not be divisible by 5.

Hence option B is correct.

Convert the ratio into integers by multiplying it by the LCM of denominators; now, the ratio becomes 40:105:36.

Let the shares held by A, B, and C be 40x, 105x, and 36x, respectively.

A, B and C hold shares for one year. Other than that, A holds 108.75% more than he holds, which is 43.5x more from the 5th month.

Now multiply shares, and the time each one holds equates to 17208.

40x$$\cdot12+105x\cdot12+36x\cdot12+43.5x\cdot8=2540\cdot x$$

2540x=17208

x=$$\frac{17208}{2504}$$

B's share is 105x*12*$$\frac{17208}{2504}$$

This is 8604

Let the cost price of an article be C. Given the profit percentage is 25%. The selling price of the article will be 1.25C. Given that the cost price of 25 articles is equal to the selling price of m articles, let's frame an equation from this.

$$25\cdot C\ =\ m\cdot1.25\ C=>m=20$$

This is a concept of weighted average; assuming that there are equal men and women, the average would be 1150. But since there are more men, the average is more than 1150. Also, the average can't go beyond 1300 and below 1000. So, option C is the

Sum to n terms in GP is $$a\cdot\frac{\left(r^n-1\right)}{r-1}$$

The sum of the first 18 terms is equal to 22 terms.

So $$a\cdot\frac{\left(r^{18}-1\right)}{r-1} = a\cdot\frac{\left(r^{24}-1\right)}{r-1}$$

This will be reduced to $$\left(r^{18}-1\right)=\ r^{22}-1$$

Reducing this we get $$r^{4\ }=1$$

Now r has 4 posibilities 1, -1, i,-i.

1 is impossible as that would lead to a zero denominator in the sum of terms formulae.

In all the other cases of r, the sum of the first four values would be zero.