SRCC GBO 2016 Question Paper

Let the quantity of hematite mined from the mine in a year = 100x

Wasted ore = 20% of 100x = 20x

Pure iron = 25% of (100x - 20x) = 20x = 80000

x = 4000

Hence, quantity of hematite mined from the mine in a year = 100x = 400000 kg

Speed of train without wagons = 24 km/h

Reduction in speed with 'n' wagons $$\propto \sqrt{n}$$

Reduction in speed with 'n' wagons $$= k\sqrt{n}$$

Speed of train with 4 wagons = 20 = 24 $$-k\sqrt{4}$$

2k = 4

k = 2

Now, 

Consider the case when the speed of train is zero. 

$$24 - 2\sqrt{n} = 0$$

$$24 = 2\sqrt{n}$$

$$12 = \sqrt{n}$$

$$n = 12^2$$

n = 144

Hence, total number of wagons will be 143 when the train cna just move. 

Case 1: Total ways when one box contains one green ball = {1}. {2}, {3}, {4}, {5}, {6} = 6 ways

Case 2: Total ways when two boxes contain one green ball = {1, 2}, {2, 3}, {3, 4}, {4, 5}, {5, 6} = 5 ways

Case 3: Total ways when three boxes contain one green ball = {1, 2, 3}, {2, 3, 4}, {3, 4, 5}, {4, 5, 6} = 4 ways

Case 4: Total ways when four boxes contain one green ball = {1, 2, 3, 4}, {2, 3, 4, 5}, {3, 4, 5, 6} = 3 ways

Case 5: Total ways when five boxes contain one green ball = {1, 2, 3, 4, 5}, {2, 3, 4, 5, 6} = 2 ways

Case 5: Total ways when all the six boxes contain one green ball = {1, 2, 3, 4, 5, 6} = 1 way

Tota; ways = 6 + 5 + 4 + 3 + 2 + 1 = 21

Since, N persons are standing on the circumference of a circle at distinct points and each possible pair of persons, not standing next to each other, sings a two-minute song one pair after the other.

Total time taken is 28 minutes when each pair sings for 2 minutes, then total time taken will be 14 minutes when each pair sings for 1 minute. 

A person standing can't form pair with himself and with his two neighbours. Hence, total possible pairs = N * (N - 3).

This also includes the pair in both clockwise and anticlockwise direction. 

Hence, the actual number of pair = $$\frac{[N(N - 3)]}{2} = 14$$

N(N - 3)=28

N = 7

Case 1: When there is 3 chairs in each row, then total rows = $$\frac{36}{3} = 12$$ (Valid)

Case 2: When there is 4 chairs in each row, then total rows = $$\frac{36}{4} = 9$$ (Valid)

Case 3: When there is 6 chairs in each row, then total rows = $$\frac{36}{6} = 6$$ (Valid)

Case 4: When there is 9 chairs in each row, then total rows = $$\frac{36}{9} = 4$$ (Valid)

Case 5: When there is 12 chairs in each row, then total rows = $$\frac{36}{12} = 3$$ (Valid)

Hence, total possible arrangements = 5

Let passing marks in the exam = 90

Marks obtained by A = 90% of 90 = 81

Marks obtained by B = $$(100 - 11\frac{1}{9})$$% of 81 = 72

Marks obtained by A and B together = 81 + 72 = 153

Marks obtained by C = $$(100 - 41\frac{3}{17})$$% of 153= 90

Hence, C will pass the exam

Since, three cards are drawn successively after replacing the previous one. 

Probability of drawing first card which is Black = $$\frac{26C1}{52C1} = \frac{26}{52} = \frac{1}{2}$$

Probability of drawing second card which is Diamond = $$\frac{13C1}{52C1} = \frac{13}{52} = \frac{1}{4}$$

Probability of drawing third card which is Ace= $$\frac{4C1}{52C1} = \frac{4}{52} = \frac{1}{13}$$

Required probability = $$\frac{1}{2} * \frac{1}{4} * \frac{1}{13} = \frac{1}{104}$$

Let boys and girls in the class are 'x' each. 

Total students = x + x = 2x

Probability of selecting two girls from the class = $$\frac{xC1}{2xC1} = \frac{7/29}$$

$$\frac{[x(x - 1)]}{[2x(x - 1)]} = \frac{7}{29}$$

29(x - 1) = 14(2x - 1)

29x - 29 = 28x - 14

x =15

Strength of class = 2x = 30

First we'll select 1 square out of total 64 squares that means we can't select second square from that row and that column. Now we're left with 7 rows and 7 columns from which we can select another square. 

Hence, total square left = 7 * 7 = 49 [From which we can select another square].

Probability to select first square = $$\frac{64C1}{64C1} = 1$$

Probability to select first square = $$\frac{49C1}{63C1} = \frac{49}{63}$$

Required probability = $$1 * \frac{49}{63}$$

$$= \frac{7}{9}$$

Let's assume the two digit number P as uv=u*10+v

Let's assume the 3 digit number Q as xyz= x*100+y*10+z

The actual multiplication result is uv*xyz=(u*10+v)*( x*100+y*10+z)

But he reversed the digits of Q. So, Q is considered as zyx = z*100+y*10+x

The result of multiplication after this operation is uv*zyx=(u*10+v)*( z*100+y*10+x)

Given the difference between these is 22770.

(u*10+v)*( z*100+y*10+x) - (u*10+v)*( x*100+y*10+z)=22770

Let's assume (u*10+v) as t.
t*( z*100+y*10+x) - t*( x*100+y*10+z)=22770

100zt+10yt+xt-100xt-10yt-zt=22770

99zt-99xt=22770

99(zt-xt)=22770

zt-xt= 230

(z-x)*t=230

z and x are two single digit numbers. So, the difference will also be a single digit number. And t is as two digit number.

Let's check the possible values of these. 

230 can be written as 2*115, 5*46, 10*23

2*115: this is not possible because t is not a 3 digit number.

5*46 is possible.

10*23 is not possible because z-x is not a two digit number.

So, the only possible value is z-x=5 and t=46

The question ask the minimum value for sum of digits of Q. The minimum value we can give for z,x is 6,1. And the minimum value of y is 0.

So, the sum is 6+1+0=7