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Arthur wanted his brother to win the tournament. He was dutiful and wanted to do all he could to help him achieve success at the tournament. He was excited about the prospects of this brother’s victory.

The given system of equations is: 

4x + 5y – 3 = 0 

kx + 15y – 9 = 0

The above equations are of the form 

a₁ x + b₁ y − c₁ = 0 

a₂ x + b₂ y − c₂ = 0 

Here, a₁ = 4, b₁ = 5, c₁ = -3 

a₂ = k, b₂ = 15, c₂ = -9 

So according to the question, 

For unique solution, the condition is 

\(\frac{a_1}{a_2}\) = \(\frac{b_1}{b_2}\) = \(\frac{c_1}{c_2}\) 

\(\frac{4}{k} = \frac{5}{15} = \frac{-3}{-9}\)

\(\frac{4}{k} = \frac{1}{3}\)

⇒ k = 12 

Hence, the given system of equations will have infinitely many solutions, if k = 12.

Correct option is (D) The value does not exist

Take |x| = X & |y| = Y

Then given system of equations converts into

3X + 5Y = 8       _________(1)

& 7X - 3Y = 48   _________(2)

Multiply equation (1) by 3 and equation (2) by 5, we get

9X + 15Y = 24        _________(3)

& 35X - 15Y = 240 _________(4)

By adding equations (3) & (4), we get

44X = 264

\(\Rightarrow\) X = \(\frac{264}{44}\) = 6

\(\therefore\) |x| = 6    \((\because X=|x|)\)

Put X = 6 into equation (1), we get

\(3\times6+5Y=8\)

\(\Rightarrow\) 5Y = 8 - 18 = -10

\(\Rightarrow\) Y = \(\frac{-10}5\) = -2

\(\therefore\) |y| = -2 which is not possible because mode never gives negative value. \((\because Y=|y|)\)

Hence, value of y does not exist.

Therefore, value of (x + y) does not exist.

Correct option is D) The value does not exist

Correct option is (A) 50

Let length and breadth of original rectangle are \(l\;and\;b\) respectively.

\(\therefore\) Area of original rectangle \(=lb\) square units

According to given condition, the new area of rectangle will be

\((l+2)(b+2)=lb+54\)

\(\Rightarrow lb+2l+2b+4=lb+54\)

\(\Rightarrow2(l+b)=54-4=50\)

\(\therefore\) Perimeter of the rectangle is

\(P=2(l+b)\) = 50 units

Correct option is A) 50

Multiplying eq2 by 2 and subtracting from eq1

⇒ - 5y + 6y = k – 24

⇒ y = k – 24

Thus, for any value of k it has a unique solution

kx + 2y = 5

3x + y = 1

Multiplying eq2 by 2 and subtracting from eq1

(k – 6)x = 3

⇒ x = 3/(k – 6)

Thus, for k ≠ 6 the system has unique solution

Let the one’s digit be ‘a’ and ten’s digit be ‘b’.

therefore no is = 10b + a

Reversed no = 10a + b

Given,

Sum of the digits of a two - digit number is 9.

Also, nine times this number is twice the number obtained by reversing the order of the digits.

⇒ a + b = 9 ----- (1)

And 9(10b + a) = 2(10a + b)

⇒ 90b + 9a = 20a + 2b

⇒ 88b = 11a

⇒ a = 8b

Substituting value of a in eq1

⇒ 8b + b = 9

⇒ 9b = 9

⇒ b = 1

Thus,

a = 8(1) = 9

Hence, no is 10(1) + 8 = 18

Correct option is (B) 73

Let required two-digit number be ab or (10a+b).

Sum of digits = a+b

According to given conditions, we have

10a+b = 9 (a+b) - 17

\(\Rightarrow\) 10a - 9a + b - 9b + 17 = 0

\(\Rightarrow\) a - 8b + 17 = 0     ______________(1)

And 10a+b = 13 (a - b) + 21

\(\Rightarrow\) 10a - 13a + b + 13b - 21 = 0

\(\Rightarrow\) -3a + 14b - 21 = 0     ______________(2)

From (1) & (2), we obtain

-3 (8b - 17) + 14b - 21 = 0

\(\Rightarrow\) -24b + 51 + 14b - 21 = 0

\(\Rightarrow\) -10b + 30 = 0

\(\Rightarrow b=\frac{30}{10}=3\)

\(\therefore\) a = 8b - 17          (From (1))

= 24 - 17 = 7

\(\therefore\) Required two-digit number is ab = 73.

Correct option is B) 73

Correct option is (A) 73

Let two-digit number be ab where b is unit's digit and a is ten's digit.

\(\therefore\) ab = 10a + b       __________(1)

According to first condition

10a + b = 9 (a+b) - 17

\(\Rightarrow\) 10a - 9a + b - 9b = -17

\(\Rightarrow\) a - 8b = -17        __________(2)

According to second condition

10a + b = 13 (a - b) + 21

\(\Rightarrow\) 10a - 13a + b + 13b = 21

\(\Rightarrow\) -3a + 14b = 21  __________(3)

Multiply equation (2) by 3, we get

3a - 24b = -51       __________(4)

By adding equations (3) & (4), we get

14b - 24b = 21 - 51

\(\Rightarrow\) -10b = -30

\(\Rightarrow b=\frac{-30}{-10}=3\)

Then from (2), we obtain

a = 8b - 17 = 24 - 17 = 7

\(\therefore ab=10\times7+3=73\)

Hence, required number is 73.

Correct option is A) 73

Let the one’s digit be ‘a’ and ten’s digit be ‘b’

Given, number consists of two digits whose sum is five.

When the digits are reversed, the number becomes greater by nine.

⇒ a+ b = 5 ----- (1)

and 10a + b – (10b + a) = 9

⇒ a – b = 1 ------ (2)

Adding (1) and (2)

Thus, 2a = 6

⇒ a = 3

∴ b = 2

Number is 23.

4x + ky + 8 = 0

2x + 2y + 2 = 0

Multiplying eq2 by 2 and subtracting from eq1

⇒ (k – 4)y + 6 = 0

⇒ y = 6/(4 – k)

Thus, for k ≠ 4 it has a unique solution

The given system of equations is: 

x + 2y – 3 = 0

5x + ky + 7 = 0

The above equations are of the form 

a₁ x + b₁ y − c₁ = 0 

a₂ x + b₂ y − c₂ = 0 

Here, a₁ = 1, b₁ = 2, c₁ = −3 

a₂ = 5, b₂ = k, c₂ = 7 

So according to the question, 

For unique solution, the condition is 

\(\frac{a_1 }{a_2}\) ≠ \(\frac{b_1}{b_2}\) 

\(\frac{1}{5} ≠ \frac{2}{k}\)

⇒ k ≠ 10 

Hence, the given system of equations will have unique solution for all real values of k other than 10.

Let the fixed charge be ‘a’ and the charge per km travelled be ‘b’

Given,

For a journey of 12 km, the charge paid is Rs 89 and for a journey of 20 km, the charge paid is Rs 145

⇒ a + 12b = 89 ----- (1)

and a + 20b = 145 ----- (2)

(1) – (2)

⇒ - 8b = - 56

⇒ b = 7

Thus,

a + 84 = 89

⇒ a = 5

For a distance of 30 km,

amount paid = distance travelled x charge per kilometre + fixed charge

= 30 × 7 + 5 = Rs. 215

Let the two digits number be 1y + x

Then, equations formed are

10y + x - 18 = 10x + y => y - x = 2 ......(i)

and x + y = 12 ......(ii)

On solving eq. (i) & (ii) we get

x = 5

and y = 7 

Hence number is 75.

Let the one’s digit be ‘a’ and ten’s digit be ‘b’.

Given, two digit number is 4 times the sum of its digits and twice the product of the digits

⇒ 10b + a = 4(a + b)

⇒ a = 2b

Also,

10b + a = 2ab

Substituting value of a.

⇒ 10b + 2b = 2 × 2b × b

⇒ b = 3

Thus, a = 6

Number is 36

Correct option is (B) 3

Given that x = 2, y = 1 is a solution of given system of equation.

Put x = 2, y = 1 in equation x + ky = 5, we obtain

2 + k = 5

\(\Rightarrow\) k = 5 - 2 = 3

Correct option is B) 3

Let the basic full fare be ‘a’, half far be ‘b’ and reservation charges be ‘r’.

Given,

Railway half ticket costs half the full fare and the reservation charge is the same on half ticket as on full ticket .

a = 2b ----- (1)

Also,

Reserved first class ticket from Mumbai to Ahmedabad costs Rs 216 and one full and one half reserved first class tickets cost Rs 327.

⇒ a + r = 216 ------ (2)

Also,

a + b + 2r = 327 ----- (3)

Substituting value of a in eq2 and eq3

⇒ 2b + r = 216 and 3b + 2r = 327

Solving the above equations, we get

r = Rs. 6 and b = Rs. 105

Thus a = Rs. 210

Let the fixed charges be Rs.x and the cost of food per day be Rs.y. 

Then as per the question 

x + 25y = 4500 ………(i) 

x + 30y = 5200 ……..(ii) 

Subtracting (i) from (ii), we get 

5y = 700 

⇒ y = 700/5 = 140 

Now, putting y = 140, we have 

x + 25 × 140 = 4500 

⇒x = 4500 – 3500 = 1000 

Hence, the fixed charges be Rs.1000 and the cost of the food per day is Rs.140.

Let the fixed charges be ‘a’ and the cost of food per day be ‘b’

Given, when a student a takes food for 20 days, he has to pay Rs 1000 as hostel charges whereas a student B, who takes food for 26 days, pays Rs 1180 as hostel charges.

⇒ a + 20b = 1000 a + 26b = 1180

Subtracting one from another

⇒ 6b = 180

⇒ b = Rs. 30

Thus,

a + 600 = 1000

⇒ a = Rs. 400

Let the one’s digit be ‘a’ and ten’s digit be ‘b’

Given, two - digit number is 4 more than 6 times the sum of its digits.

⇒ 10b + a = 6(a + b) + 4

⇒ 4b = 5a + 4 ------- (1)

Also, if 18 is subtracted from the number, the digits are reversed.

⇒ 10b + a – 18 = 10a + b

⇒ b – a = 2 ------- (2)

Multiplying eq2by 4 and subtracting from eq1

⇒ 4b – 5a – 4b + 4a = 4 – 8

⇒ a = 4

Thus, b = 6

Number is 64

The astrologer told Guru Nayak that his enemy had died long before. So he should leave his search and return home.

After taking dinner, the astrologer told his wife the reason of his being late. 

Opposite angles of a cyclic quadrilateral are supplementary

∠A + ∠C = 180°

∠B + ∠D = 180°

Given,

∠A = (2x + 4)°

∠B = (y + 3)°

∠C = (2y + 10)°

∠D = (4x - 5)°

2x + 4 + 2y + 10 = 180

⇒ x + y = 83 ------- (1)

y + 3 + 4x – 5 = 180

⇒ y + 4x = 182 -------- (2)

(1) – (2)

⇒ x – 4x = 83 – 182

⇒ x = 33

Thus,

y = 50

∠A = 2x + 4 = 70°

∠B = y + 3 = 53°

∠C = 2y + 10 = 110°

∠D = 4x – 5 = 127°

The climax of the story is when the client returns fully satisfied. The story ends in a happy mood.

He told her that a client came to him when he was ready to come home.

The astrologer felt greatly relieved on knowing that the man whom he thought he had killed was still alive. When the astrologer was young, he used to drink and quarrel with people. He had attacked a man and stabbed him and left him for dead. He ran away from his village to escape punishment and settled in a town in the guise of an astrologer.

As a strange coincidence, the same man came to him after many years to know whether he would succeed in his search of locating his enemy to take revenge. When the astrologer came to know that the man was not dead, he felt as if a great load had gone from him.

We know that, 

The sum of the opposite angles of cyclic quadrilateral should be 180°. 

And, in the cyclic quadrilateral ABCD, 

Angles ∠A and ∠C & angles ∠B and ∠D are the pairs of opposite angles.

So, ∠A + ∠C = 180° and 

∠B + ∠D = 180° 

Substituting the values given to the above two equations, we have 

For ∠A + ∠C = 180° 

⇒ ∠A = (2x + 4)° and ∠C = (2y + 10)° 

2x + 4 + 2y + 10 = 180° 

2x + 2y + 14 = 180° 

2x + 2y = 180° – 14° 

2x + 2y = 166 —— (i) 

And for, ∠B + ∠D = 180°, we have 

⇒ ∠B = (y+3)° and ∠D = (4x – 5)° 

y + 3 + 4x – 5 = 180° 

4x + y – 5 + 3 = 180° 

4x + y – 2 = 180° 

4x + y = 180° + 2° 

4x + y = 182° ——- (ii) 

Now for solving (i) and (ii), we perform 

Multiplying equation (ii) by 2 to get, 

8x + 2y = 364 —— (iii) 

And now, subtract equation (iii) from (i) to get 

-6x = -198 

x = −198/−6 

⇒ x = 33° 

Now, substituting the value of x = 33° in equation (ii) to find y 

4x + y = 182 

132 + y = 182 

y = 182 – 132 

⇒ y = 50 

Thus, calculating the angles of a cyclic quadrilateral we get: 

∠A = 2x + 4 

= 66 + 4 

= 70°

∠B = y + 3 

= 50 + 3 

= 53° 

∠C = 2y + 10 

= 100 + 10 

= 110° 

∠D = 4x – 5 

= 132 – 5 

= 127° 

Therefore, the angles of the cyclic quadrilateral ABCD are 

∠A = 70°, ∠B = 53°, ∠C = 110° and ∠D = 127°.

After harvesting farmers preserve fodder by arranging into heaps. This heap will be used for the cattle throughout the year.

1. They supply fodder from their agricultural fields. 

2. They also feed the cattle with hay, green and dry grass, oil seed cakes of ground nut.

Let the fixed charges be Rs.x and the cost of food per day be Rs.y. 

Then as per the question 

x + 25y = 4500 ………(i) 

x + 30y = 5200 ……..(ii) 

Subtracting (i) from (ii), we get 

5y = 700 ⇒ y = 700/5 = 140 

Now, putting y = 140, we have 

x + 25 × 140 = 4500 

⇒x = 4500 – 3500 = 1000 

Hence, the fixed charges be Rs.1000 and the cost of the food per day is Rs.140.

The cost of milk is decided due to fat content.

All most all the persons who won agriculture fields, rear cattle also.

ii. x + y = 37 and x – y = 11 

∴ x = 24, y = 13 

iii. x + y = 54 and x – y = 20

 ∴ x = 37, y = 17

iv. x + y = 57 and x - y = 17

∴ x = 30, y = 13

In a cyclic quadrilateral sum of opposite angles is 180°.

Given, in a cyclic quadrilateral ABCD ,

∠A = (2x + 4)°

∠B = (y + 3)°

∠C = (2y + 10)°

∠D = (4x - 5)°

∴ ∠A + ∠C = 180° and ∠B + ∠D = 180°

⇒ 2x + 4 + 2y + 10 = 180 and y + 3 + 4x – 5 = 180

⇒ x + y = 83 --------- (1) and y + 4x = 182 ---------- (2)

Subtracting eq1 from eq2.

⇒ y + 4x – x – y = 182 – 83

⇒ 3x = 99

⇒ x = 33

Substituting in eq1.

⇒ y = 50

∠A = 2 × 33 + 4 = 70°

∠B = 50 + 3 = 53°

∠C = 2 × 50 + 10 = 110°

∠D = 4 × 33 – 5 = 127°

C) by the match light

Correct Answer is : C) packing up

Let fixed charges be Rs.x and rate per km be Rs.y. 

Then as per the question 

x + 80y = 1330 ………(i) 

x + 90y = 1490 ……..(ii) 

Subtracting (i) from (ii), we get 

10y = 160 ⇒ y = 160/10 = 16 

Now, putting y = 16, we have 

x + 80 × 16 = 1330 

⇒x = 1330 – 1280 = 50 

Hence, the fixed charges be Rs.50 and the rate per km is Rs.16.

Farmers rear cocks and hens in villages. Most of these are local varieties (Natukollu). Poultry farms are of two types. One is for production of eggs (layers) and other for meat (Broilers).

Natural wild varieties grow fully in 5 to 6 years. But broilers grow fully in just 6 to 8 weeks. This happens due to genetic modifications in the hen. So genetically modified food is useful.

Sheep and goats provide him meat and wool, the dung and urine becomes good manure to the field crops.

Farmers believe that animal husbandry is part and parcel of agriculture.

This preparation is prepared by A.M. Buhari, in South Indian food industry in Chennai, in the year 1965. So this is called Chicken-65.

A hen spends 21 days to hatch its eggs.

1. 1 – B 

2 – C 

3 – A 

4 – B

2. 1 – D 

2 – A 

3 – B 

4 – B

Correct answer is (c) garden

Let the number be x.

According to the question we can write as

(1/5) x+5 = (1/4) x – 5

On rearranging

(1/5) x – (1/4) x = -5-5

(1/5) x – (1/4)x =-10

By taking L.C.M we get

(4x-5x)/20=-10

Again by transposing

x= 200

Let us consider the number as ‘x’

So,

x/5 + 5 = x/4 – 5

x/5 – x/4 = -5 – 5

By taking LCM for 5 and 4 which is 20

(4x-5x)/20 = -10

By cross-multiplying we get,

4x – 5x = -10(20)

-x = -200

x = 200

∴ The number is 200.

Let the required number be x.Then,

= (x/5) + 5 = (x/4) – 5

Transposing (x/4) to LHS and it becomes – (x/4) and 5 to RHS it becomes -5

= (x/5) – (x/4) = -5 – 5

= (4x – 5x)/20 = -10

= -x/20 = -10

Multiplying both side by (-20)

= (-x/20) × (-20) = (-10) × (-20)

= x = 200

∴ The required numbers are 200

Let fixed charge = Rs. x

and charge per kilometer = Rs. y

According to the question,

x + 110y = 690 …(i)

and x + 200y = 1050 …(ii)

On subtracting Eq. (i) from Eq. (ii), we get

x + 200y – x – 110y = 1050 – 690

⇒ 90y = 360

⇒ y = 40

On putting the value of y = 40 in Eq. (i), we get

x + 110(40) = 690

⇒ x + 440 = 690

⇒ x = 690 – 440 = 250

Hence, monthly fixed charges is Rs. 250 and charge per kilometer is Rs. 40

The boy was sitting on the platform half way up the tree. It was an April afternoon. Warm breezes had sent everyone indoors. The boy was thinking of going for a swim, when he saw a black cobra coming out of a group of cactus. It was looking for a cooler place in the garden. A mongoose also came out and went towards the cobra. They came face to face.

The Cobra knew that the 3 feet long mongoose is a fine fighter, clever and aggressive. But the cobra was also an experienced fighter. He could move with great speed and strike the mongoose. His sharp teeth were full of poison. It was a battle of champions.

The cobra hissed. His tongue darted in and out. It was6feet long. It raised its three feet high and raised its broad, spectacled hood. The mongoose was also ready to fight, its hair on the spine stood up like bristles. They would help him to prevent his body from getting bitten. A myna and a jungle crow were watching the fight. At one stage they dived towards the cobra, but they missed it. The myna went back. The crow was trying to turn around when it was struck by the cobra. It died soon. The mongoose proved too clever for the cobra and finally it was killed by the mongoose which dragged it into the bush.