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Calculation: 

Quantity I:

The logic behind the given series is as follows

5 × 0.5 + 0.5 = 3

3 × 1 + 1 = 4

 4 × 1.5 + 1.5 = 7.5

7.5 × 2 + 2 = 17

17 × 2.5 + 2.5 = ? = 45 

Quantity I = 45

Quantity II:

The logic behind the given series is as follows

2 × 1.5 + 1.5 = 4.5

4.5 × 2 + 2 = 11

11 × 2.5 + 2.5 = 30

30 × 3 + 3 = ? = 93

Quantity II = 93

∴ From the table, we can see that, Quantity I < Quantity II.

Solving for Quantity I:

Considering the given series 5, 12, 26, 47, 75, ?

The logic of above series can be explained as,

5 + 7 = 12

12 + 14 = 26

26 + 21 = 47

47 + 28 = 75

75 + 35 = ?

⇒ ? = 110

∴ The value of ? is 110

Solving for Quantity II

Considering the given series 3, 3, 6, 18, 72, ?

The logic of above series can be explained as,

3 × 1 = 3

3 × 2 = 6

6 × 3 = 18

18 × 4 = 72

72 × 5 = ?

⇒ ? = 360

∴ The value of ? is 360

∴ Quantity I < Quantity II

Concept:

• The 2 segments trapezoidal rule for integration, is exactly for integrating at the most FIRST order of the polynomial.
• The single segment trapezoidal rule is exact for at most a first-order polynomial. When it comes to the two-segment trapezoidal rule is also exact only for the same order of the polynomial, which is, a first-order polynomial.

Method of numerical integral and their order of the fitting polynomial:

Hence, the first option is correct.

Concept:

Gauss Elimination method:

• It employs both sides of the equation to be multiplied by a non-zero constant. The matrix is then reduced to Upper Triangular Matrix to get the values of the respective variables.
• The Gauss Elimination method used in linear and multilinear algebra, by a process of solving the system of simultaneous linear equation for one variable in term of the other variable and then substituting this expression to the remaining equations. Which results in a new system of equations and variables in which equations and variables are less than one to the previous one. This process is repeated until the expression is not reduced to only one variable.
• This method can also be used to find the rank of a matrix, to calculate the determinant of a matrix, and to calculate the inverse of an invertible square matrix. 

Explanation:

Floating-Point form Representation of a real number:

This representation does not reserve a specific number of bits for the integer part or the fractional part. Instead, it reserves a certain number of bits for the number (called the mantissa or significand) and a certain number of bits to say where within that number the decimal place sits (called the exponent).

Important Point:

The floating number representation of a number has two parts:

1. The first part represents a signed fixed-point number called the mantissa.
2. The second part of designates the position of the decimal (or binary) point and is called the exponent.

The fixed-point mantissa may be a fraction or an integer. Floating -point is always interpreted to represent a number in the following form: M*r^e.

Only the mantissa ‘m’ and the exponent ‘e’ are physically represented in the register (including their sign). A floating-point binary number is represented in a similar manner except that using base 2 for the exponent. A floating-point number is said to be normalized if the most significant digit of the mantissa is 1.

Given:

First card drawn = 5 to 9

Second card drawn = greater than or equal to queen but less than ace

Two cards = either between 1-5 or greater than a queen

Calculation:

Quantity I :

Probability of card being in between 5 to 9 = (5 × 4)/52 (∵ 5 -9 = 5 and 4 colors so 20)

⇒ Probability of card being in between 5 to 9 = 5/13

Probability of card being greater than or equal to queen but less than ace = (2 × 4)/51

⇒ Probability of card being greater than or equal to queen but less than ace = 8/51

Required probability = (5/13) × (8/51)

⇒ required probability = 40/663

Quantity II :

Probability of card being from 1-5 = 20/52 (∵ 5 to 1 = 5 cards so 20 cards in total)

Probability of card being from 1-5 = 5/13

Probability of card greater than queen = (2 × 4)/52 (∵ only king and ace of all colors are greater than queen)

⇒ Probability of card greater than queen = 2/13

Required probability = (5/13) + (2/13)

⇒ required probability = 7/13

∴ Quantity I < Quantity II

Given:

20% p.a. = compounded bi yearly

Difference between compound interest and simple interest = Rs 150

20% p.a. = 3 years

C.I. – S.I. = 1280

Formula used:

Net effect formula for CI = x + y + (x × y)/100

Calculation:

Quantity I :

Compound Interest (Half yearly compounded) = 10% (∵ 20/2 = 10)

⇒ Effective rate of percentage = 10 + 10 + (100/100)

⇒ Effective rate of percentage = 21%

Simple interest = 20%

Now, C.I – S.I. = Rs 150

⇒ 1% of sum = Rs 150

⇒ Sum = 150 × 100

⇒ Sum = 15,000

Quantity II:

Compound interest for three years = 20%

⇒ Effective percentage rate of first and second year = 20 + 20 + (400/100)

⇒ Effective percentage = 44%

⇒ Effective percentage of previous an third year = 44 + 20 + (880/ 100)

⇒ Effective percentage = 72.8%

Simple interest for three years = 20 × 3

⇒ Simple interest for three years = 60%

Now, (72.8 – 60%) of sum = 1280

⇒ 12.8% of sum = 1280

⇒ sum = 1280 × 100/12.8

⇒ sum of money = Rs 10,000

Given:

Profit = 425%

C.P. = Increase is 5%

S.P. =Constant

Discount = 80% on mark up

Formula used :

Profit = S.P. – C.P.

Calculation:

Quantity I :

Assume the cost price = Rs 100

⇒ Selling price = 100 + 425

⇒ Selling price = 525

⇒ Profit from new C.P. increase = 525 – (100 + (0.05 × 100))

⇒ Profit from new C.P. increase = 525 – 105

⇒ Profit from new C.P. increase = 420%

⇒ New profit margin = (420/105) × 100

⇒ New profit margin % = 400%

Quantity II :

Let the marked price be Rs 100

Selling price = 100 – (0.80 × 100)

⇒ Selling price = 20

⇒ S.P. = C.P. = 20

⇒ Percentage markup = ((100 – 20)/20) × 100

⇒ Percentage markup = 400%

∴ Quantity I = Quantity II

Given:

Quantity I : P = Rs 20,000

R = 12.5% (1/8)

T = 2 years

Quantity II : P = 18,000

Ratio = 5 : 4

Larger part at R = 20% for 3 years S.I.

Smaller part at R = 10% for two years C.I.

Formula used:

Simple interest : (P × R × T)/100

Compound interest : Amount = P (1 + (R/100))ⁿ

Where, n = Number of years, P = Principal Amount, R = Rate of Interest, A = Amount

Calculation:

Quantity I :

A = 15,000 × (1 + (1/8)) ²

⇒ A = 15,000 × (9/8) × (9/8)

⇒ A = 25312.5

⇒ I = 25312.5 – 20000

⇒ I = 5312.5

Quantity II :

P for S.I. = 5 × 18000/9

⇒ P for S.I. = 10000

⇒ S.I. = 10000 × 20 × 2/100 (∵ even though this sum is kept for three years question is asked for 2 years)

⇒ S.I. = 4000

P for C.I. = 8,000

⇒ A = 8000 × (1.10)²

⇒ A = 9680

⇒ I = 9680 – 8000

⇒ I = 1680

⇒ after two years person will earn interest of Rs 5680 (∵ 4000 + 1680 = 5680)

∴ Quantity I < Quantity II

Given:

Speed of train = 108 km/hr

Time to cross pole = 25s

At 90 km/hr time taken to cross platform of x meters = 40 seconds

Calculation:

108 km/hr = 30 m/s (∵ 108 × 5 / 18 = 30)

Quantity I:

The length of the train = 30 × 25 = 750 meters

Quantity II:

90 km/hr = 25 m/s

Let the length of the platform = x meters

⇒ (750 + x) = 25 × 40

⇒ 1000 m

⇒ x = 1000 – 750

⇒ x = 250 meters

∴ Quantity I > Quantity II

Given:

Speed of mohan = 20 km/hr initially

Speed of rohan = 60 km/hr

Formula used:

Distance = speed × time 

Calculation:

The distance travelled by Mohan in first 3 hours = 3 × 20 = 60 km 

⇒ the relative speed of X and Y = 60 – 20 = 40 km per hour 

⇒ 60 = 40 × y 

⇒ y = 1.5 hours 

⇒ It means after 1.5 hours the Rohan meets Mohan. 

Quantity I:

The total time taken by A to go and return immediately = 3 + 1.5 + 3 + 1.5

⇒ Total time taken = 9 hours 

Quantity II: 9 hours

Given:

A + B = 10 days

B + C = 12 days

A + B + C = 8 days

2/5^th of work by A = 10 days

1/2 of work by B = 8 days

Solution:

Quantity I :

2/ (A + B + C) = (1/(A + B)) + (1/(B + C)) + (1/C + A))

⇒ 2/ 8 = (1/ 10) + (1/12) + (1/x)

⇒ 1/ X = 1/4 – ((1/10) + (1/12))

⇒ 1/ X = 15/60 – (11/60)

⇒ 1/ X = 4/60

1/X = 1/15

⇒ X = 15 days

Quantity II :

A completes whole work = 10 × 5/2

⇒ A completes whole work = 25

B completes whole work = 16 (∵ 8 × 2 = 16)

⇒ 1/X = (1/25) + (1/16)

⇒ 1/X = 41/400

X = 400/41 = 9.75 days

∴ Quantity I > Quantity II

Concept:

Trapezoidal rule is given by:

\(\mathop \smallint \limits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} = \frac{{\rm{h}}}{2}\left[ {{{\rm{y}}_{\rm{o}}} + {{\rm{y}}_{\rm{n}}} + 2\left( {{{\rm{y}}_1} + {{\rm{y}}_2} + {{\rm{y}}_3}{\rm{\;}} \ldots } \right)} \right]\)

\({\rm{Number\;of\;intervals(n)}} = \frac{{{\rm{b}} - {\rm{a}}}}{{\rm{h}}}{\rm{\;}}\)

where b is the upper limit, a is the lower limit, h is the step size.

Calculation:

Given:

x    : 0 1 2

f(x) : 8 5 6

[f(x)]²:64 25 36

From the above given data n = 2, y₀ = 64, y₁ = 25, y₂ = 36, b = 2, a = 0

\(h = {(b-a)\over Number~ of ~intervals}={(2-0)\over 2}=1\)

By using the Trapezoidal rule we get:

\(\displaystyle\int_0^2 [f(x)]^2 dx={h\over2}[{y_0+y_2+2(y_1)}]={1\over 2}[64+36+2(25)]={150\over 2}=75\)

Given:

Pawan = x days

Kasim = 4x days

Formula used:

Work done in t1 and t2 days then, total work done = (1/t1) + (1/t2)

Calculation:

Quantity I: half of the work in 20 days therefore, the complete work in 20 × 2 = 40 days

⇒ (1/x) + (1/4x) = (1/40)

⇒ 5 × 40 = 4x

⇒ x = 50

Quantity II: Let P’s efficiency = a

⇒ then 25% of a = 0.25a

⇒ half of the work in 35 days then the complete work in 35 × 2 = 70 days

⇒ 0.25a efficiency he does work in 70 days

⇒ Total work = 70 × 0.25a

⇒ Total work = 17.5a units

⇒ at ‘a’ efficiency he will do work in 17.5a/a = 17.5 days

∴ Quantity I > Quantity II

Given:

Avg weight of 10 employees = 45 kg = 2 × avg weight of 5 servants

Calculation:

Weight of 10 employees = 45 × 10

⇒ Weight of 10 employees = 450 kg

Weight of 5 servants = 45 × 5 / 2

⇒ Weight of 5 servants = 112.5 kg

⇒ The sum of the weight of 10 employees and 5 servants = 450 + 112.5

⇒ 562.5 kg

Quantity I:

Required average = 562.5 / 15 ≈ 37.5 kg

Quantity II: 45 kg

∴ Quantity I < Quantity II

Given:

Cost price of 100 notebooks = Selling price of 80 Note Books

Formula used:

\({\rm{Profit\;\% }} = \frac{{{\rm{Profit}}}}{{{\rm{Cost\;price}}}} × 100\)

Profit = Selling price(S.P.) - Cost Price(C.P.)

Calculation:

Let C.P. of 100 notebooks be 100x 

S.P. of 80 notebooks be 80x

C.P. × 100x = S.P. × 80x

⇒ C.P./S.P. = 80/100 = 4/5

Profit = 5 - 4 = 1

Profit % = (1/4) × 100 = 25%

∴ The percent Gain is 25%

Concept:

Average = Sum of terms/Number of terms

Statement I.

Let the 7 consecutive even numbers be (x – 6, x – 4, x – 2, x, x + 2, x + 4, x + 6)

Now, x + x + 4 = 440

⇒ 2x +4 = 440

⇒ x = 218 

Numbers = (218 – 6, 218 – 4, 218 – 2, 218, 218 + 2, 218 + 4, 218 + 6)

⇒ Numbers = 212, 214, 216, 218, 220, 222, 224

Average = Sum of terms/Number of terms

⇒ Average = (212 + 214 + 216 + 218 + 220 + 222 + 224)/7

⇒ Average = 1526/7 = 218

So, Statement I alone is sufficient to answer the question

Statement II.

The difference of the fifth and fourth number is 8.

(x + 2) – x = 8

Here the variable is cancelled out so, we can't find value of x.

So, statement II is not sufficient

∴ The statement I alone is sufficient to answer the question but, statement II alone is not sufficient.

Given:

C.P. = Rs. 2000

Accessories = Rs. 5000

6 generators S.P. = 2500

Formula used:

S.P. = C.P. + Gain

S.P. = C.P. – Loss

loss% = (Loss/C.P.) × 100

Calculation:

Total cost price = (10 × 2000) + 5000

⇒ Total cost price = 25000

Now, 6 generators at Rs. 2500

⇒ S.P. of 6 generators = 6 × 2500

⇒ S.P. of 6 generators = 15000

Loss of 10% in whole deal

⇒ Total S.P. = Total C.P. – loss

⇒ Total S.P. = 25000 – 2500

⇒ Total S.P. = 22500

⇒ Total S.P. = S.P. of 6 + S.P. of rest 4 generators

⇒ S.P. of 4 generators = 22500 – 15000

⇒ S.P. of 4 generators = 7500

⇒ S.P. of one generator = 7500/4

⇒ S.P. of one generator = 1875

S.P. = 1875 C.P. = 25000/10

⇒ Loss = 2500 – 1875

⇒ Loss = 625

⇒ Loss% = (625/2500) × 100

⇒ Loss% = 25%

∴ Loss of 25% is incurred on remaining generators which were sold at Rs. 1875 each.

Gain:

S.P. reduced by 50%

Loss = 20% after reduction

Formula used:

S.P. = C.P. + Gain

S.P. = C.P. – Loss

Gain% = (Gain/C.P.) × 100

Calculation:

Assume initial S.P. = 100

⇒ SP reduces = 100 – 50

⇒ Reduced SP = 50

Now, on this reduced S.P. 20% loss is observed

⇒ Reduced S.P. = 50, loss = 20%

⇒ S.P. = C.P. – Loss

Assume C.P. = X

⇒ 50 = X – (0.20 × X)

⇒ 50 = 0.80 X

⇒ (50/0.8) =X

⇒ X = 62.5

Now, Initial profit percentage

⇒ Profit% = ((100 – 62.5))/62.5) × 100

⇒ Profit% = (37.5/62.5) × 100

⇒ Profit% = 0.6 × 100

∴ Initial profit% is 60%

Given:

Ratio of present age of Aman’s sister and Aman = 5 : 7

Ratio of present age of Aman and his father = 1 : 2

Calculation:

Present Age of his father = 45 – 3 = 42 years

So, present age of Aman = ½ × 42 = 21 years

As ratio of Aman’s sister and Aman = 5 : 7

Given:

Selling price = Rs. 524 when profit is earned

Selling price = Rs. 452 when loss is suffered

Formula used:

Profit = selling price – cost price

Loss = cost price – selling price

Calculation:

Let the cost price be Rs. x

According to the question:

Profit earned on selling price of Rs. 524 = loss suffered on selling price of Rs. 452

⇒ Selling price (Rs. 524) – x = x – selling price (Rs. 452)

⇒ 524 – x = x – 452

⇒ x + x = 524 + 452

⇒ 2 x = 976

⇒ x = 976/2

⇒ x = 488

∴ The cost price of the object is Rs. 488.

Given:

Ratio of present age = 4 : 3

Ratio after 15 years = 9 : 7

Calculation:

Let the age of Raman and Aman be 4x and 3x respectively

According to question

(4x + 15)/(3x + 15) = 9/7

28x + 105 = 27x + 135

x = 30

Present age of Aman = 3 × 30 = 90

⇒ Age of Aman after 3 years = 90 + 3 = 93 years

Given:

The cost price of 8 box = the selling price of 9 box

Formula Used:

Loss percentage = {(C.P – S.P)/C.P} × 100

Calculation:

Let the C.P of each box be Rs 1

C.P of 9 boxes = Rs. 9

C.P of 8 boxes = Rs. 8

S.P of 9 boxes = Rs. 8

Loss at 9 boxes = Rs. (9 – 8) = Rs. 1

Loss % = 1/9 × 100 = 100/9 % = 11.11%

Given:

A tap having a diameter d can empty a tank in 60 min

Calculation:

The area of a tap is proportional to Work done by pipe

When the diameter is double, then the area will be four times. So, will work four times faster

Hence required time taken to empty the tank = 60 × 1/4

⇒ 15 mins

∴ The required answer is 15 mins. 

Given:

Cost of 12 shirts = Selling price of 8 shirts

Formula used:

Profit% = (Profit/CP) × 100%

Calculation:

Let the cost of 12 shirts be Rs. 12

SP of 8 shirts = Rs. 12

Cost of 8 shirts = Rs. 8

Gain = Rs. (12 – 8)

⇒ Rs. 4

Gain% = (4/8) × 100%

⇒ 50%

∴ The profit percentage in this transaction is 50%

Given:

Time taken by inlet pipe to fill the tank = 4 hours

Time taken by outlet pipe to empty 3/7 of a tank = 3 hours

Formula used:

Efficiency = Total work/Time taken

Calculation:

LCM of 4 and 7 = 28 = Total work

Efficiency of inlet pipe = 28/4 = 7 work

Efficiency of outlet pipe = 28/7 = 4 work

Work done in 2 hours = (7 – 4) = 3 work

Time taken to do 27 work = (2/3) × 27 hours

⇒ 18 hours

Time taken more to complete remaining 1 work = 1/7 hour

Total time taken = 18 + (1/7) hours

⇒ 127/7 hours

∴ The total time taken to fill the tank is 127/7 hours

Given:

The quantity of mixture is 576 ml.

The ratio of milk to water is 7 : 5.

37.5% of the mixture is taken out.

Calculation:

The ratio of milk to water is 7 : 5.

The quantity of mixture is 576 ml.

Then, the quantity of milk = 7/12 × 576 = 336 ml

The quantity of water = 5/12 × 576 = 240 ml

According to the question,

37.5% of the mixture is taken out.

The mixture taken out = 37.5/100 × 576 = 216 ml

Then, the milk taken out = 37.5/100 × 336 = 126ml

The water taken out = 37.5/100 × 240 = 90 ml

Remaining milk = 336 - 126 = 210 ml

Remaining water = 240 - 90 = 150 ml

The216 ml of water added in the remaining mixture.

Then, the resultant water = 216 + 150 = 366 ml

The ratio of resultant milk to water = 210 : 366 = 35 : 61

Therefore, the ratio of milk to water in the resultant mixture is 35 : 61.

Given:

Quantity of pure milk = 80 litres 

Quantity of milk removed and replaced by water = 8 litres

Formula used:

Amount of milk left = Initial amount × [1 – (Amount took out and replaced)/Initial amount]ⁿ 

Here n is the number of times the process repeated

Calculation:

Amount of milk left = 80 × [1 – 8/80]² 

⇒ 80 × (1 - 1/10)²

⇒ 80 × (9/10)²

⇒ 80 × (81/100) = 64.8 litres

Amount of water in the final mixture = 80 – 64.8 = 15.2 litres

The ratio of milk and water in the final mixture = 64.8 ∶ 15.2 = 81 ∶ 19

∴ The ratio of milk and water in the final mixture is 81 ∶ 19

Given:

Initial quantity of mixture = 80 litres

Ratio of milk and water in the initial mixture = 2 : 3

Amount of mixture replaced with milk = 25%

Concept used:

If a fraction of mixture is removed from the mixture, then the ratio of milk and water will be same as previous.

Calculations:

Amount of mixture replaced with milk = 25% × 80

⇒ 20 litres

After removing 20 litres,

Ratio of milk and water in the remaining mixture of 60 litres will be in the ratio of 2 : 3

Let amount of milk and water be 2x and 3x litres respectively.

⇒ 2x + 3x = 60

⇒ 5x = 60

⇒ x = 12

New amount of milk = 2x

⇒ 2 × 12

⇒ 24 litres

New amount of water = 3x

⇒ 3 × 12

⇒ 36 litres

New amount of water = 36 litres

After adding 20 litres of milk in the new mixture,

New amount of milk in the final mixture = 24 + 20

⇒ 44 litres

∴ The amount of milk in the final mixture is 44 litres.

Given:

⇒ Total quantity of mixture = 63 + 77 = 140 litres

Let quantity of mixture taken out be a litres.

⇒ Percentage Quantity of water in mixture = 63/140 × 100 = 45%

⇒ Percentage quantity of alcohol in mixture = 77/140 × 100 = 55%

⇒ Quantity of water in mixture after addition =  63 - a × 45/100 + 16

⇒ Quantity of milk in mixture after addition = 77 - a × 55/100 + 4

Then,

⇒ 63 - a × 45/100 + 16 = 77 - a × 55/100 + 4

⇒ a = 20

∴ Total quantity of mixture after 20 litres of mixture taken out = 140 - 20 = 120 litres

Given:

Let quantity of water and solution be 2a litres and 3a litres respectively.

⇒ Quantity of water after replacement = 2a - 30 × 2/5 + 30 = 2a + 18

⇒ Quantity of solution after replacement = 3a - 30 × 3/5 = 3a - 18

Total quantity of mixture after replacement = 2a + 18 + 3a - 18 = 5a litres

Then,

⇒ (2a + 18)/5a × 100 = 58

⇒ 40a + 360 = 58a

⇒ a = 20

∴ Required difference = (2a + 18) - (3a - 18) = 16 litres

(B) Sodium salt of benzene sulphonic acid

Synthetic detergents are made from sodium salt of benezene sulphonic acid

(D) refraction

This phenomenon is because of refraction of light. The lines of sight intersect at a higher position than where the actual rays originated. This causes the water to appear shallower than it really is

One day Parasurama was reclining with his head on Kama’s lap when a stinging worm burrowed into Kama’s thigh. Blood began to flow and the pain was terrible, but Kama bore it without tremor lest he should disturb the master’s sleep. Parasurama awoke and saw the blood which had poured from the wound. He said: “Dear pupil, you are not a brahmana. A kshatriya alone can remain unmoved under all bodily torments”.

One day, when Parashurama was resting on Kama’s lap; it so happened that a bee stung Kama on the lower portion of his thigh. However, fearing that if he moved his legs, he would awaken Parashurama. Parashurama woke up and saw Kama bleeding. He asked Kama to tell the truth that whether he was a Brahmin or a Kshatriya. Kama addmitted that he was a Kshatriya.

I = T/α = 2/10

I = 0.2 kgm² 

And I = MK² 

K = 0.2 m

η = w/q₁ = 5.4 x 10/3.6 x 10⁹ = 15%

Q₂ = Q₁ – W 

= 36 × 10⁸ – 5.4 × 10⁸ 

= 30.6 × 10⁸ J

(B) (A) and (E) only. 

R ∝ A^1/3

V = 4/3 πR³ ∝ A

Mass ∝ A 

So density is independent of A

Correct option is (d) r²

v = \(\frac{2R^2(σ-ρ)g}{9η}\)

So, v ∝ r²

(1) Statement-1 is True, Statement-2 is True 

As w =  \(\frac{nR\triangle T}{1-\gamma}\) 

If w is negative so, \(\Delta\)T = +ve so temperature will rise.

The narrator did not want the girl travelling with him to understand that he was totally blind. So to impress her and to let her think that he was not blind, he remarks that she had an interesting face. He wanted to hide his blindness.

Ulysses drew his sword, and half resolved to thrust it with all his might in at the bosom of the sleeping monster, but wiser thought restrained him, because none but polyphemus himself could have removed that mass of stone which he had placed to guard the entrance of the cave.

The speaker caressed and nursed his emotions towards his enemy. He kept imagining that his enemy would do him harm and he lived in suffering. This increased his agony and watered his emotions of anger against his enemy. He then put on a mask of friendship towards his enemy. He pretended to be good to him and smiled at him whenever he saw him.

Compound : You were absent and we were disappointed 

Complex : We were disappointed because you were absent