Short Trick :

Let the number be 100x.

If increased by 25% then it becomes 125x.

If decreased by 30%, then becomes 70x.

Now, according to question,

125x - 70x = 22

x = 2/5

original number = 100x ⇒ 100 × 2/5 = 40

Detailed Solution :

Say A is the original number.

Number value increased by 25%

⇒ A + (0.25A) = 1.25A

Number value decreased by 30%

⇒ A – (0.30A) = 0.70A

Given that, Difference between increased and decreased value is 22.

⇒ (1.25 – 0.70) × A = 22

⇒ 0.55A = 22

⇒ A = 22/0.55 = 40

Short Trick :

Let for 100 litres he gained profit of 11.11 litres.

∴ 100 litres = 11.11 litres

for 1 litre = ?

? = 0.1111 ⇒ 1/9

Detailed Solution :

To gain $11\frac{1}{9}\mathrm{\%}$ the milkman has to mix $11\frac{1}{9}$ liter of water with 100 litres of pure milk.

If he does so, he can sell a litre of mixture [at the price of pure milk] while his investment is only 100 litre of pure milk.

∴ He is mixing $11\frac{1}{9}$ Liter water with 100 litre milk.

With 1 litre milk he has to mix litre of water.

Now, a litre of water

$=\frac{\frac{100}{9}}{100}$ litre of water

Let the total number of workers be ‘x’, then according to the question,

Skilled workers = 75% of x

Unskilled workers = (100-75)% = 25% of x,

Permanent workers = 20% of unskilled workers + 80% of skilled workers

= 20% of 25% of x + 80% of 75% of x

= x × (20/100) × (25/100) + x × (80/100) × (75/100)

= x {0.05 + 0.6} = 0.65x

According to the question, remaining workers are temporary whose number is 126,

⇒ (1 - 0.65)x = 126

⇒ 0.35x = 126

⇒ x = 126/0.35 = 360

Hence, the required number of workers is 360.

Let F be the original fraction.

As the numerator is increased by 100% and denominator is increased by 200%, the formed equation will be;

F [ (200/100) / (300/100) ]  = 4/21

F [ 2/3 ] = 4/21

F = 4/21 * 3/2

F = 2/7

Let number of cities C be T.

Number of cities B is 20% of total cities. Total number of cities is 10000.

⇒ Number of cities B = 10000 × (20/100) = 2000

⇒ Number of cities A and C = 10000 – 2000 = 8000

⇒ Number of cities A = 8000 – T

The number of cities A is 30% more than that of cities B and C together.

⇒ 8000 – T = (T + 2000) × (1 + 30/100)

⇒ 8000 – T = 1.3T + 2600

⇒ T = 5400/2.3

We see that T is not a natural number, and hence cannot be the number of cities C.

Let the total number of votes be ‘x’

25% of x did not vote

No. of polled votes = x – (25/100)x

No. of polled votes = (100 – 25)/100 × x

No. of polled votes = (75/100)x

No. of polled votes = 75%

No. of polled votes = 0.75x

Of these, 6.66% of votes were invalid

No. of valid votes = 0.75x – (6.66/100) × 0.75x

No. of valid votes = 0.75x × (100 – 6.66)/100

No. of valid votes = 0.75x × (100 - $6_3^2$)/100

No. of valid votes = 0.75x × (100 – 20/3)/100

No. of valid votes = 0.75x × (300 – 20)/300

No. of valid votes = 0.75x × 280/300

No. of valid votes = 0.25x × 280/100

No. of valid votes = 0.7x

No. of valid votes = 0.7x

Now, M + N + P = 0.7x

We know, P = 2450 valid votes

P has 40% more votes than N.

⇒ 2450 = N + (40/100)N

⇒ 2450 = (140/100)N

⇒ 2450 = (7/5)N

⇒ 2450/7 × 5 = N

N = 350 × 5

N = 1750

M has 40% of total votes

⇒ 0.4x + 1750 + 2450 = 0.7x

⇒ 4200 = 0.7x – 0.4x

⇒ 4200 = 0.3x

⇒ 4200/0.3 = x

⇒ x = 14000

Let the total distance travelled = x km

23% of the distance is travelled by bus

Thus, the remaining distance = (100 – 23)% of x = 77% of x

50% of the remaining 77% of x is travelled by train = 0.5 × 0.77 = 0.385

38.5% of the distance is travelled by train

Thus, the remaining distance = [100 – (23 + 38.5)] = 38.5%

Thus 38.5% of x is travelled by taxi = 231 km

⇒ 38.5/100 × x = 231

⇒ x = (231 × 100)/38.5

⇒ x = 600 km

∴ Distance between Hyderabad and Chennai = 600 km

Initial number of students = X

New number of students = 3X/2

First we will find Quantity A,

Quantity A:

Let the number be ‘x’.

75% of the number = 75% of x = 0.75x

According to the question,

0.75x + 75 = x – 20

⇒ 0.25x = 95

⇒ x = 95/0.25 = 380

∴ The number is 380.

Now,

Quantity B:

We know that for a G.P.-

n^th term = ar^n-1

Where, a = 1^st term, r = common ratio

⇒ 5^th term = t₅ = ar⁴ = 2

Product of its first nine terms = (a) × (ar) × (ar²) ×………× (ar⁸)

= a⁹ r^{1 + 2 + 3 + ….+ 8} = a⁹ r³⁶ = (ar⁴)⁹ = 2⁹ = 512

75% of the product of its first nine terms = (75/100) × 512 = 384

Solving for Quantity 1:

⇒ 3 × 35% of x = (8/7) x – 40

⇒ 1.05x = (8/7) x – 40

⇒ 7.35x = 8x – 280

⇒ 8x – 7.35x = 280

⇒ 0.65x = 280

⇒ x = 430.77

⇒ Quantity 1 = 430.77

Solving for Quantity 2:

⇒ 7 × 70% of x = 1800 + (6/11) x

⇒ 4.9x = 1800 + (6/11) x

⇒ 53.9x = 19800 + 6x

⇒ 53.9x – 6x = 19800

⇒ 47.9x = 19800

⇒ x = 413.36

⇒ Quantity 2 = 413.36

Quantity A:

Failed boys = (100 - 70)% of 2000 = 30% of 2000 = 600

Failed Girls = (100 - 60)%of 1600 = 40% of 1600 = 640

⇒ Percentage of failed students = (600 + 640)/(2000 + 1600) × 100 = 34.44%

Quantity B:

Total score = 140

⇒ Score made by virat by boundaries and sixes = 6 × 6 + 11 × 4 = 36 + 44 = 80

⇒ Runs made by running between the wickets = 140 - 80 = 60

⇒ Required % = 60/140 × 100 = 42.85%

∴ Quantity A < Quantity B

From statement I:

Mohan scored 158 marks in an exam and failed by 114 marks.

⇒ Minimum passing marks = 158 + 114 = 272          …(i)

Statement I alone is not sufficient to answer the question.

From statement II:

The maximum marks of the exam are 642 morethan the marks obtained by Mohan.

Statement II alone is not sufficient to answer the question.

From statement I & II together:

Maximum marks of the exam = 158 + 642 = 800

If the minimum passing percentage be x, then

$\begin{array}{l}\frac{800\times x}{100}=158+114=272\\ \Rightarrow x=\frac{272\times 100}{800}=34\end{array}$

Given, the percentage of salary, Sahil spends on food is 25% and on medicine and education is 35% and Rs 2400 on entertainment in statement(1) and the actual amount spent on food in Rs in statement (2) and the actual amount spent on medicine and education in statement (3).

Based on above statement, either statement (2) or statement (3) can be dispensed.

From statement I:

He got 60% marks on an average in hindi, chemistry and physics.

Data in statement I is not sufficient to answer the question, as the maximum marks of each subject is not given.

From statement II:

In chemistry he got 15% more than the average marks.

Same as statement I maximum marks are not given so we can say that equation II is not sufficient to answer the question.

Hence, data in statement I and statement II are not sufficient to answer the question, as the maximum marks of each subject is not given.

Let the winner got x number of votes

And his rival got y votes

Now, by statement I, we know that x = y + 300000

By statement II, we know that x = 150% of y = 150y/100

By statement III, we know that x + y = 75% of 2000000 = 1500000

Therefore by using any of the 2 statements(I or II) and third statement, we can get 2 different equations in x and y and hence find their values.

Here x = 900000 and y = 600000