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Given:

18 women corporators out of 60 in a municipality

Calculation:

Let be assume p% women are corporators

⇒ p = (18/60) × 100 = 30%

∴ The required result will be 30%.

Given:

4.5 kg is 18 gm

Formula used:

Percentage = (Part value/Whole value) × 100

Calculation:

Part value is 18gm

The whole value is 4500g

⇒ (18/4500) × 100

⇒ 0.4%.

∴ 18 is 0.4% of 4.5kg.

Calculation:

Let the width of the rectangle = x cm

Then, 300% of x = 63

⇒ 300/100 × x = 63

⇒ x = 63 × 100/300

⇒ x = 21

Perimeter of the rectangle = 2(length + width)

⇒ 2(63 + 21)

⇒ 168 cm

∴ Perimeter of the square = (168 + 8) = 176 cm

The side of the square = 176/4

⇒ 44

The area of the square playground without path = (44)2 sq. cm

⇒ 1936 sq. cm

The area of the square playground with path = (44 + 20)2 sq. cm

⇒ 4096 sq. cm

The area of the path = 4096 – 1936

⇒ 2160 sq. cm

The total cost of constructing the path = 2160 × 25

⇒ 54000

The total cost of constructing the path at the rate of Rs. 25 per sq. cm is 54,000.

2400

Calculation:

Let us suppose there are X employees in company. 

If 48 Executive get promoted to Sr. Executive, there will be

22% of total employees who are Sr. Executive instead of 20%

(1/5th of total employees = 20% of total employees)

⇒ 48 represents 2% employees of ALPHA.

48 = X × (2/100)

X = 4800/2

X = 2400

∴ There are 2400 employees in the organization.

Given:

Invalid Votes = 30%

contestant A got 65.6% of valid votes and

contestant B got 34.4% of valid votes.

Total population = 7500.

Calculation:

Total no. of Votes = 7500

Total no. of Valid Votes = 0.7 × 7500

⇒  5250

No. of valid votes which A received = (65.6/100) × 5250 = 3444

No. of valid votes which B has got = (34.4/100) × 5250 = 1806

∴ A won to B by = (3444 - 1806)/3444 × 100

= 47.56%

∴ A will win by 47.56% votes.

Given:

Ram gets 40% of the total votes.

Ranu gets 60% of the total votes.

Difference of their votes = 3500.

Calculation:

Let total number of votes be x.

Votes of Ram = 40% of x = 0.4x

Votes of Ranu = 60% of x = 0.6x

Difference = 0.6x – 0.4x 

⇒ 3500 = 0.2x

⇒ x = 3500/0.2

⇒ x = 17500

∴ Total number of votes are 17500.

Given:

Cost of book = Rs. 50

Increment = Rs. 25

Calculation:

Rate of increase = (25/50) × 100%

⇒ 50%

∴ The rate of increase is 50%

Given:

The ratio of income of A and B = 11 : 9

The ratio of expenditure of A and B = 17 : 15

Saving of A = 2500

Saving of B = 1500

Formula used:

Income = Expenditure + Saving

Calculation:

Let the income of A and B is Rs. 11x and Rs. 9x respectively.

Then, (11x – 2500)/(9x – 1500) = 17/15

⇒ 165x – 37500 = 153x - 25500

⇒ 165x – 153x = 37500 – 25500

⇒ 12x = 12000

⇒ x = 1000

Income of B = 9 × 1000 = 9000

∴ The income of B = 9000

Given:

At point A, 1/6th of the passengers alight and 10 boards of the bus.

At point B, 1/5th of the passengers alight and 3 boards the bus.

At point C which is the last stop, all the 55 passengers alight.

Calculations:

Let total number of passenger be x.

At point A,1/6th of the passengers alight and 10 boards of the bus,

Number of passengers in the bus = x - x/6 + 10

⇒ 5x/6 + 10

At point B, 1/5th of the passengers alight and 3 boards the bus,

Now, Number of passengers in the bus = (5x/6 + 10) - {(1/5)(5x/6 + 10)} + 3

⇒ (5x/6 + 10)(1 - 1/5) + 3

⇒ (5x/6 + 10)(4/5) + 3

⇒ (4x/6 + 8) + 3

⇒ 4x/6 + 11

⇒ 2x/3 + 11

At point C, 55 passengers alight the bus,

⇒ 2x/3 + 11 = 55

⇒ 2x/3 = 44

⇒ x = 132/2

⇒ x = 66

∴ The capacity of the bus is 66.

Given:

A is 25% more than B.

Formula Used:

Percentage = \(\frac{{increased\;value}}{{value\;of\;A\;}} \times 100\)

Calculation:

Let the value of B be x.

Then the value of A be \(x \times \frac{{100 + 25}}{{100}}\) = 1.25x

B less than A = 1.25x - x = 0.25x

% B less than A = \(\frac{{0.25x}}{{1.25x}} \times 100\) = 20%

Therefore, B less than A by 20%.

GIven:

A number, on subtracting 24 from it, reduces to 60% of itself

Calculation:

If the number be x, then

x - 24 = 60% of x

⇒ x - 24 = 60x/100

⇒ x - 24 = 3x/5

⇒ 5x - 120 = 3x

⇒ 2x = 120

⇒ x = 60

85% of x = (85/100) × 60 = 51

∴ 85% of the number is 51

Given:

44% of the students are females

Number of male is 42

Calculation:

Total number of students in a class = 100%

Females = 44%

⇒ Males = 100% – 44%

⇒ Males = 56%

⇒ 56% = 42

⇒ 1% = 42/56

⇒ 100% = 42/56 × 100

⇒ 100% = 0.75 × 100

⇒ 100% = 75

∴ Total number of students in a class is 75.

GIVEN:

48% of X = 336

CALCULATION:

48/100 × X = 336

X = (336 × 100)/48  

X = 700

Given:

A = 148%B

C = 160% of 2(A – B)

Calculation:

According to question

⇒ A = 148/100 × B

⇒ A = 1.48B      ----(1)

⇒ C = 160/100 × 2(A – B)

⇒ C = 3.2(A – B)

⇒ C = 3.2A – 3.2A/1.48

⇒ C = (4.736A – 3.2A)/1.48

⇒ C = 1.536A/1.48

⇒ C = 1.038A

⇒ C/A = 1.038/1

% of C more than A = (1.038 – 1)/1 × 100

⇒ 3.8%

∴ C more than A by 3.8%

Given

Richa invests 5,59,968 in mutual fund. Which is 19% of annual income of Richa

Calculation

Let annual income income of Richa be 100%

∴ 19% = Rs.5,59,968

⇒ 100% = Rs.2,947,200

⇒  monthly income of Richa = Rs.2947200/12

⇒ Rs.2,45,600

Given:

Total population of village = 8000

Number of people who know coding = 30%

Number of people learning coding = 1/4

Formula used:

Total percentage of non-coders = 100 – total number of coders

Calculations:

Here, people studying coding won’t be considered as non-coders

Total number of coders = 30% + ((1/4) × 100)

Total number of coders = 30% + 25%

Total number of coders = 55%

Total number of non-coders = 100% – 55% = 45%

Total number of non coders = (45/100) × 8000

Total number of non-coders = 3600

∴ The total number of non-coders is 3,600

Given:

Proportion of water by weight in gin and water mixture = 75%

Mixture = 60 gm

Water added in 60 gm mixture = 15 gm

Calculations:

Weight of water in the mixture of 60 gm = (75/100) × 60 = 45 gm

Weight of water in the mixture of 45 gm = 45 + 15 = 60 gm

New total mixture = 60 + 15 = 75

New Percentage of water = (60/75) × 100 = 80%

Given:

Increase in income = 28.56%

Deduction in Income = 33.33%

Difference between increased income and present income = Rs. 3300

Concept Used:

In this type of questions, Percentage values should be changed into Fraction values.

Formula Used:

To change given % in fraction, we first convert the % into mixed fraction value.

\({\rm{Fraction\;value}} = \frac{{{\rm{Percentage\;given}}}}{{100}}\)

Representation done by fraction values will be taken as,

\({\rm{Fraction\;value}} = \frac{{{\rm{Increased\;value}}}}{{{\rm{Initial\;value}}}}\)

\({\rm{Percentage\;less}} = \frac{{{\rm{Difference\;between\;Initial\;and\;final\;value}}}}{{{\rm{Initial\;Value}}}}\; \times 100\)

Calculation:

Using Formula and concept,

\(\begin{array}{l} 28.56{\rm{\% }} = 28\frac{4}{7} = \frac{{200}}{7}\% \\ 33.33{\rm{\% }} = 33\frac{1}{3} = \frac{{100}}{3}\% \\ {\rm{Fraction\;value\;of\;}}28.56{\rm{\% }} = \frac{{\frac{{200}}{7}}}{{100}} = \frac{{200}}{{700}} = \frac{2}{7}\\ {\rm{Fraction\;Value\;of\;}}33.33{\rm{\% }} = \frac{{\frac{{100}}{3}}}{{100}} = \frac{{100}}{{300}} = \frac{1}{3}\; \end{array}\)

Suppose Income of Sumit be 7 Units then after increase of 2 Units

New Income = 7 + 2 = 9 Units

Deduction in salary in the ratio of 1/3

\({\rm{deducted\;amount}} = \frac{1}{3}\; \times 9 = 3\;{\rm{Units}}\)

New Salary = 9 – 3 = 6 Units

Difference in Initial and final Salary = 7 - 6 = 1 Unit

\({\rm{Final\;SalaryPercentage\;less\;than\;the\;Initial\;salary}} = \frac{1}{7}\; \times 100 = 14.28\% \)

∴ Final Salary of Sumit is 14.28% less than the Initial Salary.

Given:

Mahi’s salary = (5/6)^th of Shekhar’s salary

Money spent by both = Rs. 2000

The Ratio of Mahi and Shekhar’s savings = 4 : 5.

Calculations:

Let Shekhar’s salary be x

Mahi’s salary = 5x/6

According to the question,

⇒ (5x/6) – 2000 ∶ (x – 2000) = 4 ∶ 5

⇒ 5 ((5x/6) – 2000)) = 4(x – 2000)

⇒ (25x/6) – 10000 = 4x – 8000

⇒ (25x/6) – 4x = 10000 – 8000

⇒ (x/6) = 2000

⇒ x = Rs. 12000

Mahi’s salary = (5/6) × 12000 = 10000

∴ Mahi’s salary is Rs. 10,000 and Shekhar’s salary is Rs. 12,000.

(c) 98% of 250 = 245 & 32 % of 750 = 240

∴ 32 of 750 < 98% of 250.

(e) No. of candidate selected = 855 × 20% = 171

Given:

An outlet gives flat 30% discount to the customers if they buy goods for more than Rs. 5000, and discount of 20% if they buy goods for less than Rs. 5000.

The worth of goods Avi bought = Rs. 4800.

Discount for Avi = Discount for Bhavi – 1068

Calculation:

Let the worth of goods Bhavi bought be x.

Discount for Avi = Discount for Bhavi – 1068

⇒ 20% of 4800 = 30% of x – 1068

⇒ 960 + 1068 = 0.3x

⇒ 0.3x = 2028

⇒ x = 6760

∴ The worth of goods Bhavi bought is Rs. 6760.

Let the number be x. 

Then, 4*x/5 –(65% of x) = 21 

4x/5 –65x/100 = 21 

5x = 2100 

x = 140.

Given:

50% of a certain number is equal to \(\frac{3}{4}^{th}\) of another number.

Calculation:

Let the numbers be x and y.

According to the question:

(50% of x) = [(3/4) × y]

⇒ 50x/100 = 3y/4

⇒ x/2 = 3y/4

⇒ x/y = 3/2

⇒ x : y = 3 : 2

∴ The ratio between the numbers are 3 : 2.

(d) Dhreeu’s monthly salary = 6,00,000 /12 = Rs.50,000

Surya’s monthly salary = 50,000 x 40 /100 = Rs.20,000

Pranab’s monthly salary = 20,000 x 80 /25 =Rs.64,000

Calculation:

Let the number be x.

60% of the number = 60% × x = 0.6x

60 subtracted from 60% of the number = 0.6x - 60 = 60

⇒ 0.6x = 120

⇒ x = 200

∴ The number is 200.

Given:

M% of x = y

N% of y = x

Calculation:

Substituting the value of y in second equation,

N% of (M% of x) = x

⇒ (N/100) × (M/100) × x = x

⇒ N × M = 10,000

∴ The relation NM = 10,000 is the solution.

Given:

Per day chocolate eats by Himanshu = 2 chocolates

Per day chocolate give to Chesta = 1 chocolate 

After 15 days remaining chocolates in the box = 15 

Calculations:

Total chocolate which was eaten by Himanshu in 15 days = 2 × 15 = 30 

Total chocolate which was given to Chesta by Himanshu in 15 days = 1 × 15 = 15 

Total chocolate in the box = 30 + 15 + 15 = 60 

Required % = (15/60) × 100 = 25% 

∴ Himanshu gave 25% chocolate to Chesta.

Given: 

Rotten apples = 10%

Remaining sold good apples = 85%

good apples = 405

Calculation:

let he had originally be X apples.

According to the question:

∴ Rotten apples = (X × 10%) =10X/100

⇒ Rotten apples = X/10

Again,

sold apples = 85% 0f (X – X/10)

⇒ [(85/100) × (9X/10)] = 153X/200

Now,

(X/10 + 153X/200 + 405) = X

⇒ (20X + 153X + 81000)/200 = X

⇒ 173X + 81000 = 200X

⇒ 200X – 173X = 81000

⇒ 27X = 81000

⇒ X = 81000/27

⇒ X = 3000

∴ He originally had 3000 apples.