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Given:

Percentage of salary decided to be donated to orphanage = 15%

Actual donation is Rs. 2896 which is 90% of what was dedided earlier

Calculation:

Let Shyam's salary be Rs. x.

Now, according to the question,

90% of 15% of x = Rs. 2896

⇒ (9/10) × (3/20) × x = Rs. 2896

⇒ x = Rs. 21,451.85 ≈ Rs. 21,452

∴ The salary of Shyam is Rs. 21,452

Let the initial amount with Mr.jones be Rs.x then, 

Money given to wife= Rs.(40/100)x=Rs.2x/5.Balance=Rs(x-(2x/5)=Rs.3x/5. 

Money given to 3 sons= Rs(3X((20/200) X (3x/5)) = Rs.9x/5. 

Balance = Rs.((3x/5) – (9x/25))=Rs.6x/25. 

Amount deposited in bank= Rs(1/2 X 6x/25)=Rs.3x/25. 

Therefore 3x/25=12000 

x= ((12000 x 35)/3)=100000 

So Mr.Jones initially had Rs.1,00,000 with him.

Given:

The value of a machine depreciates at the rate of 25% each year.

Concept Used:

Price after depreciation = Original price × (100 - depreciation %)/100

Calculation:

Let value of the machine at the end of first year be Rs.a

The value of the machine at the end of second year = a × 75/100

The value of the machine at the end of third year = a × 75/100 × 75/100

Accordingly,

a × 75/100 - a × 75/100 × 75/100 = 24000

⇒ a = 128000

∴ The value of machine at the end of first year is Rs.128000.

Given:

The value of a machine depreciates by 30% every year.

Value of the machine at the end of 2 years would be Rs.34,300.

Concepts used:

The present value of machine = Value of machine at the end of 2 years/(1 – Rate of depreciation)² 

Calculation:

Let the present value of machine be Rs. Z.

The present value of machine = Value of machine at the end of 2 years/(1 - Rate of depreciation)²

⇒ Z = Rs. 34,300/(1 – 0.30)² 

⇒ Z = Rs. 34,300/(0.70)² 

⇒ Z = Rs. 70,000.

∴ The present value of the machine is Rs.70,000.

Alternate method:

Let the present value be Rs. x

After 1st year, value = Rs. 70x/100

After 2nd year, it will be = Rs. {(70x/100) × (70/100)} = Rs. (0.7 × 0.7 × x)

⇒ (0.7 × 0.7 × x) = 34300

⇒ x = 34300/(0.7 × 0.7)

⇒ x = 70000

∴ The present value of the machine is Rs.70,000.

Formula used

Value of a commodity after 'n' years with depreciation rate 'r'% p.a. is

\(Actual\ value \times (1 - {r\over100})^n\)

Given

Cost of generator = Rs.60,000

Rate of depreciation = 5% p.a.

Calculation

The Value of the generator after two years will be,

\(Actual\ value \times (1 - {r\over100})^n\)

\(⇒ 60000 \times (1 - {5\over100})^2\)

\(⇒ 60000 \times ({19\over20})^2\)

⇒ Rs. 54150

Given: 

Total people = 3000, only 70% have the saving habit,

Calculation: 

The number people which have saving habit = 3000 ×  70%

⇒ 2100

Now, 

Out of saving habit people, percentage of share holders is

⇒ 100 - 30 - 38

⇒ 32%

So, 

The number of share holders are 

⇒ 2100 × 32%

⇒ 21 × 32

⇒ 672

∴ The required number of share holders are 672. 

Solution:

Given data:

Present value of motorcycle = Rs. 75,000

Rate of depreciation = 4%

Time for depreciation = 2 years

Concept used:

Final value = Initial value × [(100 – Rate of depreciation)/100]ⁿ

Where, n = no. of years

Calculation:

Final value = Initial value × [(100 – Rate of depreciation)/100]ⁿ

⇒ 75,000 × [(100 – 4)/100]²

⇒ 75,000 × (96/100)²

⇒ 75,000 × (24/25)²

⇒ (75,000 × 576)/625

⇒ 69,120

∴ The value of the motorcycle after 2 years will be Rs. 69,120.

Given:

Water tax is increased by 20% but its consumption is decreased by 20%.

Calculations:

Let the expenditure of water consumption be Rs. y and the tax imposed is x%.

so the tax needs to be paid = Rs. (x/100) × y = Rs. xy/100

Now new tax = (x × 120/100)% = (6x/5)%

and new expenditure = Rs. y × 80/100 = Rs. 4y/5

new tax = Rs. (6x/500) × 4y/5 = Rs. 24xy/2500

so change in tax = xy/100 - 24xy/2500 = xy/2500

then percentage change = (change/original) × 100

⇒ \(\frac{\frac{xy}{2500}}{\frac{xy}{100}}\ ×\ 100\ = \) (1/25) × 100 = 4% decrease (since new tax is less than original tax)

∴ The correct answer is 4%.

Short Tricks:

1.2 × 0.8 = 0.96 = 96 % (4 % decrease from original)

Given:

Increment in the number = 25%

Decrement in the number = 30%

Formula Used:

Net increase/decrease  = Increase% – Decrease% - (Increase %× Decrease%)/100

Calculation:

Net increase/decrease  = Increase% – Decrease% - (Increase %× Decrease%)/100

⇒ 25 – 30 – (30 × 25)/100

⇒ - 5 – 7.5%

⇒ - 12.5%

∴ Net decrement is 12.5%.

The correct option is 3 i.e. Decrement 12.5%

Given:

Income of the person = Rs. 6400

Percentage increase every year = 10%

Time = 2 years

Formula used:

If there is increase in x% then then the value after the increment = Initial value × (100 + x)/100

Calculation:

Income after one year = Initial income × (100 + 10)/100

⇒ Income after one year = 6400 × 11/10 = 7040

Income after second year = 7040 × (100 + 10)/100 = 7744

∴ Income after two years will be Rs. 7744

Given 

Total Questions = 70 

Formula Used 

Percentage = (Actual/Total) × 100 

Calculation 

Question correctly by anita as follow:

⇒ Airthmetic = 70% of 10 = 7 

⇒ Algebra = 40% of 30 = 12 

⇒ Geometry = 60% of 30 = 18 

⇒ Total correct Question by anita = 37 

⇒ For pass no. of Question should be correct = 60% of 70 = 42 

⇒ Number of more questions she would have to answer correctly to earn a 60% passing marks = 42 - 37 = 5 

∴ Number of more questions she would have to answer correctly to earn a 60% passing marks is 5.

Given:

Marks of A = Marks of B + 20% marks of B

Calculation:

Let the B’s marks is x

A’s marks = 20% of x + x

⇒ 120x/100

Required percentage = {[(120x/100) - x]/[115x/100]} × 100

⇒ {[(20x/100)]/[(120x/100)]} × 100

⇒ 16.66%

Alternate method:

If A is x% more than B, then B is {[(x)/(100 + x)] × 100}% less than A.

⇒ [(20/120) × 100]%

⇒ 16.66%

Given:

Height in 2014 = 160 cm

Height increased in 2015 = 10%

Calculation:

Height in 2015 = 160 + 10% of 160

⇒ 176 cm

Given:

Quantity of water added = 18 litres

Quantity of initial mixture = 15 litres

Percentage of spirit in the initial solution = 60%

Calculation:

Quantity of spirit in the initial solution of spirit and water = 15 × (60/100) = 9

Only water is added due to which quantity of spirit will be the same

Quantity of spirit in the final solution = 9 litre

Concentration of spirit in the final solution = {9/(18 + 15)} × 100 = 300/11 = \(27\frac{3}{{11}}\)%

∴ The concentration of spirit in the final solution is \(27\frac{3}{{11}}\)%

Given:

Initial consumption of mustard oil = 12 kg

Percentage increase in the price of mustard oil = 10%

Percentage increment in price = 5%

Formula used:

Expenditure = Price × Consumption

Calculation:

Let the initial price of mustard oil be x

Expenditure on mustard oil = 12x

⇒ After increment expenditure on mustard oil = (105/100) × 12x = (63/5)x

⇒ Price of mustard oil = (110/100) × x = (11/10)x

⇒ Consumption after increment = {(63/5)x}/{(11/10)x}

⇒ Consumption after increment = {(63 × 10)/(5 × 11)} = 126/11 = \(11\frac{5}{{11}}\;kg\)

∴ The new consumption is \(11\frac{5}{{11}}\;kg\)

Given:

Total number of workers in the factory = 6000

Percentage of male workers in the factory = 80%

Percentage of male workers at least 55 years old out of total male workers = 19%

Concept used:

If x% of y

We can write (x/100) of y

Calculation:

Total number of male workers in the factory = 6000 × (80/100) = 4800

⇒ Total number of male workers at least 55 years old = 4800 × (19/100) = 912

⇒ Total number of male workers below 55 years old = 4800 – 912 = 3888

∴ The number of male workers below 55 years is 3888.

Given:

Ratio of male to female = 3 ∶ 7

Children % among male = 25%

Children % among female = 30%

Number of adult males = 18000

Calculation:

Let the number of males and females be 3x and 7x respectively

Total population = 3x + 7x = 10x

Percentage of adult males = (100 - 25)% = 75%

Adult males = 75% of 3x = 9x/4

⇒ 18000 = 9x/4

⇒ 9x = 18000 × 4 = 72000

⇒ x = 8000

Total population = 10x = 10 × 8000 = 80000

∴ Total population of the city is 80,000

Given that:

A is 50% greater than B

and B is 60% less than C

Hint:

• x% of 100 = x and
• less or more is always from 100%

∴ 60% less than C = (100 - 60) % of C = 40% of C

Calculations:

Let B = 100

∴ A = 50% greater then B

⇒ A = 100 + 50 = 150

∴ B = 60% less then C = 40% of C

⇒ 100 = (40/100) × C

⇒ C = (100 × 100)/40

 ⇒ C = 250

∴ ratio of A to C is 150: 250

 ⇒A: C = 3: 5

Given:

60% of A = 50% of B

Calculations:

According to the question,

A × 60/100 = B × 50/100

B = 6/5 × A 

So, B = (6/5) × 100 = 120% of A

∴ the correct answer is 120%.

Calculation:

Let previously the price of wheat be Rs.a

New price of wheat = a × (117.5/100) = Rs.4.7a/4

Quantity of wheat initially bought be m.

New quantity of wheat bought = 90m/100 = 9m/10

Previously total price of wheat was am

Then,

Amount spent increase = (4.7m/4 × 9m/10 - am)/am × 100

⇒ 5.75%

∴ 5.75% increase in amount.

Given:

Ramesh got 304 marks out of 400

Calculation:

⇒ The gain percentage of Ramesh = (304/400) × 100 = 76%

⇒ The gain percentage of Govind = (385/500) × 100 = 77%

⇒ The gain percentage of Reena = (273/300) × 100 = 91%

⇒ The gain percentage of Veena = (164/200) × 100 = 82%

∴ The required result will be "Reena".

Given:

A is 50% of C

B is 75% of C

Calculation:

A = 50% of C

⇒ A = (50/100) × C

⇒ A/C = 1/2

⇒A : C = 1 : 2      ----(1)

Again, 

B = 75% of C

⇒ B = (75/100) × C

⇒ B/C = 3/4

⇒ B : C = 3 : 4      ----(2)

Now, Multiply by 2 in eq.(1),

A : C = 1 : 2   

⇒ A : C = (1 : 2) × 2

⇒ A : C = 2 : 4

A : B : C = 2 : 3 : 4

Let A be 2x

B be 3x

And C be 4x.

According to the questions:

2x = 3x of y%

⇒ (2x/3x) × 100 = y

⇒ y = 200/3 

⇒  y = 66.66%

∴ A is 66.66% percentage of B.

Given:

The price of fridge = The price of TV + Rs. 9000

SP of TV = 70% higher than CP of TV

SP of fridge = 40% of CP of fridge

Sum of the SP of both the items = Rs. 28800

Calculation:

Let the CP of TV and fridge be x and (x + 9000) respectively

SP of fridge = 40% of CP of fridge

⇒ SP of fridge = 0.4(x + 9000)

SP of TV = 70% higher than CP of TV

⇒ SP of TV = CP of TV + 70% of CP

⇒ SP of TV = 1.7x

The sum of the SP = 28800

⇒ 1.7x + 0.4(x + 9000) = 28800

⇒ 2.1x = 28800 – 3600

⇒ 2.1x = 25200

⇒ x = 12000

SP of fridge = 0.4(x + 9000)

⇒ 0.4(12000 + 9000)

⇒ 8400

∴ The selling price of fridge is Rs. 8400.

Given:

Population of town = 12,000

Percentage increase in population = 15%

Concept used:

Population of the city in n years = initial population × (1 + r/100)ⁿ

Calculation:

Population of the city in two years = 12,000 × (1 + 15/100)² = 15,870

∴ The population of the city in two years is 15,870.

Calculation:

Two-third of 225 = (2/3) × 225

⇒ 150

 Required percentage = (150/450) × 100

⇒ 100/3%

∴ The required percentage is 100/3%

Let x% of 25 = 2.125. 

Then , (x/100)*25 = 2.125 

X = (2.125 * 4) = 8.5.

Let 0.25% of x = 0.04. 

Then , 0.25*x/100 = 0.04 

X= [(0.04*100)/0.25] = 16.

Calculation:

9% of 50 = 5x/2

 ⇒ 9/2 = 5x/2

⇒ 9 = 5x

 ⇒ 1.8 = x

∴ The value of x is 1.8

Let 9% of x =6.3. 

Then , 9*x/100 = 6.3 

X = [(6.3*100)/9] =70.

Given:

Radioactivity of the atom = 35 curie

Decay rate = 10% of present reactivity

Concept:

Since, the candle loses its length after 40 minutes seconds, so it will lose its reactivity 3 times in 2 hours. Use successive decrease percentage formula to calculate the reactivity after 2 hours.

Calculation:

% decrease in reactivity after 40 minutes;

= -10 -10 + (100/100)

= -19%

% decrease after 80 seconds;

= - 19 – 10 + (19 × 10)/100

= -29 + 1.9

= 27.1%

% decrease after 120 seconds;

= - 27.1 – 10 + (27.1 × 10)/100

= - 37.1 + 2.71

= 34.39%

∴ % decrease in radioactivity after 120 minutes = 34.39% 

Radioactivity after 120 minutes = 100% - 34.39% = 65.61% of 35 curie

Given:

% of uncast vote = 20%

Votes got by 1^st Candidate = 20% of cast vote

Votes got by 2^nd Candidate = 30% of the remaining vote

Votes got by 3^rd Candidate = Total cast vote – (Votes of 1^st Candidate + votes of 2^nd Candidate)

Concept:

Consider the total votes to be 100x. Then proceed.

Calculation:

Let total number of voters be 100x

Cast vote = 80% of 100x = 80x

Vote received by A = 20% of 80x votes = 16x

Remaining votes = (80x – 16x) = 64x

Votes received by B = 30% of 64x votes = 19.2x votes

Remaining votes = (64x – 19.2x) = 44.8x votes

⇒ Votes received by 3^rd candidate = 44.8x

Votes received by 3^rd candidate – Votes received by 1^st candidate = 1152

⇒ 44.8x – 16x = 1152

⇒ 28.8x = 1152

⇒ x = 40

Total number of voters in the voting list = 100x = 100 × 40 = 4000

∴ The total number of voters in the voting list is 4000.

Given:

15% of the number is 30

Calculation:

Let be assume the number is p

⇒ p × (15/100) = 30

⇒ p = 200

⇒ Three times of the number = 3p = 3 × 200 = 600

∴ The required result will be 600.