Ratio and Proportion

Here's another way to calculate C's share:

C's share = C's ratio * (total amount / sum of ratios)
= 5 * (1260 / 10)
= 5 * 126
= 630

To find the ratio of a: b: c, you can follow these steps:

Find the ratio of a to b: a: b = 3: 4
Find the ratio of b to c: b: c = 2: 5
Combine the two ratios to find the ratio of a: b: c: a: b: c = 3: 4: 10
This gives you the ratio of a to b to c as 3: 4: 10.

To find the ratio of A: B: C: D, you can follow these steps:

Find the ratio of A to B: A: B = 1: 2
Find the ratio of B to C: B: C = 3: 2
Find the ratio of C to D: C: D = 1: 3
Multiply all the ratios together: A: B: C: D = (131): (231): (221): (223) = 3: 6: 4: 12
This gives you the ratio of A to B to C to D as 3: 6: 4: 12.

To find the ratio of A: B: C: D, you can follow these steps:

Find the ratio of A to B: A: B = 1: 2
Find the ratio of B to C: B: C = 3: 3
Find the ratio of C to D: C: D = 2: 3
Multiply all the ratios together: A: B: C: D = (132): (232): (212): (213) = 6: 12: 4: 6
This gives you the ratio of A to B to C to D as 6: 12: 4: 6.

If the sum of the ratios is 14 and the total amount is 5600, you can find the share of each quantity by multiplying each ratio by 400, as you did. 

This gives you the shares of A, B, C, and D as 1200, 2400, 800, and 1200, respectively.

The share of (A+C) is the sum of the shares of A and C, which is 2000. 

The share of (B+C) is the sum of the shares of B and C, which is 3200.

The total salary of males: the total salary of females = 6:5
The salary of each male: salary of each female = 2:3

To find the number of men and women, divide the total salary of males and females by salary of each male and female.
i.e., 6/2: 5/3
Or, 18: 10 = 9: 5
So, the ratio of the number of males and females = 9:5

Let X be the value of 1 unit in terms of the currency being used.
Then, the initial expenditure on salary is 12 units * X = 12X.

We are given that 2 units = 12000, so we can set up the following equation:
2X = 12000.
Solving this equation gives us X = 6000.

Substituting this value back into the equation for the initial expenditure on salary, we find that the initial expenditure on salary is 12 units * 6000 = 72000.

Therefore, the initial expenditure on salary is 72000.

Let X be the salary of A one year ago.
Then, the salary of B one year ago is 2X.

The salary of A now is 2X * 3/2 = 3X.
The salary of B now is 2X * 4/3 = 8/3X.

The total present salary of A and B is 3X + 8/3X = 21500.

Solving this equation gives us X = 3000.

Substituting this value back into the equation for the salary of B now, we find that the salary of B is 8/3 * 3000 = 8000.

Therefore, the salary of B is 8000.

Let X be the income of A.
Then, the income of B is 4X.

The expenditure of A is 2X * 2/3 = 4/3X.
The expenditure of B is 4X * 3/3 = 4X.

We are given that the saving in each case is 200, so we can set up the following equations:
X - 4/3X = 200
4X - 4X = 400

Solving these equations gives us X = 200.

Substituting this value back into the equations for the income of A and B, we find that the income of A is 3 * 200 = 600 and the income of B is 4 * 200 = 800.

Therefore, the income of A is 600 and the income of B is 800.

Let X be the total income of Pervez, Sunny, and Ashu.
Then, Pervez's salary is X * 16/37.
Sunny's salary is X * 12/37.
Ashu's salary is X * 9/37.

We are given that Pervez's saving is 20% of his income, Sunny's saving is 25% of his income, and Ashu's saving is 40% of his income, so we can set up the following equations:
X * 16/37 * 20/100 = X * 16/37 * 1/5 = X * 16/185 = X * 16/185 * 100/100 = X * 16/185 * 1
X * 12/37 * 25/100 = X * 12/37 * 1/4 = X * 12/148 = X * 12/148 * 100/100 = X * 12/148 * 1
X * 9/37 * 40/100 = X * 9/37 * 2/5 = X * 9/74 = X * 9/74 * 100/100 = X * 9/74 * 1

Adding these equations gives us:
X = 1530.

Substituting this value back into the equation for Sunny's salary, we find that Sunny's salary is 1530 * 12/37 = 480.

Therefore, Sunny's salary is 480.

Let the income of Pervez: Sunny: Ashu = 3x, 7x, 4x
The income of Pervez = 3x = 2400 (given in the question)
That means x = 2400/3 = 800
Now, the income of Sunny = 7x = 7*800 = 5600
Similarly the income of Ashu = 4x = 4*800 = 3200

Now,
Their expenditure is in the ratio of 4: 3: 5
So, let their expenditure is 4y, 3y, 5y
The expenditure of Pervez = income of Pervez - saving of Pervez
Or, expenditure of Pervez = 4y = 2400-300 = 2100
Or, y = 2100/4 = 525
Here 4y comes from expenditure's ratio
Similarly, the expenditure of Sunny = 3y = 3* 525 = 1575
The expenditure of Ashu = 5y = 5*525 = 2625

Saving's of Ashu = Income of Ashu - Expenditure of Ashu
Saving's of Ashu = 3200- 2625 = 575

That's correct! The income ratio of Pervez, Sunny, and Ashu is 3:7:4 and their expenditure ratio is 4:3:5. You correctly calculated that Pervez has an income of 2400 and expenditure of 2100, Sunny has an income of 5600 and expenditure of 1575, and Ashu has an income of 3200 and expenditure of 2625. That means Ashu has a saving of 575. Great work!

Let the First-year income = 2x
And, Second-year income = 3x
But ATQ, the second-year income = 45000
So, x = 45000/3 = 15000
Then, first-year income = 2*15000=30000

Similarly,
Let the first-year expenditure = 5y
The first-year expenditure =25000
So, y = 25000/5 = 5000
Second-year expenditure = 9y= 9* 5000 = 45000

Apply formula:

Income = expenditure + savings
Or, savings = income - expenditure
Or, first-year savings = 30000- 25000 = 5000
Similarly,
Second-year savings = 45000- 45000 = 0

Now, total saving in two years = first-year savings + second-year savings
Or, Total saving = 5000+0 = 5000.

Here is the equation version of the solution:

First year income: 2x
First year expenditure: 5y
First year savings: 2x - 5y = 5000
Second year income: 3x
Second year expenditure: 9y
Second year savings: 3x - 9y = 0
Total saving: 5000 + 0 = 5000

Where x and y are variables representing the first year income and expenditure, respectively.

Total income = total days * per day salary

Let per day salary of Pervez = 4
Pervez can complete a piece of work in 30 days and his per day salary is 4
So, the total income of Pervez = 30* 4 = 120
Let per day salary of Sunny = 3

Similarly, Sunny can complete the same work in 50 days and his per day salary is 3
So, the total income of Sunny = 50 * 3 = 150

Let per day salary of Ashu Bhati = 2
Ashu Bhati can complete the same work in 40 days and his per day salary is 2
So, the total income of Ashu Bhati = 40* 2 = 80

Or, the ratio of total income of Pervez: Sunny: Ashu Bhati = 120: 150: 80 = 12: 15: 8
That means total income of Pervez = 12, but according to the question it is 144.
On multiplying 12 by 12, we get the original value.
So, multiply each by 12.
Hence, the total income of Pervez = 144
Total income of Sunny = 15*12 = 180
Total income of Ashu Bhati = 8*12 = 96

Distance covered in the ratio T: B: C = 4: 3: 2
Fair ratio per km. T: B: C = 1: 2: 4
So, the ratio of total fair T: B: C = 4: 6: 8

Sum of the ratio of total fair = 18
But ATQ, it is 720, so multiply 18 by 40.
Now, multiply each and every ratio with 40.
The total expenditure as fair on a train = 4*40=160

Given ratio = 3: 2: 1
Or, 3x: 2x: x
The initial price= (6x)2 = 36x2
After broke down the price = (3x)2: (2x)2: x2 = 9x2: 4x2: x2 = 14x2
After breakdown, a loss of Rs. 4620 occurs.
i.e., loss = initial price - final price
Loss = 36x2 - 14x2 = 22x2
22x2 = 4620
Or, x2 =4620/22 = 210
The initial price of silver-biscuit = 36*210 = 7560

As per the question, B is inversely proportional to A3
i.e., B α 1/A3
Or, B =k/ A3
Given that, B=3, A=2
So, 3 = k/ 8, or k=24
Given that, B = 8/9
So, 8/9 = 24/A3
A3 = (9*24)/8 = 27
So, A = 3√ 27 = 3

The share of A: B is 3: 4
The share of B: C is 2: 3

Note: Whenever such form is given, multiply a to b, then b to b, and then b to c.

i.e., A: B: C = 3*2: 4*2: 4*3
Or, A: B: C = 6: 8: 12
Or, A: B: C = 3: 4: 6
Sun of ratios = 13
Now, the share of B = [4/13] * 7800 = 2400
Share of C = [6/13]* 7800 = 3600

The difference between the share of B and C = 3600- 2400 = 1200

If the ratio of coins = 4: 6: 9
That means if Rs 5 coins are 4, Rs 2 coins are 6, and then Rs 1 coins are 9.

According to the given ratio, the ratio of amounts = 5*4: 6*2: 9*1 = 20: 12: 9
The sum of the ratios of the amounts = 20+12+9 = Rs 41
But ATQ, it is Rs 410, which means multiply each ratio by 10
i.e., new ratio = 40: 60: 90
Now, 40*5: 60*2: 90*1 = 200: 120: 90

The total amount in the form of two rupees coins = 120
So, the two rupees coins = 120/2= 60

Amount of copper in alloy = [copper ratio/ sum of ratios]* total quantity of alloy

Copper = {4/7}* 63 = 36 kg
Similarly, the amount of zinc in alloy = [3/7] * 63 = 27 kg
Let the extracted copper from alloy = x kg
Remaining copper in alloy = 36-x
The new ratio = 10: 9
i.e., 36-x: 27 = 10: 9
36-x: 3 = 10: 1
36-x = 10* 3
36-x = 30
x = 36-30 = 6 kg.
Hence, the extracted amount of copper is 6 kg.

The ratio of land: water = 1: 2
In the northern hemisphere the ration of land: water = 2: 3

Note: Earth is divided equally into two hemispheres called northern hemisphere and southern hemisphere.

i.e., the northern hemisphere is 50% of the total earth.

We can say southern hemisphere = total area- northern hemisphere.

To make the northern hemisphere 50% of the total area
Multiply the ratio of the earth by 10 and northern hemisphere by 3
Now, earth's ratio Land: water = 1*10: 2*10 = 10: 20............... (i)
Northern hemisphere's ratio = 2*3: 3*3 = 6: 9...................... (ii)

Subtract equation i by ii

Hence, southern hemisphere = 10- 6: 20-9 = 4: 11

Let the quantity of vessels A and B in the ratio x: y to form a mixture of water and milk in the ratio 4: 5.
Or, Milk: water = 5: 4.
From vessel A, milk: water = 4x/9: 5x/9
From vessel B, milk: water = 5y/6: 1y/6

Now, ATQ quantities of the mixture should be taken from A and B.
Ratio of milk: water = [4x/9 +5y/6]: [5x/9+ 1y/6] = 5/4
Take the LCM of 9 and 6 = 54
Or, [24x + 45y]/54: [30x+9y]/54 = 5/4
Or, [8x + 15 y]: [10x + 3y] = 5/4
Or, [8x + 15y] * 4 = [10x + 3y] * 5
Or, (60-15) y = (50-32) x
Or, 45y = 18x
Or, 5y = 2x
Or, x: y = 5: 2
Hence, the quantity of mixture should be taken from A and B in the ratio 5: 2.

 (9*12)/3 = 36

 Total no's of radios can be placed in the box 

(25 * 42 * 60) / (7 *6 * 5) 

= 6300 / 210 

= 300

 Let the ages of Kunal and sagar be x and y respectively.
According to the equation, (x-6)/(y-6) = 6/5
=> (x+4)/(y+4) = 11/10
After solving we get y = 16

 Let 40% of A = 2/3 B. Then,
40A/100 = 2B/3 => 2A/5 = 2B/3
A/B = (2/3 * 5/2) = 5/3
Hence, A : B = 5 : 3

Required number of trees = 24/36 * 42 = 28

1/3 : 1/2 : 1/1 = 2 : 3 : 6

The savings of Chetan and Dinesh are 3x – 5y and 4x – 7y respectively.
3x – 5y = 200 --- (1)
4x – 7y = 200 --- (2)
Multiplying (1) by 7 and (2) by 5 and subtracting the resultant equation (2) from resultant equation (1), we get x = 400.
The incomes of Chetan and Dinesh are 3x = Rs.1200 and 4x = Rs.1600 respectively.

4/15 A = 2/5 B
=>A = (2/5 * 15/4) B
=>A = 3/2 B
=>A/B = 3/2
A : B = 3 : 2
Therefore B's share = Rs. (1210 * 2/5) = Rs. 484

 Let the third number be x
Then, first number = 120% of x = 120x/100 = 6x/5
Second number = 150% of x = 150x/100 = 3x/2
Ratio of first two numbers = 6x/5 : 3x/2 = 12x : 15x = 4 : 5

Let the soldiers in the two armies be 10X, 3X and losses be 20Y, 3Y,

Then we have,

survivors in the larger army

10X - 20Y = 24000 ...(i)

Likewise survivors in the smaller army

And 3X - 3Y = 24000 x 13/40 = 7800 -----(ii)

Solving, we have 10X = 28000, 3X = 8400

If X be the required number, then
(45 x X)1/2 = 3 x (5 x 22)1/2
45X = 9 x 110
X = 22

The first alloy is prepared by mixing metals in proportions 7 : 2,
Then the weight of the first alloy will be = 7x + 2x
Assume ’x’ = 1Kg
The weight of the first alloy will be 9kg
The second alloy is prepared by mixing metals in proportions 7 : 11,
The weight of the second alloy will be 18kg
If equal quantities of the two alloys are melted to form a third alloy,
Then 18Kg of first alloy should be used, then its ratio will become 14 : 4,
Now, the proportion of argentum and brass in third alloy = 7 + 14 : 11 + 4 = 21 : 15 = 7 : 5

Let X’s income be 3k then Y’s income is 4k.
Let X’s expenditure be 4g then Y’s expenditure is 5g.
But 3k = 9/10 (5g) or k = 3/2g
X’s saving/Y’s saving = (3k - 4g)/(4k - 5g)
[3 (3 / 2g) - 4g]/[ 4(3 / 2g) - 5g] = 1/2
X’s saving : Y’s saving = 1 : 2

Let number of one rupee coins = 3K

Number of 50 paise coins = 4K

Number of 25 paise coins = 4K

Therefore 3K + 0.50 x 4K + 0.25 × 4K = 216

=> 3K + 2K + 1K = 216

=> K = 36

Therefore number of 50 paise coins = 4K = 144

Let number of boys = x
Let number of girls = y
Therefore x/y = B and y/x = G
3(B + G) = 3[(x/y) + (y/x)] = 3[(x2 + y2)/xy] > 3If the boy-to-girl ratio in a class is B and the girl-to-boy ratio is G, then 3 (B + G) equals?

(A/3) = (B/4) = (C/5) = K

=> A = 3K, B = 4K, C = 5K

Therefore, A : B : C = 3K : 4K : 5K = 3 : 4 : 5

Let x and y be the two numbers
Therefore, (x/y) = (2/3),
(x + 8)/(y + 8) = 3/4 => x = 16, y = 24

New ratio will be 3 x (150/100) : 5 x (160/100) : 7 x (150/100) = (9/2) : 8 : (21/2) = 9 : 16 : 21

(2a + 3b)/(5a - 3b) = 23/5
=> 10a + 15b = 115a - 69b
=> 105a = 84b => 5a = 4b
a/b = 4/5

Given, y/(x - z) = (y + x)/z
yz = xy + x2- yz - xz .. (1)
Also (x/y) = (y/(x - z)) => x2 - xz = y2 .. (2)
Using (1) and (2), we get yz = xy - yz + y2
2yz = xy + y2
2z = x + y
Only option (c) satisfies the equation