Shortcut - Ratio and Proportion
Splitting a number in the given ratio
1. Divide 720 in the ratio 2: 3.
Solution 1:Total parts = 2 + 3 = 5
First number = (2/5) 720 = 288
Second number = (3/5) 720 = 432
2x + 3x = 720;
5x = 720;
x = 144
2x = 2(144) = 288;
3x = 3(144) = 432
2. Divide 720 in the ratio 2: 3: 4.
Solution 1:Total parts = 2 + 3 + 4 = 9
First number = (2/9) 720 = 160;
Second number = (3/9) 720 = 240;
Third
number = (4/9)720 = 320
2x + 3x + 4x= 720;
9x = 720;
x = 80
2x = 2(80) = 160;
3x = 3(80) = 240;
4x = 4(80) = 320
Shortcut– Ratio and Proportion
Direct Proportion:
When two parameters are in direct proportion if one parameter
increases the other one will also increase and if one parameter decreases
the other one will also decrease.
If the parameters A and B are in direct proportion they will satisfy the
below equation.
$\dfrac{A_{1}}{A_{2}}=\dfrac{B_{1}}{B_{2}}$
Question:
Price of diamond is directly proportional to its weight. If 2 grams of diamond costs $ 45,000 then what is the price of a diamond that weighs 6 grams?
Answer:Weight of diamond in first case, $A_{1} $=2
Weight of diamond in second case, $A_{2} $ =6
Price of diamond in first case, $ B_{1} $= 45000
Price of diamond in second case, $B_{2} $=?
According to the equation,
2/6 = 45000/ $B_{2} $
$B_{2} $ = 45000 x (6/2)
$B_{2} $ = 135,000
Shortcut– Ratio and Proportion
Inverse Proportion:
When two parameters are in inverse proportion if one parameter
increases the other one will decrease and if one parameter decreases the
other one will decrease.
If the parameters A and B are in inverse proportion they will
satisfy the below equation.
$\dfrac{A_{1}}{A_{2}}=\dfrac{B_{2}}{B_{1}}$
Question:
Mileage and Engine capacity are inversely proportional. A bike with 100cc engine capacity gives a mileage of 80 km. What will be the mileage given by a bike with 150cc engine capacity?
Answer:Engine capacity of first bike, $A_{1}= 100$
Engine capacity of second bike,$ A_{2}= 150
$
Mileage of first bike,$ B_{1}= 80$
Mileage of second bike,$B_{2} =?$
According to the equation,
$100/150 =B_{2}/80$
$B_{2} = 100(80)/150$
$B_{2}= 53.33 km$
Shortcut– Ratio and Proportion
Finding A: C from A: B and B: C
Question:
The ratio between salary of A and B is 4: 5. The ratio between salary of B and C is 3: 4. Find the salary of C if A is earning $3600.
Answer:
A: C = 12: 20 = 3: 5
A/C = 3/5
3600/C = 3/5;
C = 6000
Shortcut – Ratio and Proportion
Finding A: D from A: B, B: C and C: D
Question:
Ratio between salary of A and B is 3: 5, B and C is 4: 5, C and D is 6: 7. If the salary of A is $ 7200, find the salary of D.
Answer:A: D = 72: 175
7200/D = 72/175
=17500
Shortcut– Ratio and Proportion
Usage of Common Factor x
Question:
Two numbers are in the ratio 4: 5. Sum of their squares is 1025. Find the numbers.
Answer:Substitute the common factor x to the ratio 4: 5
Assume the actual numbers as 4x and 5x
Given,
$(4x) ^{2} + (5x)^{ 2} = 1025$
$16x^{2} + 25x^{2} = 1025;$
$41x^{2} = 1025$
$x^{2}= 25; $ x =5
Substitute x = 5 in 4x and 5x to find the numbers
The numbers are, 20 and 25.
Question:
Two numbers are in the ratio 3: 2. Cube of their difference is 125. Find the numbers.
Answer:$(3x – 2x) ^{3} = 125$
$x ^{3} = 125$
x=5
The numbers are 15 and 10.
Shortcut – Ratio and Proportion
Finding number of Coins in a bag
Question:
A bag contains 5 paise, 10 paise and 20 paise coins in the ratio 2:4:5. Total amount in the bag is Rs. 4.50. How many coins are there in 20 paise?
Answer:T = (2 x 5) + (4 x 10) + (5 x 20)
= 150 paise
= Rs. 1.50
X = 4.50/1.50
X=3
Quantity of 20 paise coins
=5x3
= 15 coins.
Shortcut– Ratio and Proportion
Adding and removing quantities
Question:
Container 1 has milk and water in the ratio 2 : 5. After adding 4 liters of pure water from container 2, the ratio between milk and water in container 1 became 1 : 3. Find the quantity of milk in the container.
Answer:
Let us assume the actual quantity of milk and water as 2x and 5x.
New quantity of water = 5x + 4
2x/(5x + 4) = 1/3
6x = 5x + 4
x=4
Quantity of milk = 2x
= 2(4)
= 8 liters.
Question:
If a:b=2:3 and b:c=3:4 Find a:b:c=?
Solution:a:b=2:3
b:c=3:4
Tricks:
a:b:c=(a *b) :(b *b) :(b:c)
Therefore,a:b:c=6:9:12
that is,a:b:c=2:3:4
Question:
x:y=3:4 and y:z=8:9 Find x:z=?
Solution:x:y=3:4
y:z=8:9
x:y:z=(x*y):(y*y):(y*z)
Therefore,x:y:z=(3 * 8):(8 * 4) :( 4 * 9)
24:32:36
That is,x:y:z=6:8:9
In question they asked to find x:z=6:9 or 2:3
Question:
If A:B=2:3,B: C=4:5 and C:D=6:7 .Find A:B:C:D
Solution:A:B=2:3
B: C=4:5
C:D=6:7
A:B:C:D=(ABC):(BBC):(BCC):(BCD)
=(2*4*6) :(3*4*6) :(3*5*6):(3*5*7)
A:B:C:D=48:72:90:105
Question:
Divide 600 in the rate of 2:3
Solution:2x + 3x =600
5x=600
x=120
Therefore,2x=>2 *120=240
3x=>3*120=360
Question:
If $\dfrac{A}{2}=\dfrac{B}{3}=\dfrac{C}{5}$ then A:B:C=?
Solution:$\dfrac{A}{2}=\dfrac{B}{3}=\dfrac{C}{5}=x$
$\dfrac{A}{2}=x$
A=2x
B=3x
C=5x
Therefore,A:B:C=2x:3x:5x=>2:3:5
Question:
If 3A=4B=5C then Find A:B:C=?
Solution: 3A=4B=5C =x
3A=x =>$\dfrac{x}{3}$
4B=x=>$\dfrac{x}{4}$
5C =x=>$\dfrac{x}{5}$
Therefore,A:B:C=>$\dfrac{x}{3}:\dfrac{x}{4}:\dfrac{x}{5}$
Here L.C.M is 60
then,60($\dfrac{x}{3}):60(\dfrac{x}{4}):60(\dfrac{x}{5}$)
by cancelling finally we get,
A:B:C=20:15:12
Question:
If 10% of x = 30% of y then x:y=?
Solution:$\dfrac{10}{100}\times x$ =$\dfrac{30}{100}\times y$
by cancelling we get ,1x=3y
$\dfrac{x}{y}$=$\dfrac{3}{1}$
x:y=>3:1
Question:
If 2:x:: 3:4 then x is equal to ?
Solution:a:b::c:d
Products of extremes=Products of means
$2 \times 4$= $3\times x$
8=3x
$x=\dfrac{8}{3}$
=2.67
Therefore,2 :2.67 : : 3: 4
Question:
What is the inverse ratio of 3:2:1=?
Solution:3:2:1=>$\dfrac{1}{3}$:$\dfrac{1}{2}$:$\dfrac{1}{1}$
Here LCM is 6
then,6($\dfrac{1}{3})$:6($\dfrac{1}{2}$):6($\dfrac{1}{1}$)
by cancelling we get,2:3:6
Question:
Fourth proportional to 5,8,15
Solution:a:b::c:d
5:8::15:x
Products of extremes=Products of means
$5 \times x= 8 \times 15$
x=24
Therefore,4th proportional =24.