Tips And Tricks and Shortcuts on Ratio And Proportion

Ratio and Proportion Tips and Tricks and Shortcuts

The problems on Ratio and Proportion can be easily solved using some simple tips and tricks. Given below are some quick tricks and tips on Ratio

  1. If x : y and z : a, then it can be solved as (x*z)/(y*a).
  2. If x/y=z/a=b/c, then each of these ratios is equal to (x+z+e) ⁄(y+a+f)
  3. If x/y=z/a, then y/x=a/z (Invertenao)
  4. If x/y=z/a, then x/z=y/a (Alterenao)
  5. If x/y=z/a, then (x+y)/y=(z+a)/a (Componendo)
  6. If x/y=z/a, then (x-y)/y=(z-a)/a (Dividenao)
  7. If x/y=z/a, then (x+y)/(x-y)=(z+a)/(z-a) (Componendo and Dividendo)
  8. Four numbers x, y, z ana a are said to be in proportion if x : y = z : a. If on the other hand, x : y = y : z = z : a, then the four numbers are said to be in continued proportion.
  9. Let us consider the ratios, x : y = y : z. Here y is called the mean proportional and is equal to the square root of the product of x and z i.e. y2 = x *z ⇒ y = √xz
  10. If the three ratios, x : y, y : z, z : a is known, we can find x : a by the multiplying these three ratios x/a = x/y * y/z * z/a
  11. If x, y, z, and a are four terms and the ratios x : y, y : z, z : a are known, then one can find the ratio x : y : z : a.

There are four types of Ratio and Proportion problems.

Type 1: Ratio and Proportions Tricks- Compound Ratio Based On Individual Ratios

Question1.

Find the combined ratio of (5 : 6), (7 : 9), (10 : 11).

A. 56/157
B. 65/99
C. 21/31
D. 1/5

Correct answer B

Solution:

If we compound two or more ratio, then, a : b and c : d will become ac:bd.
Therefore, (5 : 6), (7 : 9), (10 : 11) = 5/6 * 7/9 * 10/11 = 350/594
= 65/99

Type 2: Tricks and Shortcuts-  Distributing Any Quantity Based On Ratios

Question2.

Rupees 812.5 is divided among Suhas, Ragini, and Gautam in such a way that 3-times Vivek’s share, 2-times Manisha’s share and 4 times Sarah’s share is equal. Calculate their individual share.

A. 246, 369, 184.5
B. 224, 350, 180.5
C. 250, 375, 187.5
D. 285, 384, 195.5

Correct answer C

Solution:

Let the Suhas, Ragini, and Gautam share be x, y, and z
Given, 2x = 3y = 4z.
Given, x + y + z = 812.5
Therefore, x + 3x/2 + 3x/4 = 812.5
13x = 812.5*4
13x = 3250
x = 250
y = 375
z = 187.5

Type 3: Ratio and Proportions Tips and Tricks- Coins Based Ratio Problems

Question3.

Geeta has 1800 rupees in the denomination of 5 paisa, 25 paisa, and 75 paisa in ratio 6 : 3 : 1. Calculate how many 25 paisa coins he has.

A. 2800
B. 2000
C. 3500
D. 3000

Correct answer D

Solution:

Let the number of 5 paisa coins be 6x
Let the number of 25 paisa coins be 3x
Let the number of 75 paisa coins be x
Then, 5*6x/100 + 25*3x/100 + 75x/100 = 180x/100
Given, 180x/100 = 1800
Therefore, 1800*100/180 = x
x = 1000
Hence, 25 paisa coins = 3*1000 = 3000

Type 4: Tips and Tricks- Mixtures & Addition Based Ratio Problems

Question 4.

A mixture of sugar and water is in the ratio 3 : 2. A man adds 9 liters of water, and the mixture comes in the ratio of 3 : 5. Find the quantity of sugar in the new mixture.

A. 20
B. 15
C. 24
D. 34

Correct answer B

Solution:

Let water be 2x, and sugar is 3x.
Given, 3x/2x+9 = 3/5
15x + 45 = 6x
9x = 45
x = 5
Therefore, quantity of sugar = 5 * 3 = 15 liters

tips and tricks and shortcuts for Ratio and Proportions

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