* The ratio of A:B:C
1) On a tree planting day, class A planted 2/5 the number of trees class B planted. Class C planted thrice as many trees as class B. Class A planted 48 trees. How many trees did the three classes plant altogether?
Important note : This is very simple sum involving three quantities ratio. Only thing important to understand is the sentence, Class A planted 2/5 the number of trees class B planted. That means if Class B planted 5u trees so Class A planted 2u trees. 

Note that Class C Planted thrice as many trees as Class B. If Class B planted 5u trees that means Class C planted 5 × 3 = 15u trees. 
Class A planted 48 trees but same time we can see that Class A planted 2u trees that means
2u = 48
u = 24
Altogether they have planted : 15u + 5u + 2u = 22uTrees
22u = 22 × 24
= 528 Trees
Ans: Three classes planted 528 Trees Altogether
2) Kim, Sima and Nadia had to sell some cookies for school charity. Each cookie costs $3. Kim sold ⅓ of the cookies. Sima and Nadia sold the remaining in the ratio of 2:3 . Kim sold 17 cookies more than Sima how much money did they raise altogether?
Important note : In this sum it is given that Kim sold ⅓ of the stickers, that means there were total 3u stickers and Kim sold 1u of the stickers. It is also given that Sima and Nadia sold remaining stickers[3u(Total)1u(Kim sold)= 2u] in the ratio of 2:3. 

Here we have used equivalent Ratio. Remaining 2u are distributed in Sima and Nadia's 2:3 ratio that means 5 parts. 
It is given that Kim sold 17 more stickers than Sima
Kim sold 5u stickers while Sima sold 4u stickers that means
5u  4u = 17
u = 17
Altogether they sold : 5u + 4u + 6u = 15u
15u = 15 × 17 —› 255 Stickers
Each stickers cost $3 that means:
255 × $3 = $765
Ans : They collected $765 altogether
3) Dev, Jack and Ahmad shared some stickers. The ratio of Dev's stickers to Jack's stickers is 3:8 while the ratio of Jack's stickers to Ahmad's stickers is 3:5. If Dev has 27 stickers, then find the number of stickers that Ahmad has.
Important note : This is very simple sum involving three quantities ratio. To solve this sum, only thing we have to do is to use equivalent ratio. 

We have been given two relations. First is between Dev and Jack, Second is between Jack and Ahmad. Jack is common in both situations that's why we are using his ratios to find equivalent ratio. We have multiplied Dev and Jack's ratios with 3 while Jack and Ahmad's ratios with 8. But in both situations Jack's ratios will remain 24 that is common. 
It is given that Dev had 27 stickers but we have found out that Dev has 9u stickers. That's why 9u = 27 —› u = 3
Ahmad has 40u Stickers. That means :
40u = 40 × 3 —› 120 Stickers
Ans : Ahmad had 120 Stickers
4) In Miss Li's orchard, the number of orange trees is 1/2 the number of mango trees. the number of chicku trees is 3/4 that of orange trees.
a) Find the ratio of mango trees to chicku trees
b) If Miss Li plants 50 more chicku trees, the ratio of mango trees to chicku trees will become 2:1. Find the number of mango trees in the end.
Important note : This is very simple sum involving three quantities ratio. To solve this sum, only thing we have to do is to use equivalent ratio. The way we used in previous sum. Here in this sum Orange is common in both relations so we will use It's both ratios to find equivalent ratio. 
Orange  :  Mango  :  Chicku 
1  2  
× 4  × 4  
4  3  
× 1  × 1  
4  :  8  :  3 
a) Ratio of the number of Mango trees to the number of Chicku trees is –› 8 : 3
b) If Mr. Li plants 50 more Chicku trees ratio of mango trees to chicku trees will become 2 :1

Here we will use unit and parts method. We will use units for before situation and parts for after situation. Here, the number of mango trees does not change but if 50 more chicku trees added their ratio changes and we will use that fact to solve our sum. So in this part of sum we will use skill 3 : an unchanged quantities . 
From table we can see that number of mango trees remain constant that's why, 8u = 2p –› 4u = 1p
Same way from table we can say that 3u + 50 = 1p
Here we can replace 1p's value because 1p = 4u
We will get , 3u + 50 = 4u
That we can solve for u
u = 50
In Mr. Li's orchard he has 8u mango trees that means –›
8 × 50 = 400 Mango trees
Ans :a) 8 : 3
b) There are 400 Mango trees in Mr. Li's Orchard
5) Alka collected some 10¢, 20¢ and 50¢ coins, the ratio of number of 10¢ coins to the number of 20¢ coins is 5:3. The ratio of number of 20¢ coins to the number of 50¢ coins is 4:3.
a) Find the ratio of number of 10¢ coins to the number of 20¢ coins to the number of 50¢ coins.
b) Alka spent 32 10¢ coins and added 14 50¢ coins and the ratio 0f number of 10¢ coins to number of 50¢ coins became 1:4. Find the value of money she had at first.
Important note : This is very simple sum involving three quantities ratio. To solve this sum, only thing we have to do is to use equivalent ratio. The way we used in previous sum. Here in this sum number of 20¢ coin is common in both relations so we will use It's both ratios to find equivalent ratio. 
10¢  :  20¢  :  50¢ 
5  3  
× 4  × 4  
4  3  
× 3  × 3  
20  :  12  :  9 
a) Ratio of number of 10¢ coins to the number of 20¢ coins to the number of 50¢ coins. –› 20 : 12 : 9
b) If Alka spend 32 10¢ coins and add 14 50¢ coins, New ratio of number of 10¢ coins to the number of 50¢ coins will be 1 : 4

Here we will use unit and parts method. We will use units for before situation and parts for after situation. Here,both quantities changes. So in this part of sum we will use skill 2 : changing quantities. 
From table we can see that
20u  32 = 1p –› ( Multiply this equation by 4 to get 4p so we can replace 4p's value in equation 2)
80u  128 = 4p
Same way from table we can say that
9u + 14 = 4p
Here we can replace 4p's value
We will get ,
9u + 14 = 80u  128
80u  9u = 128 + 14
71u = 142
u = 2
10¢ coins —› 20u = 20 × 2 —› 40 × 10¢ = 400¢
20¢ coins —› 12u = 12 × 2 —› 24 × 20¢ = 480¢
50¢ coins —› 9u = 9 × 2 —› 18 × 50¢ = 900¢
Total —› 400¢ + 480¢ +900¢ = 1780¢ = $17.80
Ans :a) 20 :12 :9
b) At first Alka had $17.80
6) For his school's fund raising day, Vandy sold 5/8 as many badges as Shan, while the ratio of badges sold by Mark to those sold by Vandy was 3:4. If mark sold 136 less badges than Shan,
a) How many badges did Mark and Shan sell altogether?
b) How many more badges should Mark have sold so that the ratio of his sold badges to Vandy's sold badges would have become 3:2
Important note : In this sum Vandy is common in both relation. That's why we will use Vandy's ratio to find out Equivalent ratio. 
Vandy  :  Shan  :  Mark 
5  8  
× 4  × 4  
4  3  
× 5  × 5  
20  :  32  :  15 
It is given that Mark sold 136 less badges than shan, that means :
32u  15u = 136
17u = 136
u = 8
a) Altogether, Shan and Mark sold badges :
32u + 15u —› 47u = 47 × 8 —› 376 Badges

From first part of the sum we found out that Vandy sold 20u badges and Mark sold 15u badges. But in this part of sum it is given that , What if Mark would have sold more badges than 15u. So how many more badges would he have sold that the ratio of his sold badges to that of vandy's sold badges would have become 3 : 2 . Most important thing to take note is here Vandy's number of badges remain constant. Here, we are using skill 3: An unchanged quantities. 
We have been given ratio of before and after situations and we know Vandy's number of badges remain constant so we can use equivalent ratio here.
From table it is cleared that if Mark would have sold 30u of badges instead of 15u of badges than the ratio of his sold badges to that of Vandy's sold badges would have become 3 : 2
30u  15u = 15u —› 15 × 8 = 120 Badges
Ans :a) Mark and Shan sold 376 Badges together.
b) Mark should have sold 120 badges more
7) At a hyper mart, a salesperson sold 4 times as many Washing machines as Dryers and additional 8 Washing machines.
Later in the day, the number of Dishwashers that were sold was 1/8 of the number of Washing machines that were sold and additional 8 Dishwashers.
In the end of the day, the salesperson realized he has sold 4 less Dishwashers than Washing machines. Find the total number of items he sold that day.
Important note : In this sum Washing Machine is common in both relation. But this is not simple sum of ratio because here additional number of products also given. That's why we will use unit and part method but here for Washing machine and dryer's relation we will use unit while for Washing machine and Dish washer's relation we will use part. Below given table will explain all situation clearly. 

Point 1: Sales person sold 4 times as many WM as D and an additional 8 WM. That means if he sold 1u dryer than 4u + 8 WM would have been sold.
Point 2: DW sold was 1/8 of the number of WM and an additional 8 DW. That means, if he sold 8p WM than 1p + 8 DW would have been sold.
Point 3: He sold 4 less DW than dryer. That means if he sold 1u dryer than he would have sold 1u  4 DW.

From table we can say that 4u + 8 = 8p
And, 1p + 8 = 1u  4
Simplify both equation so will we get u + 2 = 2p (divide 1st equation by 2)
And, 1u  12 = 1p Multiply this equation by 2 so we will get
2u  24 = 2p
We can replace 2p's value
u + 2 = 2u  24
u = 26
Total electric Item has been sold : 4u + 8 + 1u + 1u  4 = 6u + 4
= (6 × 26) + 4
= 156 + 4
=160 Electric items
Ans :a) In entire day, salesperson sold 160 Electric items
8) Dick had a total of 1228 Rose, Lily and Jasmine flowers for sale.
The ratio of the number of Rose to Lily was 5:2.
After Dick sold 2/3 of the Lilys and 1/4 of the Jasmines, and none of the rose, he had 931 flowers left. How many jasmine flowers did Dick have at first?
Important note : In this sum Common relation is not given. But instead only relation between Rose and Lily is given while for Jasmin only his sold portion is given so lets use unit for Rose and Lily, and for our convenience we will use 4p for Jasmine because it is given that 1/4 of Jasmine sold that means 1u of Jasmine sold. One more thing given is total number of flowers so we can use that to solve the sum. 

Here Total number of Flowers have been given for both, before and after, situations. For our convenience we have multiplied Rose and Lily's ratio with 3 because it is given that Dick sold 2/3 of the Lilys. So initially he had 6u Lilys and after selling 2/3 Lilys he is left with 2u Lilys. 
Initially Dick had total 1228 Flowers that means :
15u + 6u + 4p = 1228 —› 21u + 4p =1228 —› eq (1)
After selling Lily and Jasmine he had total 931 Flowers
15u + 2u + 3p = 931 —› 17u + 3p = 931 —› eq(2)
Now multiply eq(1) by 3 and eq (2) by 4 so we will get 12p in both equations that is common factor between both equations, so we can replace 4p's value in one equation and solve for unit.
63u + 12p = 3684 —› eq(3)
68u + 12p = 3724 —› eq (4)
68u + 3684  63u = 3724
5u = 40
u = 8
To find out number of Jasmine flower Dick had we have to find 4p. That we can find replacing u's value in equation (1)
(21 × 8) + 4p = 1228
168 + 4p = 1228
4p = 1060 Jasmine flowers
Ans : Dick had 1060 Jasmine flowers at first.
9) In a fun fair, each woman was given 4 coupons, each man was given 2 coupons and each child was given 5 coupons.
2/5 of the people at the gathering were children. The number of women was 5/7 the number of men.
If 1702 coupons were given out, how many adults were there altogether?
Important note : In this sum two most important things have been given and those are, coupons each person receive and total number of coupon given. We will use these informations to solve our sum.
Another thing given is, 2/5 of the people were children, that means 2u were children while 3u were adult. Same time it has been given that number of women were 5/7 of men that means among Adult, 5u were Women and 7u were Men. So we will use equivalent ratio here because men and women together make 12u. In another there were 12u Adult. 

To bring equivalent ratio we have multiply Adult and children's ratio by 4. and to solve sum we have multiply those ratios by the coupon given to each of them so we will get total number of coupon given in unit.

74u = 1702
u = 23
There were 12u Adult in gathering that means :
12 × 23 = 276 Adults
Ans : There were 276 Adult in gathering
10) In a book fair, ^{2}⁄_{5} of the books and a additional 20 books were Thriller.
¼ of the remaining and another 15 books were Romance.
The rest were comic. Given that there were 735 comic books, how many books were there altogether?
Important note : This is very simple sum of fraction. Total number of comic books has been given using that we will find out number of Thriller and Romance books . 
Total 
Thriller 
Romance 
Comics 
^{2}⁄_{5} + 20  [(^{3}⁄_{5}  20)×(¼)]+15  735  
If T= 20u 
20u × ^{2}⁄_{5 }  [(12u  20) × ¼] +15  (12u  20)  (3u + 10) 
8u + 20  (3u  5) + 15  9u  30  
3u + 10 
Now Explanation of above given table
> It is given that ^{2}⁄_{5} of the books and additional 20 books were thriller that means thriller books are ^{2}⁄_{5} +20
> ¼ of the remaining and another 15 books were romance. That means if total we have ^{5}⁄_{5} books out of it ^{2}⁄_{5} + 20 are Thriller so remaining books are ( ^{3}⁄_{5}  20) .
> To find out ¼ of ^{ 3}⁄_{5} we have to multiply them.
> Last it is given that rest of the books were comics.
To simplify above given relation we have used 20u as total books.
^{2}⁄_{5} of 20u = ^{2}⁄_{5} × 20u > 8u ,
that means Thriller books = 8u +20
Now remain books are 20u  (8u + 20) > 20u  8u 20 = 12u  20
¼ of remaining books = (12u  20) × ¼ > 3u  5
That means Romance books are 3u  5 + 15 = 3u + 10
Now rest of the books means :
Total  Thriller  Romance = Comics
20u  (8u + 20)  (3u + 10) = 20u  8u  20  3u 10 = 9u  30
that means romance books are = 9u  30
But we have been given that there were 735 comic books
That means : 9u  30 = 735 , that we can solve for unit and we will get unit value
9u  30 =735
9u = 765
u = 85
We have to find out Total books , and total books are 20u
20u = 20 × 85
= 1700
Ans : There are 1700 total books in book fair