Ten Most Important Box Problems in reasoning for Competitive Exams

Ten Most Important box problems in reasoning for SSC CGL,  SSC CHSL , RRB NTPC and other similar competitive exams. 

                                        Problem # 1

Multiplying the numbers in first column with numbers in second column , Its product must be equal to the multiplication of  numbers in third column and fourth column. 

Since all the numbers in third and fourth columns are given therefore we have to start from here.

 In 3rd  and 4th columns

1 × 24 = 24

8 ×  3  = 24

2 × 12 = 24

12 × 2 = 24

In 1st and 2nd columns 

6 × 4 = 24

4 × 6 = 24

3 × 8 = 24

 Similarly  ? × 1 = 24

Hence ? = 24

Hence option (D) is right option

               Problem # 2

Carefully study all the numbers in 1st row which are 3 ,8, 13, 18, ? , Now take the difference of two successive numbers, this is 5 in all the cases.

18 - 13 = 5 

13 - 8 = 5 

8 - 3 = 5

so  ? - 18 = 5

⇒ ? = 5 + 18 

⇒ ? = 23

Now study  all the numbers in 2nd row which are 2 ,11, 20, 29, ? , Now take the difference of two successive numbers, this is 9 in all the cases. 

Because  29 - 20 = 9

20 - 11 = 9

11 - 2 = 9

So ? - 29 = 9

⇒ ? = 9 + 29

⇒ ? =  38

Hence option (C) is right option

                                                                     Problem # 3

If we multiply 1st column with  the square root of 3rd column ,then we shall get the 2nd column.

7 × √4 = 7 × 2 = 14 (1st element in 2nd column)

4 × √9 = 4 × 3 = 12 ( 2nd element in 2nd column)

6  × √?   = 24 

 √?   =  4 

 √?   = 4 ,  squaring both sides

 ?    = 16   ( 3rd element in 2nd column)

Hence option (A) is right option

  Problem # 4

Taking the sum of squares of all the elements of 1st three rows to get the element in fourth row of respective column.

4² + 2² + 3² = 16 + 4 + 9 = 29          (Last number in 1st column)

 5² +  1² +  6² = 25 + 1 + 36 = 62        (Last number in 2nd column)

 1² + 2² + 3² = 1 + 4 + 9 = 14           (Last number in 3rd column)

5² + 5² + 2² = 25 + 25 + 4 = 54  ,The value of question Mark (Last number in 4th column)

Hence option (A) is right option

                                                                     Problem # 5

Look at the numbers in fourth row, these are large as compare to all other numbers.  All the numbers in fourth row are sum of 2nd row and product of 1st row and 3rd row .

( 7 × 3 )  +  ( 8 ) = 21 + 8 =  29  ( Last number in 1st column)

( 4 × 3 )  +  ( 7 ) = 12 + 7 =  19  ( Last number in 2nd column)

( 5 × ? )  +  ( 6 ) = 5? + 6 =  31  ( Last number in 3rd column)

⇒ 5? =  31 - 6 

⇒ 5? =  25

⇒ ? =  25/5

⇒ ? =  5

Hence option (D) is right option

                

 Problem # 6

 In this box problem the difference of the numbers in 1st and 2nd columns is equal to the number in 3rd column, while the sum of the numbers in 1st and 2nd columns is equal to the number in 4th column.

5  -  2 =  3 ( Number in 3rd column of 1st row )

6  -  3 =  3 ( Number in 3rd column of 2nd row  )

4  -  1 =  3 ( Number in 3rd column of 3rd row )

8  -  1 =  7 ( Number in 3rd column of 4th row )

5  +  2 =  7 ( Number in 4th column of 1st row )

6  +  3 =  9 ( Number in 4th column of 1st row  ) 

4  +  1 =  5 ( Number in 4th column of 1st row  )

8  +  1 =  9 ( Number in 4th column of 1st row   ) ,

 Therefore this will be the value of  question mark  "? "

Hence option (D) is right option

                                              Problem # 7

In this box problem every number in 2nd column is the sum of 1st and 3rd column.

481 + 365 = 846 

655 + 184 = 839

297 + 492 =  789 (The value of question mark)

Hence option (B) is right option

                                                        Problem # 8

In this box problem the difference of sum of last two columns and 1st two columns in every row is one.

( 10 + 5 ) - ( 6 + 8 )  = 15 - 14 = 1 ( 1st Row)

( 8 + 4 ) - ( 9 + 2 )   = 12  - 11 = 1  ( 2nd Row)

( 7 + 8 ) - ( 11 + 3 )  = 15 - 14 = 1 ( 3rd Row)

( ? + 2 ) - ( 5  + 8 )  = ( ? + 2 ) - ( 13) = 1 ( 4th Row)

So in order to get the difference equal to one we  have to put  ? = 12,

so that ( 12 + 2 ) - ( 5  + 8 )  = 14 -  13 = 1 ( 4th Row)

Hence option (B) is right option

Also read these posts on Reasoning

1.Reasoning for competive exams
2 .Box and circle reasoning
3 . Reasoning for bank exams
4. Ten Tricky logical reasoning
5. Missing  number series questions
6. Reasoning questions with answers
7. Circle Reasoning
8.  Box Problems
9. 15 Questions Circle Problems 
10.  SSC CGL Reasoning

                                                                      Problem # 9

In this box problem every number in fourth row of every column is the sum of  Multiplication of  all the numbers in every column and sum of all the numbers in every column. 

In 1st column ( 3 × 4 ×  5 ) + ( 3 + 4 + 5 ) = 60 + 12 = 72

In 2nd column ( 2 ×  5 × 6 ) + ( 2 + 5 + 6 ) = 60 +13 = 73

In 3rd column ( 5 × 9 × 1 ) + ( 5 + 9 + 1 ) = 45 + 15 = 60 ( The value of question mark )

Hence option (D) is right option

                                              Problem # 10

Multiplying the numbers in first column with numbers in second column , Its product must be equal to the multiplication of  numbers in third column and fourth column. 

Since all the numbers in 1st and  2nd columns are given therefore we have to start from here.

 In 1st and 2nd columns 

6  ×  8 =  48

4 ×  12  = 48

3 ×  16  = 48

8  ×  6  =  48

 In 3rd  and 4th columns

1 × 48 = 48

8  ×  6  =  48

16 × 3 =  48

 Similarly multiplication of  2  × ? = 48

Hence ? =  24

Hence option (D) is right option

Also reads these posts on Reasoning
ReasoningProblems #11  
ReasoningAnalogy #10
Reasoning Questions #9    
Circle Reasoning #8
TenBox-Problems #7   
TenReasoning problems #6
Ten-Important-Problem#5  
Ten--Tricky-Puzzles #4
Twelve-Figures-Problems#3   
Ten-Important-Reasoning#2     
Picture-Reasoning#1

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15 Most Important Questions of Reasoning in Circle Problems

15 Missing term in circle reasoning with solution have been discussed in this post . These types of problems are very helpful for cracking competitive exams like ssc cgl, ssc chsl , RRB NTPC, Group D and various Bank exams and many other similar exams.

15 Most Important circle Problems of Reasoning for different competitive Exams

Problem # 1

Add all the numbers in every circle except middle number then divide it with 3 to get middle number.

{ 13 + 10 + 7 + 9  } / 3 = 39 / 3 = 13 ( Middle number in 1st picture)

{ 27 + 23 + 5 + 8  } / 3 = 63 / 3 = 21 ( Middle number in 2nd picture)

{ 35 + 44 + 9 + 8  } / 3 = 96 / 3 = 32 ( Middle number in 3rd picture)

Hence option (1) is right answer.

Problem # 2

Multiply both the numbers in the second line then divide it with difference of the numbers in the 1st line of every figure. In last step add "7" to it to get the number in the middle of every figure.

{ ( 9 * 6 ) / ( 13 -10 ) } + 7 = ( 54 / 3 )  + 7 =  18 + 7 = 25 ( Middle number in 1st picture )

{ ( 8 * 3 ) / ( 27 - 23) } + 7 = ( 24 / 4 )  + 7 =  6 + 7 =  13 ( Middle number in 2nd picture )

{ ( 11 * 9 ) / ( 44 - 35 ) } + 7 = ( 99 / 9 )  + 7 =  11 + 7 = 18 ( Middle number in 3rd picture ).

Hence option (4) is right answer.

Problem # 3

This figure consist of four sectors and every sector consists of three numbers. Every number in the inner part of each sector is H C F ( Highest Common Factor ) of two other numbers, which are in outer part of each sector.

HCF of 12 and 27 = 3
HCF of 16 and 24 = 8
HCF of 2 and 3 = 1                            

Similarly HCF of 4 and 8  = 4  ( The value of question mark )

Hence option (1) is right answer.

Problem # 4

This figure consist of four sectors and every sector consists of three numbers. Every number in the inner part of each sector is HCF ( Highest Common Factor ) of two other numbers which are in outer part of each sector.

  H C F of 12 and 15 = 3 

   H C F of   6 and 4 = 2 

   H C F of 3 and  2 = 1 

  H C F of 12 and 6 = 6 ( The value of question mark)

Hence option (4) is right answer.

Problem # 5

This figure consist of four sectors and every sector consists of three numbers. Every number in  the inner part  is LCM ( Lowest Common Multiple ) of two other numbers which are in outer part of each sector.

LCM of 12 and 15 = 60 

LCM  of  9 and 6 = 18

LCM  of  3 and 2 = 6 

LCM of 8 and 6 = 24  (The value of question mark)

Hence option (4) is right answer.

 Problem # 6

All the Numbers except one in both the circles are written in same pattern . If we multiply all the three digits of any number in both the circles then we shall get same result except in one number in both the circles. And that different result will be the odd one out in this problem.

In 1st circle

The product of digits of number 262 = 2 * 6 * 2 = 24

The product of digits of number 226 = 2 * 2 * 6 = 24

The product of digits of number 423 = 4 * 2 * 3 = 24

The product of digits of number 333 = 3 * 3 * 3 = 27 ( Odd one out)

The product of digits of number 342 = 3 * 4 * 2 = 24

In 2nd circle

The product of digits of number 562 = 5 * 6 * 2 = 60

The product of digits of number 345 = 3 * 4 * 5 = 60

The product of digits of number 543 = 5 * 4 * 3 = 60

The product of digits of number 256 = 2 * 5 * 6 = 60 

The product of digits of number 452 = 4 * 5 * 2 = 40 ( Odd one out)

So from both these circles two results 27 and 40 of numbers 333 and 452 are different from other.

Hence option (3) is right answer.

Problem # 7

All the Numbers except one in both the circles are written in same pattern . If we add all the three digits of any number in both the circles then we shall get same result except in one number in both the circles. And that different result will be the odd one out in this problem.

In 1st circle

The sum of digits of number 163 = 1+ 6 + 3 = 10

The sum of digits of number 145 = 1 + 4 + 5 = 10

The sum of digits of number 334 = 3 + 3 + 4 = 10

The sum of digits of number 441 = 4 + 4 + 1 = 9  ( Odd one out)

The sum of digits of number 343 = 3 + 4 + 3  = 10

In 2nd circle

The sum of digits of number 414 =  4 + 1 + 4 =  9

The sum of digits of number 333 = 3 +  3 + 3 =  9

The sum of digits of number 444 = 4 + 4  + 4 =  12  ( Odd one out). 

The sum of digits of number 504 = 5 + 0 + 4 =  9 

The sum of digits of number 108 = 1 + 0 + 8 = 9 

So from both these circles two results 9 and 12 of numbers 441 and 444 are different from other.

Hence option (2) is correct answer.

 

Problem # 8

In this circle all the numbers are written in a line and the sum of each line is 25. Starting from the line which is right side of ? ( Question mark),

 4 + 11 + 3 + 7 = 25

10 + 1 + 9 + 5 = 25

11 + 6 + 2 + 6 = 25

? + 5 + 5 + 8  = 25

⇒   ? + 18 = 25 

? = 25 - 18 = 7

Hence option (1) is correct answer

Also read these posts on Reasoning

1.Reasoning for competive exams
2 .Box and circle reasoning
3 . Reasoning for bank exams
4. Ten Tricky logical reasoning
5. Missing  number series questions
6. Reasoning questions with answers
7. Circle Reasoning
8.  Box Problems
9. 15 Questions Circle Problems 
10.  SSC CGL Reasoning

 Problem # 9

Here 1st circle consist of four numbers and if we divide sum of all the digits in its outer part with total numbers then result will be the middle number.

( 3 + 8 + 7 + 2  )/4 =  20/4 = 5 ( Middle number)

2nd circle consist of five numbers and if we divide sum of all the digits in its outer part with total numbers then result will be the middle number. 

( 6 + 4 + 8 + 5 + 7 )/5 = 30/5 =6 ( Middle number)

3rd circle consist of six numbers and if we divide sum of all the digits in its outer part with total numbers then result will be  the middle number. 

(  9 + 7 + 8 + 9 + 7 + 8   )/6 = 48/6 =8 ( Middle number)

Hence option (2) is correct answer

Problem # 10

This figure consist of four sectors and every sector consists of three numbers. Every number in  the inner part  is cube of difference of two other numbers which are in outer part of each sector.

Cube of ( 8 - 5 )  =  3³ = 27  

Cube of  ( 23 - 19 ) = 4³ = 64

Cube of ( 3 - 2 ) =   1³ = 1

Cube of  ( 26 - 24 ) = 2³ = 8 ( The value of question mark)

Hence option (4) is correct answer

Problem # 11

This figure consist of four sectors and every sector consists of three numbers. Every number in  the inner part  is  Four times of square of difference of two other numbers which are in outer part of each sector.

4 × Square of ( 11 - 8 )  =  4 × (3)²  =  4 × 9 = 36  

4 × Square of ( 5 - 5 )  =  4 × (0)²  =  4 ×  =  0 

4 × Square of ( 7 - 3 )  =  4 × (4)²  =  4 × 16 = 64  

4 × Square of ( 8 - 2 )  =  4 × (6)²  =  4 × 36 = 144 ( The value of question mark)

Hence option (1) is correct answer

Problem # 12

This figure consist of four sectors and every sector consists of three numbers. Every number in  the inner part  is cube of difference of two other numbers which are in outer part of each sector.

Cube of (11 - 8 )  =  3³ = 27  

Cube of  ( 5 - 4 ) =   1³ = 1

Cube of ( 7 - 3 ) =    4³ = 64

Cube of  ( 8- 2 )  =  6³ = 216 ( The value of question mark)

Hence option (1) is correct answer

Problem # 13

This figure consist of four sectors and every sector consists of three numbers. Every number in  the inner part  is five times of  product of two other numbers which are in outer part of each sector.

( 6 × 3 ) × 5 = 18 × 5 = 90

( 2 × 3 ) × 5 =  6  × 5 = 30

( 4 × 2 ) × 5 =  8  × 5 = 40

( 5 × 4 ) × 5 = 20 × 5 = 100 ( The value of question marks ) . 

Hence option (1) is right answer.

Problem # 14

This figure consist of four sectors and every sector consists of three numbers. Every number in  the inner part  is the sum  of  two other numbers which are in outer part of opposite sector to these numbers.

 24 + 11 = 35 ( The number opposite to both 24 and 11 ) . 

13 + 26 = 39 ( The number opposite to both 246and 13 ) . 

9  +  25 = 34 ( The number opposite to both 9 and 25 ) . 

7 + ? = 35 ( The number opposite to both 7 and ? ) . 

So if we put ? = 28 then we shall have total equal to 35.

Hence option (2) is right answer

Problem # 15

All the Numbers except one in both the circles are written in same pattern . If we multiply all the three digits of any number in both the circles then we shall get same result except in one number in both the circles. And that different result will be the odd one out in this problem.

In 1st circle

The product of digits of number 145 = 1 *  4 *  5  = 20

The product of digits of number 451 = 4 * 5 * 1 =  20

The product of digits of number 225  = 2 * 2 * 5 = 20

The product of digits of number 541 = 5 * 4 * 1  = 20 

The product of digits of number 255 = 2 * 5 * 5 = 50 (Odd one out)

In 2nd circle

The product of digits of number 414 = 4 * 1 * 4  = 16

The product of digits of number 444 = 4 *  4  * 4  = 64  (Odd one out)

The product of digits of number 224 = 2 * 2  * 4  = 16

The product of digits of number 441 = 4  * 4 * 1  = 16 

The product of digits of number 128 = 1 * 2 * 8  =  16  

So from both these circles two results 50 and 64 of numbers 255 and 444 are different from other.

Hence option (3) is right answer.

Also Reads following posts on Reasoning

ReasoningProblems #11  
ReasoningAnalogy #10
Reasoning Questions #9    
Circle Reasoning #8
TenBox-Problems #7   
TenReasoning problems #6
Ten-Important-Problem#5  
Ten--Tricky-Puzzles #4
Twelve-Figures-Problems#3   
Ten-Important-Reasoning#2     
Picture-Reasonin#1

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Twelve Important Tricky Reasonining Problems for SSC CGL ,SSC CHSL And RRB NTPC Exams

12 Most important problems of  tricky reasoning problems which includes box problems , triangles problems and star problems for competitive exams like SSC CGL ,SSC CHSL and RRB NTPC etc have been included in this post. 

12 Important Tricky Reasoning Problems for Reasoning Various Exams

 Problem # 1

Adding all the numbers which are forming triangle in corner like this

Top Right corner

10 + 13 + 9 = 32 Top Right corner

Top Left corner

15 + 10 + 7 = 32  Top left  corner

Bottom Left corner

 7 +  17 +  8 = 32  Bottom left  corner

Bottom Right corner

Similarly 8 + 9 + ? = 32 means Question mark "?"  will take the value 15. So

Option (1) is correct Option.

Problem # 2

Add all the numbers from corner which are forming rhombus and square respectively , because sum of all the numbers in both rhombus and square in corner position are equal to 34.

Rhombus

7 + 10 + 9 + 8 = 34

Square

15 + 5 + ? + 6 = 34 . 

26 + ? = 34

? = 34 - 26

? = 8

This means question mark ? will have  8 value. 

So Option (2) is correct Option.  

Problem # 3

Add all the three numbers in every small square then divide this sum with the fourth number in same square to get result every time 10.

Top Left Square 

 ( 9 + 4  + 7 )/2 = 20 /2 = 10

Top Right Square

 ( 9 + 13  + 8 )/3 = 30 /3 = 10

Bottom Left Square

( 9 + 16 + 15 )/4 = 40/4 = 10

Similarly in 

Bottom Right Square

 (6 + ? +3 )/1 = 10/1 = 10. If we take the value of question mark "?"  equal to 1 then we shall get total equal to 10 ,

Hence  Option (1) is correct Option. 

Problem # 4

 

In all the picture add both the numbers in 1st row and then umbers in 2nd row and left number in last row, then take the difference of these sum to get the number in right side of 3rd row.

In 1st row 9 + 8 = 17,  4 + 9 = 13  then 17 - 13 = 4 

In 2nd row  11 + 5 = 16 , 10 + 3 = 13 then 16 - 13 = 3 , Similarly

in 3rd row  7 + 16 = 23  , 6 + 12 = 18 then 23 - 18 = 5 .

Hence option (1) is correct option.

 Problem # 5

There are five squares in this picture. To solve this problem or to find the value of question mark , sum of all the numbers in each square when divided by the number in the corner must be equal to 6 in each four figure. In  the 

Top Left square

    (  3 + 2 +  7 ) /2  =  12/2 = 6

Top Right square

   (  9 + 5 +  4) /3  =  20/3 = 6

Bottom Left square

    (  9 + 7 +  8 )/4  =  24/4 = 6

Bottom Right square

    (  3  + ? +  1 )/1  = (4 + ?)/1= 6.

4 + ? = 6

? = 6 - 4

? = 2

This means if we put ? = 2, Only then we shall 6 as answer.

Hence option (3) is correct option. 

Problem # 6

 

To solve this problem divide each number in the second row with 6 , then subtract one from the answer so obtained to get the number in 1st row.

(78/6) -1 => 13 - 1 = 12 ( The  1st number in the 1st row)

(60/6 ) -1 => 10 - 1 = 9 ( The 2nd number in the 1st row)

(24/6) -1 =>  4 - 1 =  3 ( The 3rd number in the 1st row)

( ?/6 - 1 ) -1  = >10 ( The 4th number in the 1st row). that is if we put ? = 66 then upon dividing 66 with 6 we shall get 11, And after subtracting 11 from 1 , we shall get the answer.

Hence option (C) is correct option. 

 

Problem # 7

 

Take the square root of  difference of  21 and 12 to get middle number 3. Similarly take the square root of  difference of  15  and 6 to get middle number 3 in 1st picture.

Similarly take the square root of  difference of  29 and 4 to get middle number 5. Similarly take the square root of  difference of  36  and ? to get middle number 5 in 2nd picture.

In 1st picture

√(21 - 12 ) = √9 = 3 

√(15 - 6 ) = √9 = 3 

In 2nd picture

√(29 - 4 ) = √25 = 5 

√(36 - ? ) = √25 = 5 , If we put ? = 11 then we can solve this problem.

Hence option (4) is correct option. 

  Problem # 8

1st Picture

  (10)² - 16  = 100 - 16 = 84 in the same way

 (12)² - 60 = 144 - 60 = 84

2nd Picture

(9)² - 13  = 81 - 13 = 68 in the same way 

(?)² - 53  = 68

(?)² = 68 - 53

(?)² = 121

? = 11

so ? will take the value 11

Hence option (1) is correct option.

Also read these posts on Reasoning

1.Reasoning for competive exams
2 .Box and circle reasoning
3 . Reasoning for bank exams
4. Ten Tricky logical reasoning
5. Missing  number series questions
6. Reasoning questions with answers
7. Circle Reasoning
8.  Box Problems
9. 15 Questions Circle Problems 
10.  SSC CGL Reasoning

 

            Problem # 9

Sum of all the numbers except middle number in 1st picture is equal to the middle number in 2nd picture.

5 + 2 + 7 + 5 = 19 (Middle number in 2nd picture).

Sum of all the numbers except middle number in 2nd picture is equal to the middle number in 3rd picture.

6 + 11 + 2 + 4 = 19 (Middle number in 3rd picture).

Sum of all the numbers except middle number in 2nd picture is equal to the middle number in 1st  picture.

6 + 1 + 6 + ? = 15 (Middle number in 1st picture).

To make toal 15 in 1st picture ? will have to take 2 value in 3rd picture.

Hence option (4) is correct option.

      Problem # 10

Take the sum of squares of each number in the corner positions in each picture to get number the middle positions.

In 1st Picture 

   9² + 8² + 6² + 7² =  81 + 64 + 36 +  49 = 230

In 2nd Picture 

 6² +  7² +  4² +  3² =  36 + 49 + 16 +  9 = 110

In 3rd Picture 

9² + 6² + 4² +  5² =  81 + 36 + 16 +  25 = 158

 Therefore ? will take the value equal to 158

Hence option (3) is correct option.

   Problem # 11

Add all those numbers which are above central number then take the difference of this sum from the addition  of those numbers which are below the central number .

1st figure  ( 4 + 2  + 7 ) -  ( 3 + 1 ) =  13 - 4  = 9 middle number

2nd figure  ( 3 + 3  + 5 ) -  ( 4 + 2 ) =  11 - 6  = 5 middle number

3rd figure  ( 6 + 9  + 2 ) -  ( 4 + 3 ) =  17 - 7  = 10 middle number

Hence option (1) is correct option.

 Problem # 12

In this problem all the numbers written in 1st figure are related to 2nd figure and all the numbers written in 2nd figure are related to 3rd figure , similarly all the numbers written in 3rd figure are related to 4th figure .

i.e. number 4 in 1st figure is related to 10 in 2nd figure and number 10 in 2nd figure is related to 22 in 3rd figure. similarly number 22 in 3rd figure is related to 46 in 4th figure.

4 =  1st number in 1st figure

(4   ×  2 ) + 2 = 10 ( 1st number in 2nd figure ) 

(10  ×  2) + 2  = 22 ( 1st number in 3rd figure )

(22   ×  2) + 2 = 46 ( 1st number in 4th figure )

1 =  2nd number in 1st figure

(1  ×  2 ) + 2 = 4 ( 2nd number in 2nd figure ) 

(4  ×  2) + 2 = 10 ( 2nd number in 3rd figure ) 

(10  ×  2) + 2  = 22 ( 2nd number in 4th figure ) 

3 = Bottom number in 1st figure

 (3  ×  2) + 2 = 8  ( bottom number in 2nd figure )

(8 × 2) + 2 = 18 ( bottom number in 3rd figure )

(18 × 2) + 2 = 38  (The value of question mark)

Hence option (1) is correct option.

 Also Reads these posts on Reasoning

Reasoning Problems #11  
Reasoning Analogy #10
Reasoning Questions #9    
Circle Reasoning #8
Ten Box-Problems #7   
Ten Reasoning problems #6
Ten-Important-Problem#5  
Ten--Tricky-Puzzles #4
Twelve-Figures-Problems#3  
Ten-Important-Reasoning#2   
Picture-Reasoning#1

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