Can You See The Pattern - Class 5 Maths - Extra Questions

Solve the magic square.

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The Concept of Solving a Magic Square is that the sum of all numbers in each $$row, column $$ and $$diagonal $$ are $$equal$$ to $$15$$

Now, 

$$ 4$$let $$x$$  $$2$$
let $$y$$  $$5$$$$7 $$
$$8 $$$$1$$ let $$z$$ 

Now, in first row 
$$Sum=4+x+2=15\Rightarrow x=9$$

Now in second Row
$$Sum=y+5+7=15\Rightarrow y=3$$

Now in third Row
$$Sum=8+1+z=15\Rightarrow z=6$$

Therfore Required magic Square is

$$ 4$$$$9$$  $$2$$
$$3$$  $$5$$$$7 $$
$$8 $$$$1$$ $$6$$ 



Draw a magic triangle whose magic number's sum is $$11$$.

Consider the given question,

We take numbers are $$1$$ , $$2$$, $$3$$,$$4$$, $$5$$ and $$6$$

We put the numbers as that sum of every side is equal to be $$11$$.
$$6+1+4=11$$
$$4+5+2=11$$
$$6+3+2=11$$


1000206_1070243_ans_c5bb2f1edf434590943248c4ccb3ab52.jpg


Observe the toothpick pattern given below:
Would make sense to join the points on this graph Explain.
1795231_ff5c4a34b5cb4f739d1da3bfed4eb432.png

Yes, it makes sense to join the points on the graph as  the points join together form a straight line y = 3x + 1 .
1882955_1795231_ans_9da6ed84ad074bb2ae09ac5a2c3173e0.jpg


Fill in the blank spaces by integers of the given magic square so that the sum of the integers in each row, each column and each diagonal is $$-6$$.
$$-1$$
$$3$$$$-2$$

Sum of integers of the magic square is $$-6$$ in each row, column and each diagonal as shown here:
$$-1$$    $$-9$$    $$\ \ 4$$
$$\ \ 3$$    $$-2$$    $$-7$$
$$-8$$    $$\ \ 5$$    $$-3$$


Using the numbers from $$9$$ to $$17$$, construct a $$3\times 3$$ magic square. What is the magic sum here? What relation is there between the magic sum and the number in the central cell?

Using the numbers from $$9$$ to $$17$$, a $$3\times 3$$ magic square is constructed as shown above:

Considering the above drawn magic square, the magic sum is $$13\times 3=39$$.

Hence, the relation between the magic sum and the number in the central cell is that if we multiply the middle number to $$3$$ (middle number$$\times 3$$), then we get the magic sum.

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Find the value of magic square.

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The Concept of Solving a Magic Square is that the sum of all numbers in each $$row, column $$ and $$diagonal $$ are $$equal$$ to $$15$$

Now, 

$$ 7$$let $$x$$  $$17$$
let $$y$$  $$5$$let $$z$$
let $$a$$let $$b$$ let $$c$$ 

Now, in first row 
$$Sum=7+x+17=15\Rightarrow x=-9$$

Now in second column
$$Sum=-9+5+b=15\Rightarrow b=19$$

Now in first Diagonal
$$Sum=7+5+c=15\Rightarrow c=3$$

Now in Second Diagonal 
$$Sum=17+5+a=15\Rightarrow a=-7$$

Now in first column
$$Sum=7+y-7=15\Rightarrow y=15$$

Now in third column
$$Sum=17+z+3=15\Rightarrow z=-5$$


Therfore Required magic Square is

$$ 7$$$$-9$$  $$17$$
$$15$$  $$5$$$$-5$$
$$-7$$$$19$$ $$3$$


Solve the magic square.

454875.PNG

Let the sum of each row, column and diagonal be x.
So, we have
19 + 16 + g = x 
35 + g = x 
g = x - 35 --(1)

Hence, i = x - (x-35) - 14 = 21

So, the sum of the left diagonal is 60.
Hence, g = 25
b = 26
f = 24

475685_454875_ans.png


Solve the magic square.

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The rows, columns and diagonals must add up to $$36$$.
$$1\times 2 = 19$$  [first row, second column]
$$2\times 3 = 9$$  [second row, third column]
$$2\times 2 = 12$$

472369_454868_ans.png


Complete the magic square.

454880.PNG

The Concept of Solving a Magic Square is that the sum of all numbers in each $$row, column $$ and $$diagonal $$ are $$equal$$ to $$15$$

Now, 

let $$ x$$$$-8$$  $$18$$
let $$y$$ $$4$$let $$z$$
$$-10$$let $$a$$ let $$b$$ 

Now, in first row 
$$Sum=18-8+x=15\Rightarrow x=5$$

Now in first column
$$Sum=x+y-10=15\Rightarrow y=20$$

Now in second Row
$$Sum=20+4+z=15\Rightarrow z=-9$$

Now in second column 
$$Sum=-8+4+a=15\Rightarrow a=19$$

Now in third column 
$$Sum=18-9+b=15\Rightarrow b=6$$

 
Therfore Required magic Square is

$$5$$$$-8$$  $$18$$
$$20$$  $$4$$$$-9$$
$$-10$$$$19$$ $$6$$ 


Solve the magic square.

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Sum of any row, column, or diagonal is 9.
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In a row of boys, Srinath is $${7}^{th}$$ from the left and Venkat is $${12}^{th}$$ from the right. If they interchange their positions, Srinath becomes $${22}^{nd}$$ from the left. How many boys are there in the row?

Srinath is $${7}^{th}$$ from the left and Venkat is $${12}^{th}$$ from the right.
Now ,If they interchange their positions, Srinath becomes $${22}^{nd}$$ from the left.
So, Srinath is $$22$$ from left and $$12$$  from right.
So, he has $$21$$ boys before him and $$11$$ boys after him.
So, total number of boys $$=$$ number of boys before him $$+$$Number of boys after him $$+$$ Srinath
$$= 21+11+1=33$$
Hence there are $$33$$ boys


In a row of plants, a plant is $$16^{th}$$ from either end of the row. How many trees are there in the row ?

A plant is $$16^{th}$$ from either end of the row.

So, it have $$15$$ plants in front and $$15$$ plants behind.

Thus, number of plants $$=15+15+1$$   [Including the mentioned plant]
                                        $$=31$$

Hence, number of plant is $$31$$.


$$\square +\square +\square =30$$
Fill the boxes using $$(1, 3, 5, 7, 9, 11, 13, 15)$$ You can also repeat the numbers.


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In a "magic square", the sum of the numbers in each row, in each column and along the diagonals is the same. Is this a magic square?
$$\dfrac {4}{11}$$$$\dfrac {9}{11}$$$$\dfrac {2}{11}$$
$$\dfrac {3}{11}$$$$\dfrac {5}{11}$$$$\dfrac {7}{11}$$
$$\dfrac {8}{11}$$$$\dfrac {1}{11}$$$$\dfrac {6}{11}$$
(Along the first row $$\dfrac {4}{11} + \dfrac {9}{11} + \dfrac {2}{11} = \dfrac {15}{11})$$.

Sum of first row $$= \dfrac {4}{11} + \dfrac {9}{11} + \dfrac {2}{11} = \dfrac {15}{11}$$ [Given]

Sum of second row $$= \dfrac {3}{11} + \dfrac {5}{11} + \dfrac {7}{11} + \dfrac {3 + 5 + 7}{11} = \dfrac {15}{11}$$

Sum of third row $$= \dfrac {8}{11} + \dfrac {1}{11} + \dfrac {6}{11} = \dfrac {8 + 1 + 6}{11} = \dfrac {15}{11}$$

Sum of first column $$= \dfrac {4}{11} + \dfrac {3}{11} + \dfrac {8}{11} = \dfrac {4 + 3 + 8}{11} = \dfrac {15}{11}$$

Sum of second column $$= \dfrac {9}{11} + \dfrac {5}{11} + \dfrac {1}{11} = \dfrac {9 + 5 + 1}{11} = \dfrac {15}{11}$$

Sum of third colum $$= \dfrac {2}{11} + \dfrac {7}{11} + \dfrac {6}{11} = \dfrac {2 + 7 + 6}{11} = \dfrac {15}{11}$$

Sum of first diagonal (left to right) $$= \dfrac {4}{11} + \dfrac {5}{11} + \dfrac {6}{11} = \dfrac {4 + 5 + 6}{11} = \dfrac {15}{11}$$

Sum of second diagonal (left to right) $$= \dfrac {2}{11} + \dfrac {5}{11} + \dfrac {8}{11} + \dfrac {2 + 5 + 8}{11} = \dfrac {15}{11}$$

Since the sum of fractions in each row, in each column and along the diagonals are the same, therefore it's a magic square.


In a "magic square", the sum of the numbers in each row, in each column and along the diagonal is the same. Is this a magic square ?
$$\dfrac{4}{11}$$$$\dfrac{9}{11}$$$$\dfrac{2}{11}$$
$$\dfrac{3}{11}$$$$\dfrac{5}{11}$$$$\dfrac{7}{11}$$
$$\dfrac{8}{11}$$$$\dfrac{1}{11}$$$$\dfrac{6}{11}$$


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Complete the following magic square by supplying the missing numbers: 
21516
912
710
1417


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Fill in the blanks to complete the following number triangle:
1808036_3a928ecd920b4e78b97a262e01870ac4.PNG

here is the complete triangle :
1715214_1808036_ans_b7fc7debd1f04a9e8bf1030e4fe9afde.png


In two-digit numbers, how many times does the digit $$5$$ occur in the unit's place?

$$15,25,35,45,55,65,75,85,95$$ are the two digit numbers which have 5 at units place. Total numbers are $$9.$$


A magic square is an array of numbers having the same number of rows and columns and the sum of numbers in each row, column or diagonal being the same. Fill in the blank cells of the following magic squares: 
2261320
101219
9111825
15172426
16714


1872309_1658484_ans_3891780569784f76884737707a27b1ed.jpg


A magic square is an array of numbers having the same number of rows and columns and the sum of numbers in each row, column or diagonal being the same. Fill in the blank cells of the following magic squares: 
813
12
11


1872307_1658483_ans_18ccdd5fb9be4ec39745fd62ab6af61c.jpg


The given figure represents a weighing balance. The weights of some objects in the balance are given.
Find the weight of each square and the circle.
1790097_59ff99225f7f42a39763305fd1795902.png

Since weight is balanced 
$$\implies$$ Weight on L.H.S $$=$$ Weight of R.H.S
$$\implies 20+20=14+4+x$$    (suppose $$x$$ is the weight of circle)
$$\implies 40=18+x$$
$$\implies x=40-18=22\text{ kg}$$
$$\therefore$$ the weight of the circle is $$22\text{ kg}$$


Observe the toothpick pattern given below:
Imagine that this pattern continues. Complete the table to show the number of toothpicks in the first six terms.
1795225_c2beb0f87a0746878231ec52f3ca8ccd.png

By observing the given pattern, we can say that

1 →  4
2 →  4 + 3 = 7
3 →  7 + 3 = 10
4 →  10 + 3 = 13
5 →  13 + 3 = 16
6 →  16 + 3 = 19

In tabular form it can be represented as :

 Pattern 12 3 4  56 
 Toothpicks 47 10 13 16 19 




Complete the magic square given alongside using number $$ 0,1,2,3,........,15 $$ (only once),so that sum along each row, column and diagonal is $$ 30.$$
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Given :
In the magic square , the use of number $$ 0 , 1 , 2 , 3 ............, 15 $$
So only once the sum along each row, column and diagonal is $$ 30 $$.
Some numbers are already filled .
$$ 3+ 8 +4 = 15 + 15 = 30 $$
$$ 3 +12 = 5 + 10 = 30 $$
$$ 3 + 14 + 0 = 17 + 13 = 30 $$
$$ 0+ 6 = 9 + 15 = 30 $$
$$ 0+ 7 + 12 = 19 +11 = 30 $$
$$ 15 + 1 + 12 = 28 + 2 = 30 $$
$$ 8 + 6 + 11 = 25 + 5 = 30 $$
$$ 13 + 6 +1 = 20  +10 = 30 $$
$$ 14 + 5 + 2 = 21 + 9 = 30 $$


In two-digit numbers, how many times does the digit $$5$$ occur in the ten's place.

$$50,51,52,53,54,55,56,57,58,59$$ are the numbers which have 5 at ten's place. Total numbers are 10.


Solve the following riddles, you may yourself construct such riddles.
Who am I?
Go round a square counting every corner Thrice and no more! Add the count to me To get exactly thirty-four!

There are $$4$$ comers in a square
Thrice the number of comers in the square will be $$3 \times 4 = 12$$
When this result i,e $$12$$, is added to the number if Comes to be $$34$$. therefore the number will be the difference of $$34$$ and $$12$$ i.e $$34 - 12 = 22$$.


Complete the following magic squares:

1819450_df53f6eb075b46e4b044f0d2fe54f0d1.png

Sum for row-wise is as follows:
$$6 + 7 + 2 = 15$$,
$$1 + 5 + 9 = 15$$,
$$8 + 3 + 4 = 15$$.
Sum for column wise is as follows:
$$6 + 1 + 8 = 15$$,
$$7 + 5 + 3 = 15$$,
$$2 + 9 + 4 = 15$$.
Sum for diagonal wise is as follows:
$$6 + 5 + 4 = 15$$,
$$2 + 5 + 8 = 15$$
Hence, the magic square is as shown above

1790167_1819450_ans_5578d34105ef43ab9060ccd823055bb2.png


Using the numbers from 9 to 17 construct a $$ 3 \times 3 $$ magic square. What is the Magic sum here? What relation is there between the magic sum and the number in the central cell?

Magic sum is $$39$$
Central Number is $$13$$

we have $$ \Rightarrow 39=3 \times 13 $$
Thus, the magic sum is 3 times the number in central cell

1961057_1871173_ans_09d0a7413c044bb9bae77c6c3c3eaf00.png


Construct a $$ 3 \times 3 $$ magic square using all odd numbers from 1 to 17.

As shown in figure, magic sum is $$27$$

Number in the central cell is $$9$$

$$9\times 3=27$$

Thus, magic sum is three times number in the central cell

1961136_1871178_ans_b7633c633bf44a918876f33d2aebe0ac.png


Starting with the middle cell in the bottom row of the square and "using numbers from 1 to 9 construct a $$ 3 \times 3 $$ magic square. 

Figure shows $$3\times 3$$ magic square in which magic sum is $$15$$ and central cell contains $$5$$

We know that $$5\times 3=15$$
Thus, magic sum is 3 times the number in central cell.

1961082_1871175_ans_a0f72c65fdb24366a3ce1cb8915a3beb.png


Complete the following magic square:

1819454_2d2e9d066653442db1b92db52eb2f663.png

Row wise sum is as follows:
$$16 + 2 + 12 = 30$$,
$$6 + 10 + 14 = 30$$,
$$8 + 18 + 4 = 30$$.
Column wise sum is as follows:
$$16 + 6 + 8 = 30$$,
$$2 + 10 + 18 = 30$$,
$$12 + 14 + 4 = 30$$.
Diagonal wise sum is as follows:
$$16 + 10 + 4 = 30$$,
$$12 + 10 + 8 = 30$$.
Hence, the magic square is as shown above

1790158_1819454_ans_8c7022578cc04552801aca899948c8bd.png


Complete the following magic square:

1819452_3916e5a4f23b456f9becec2e6db1ce65.png

Row wise sum is as follows:
$$4 + 9 + 8 = 21$$,
$$11 + 7 + 3 = 21$$,
$$6 + 5 + 10 = 21$$.
Column wise sum is as follows:
$$4 + 11 + 6 = 21$$,
$$9 + 7 + 5 = 21$$,
$$8 + 3 + 10 = 21$$.
Diagonal wise sum is as follows:
$$4 + 7 + 10 = 21$$,
$$8 + 7 + 6 = 21$$.
Hence, the magic square is

1790162_1819452_ans_be2c168b396b44189f7ace6d9f3f3816.png


Complete the adjoining magic square (Hint. In a $$ 3 \times 3 $$ magic square the magic sum is $$$$ times the central number)
1871222_9e1a47b50e12448497f3141743b75ce9.png

Central number is $$7$$. 

The magic sum is $$3\times 7=21$$

1961191_1871222_ans_40f8d02a987a48d5830ab58bbfd843e4.png


Construct a $$ 5 \times 5 $$ magic square using all even numbers from 1 to 50.

As shown in the figure,
magic sum is $$130$$

Number in the central cell is $$26$$

$$26\times 5=120$$

Thus, magic sum is five times the number in central cell

1961137_1871182_ans_550da74eec134e8995585d6638e7ca82.png


What relation is there between the magic sum and the number in the central cell?

1) If we take sum of numbers in any row, in any column and in any diagonal, it comes up to be $$27$$

2) Number in the central cell is $$9$$

3) $$9\times 3=27$$
Thus, we can say that magic sum is three times the number in the central cell

1961051_1871166_ans_14aeb55f1ff64284a60c6328a4807236.png


Write an algorithm to check if a number is even or odd, using both pseudocode and flowchart.

pseudocode:
take Number
remainder=Number%2
if remainder==0;
   print even
else
   print odd

1945804_1940775_ans_e139d4ff322c4dc6972f70bda370505c.png


Now use number $$1 - 6$$ to make your own magic triangle.
Rule : Numbers on each side must add up to $$10$$.
1985496_ea6db0183ef94f36ae6cca99585ac52c.jpg




Starting with the middle cell in the bottom row of the square and using numbers from $$1$$ to $$9$$, construct a $$3\times 3$$ magic square

Using the numbers from $$1$$ to $$9$$, a $$3\times 3$$ magic square is constructed as shown above:

Considering the above drawn magic square, the magic sum is $$5\times 3=15$$.



620394_557477_ans.png


Construct a $$3\times 3$$ magic square using all odd number from $$1$$ to $$17$$

Using all the odd numbers from $$1$$ to $$17$$, a $$3\times 3$$ magic square is constructed as shown above:

Considering the above drawn magic square, the magic sum is $$9\times 3=27$$.


620401_557478_ans.png


Using the numbers from $$5$$ to $$13$$, construct a $$3\times 3$$ magic square. What is the magic sum here? What relation is there between the magic sum and the number in the central cell?

Using the numbers from $$5$$ to $$13$$, a $$3\times 3$$ magic square is constructed as shown above:

Considering the above drawn magic square, the magic sum is $$9\times 3=27$$.

Hence, the relation between the magic sum and the number in the central cell is that if we multiply the middle number to $$3$$ (middle number$$\times 3$$), then we get the magic sum.

620386_557475_ans.png


Construct a $$5\times 5$$ magic square using all even number from $$1$$ to $$50$$

Using all the even numbers from $$1$$ to $$50$$, a $$3\times 3$$ magic square is constructed as shown above:

Considering the above drawn magic square, the magic sum is $$26\times 5=130$$.


620403_557479_ans.png


Replace each $$\ast$$ by the correct digit of the above:
1778081_59d05d4c9666428ab05389a49cf3699c.png


1678839_1778081_ans_2b0205d8564240fd9b1081621cb9dd33.jpg


Replace each $$\ast$$ by the correct digit of the above:
1778079_e93069f91449470e9efbd7ca6378ee32.png


1678828_1778079_ans_d2f2b61472684d8bb233432047847a74.jpg


Class 5 Maths Extra Questions