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Answer 1

$\sf{\large {\underline {1. \:Diagram}}}$ :

One cone has 3 sides.

3 × 3 × 3 = 27

$\sf{\large {\underline {2. \:Diagram}}}$:

2 cone with 2 ice - creams ( white ) over it.

1 cone = 3, 1 ice = 2

( 3 + 2 ) + ( 3 + 2) = 10

$\sf{\large {\underline {3. \:Diagram}}}$ :

1 cone ( 3) + 2 White ices ( 2 + 2) and 1 pink ice ( 1 )

( 3 + 2 + 2 ) + ( 3 + 1 ) = 7 + 4 = 11

$\sf{\large {\underline {4. \:Diagram}}}$ :

( 3 + 2 ) + 3 + ( 3 + 2 + 1 +1 ) = 5 + 3 + 7 = 15

$\sf{\large {So, \:Cone \:=\: 3, \:White\: ice \:=\: 2,\: Pink \:Ice\: = \:1. }}$

Answer 2

Answer:

3

2

1

Step-by-step explanation:

Let us solve the given problem algebraically .

Let the value of the ice cream cone be x .

In the first figure , there are 3 cones multiplied with each other .

The value of the cone if multiplied thrice with the same value gives 27 .

We can write the above statement algebraically :

x³ = 27

Take cube root both sides :

⇒ x = 3

The value of x is 3 .

Next figure says two cones and they are filled with same quantities of ice cream .

Let the quantity of ice cream in each be y .

x + x + y + y = 10

⇒ 2 x + 2 y = 10

⇒ 2 ( x + y ) = 10

⇒ x + y = 5

x was calculated as 3 and hence :

3 + y = 5

⇒ y = 5 - 3

⇒ y = 2

The value of y is 2 .

Next step there are 2 y , 2 x and 1 different coloured quanitity of ice cream .

Let that be z .

2 y + 2 x + z = 11

⇒ 2( 2 ) + 2( 3 ) + z = 11

⇒ 4 + 6 + z = 11

⇒ z = 11 - 10

⇒ z = 1

The value of z is 1 .

Verification :-

Last figure has 3 x , 2 y and 2 z .

3 x = 3 × 3 = 9

2 y = 2 × 2 = 4

2 z = 2 × 1 = 2

Adding all :

3 x + 2 y + 2 z = 15

Hence it is verified .