CBSE BOARD XII, asked by TheKnowledge, 1 year ago

Hi , here is a Riddle

use your brain


don't spam


thanks !!


@ Ranjan Kumar




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gaurav2013c: @ranjankumar12
VAISHVItheBEATboxer: pakka 120
gaurav2013c: Bhai I can explain
gaurav2013c: 38 is correct
rajc7: okk
rajc7: good
gaurav2013c: i can, if the you want

Answers

Answered by gaurav2013c
34
In the first equation, there are three polygons.

1.) Hexagon

2.) Pentagon

3.) Quadrilateral

In equation (1)

Value of 1 figure = 15 ( because 3 times of one figure = 45)

Value of Hexagon = 6 ( it has 6 sides)

Value of Pentagon = 5 ( it has 5 sides)

Value of quadrilateral = 4 ( it has 4 sides)


[ you can add these for confirmation]


Now,

In second equation,

4 × Banana + 4 × Banana + 1 × Polygon = 23

=> 8 × Banana + 15 = 23

=> 8 × Banana = 8

=> 1 Banana = 1

Value of 1 banana = 1


Now,

In third equation,

4 × Banana + Clock with time 3:00 + Clock with time = 10

=> 4 × 1 + 2 × Clock with time 3:00 = 10

=> 4 + 2 × Clock with time 3:00 = 10

=> 2 × Clock with time 3:00 = 6

=> Clock with time 3 :00 = 3

Value of Clock with time 2:00 = 2



Now,


Put all the values in 4th

Clock with time 2:00 + 3 × banana + [ 3× Banana × ( Hexagon + Pentagon)]

= 2 + 3×1 + [ 3× 1 × ( 6 + 5)]

= 2 + 3 + [ 3 × 11]

= 5 + 33

= 38


gaurav2013c: Koi nhi bhai
gaurav2013c: kabhi bhi
supriya7402: answer is 88
supriya7402: sorry answer is 38 itself...
supriya7402: u know how,, 2+3+3×11=38
prachi8578: hello
prachi8578: haaaaa gaurav
Answered by Anonymous
7
Hey mate.......

here's ur answer......

Hope it helps ☺️❤️
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