Quadrilateral Based Problems For IBPS

Dear Reader,
Below are four problems on the angles of quadrilateral with simple arithmetic calculations.

Question 1

If 12:3:10:5 is the ratio of angles of a quadrilateral, then the difference between the largest and second largest angles of the quadrilateral is:

a) 30 b) 62 c) 24 d) 35

Answer : c)24

Solution :

Given that the ratio of the angles of the quadrilateral is 12:3:10:5.
Then the angles of the quadrilateral are 12x, 3x, 10x and 5x.
We know that the sum of the angles of any quadrilateral is 360.
Therefore, 12x + 3x + 10x + 5x = 360
30x = 360
x = 12

Then the angles are 144, 36, 120 and 60.
Here the largest and the second largest are 144 and 120.
Then the required answer is 144 - 120 = 24.

Question 2

If 48 is one of the angles of a quadrilateral and the ratio of the other three angles is 11:24:43 then what is the sum of the least two angles?

a) 48 b) 92 c) 72 d) 24

Answer : b)92

Solution :

Given that the ratio of three angles of a quadrilateral is 11:24:43 and the fourth angle is 48.
Then the angles of the quadrilateral are 11x, 24x, 43x and 48.
We know the sum of the angles is 360.
i.e, 11x + 24x + 43x + 48 = 360
78x + 48 = 360
78x = 312
x = 4

Then the angles are 44, 96, 172 and 48.
The least two angles are 44 and 48.
Then the required sum is 44 + 48 = 92.
Hence the answer is 92.

Question 3

The angles of a quadrilateral are 37X + 25, 23X - 19, 17X - 6 and 15X - 8 then which will be the largest one?

a) 180 b) 168 c) 98 d) 173

Answer : d) 173

Solution :

Given that the angles of a quadrilateral are 37X + 25, 23X - 19, 17X - 6 and 15X - 8
We know that, 37X+25 + 23X-19 + 17X-6 + 15X-8 = 360
92X - 8 = 360
92X = 368
X = 368 / 92 = 4
Then the angles are 173, 73, 62 and 52
The largest angle is 173.

Question 4

If 8X+8, 14X+14, 43X+43 and 28X+16 are the angles of a quadrilateral then the difference between the largest and the least angle is:

a) 140 b) 68 c) 56 d) 100

Answer : a)140

Solution :

Given that the angles of a quadrilateral are 8X+8, 14X+14, 43X+43 and 28X+16
Then, 8X+8 + 14X+14 + 43X+43 + 28X+16 = 360
93X + 81 = 360
93X = 279
X = 279/93 = 3
Then the angles are 32, 56, 100 and 172
The required difference = 172 - 32 = 140


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