Data Interpretation Practice Questions – Set 95

Mentor for Bank Exams
Data Interpretation Practice Questions – Set 95
Directions (1 – 5): Below is given a table that shows the details of some two dimensional figures. Based on this information, answer the questions that follow.

1. Find the length of shorter diagonal of rhombus.(in cm)
(a)  30
(b)  40
(c)  50
(d)  80
(e)  60
2. Find the area of equilateral triangle.(in square cm)
(a)  561.18
(b)  582.82
(c)  602.74
(d)  637.19
(e)  551.53
3. What is the value of N if square and rhombus have same area?
(a)  20
(b)  22.5
(c)  24.5
(d)  25.75
(e)  28.5
4. What is the value of P if rectangle and rhombus have same area?
(a)  15.25
(b)  16.65
(c)  17.65
(d)  18.35
(e)  19.95
5. What is the value of Q if rectangle and equilateral triangle have same perimeter?
(a)  340
(b)  510
(c)  680
(d)  750
(e)  600
Directions (6 – 10): Study the following table carefully to answer the following questions.
6. What is the cost of flooring of A?
(a)  Rs. 4000
(b)  Rs. 4600
(c)  Rs. 4800
(d)  Rs. 5000
(e)  Rs. 4400
7. What is the difference between the cost of fencing of C and that of B?
(a)  Rs. 180
(b)  Rs. 120
(c)  Rs. 240
(d)  Rs. 360
(e)  Rs. 480
8. What is the ratio of the cost of flooring to that of fencing of field D?
(a)  4 : 1
(b)  6 : 1
(c)  8 : 1
(d)  9 : 1
(e)  5 : 1
9. The cost of fencing of field E is approximately what per cent of the cost of flooring of field C?
(a)  10.5%
(b)  19.46%
(c)  18.71%
(d)  15.36%
(e)  13.82%
10. The cost of fencing of field C is what per cent of the cost of fencing of field D?
(a)  87.54%
(b)  67.5%
(c)  72.13%
(d)  54.36%
(e)  46.5%


Solutions:
1. A) Diagonals of a rhombus bisect each other perpendicularly.
If shorter diagonal of rhombus is T cm,
(T/2)2 + (40/2)2 = 252
T2 + 1600 = 2500
T2 = 900
Shorter diagonal = T = 30 cm

2. A) Area of equilateral triangle = (√3/4) × 36 × 36 square cm = 561.18 square cm

3. C) Diagonals of a rhombus bisect each other perpendicularly.
If shorter diagonal of rhombus is T cm,
(T/2)2 + (40/2)2 = 252
T2 + 1600 = 2500
T2 = 900
T = 30 cm
Area of rhombus = Half the product of diagonals = (1/2) × 30 × 40 = 600 square cm
Area of square = 600 = N × N
N = 600 = 24.5

4. C) Diagonals of a rhombus bisect each other perpendicularly.
If shorter diagonal of rhombus is T cm,
(T/2)2 + (40/2)2 = 252
T2 + 1600 = 2500
T2 = 900
T = 30 cm
Area of rhombus = Half the product of diagonals = (1/2) × 30 × 40 = 600 square cm
Area of rectangle = 600 = 34 × P

5. C) Perimeter of equilateral triangle = 3 × 36 cm = 108 cm
2(34 + P) = 108
P = 54 34 = 20
Area of rectangle = Q = 34 × P = 34 × 20 = 680

6. C) Area of triangle is given by: ½(base × height)
½ (16 × 12)
96 meter square.
cost of flooring of A = 50 Rs./square meter
Cost of flooring A = 50 × 96 = 4800 Rs

7. A) Perimeter of B = 2(10 + 20) = 60 m
cost of fencing of B = 60 × 15 = 900 Rs.
Perimeter of C = 4 × 15 = 60 m
So, cost of fencing of C = 60 × 18 = 1080 Rs.
Required difference = 1080 - 900 = 180 Rs

8. D)  Cost of flooring of D = 20 × 12 × 60 = 14400
 Cost of fencing of field D = 2 (20 + 12) × 25 = 1600
Required Ratio = 14400 : 1600 = 9 : 1

9. D) Perimeter of E = 2 × 22/7 × 10 = 440/7 m
Cost of Fencing of E = 440/7 × 22 = 1382.85
Area of C = 152 = 225m2
Cost of flooring of C = 225 × 40 = 9000
Required % = 1382.85 × 100/9000 = 15.36%

10. B) Fencing cost of C = 1080
Fencing cost of D = 1600
Required % = (1080 × 100)/1600 = 67.5%