__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}= 25^{2}
⇒ T

^{2}+ 1600 = 2500
⇒ T

^{2}= 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}= 25^{2}
⇒ T

^{2}+ 1600 = 2500
⇒ T

^{2}= 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}= 25^{2}
⇒ T

^{2}+ 1600 = 2500
⇒ T

^{2}= 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 = 15

^{2}= 225m^{2}
⇒ 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%