**Q1. The parallel sides of a trapezium field are 12 m and 20 m. If distance between the parallel sides is 8 m, find the area of the field.**

**(a) 128 sq m**

(b) 132 sq m

(c) 130 sq m

(d) 135 sq m

**(Area of trapezium =1/2 (sum of parallel sides) × Distance between them**

**=1/2×(12+20)×8**

**=1/2×32×8=128 sq cm )**

**Q2. If side of a square is doubled, how many times its area will be increased?**

(a) 5 times

(b) 2 times

**(c) 4 times**

(d) 3 times

**( Let original side of the square be a.**

**Then, original area = a^2**

**New side = 2a**

**New area =(2a)^2=4a^2**

**Thus, area will be 4 times the original area. )**

**Q3. Area of a rhombus is 256 cm^2. One of the diagonal is half of the other diagonal. The sum of the diagonals is-**

(a) 38 cm

**(b) 48 cm**

(c) 28 cm

(d) 56 cm

**( Let both the diagonals be x and 2x**

**Then, 1/2*x*2x=256**

**=x^2=256**

**x=16**

**∴ 2x = 32**

**Hence, sum of diagonals**

**x + 2x = 16 × 32 = 48 cm )**

**Q4. The length of a rope by which a buffalo must be gathered so that may be able to graze a grassy area of 2464 sq m is-**

(a) 35 m

(b) 27 m

(c) 24 m

**(d) 28 m**

**( Let length of rope be r.**

**Then area =πr^2=2464**

**r^2=2464/22×7=784**

**r = 28 m)**

**Q5. The inner and outer radii of a circular track are respectively 21 m and 28 m. the cost of leveling the track at Rs. 5 per sq m is-**

(a) Rs. 1078

(b) Rs. 2156

(c) Rs. 4312

**(d) Rs. 5390**

**(Area of circular track**

**=π(r1^2〗_1-r2^2 )**

**=π(28^2-21^2 )**

**=π(784-441)**

**=22/7×343=1078 sq m**

**Total cost =1078×5 = Rs. 5390 )**

**Q6. A brick measures 20 cm × 10 cm × 7.5 cm. How many bricks will be required for a wall 20 m × 2 m × 0.75 m?**

(a) 10000

(b) 15000

(c) 18000

**(d) 20000**

**(Number of bricks =(Total volume of a wall)/(Volume of one brick )**

**=(20 ×2 × 0.75 × 100 × 100 × 100)/(20 × 10 × 7.5)=20000 )**

**Q7. The capacity of cylindrical tank is 1848 m3 and diameter is 14 m. Determine the depth of the tank.**

(a) 10 m

**(b) 12 m**

(c) 14 m

(d) 16 m

**( Let depth of the cylindrical tank = h m**

**We know that,**

**Volume =πr^2 h**

**1848=22/7×7×7×h**

**=h=1848/154=12 m)**

**Q8. If diameter of a circle is decreased by 5%, by what per cent its area will be decreased?**

(a) 10.25%

**(b) 9.75%**

(c) 9%

(d) 10.36%

**(Here, a = 5% (we took – ve sign because of decrease)**

**=(2a+a^2/100)%=(-10+25/100)%**

**=(-10+0.25)%=9.75%**

**( – ve sign shows that there is a decrease)**

**Q9. The sides of a triangle are in the ratio 12 : 15 : 20. If its perimeter be 94 cm, find the length of smallest side of the triangle.**

(a) 18 cm

(b) 22.5 cm

**(c) 24 cm**

(d) 27 cm

**(Let sides of the triangle are 12x , 15x and 20x.**

**Then,**

**12x + 15x + 20x = 94**

**= 47x = 94**

**x = 2**

**Thus, the smallest side = 12x**

**= 12 × 2 = 24 cm)**

**Q10. The side of a square is 5 cm which is 13 cm less than the diameter of a circle. What is the approximate area of the circle?**

(a) 245 sq cm

**(b) 255 sq m**

(c) 235 sq m

(d) 265 sq m

**( Diameter of the circle = 5 + 13 = 18 cm**

**∴ Radius of circle = 18/2= 9 cm**

**∴ Area of circle =π(9)^2=22/7×81**

**≈255 sq m )**