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 )
