Q1. In a rectangle the ratio of the length and breadth is 3:2. If each of the length and breadth is increased by 4m their ratio becomes 10:7. The area of the original rectangle in m² is?

A. 846 B. 864 C. 840 D. 876

Answer: B
Explanation: [latex] [3x + \frac {4}{2x + 4}] = \frac {10}{7} [/latex]
[latex]x = 12 [/latex]
Area of the original rectangle = [latex] 3x \times 2x = 6x² [/latex]
Area of the original rectangle = [latex] 6 \times 144 = 864 m² [/latex]
Q2. The perimeter of a rectangle and a square is 160 cm each. If the difference between their areas is 500 cm. If the area of the rectangle is less than that of a Square then find the area of the rectangle?

A. 1800 cm² B. 1500 cm² C. 1100 cm² D. 1600 cm²

Answer: C
Explanation: Perimeter of rectangle = Perimeter of Square = 160
4a = [latex] 160 \Rightarrow a [/latex] = 40
Area of square = 1600
1600 – lb = 500
lb = 1100 cm²
Q3. Circumference of a circle A is [latex] \frac {22}{7} [/latex] times perimeter of a square. Area of the square is 441 cm². What is the area of another circle B whose diameter is half the radius of the circle A(in cm²)?

A. 354.5 B. 346.5 C. 316.5 D. 312.5

Answer: B
Explanation: Area = 441 cm²
a = 21 cm
Perimeter of Square = [latex] 4 \times 21 [/latex]
Circumference of a Circle = [latex] 4 \times 21 \times \frac {22}{7} [/latex]
2 [latex] \pi r = 4 \times 3 \times 22 [/latex]
r = [latex] \frac {12 \times 22 \times 7}{2} \times 22 [/latex] = 42 cm
Radius of Circle B = [latex] \frac {42}{4} [/latex] = 10.5 cm
Area of Circle = [latex] \pi r² = \frac {22}{7} \times 10.5 \times 10.5 [/latex] = 346.5 cm²
Q4. The area of a rectangle is equal to the area of a square whose diagonal is [latex] 12 \sqrt {2} [/latex] meter. The difference between the length and the breadth of the rectangle is 7 meter. What is the perimeter of a rectangle ? (in meter).

A. 68 meter B. 50 meter C. 62 metre D. 64 metre

Answer: B
Explanation: d = [latex] a \sqrt {2} [/latex]
[latex] 12 \sqrt {2} = a \sqrt {2} [/latex]
a = 12
[latex] l \times b [/latex] = a² = (12²) = 144
l – b = 7
l = b + 7
[latex] (b + 7) \times (b) [/latex]= 144
b² + 7b – 144 = 0
b = 9
l = 16
2(l + b) = 2(16 + 9) = 50 m
Q5. The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and breadth by 2 units, then the area is increased by 67 square units. Find the area of the rectangle?

A. 159 m B. 179 m C. 147 m D. 153 m

Answer: D
Explanation: Length = [latex] x [/latex]
Breadth = [latex] y [/latex]
[latex] xy – (x-5)(y+3) [/latex] = 9
[latex] 3x – 5y – 6 [/latex]= 0 ................(i)
[latex] (x+3)(y+2) – xy [/latex] = 67
[latex] 2x + 3y -61 [/latex] = 0 ................(ii)
solving (i) and (ii)
[latex] x = 17 m [/latex]
[latex] y = 9 m [/latex]
Area of the Rectangle = 153 m

Q1. The length of a plot is four times its breath. A playground measuring 900 square meters occupies one fourth of the total area of a plot. What is the length of the plot in meter.?

A. 150 B. 130 C. 160 D. 120

Answer: D
Explanation: Area of the plot = [latex] (4 \times 900) m² [/latex]
= [latex] 3600 m² [/latex]
Breadth = [latex] y [/latex] meter
Length = [latex] 4y[/latex] meter
Now area = [latex] 4y \times y = 3600 m² [/latex]
[latex] \Rightarrow y² = 900 m² [/latex]
[latex] \Rightarrow y = 30 m [/latex]
∴ Length of plot = [latex] 4y = 120 m [/latex]
Q2. The sum of the radius and height of a cylinder is 19 m. The total surface area of the cylinder is 1672 m², what is the radius and height of the cylinder? (in m)

A. 12, 7 B. 11, 8 C. 13, 6 D. 14, 5

Answer: D
Explanation: r + h = 19 m
2 [latex] \pi r(r + h) [/latex] = 1672
r = [latex] 1672 \times \frac {7} {2} \times 22 \times 19 [/latex] = 14
r = 14
h = 5
Q3. If each side pair of opposite sides of a square is increased by 20 m, the ratio of the length and breadth of the rectangular so formed becomes 5:3. The area of the old square is?

A. 990 m² B. 900 m² C. 930 m² D. 945 m²

Answer: B
Explanation: [latex] \frac {(x+20)} {x} = \frac {5} {3} [/latex]
[latex] 3x + 60 = 5x [/latex]
[latex] x = 30 m [/latex]
Area = [latex] 900 m² [/latex]
Q4. The length of a park is four times its breadth. A playground whose area is 1024 m² covers 1/4th part of the park. The length of the park is?

A. 118 m B. 142 m C. 124 m D. 128 m

Answer: D
Explanation: l = 4b
Area of the park = [latex] 4 \times 1024 m² [/latex]
[latex] l \times b = 4 \times 1024 [/latex]
[latex] l times \frac {l}{4} = 4 \times 4 \times 1024 [/latex]
[latex] l² = 1024 \times 16 [/latex]
[latex] l = 32 \times 4 = 128 m [/latex]
Q5. The width of a rectangular piece of land is [latex] \frac {1}{{4}{th}} [/latex] of its length. If the perimeter of the piece of land is 320m its length is?

A. 140 m B. 128 m C. 120 m D. 156 m

Answer: B
Explanation: length = [latex] l [/latex]
breadth = [latex] \frac {l}{4} [/latex]
[latex] 2(l + b) [/latex] = 320
[latex] 2(l + \frac {l}{4}) [/latex] = 320
[latex] l = 320 \times \frac {4}{10} [/latex] = 128m

Q1. A deer and a rabbit can complete a full round on a circular track in 9 minutes and 5 minutes respectively. P, Q, R, and S are the four consecutive points on the circular track which are equidistant from each other. P is opposite to R and Q is opposite to S. After how many minutes will they meet together for the first time at the starting point when both have started simultaneously from the same point in the same direction?

A. 15 minutes B. 25 minutes C. 35 minutes D. 45 minutes

Answer: D
Explanation: Time is taken by a deer to complete one round = 9 minutes
Time is taken by a rabbit to complete one round = 5 minutes
They meet together for the first time at the starting point = LCM of 9 and 5 = 45 minutes.
Q2. A deer and a rabbit can complete a full round on a circular track in 9 minutes and 5 minutes respectively. P, Q, R, and S are the four consecutive points on the circular track which are equidistant from each other. P is opposite to R and Q is opposite to S. After how many minutes will they meet together for the first time when both have started simultaneously from the same point in the same direction(in min)?

A. [latex] \frac {15}{4} [/latex] B. [latex] \frac {45}{4} [/latex] C. [latex] \frac {35}{4} [/latex] D. [latex] \frac {25}{4} [/latex]

Answer: B
Explanation: Circumference of the track = LCM of 9 and 5 = 45 m.
The ratio of time of deer and rabbit = 9 : 5
The ratio of the speed of deer and rabbit = 5 : 9
Relative Speed = 4 m/min
They meet together for the first time at the starting point = [latex] \frac {45}{4} [/latex]min
Q3. A cylindrical cistern whose diameter is 14 cm is partly filled with water. If a rectangular block of iron 22 cm in length, 14 cm in breadth and 7 cm in thickness is wholly immersed in water, by how many centimeters will the water level rise?

A. 10 cm B. 14 cm C. 12 cm D. 15 cm

Answer: B
Explanation: Volume of the block = [latex] 22 \times 14 \times 7 [/latex]
Radius of the cistern = [latex] \frac {14}{2} = 7 [/latex]
Volume of the Cylinder = [latex] \frac {22}{7} \times R2 \times h [/latex]
[latex] \frac {22}{7} \times R2 \times h = \frac {22}{7} \times 7 \times 7 \times h [/latex]
[latex] \frac {22}{7} \times 7 \times 7 \times h = 22 \times 14 \times 7 \Rightarrow h [/latex]= 14
Q4. A well with 28 m inside diameter is dug out 18 m deep. The earth taken out of it has been evenly spread all around it to a width of 21 m to form an embankment. Find the height of the embankment.

A. 3 m B. 8 m C. 9 m D. 6 m

Answer: A
Explanation: [latex] \frac {22}{7[(R2) – (r2)]} \times h = \frac {22}{7(7 \times 7 \times 18)} [/latex]
[(352) – (72)] h = [latex] 14 \times 14 \times 18 [/latex]
[latex] (42 \times 28) h = 14 \times 14 \times 18 [/latex]
h = 3 m
Q5. The radii of two cylinders are in the ratio 4:5 and their heights are in the ratio 5 : 7, What is the ratio of their curved surface areas?

A. 2 : 5 B. 4 : 7 C. 7 : 4 D. 2 : 3

Answer: B
Explanation: [latex] \frac {2 \pi r1 h1}{2 \pi r2 h2} = [\frac {4}{5} \times {5}{7}] [/latex] = 4 : 7