IUC AcademyHome961. The area of a rectangle is thrice that of a square. If the length of the rectangle is 40 cm and its breadth is 3 2 times that of the side of the square, then the side of the square is
- A. 15 cm
- B. 20 cm
- C. 30 cm
- D. 60 cm
Answer: 20 cm
Mensuration MCQ Practice
Practice question set with answers. Use deep links to save questions.
Quiz App / Mensuration
1059 questions loaded. Showing page 13 of 14 for fast crawlable revision and deep linking.
962. The perimeter of a square is 48 cm. The area of a rectangle is 4 cm2 less than the area of the square. If the length of the rectangle is 14 cm, then its pe-rimeter is
- A. 24 cm
- B. 48 cm
- C. 50 cm
- D. 54 cm
Answer: 48 cm
Explanation: Side of the square = 12 cm Area of rectangle = [(12 × 12) – 4] cm2 = 140 cm2. ∴ Breadth = Area 140 = = 10 cm. Length 14 Hence, perimeter = 2(l + b) = 2 (14 + 10) cm = 48 cm.
Explanation: Side of the square = 12 cm Area of rectangle = [(12 × 12) – 4] cm2 = 140 cm2. ∴ Breadth = Area 140 = = 10 cm. Length 14 Hence, perimeter = 2(l + b) = 2 (14 + 10) cm = 48 cm.
963. The area of the shaded portion is
- A. 10 sq cm
- B. 14 sq cm
- C. 21 sq cm
- D. 25 sq cm
Answer: 21 sq cm
Explanation: Required area = [(2 × 3) + (3 × 3) + (2 × 3)] cm2 = (6 + 9 + 6) cm2 = 21 cm2. 2cm 3cm 2cm 2cm 3cm 3cm 2cm 3cm
Explanation: Required area = [(2 × 3) + (3 × 3) + (2 × 3)] cm2 = (6 + 9 + 6) cm2 = 21 cm2. 2cm 3cm 2cm 2cm 3cm 3cm 2cm 3cm
964. Perimeter of a rectangular field is 160 metres and the difference between its two adjacent sides is 48 metres. The side of a square field, having the same area as that of the rectangle, is
- A. 4 m
- B. 8 m
- C. 16 m
- D. 32 m
Answer: 32 m
Explanation: Let the sides of the rectangle be x metres and (x + 48) metres. Then, 2(x + x + 48) = 160 ⇒ 4x + 96 = 160 ⇒ 4x = 64 ⇒ x = 16. So, sides of the rectangle are 16 m and 64 m. Area of the rectangle = (16 × 64) m2 = 1024 m2. Area of the square = 1024 m2. ∴ Side of the square = 1024 m = 32 m.
Explanation: Let the sides of the rectangle be x metres and (x + 48) metres. Then, 2(x + x + 48) = 160 ⇒ 4x + 96 = 160 ⇒ 4x = 64 ⇒ x = 16. So, sides of the rectangle are 16 m and 64 m. Area of the rectangle = (16 × 64) m2 = 1024 m2. Area of the square = 1024 m2. ∴ Side of the square = 1024 m = 32 m.
965. A 7 m long partition divides a rectangular room into two equal square rooms. What is the area of the whole room in square metres?
- A. 49
- B. 147
- C. 196
- D. None of these
Answer: None of these
Explanation: Length of the room = (7 + 7) m = 14 m. Breadth of the room = 7 m. ∴ Area of the room = (14 × 7) m2 = 98 m2.
Explanation: Length of the room = (7 + 7) m = 14 m. Breadth of the room = 7 m. ∴ Area of the room = (14 × 7) m2 = 98 m2.
966. What is the minimum number of identical square tiles required to tile a floor of length 6 m 24 cm and width 4 m 80 cm?
- A. 122
- B. 130
- C. 148
- D. 165
Answer: 130
967. A room is 1 12 4 m long and 7 m wide. The maximum length of a square tile to fill the floor of the room with whole number of tiles should be
- A. 125 cm
- B. 150 cm
- C. 175 cm
- D. 200 cm
Answer: 175 cm
Explanation: Length of largest tile = H.C.F. of 1 12 m and 7 m 4 = H.C.F. of 12.25 m and 7 m = H.C.F. of 1225 cm and 700 cm = 175 cm.
Explanation: Length of largest tile = H.C.F. of 1 12 m and 7 m 4 = H.C.F. of 12.25 m and 7 m = H.C.F. of 1225 cm and 700 cm = 175 cm.
968. Three plots having areas 110, 130 and 190 square metres are to be subdivided into flower beds of equal size. If the breadth of a bed is 2 metre, the maximum length of a bed can be
- A. 5 m
- B. 11 m
- C. 13 m
- D. 19 m
Answer: 5 m
969. A big rectangular plot of area 4320 m2 is divided into 3 square-shaped smaller plots by fencing parallel to the smaller side of the plot. However some area of land was still left as a square could not be formed. So, 3 more square-shaped plots were formed by fencing parallel to the longer side of the original plot such that no area of the plot was left surplus. What are the dimensions of the original plot?
- A. 160 m × 27 m
- B. 240 m × 18 m
- C. 120 m × 36 m
- D. 135 m × 32 m
Answer: 120 m × 36 m
970. The length and breadth of the floor of the room are 20 feet and 10 feet respectively. Square tiles of 2 feet length of different colours are to be laid on the floor. Black tiles are laid in the first row on all sides. If white tiles are laid in the one-third of the remaining and blue tiles in the rest, how many blue tiles will be there?
- A. 16
- B. 24
- C. 32
- D. 48
Answer: 16
971. How many squares with side 1 2 inch long are needed to cover a rectangle that is 4 feet long and 6 feet wide?
- A. 24
- B. 96
- C. 3456
- D. 13824
Answer: 13824
972. 50 square stone slabs of equal size were needed to cover a floor area of 72 sq. m. The length of each stone slab is
- A. 102 cm
- B. 120 cm
- C. 201 cm
- D. 210 cm
Answer: 120 cm
973. How many marble slabs of 20 cm × 30 cm are needed to cover the floor of a square room with a 3 m side?
- A. 100
- B. 150
- C. 225
- D. 250
Answer: 250
974. A coaching institute wants to execute tiling work for one of its teaching halls 60 m long and 40 m wide with a square tile of 0.4 m side. If each tile costs ₹ 5, the total cost of tiles would be
- A. ₹ 60000
- B. ₹ 65000
- C. ₹ 70000
- D. ₹ 75000
Answer: ₹ 75000
975. Total area of 64 small squares of a chessboard is 400 sq. cm. There is 3 cm wide border around the chess board. What is the length of the side of the chessboard?
- A. 17 cm
- B. 20 cm
- C. 23 cm
- D. 26 cm
Answer: 26 cm
976. The perimeters of five squares are 24 cm, 32 cm, 40 cm, 76 cm and 80 cm respectively. The perimeter of another square equal in area to the sum of the areas of these squares is
- A. 31 cm
- B. 62 cm
- C. 124 cm
- D. 961 cm
Answer: 124 cm
977. ABJH, JDEF, ACEG are squares. BC = 3. AB Area BCDJ =? Area HJFG
- A. 1 9
- B. 13
- C. 1
- D. 3
Answer: 1
978. The following squares represent the monthly incomes of two families A B 2 3 (P.C.S., 2008) If the monthly income of family A is ₹ 40000, the monthly income of family B is
- A. ₹ 50000
- B. ₹ 60000
- C. ₹ 90000
- D. ₹ 120000
Answer: ₹ 90000
979. The perimeter of a square and a rectangle is the same. If the rectangle is 12 cm by 10 cm, then by what percentage is the area of the square more than that of the rectangle?
- A. 2 3
- B. 1
- C. 1 1 3
- D. None of these
Answer: None of these
980. Cultivating a square field at ₹ 685 per hectare costs ₹ 6165. What would it cost to fence it at ₹ 48.75 per metre?
- A. ₹ 23400
- B. ₹ 52650
- C. ₹ 58500
- D. ₹ 117000
Answer: ₹ 58500
981. ABCD is a square and AEFG is a rectangle. Area of each of them is 36 sq. m. E is the mid-point of AB. The perimeter of the rectangle AEFG is
- A. 12 m
- B. 18 m
- C. 30 m
- D. 36 m
Answer: 30 m
982. The area of a square is 1024 sq. cm. What is the ratio of the length to the breadth of a rectangle whose length is twice the side of the square and breadth is 12 cm less than the side of this square?
- A. 5 : 18 / 5: 18
- B. 16 : 7 / 16: 7
- C. 14 : 5 / 14: 5
- D. None of these
Answer: None of these
Explanation: Area of square = 1024 cm 32 cm. = Length of rectangle = (2 × 32) cm = 64 cm. Breadth of rectangle = (32 – 12) cm = 20 cm. ∴ Required ratio = 64 : 20 = 16 : 5.
Explanation: Area of square = 1024 cm 32 cm. = Length of rectangle = (2 × 32) cm = 64 cm. Breadth of rectangle = (32 – 12) cm = 20 cm. ∴ Required ratio = 64 : 20 = 16 : 5.
983. The area of a square is three-fifths the area of a rectangle. The length of the rectangle is 25 cm and its breadth is 10 cm less than its length. What is the perimeter of the square?
- A. 44 cm
- B. 60 cm
- C. 80 cm
- D. Cannot be determined
Answer: 60 cm
984. The length of the side of a square whose area is four times the area of a square with side 25 m is
- A. 12.5 m
- B. 50 m
- C. 100 m
- D. 125 m
Answer: 50 m
Explanation: Area of given square = (25 × 25) m2 = 625 m2. Area of new square = (625 × 4) m2 = 2500 m2.. ∴ Side of new square = 2500 m 50 m. =
Explanation: Area of given square = (25 × 25) m2 = 625 m2. Area of new square = (625 × 4) m2 = 2500 m2.. ∴ Side of new square = 2500 m 50 m. =
985. The area of a square with perimeter 48 cm is
- A. 144 sq. cm
- B. 156 sq. cm
- C. 170 sq. cm
- D. 175 sq. cm
Answer: 144 sq. cm
986. A housing society has been allotted a square piece of land measuring 2550.25 sq. m. What is the side of the plot?
- A. 50.25 m
- B. 50.5 m
- C. 50.65 m
- D. None of these
Answer: 50.5 m
987. The dimensions of a rectangle are 51 m and 49 m respectively while side of a square is 50 m. Which of the following statements is correct?
- A. Diagonals of the square and the rectangle are equal.
- B. Diagonals of both the geometrical figures intersect at right angles.
- C. The perimeters of both the geometrical figures are equal.
- D. Both the geometrical figures are of the same area.
Answer: The perimeters of both the geometrical figures are equal.
988. A garden is 24 m long and 14 m wide. There is a path 1 m wide outside the garden along its sides. If the path is to be constructed with square marble tiles 20 cm × 20 cm, the number of tiles required to cover the path is
- A. 200
- B. 1800
- C. 2000
- D. 2150
Answer: 2000
989. A rectangular park 60 m long and 40 m wide has two concrete crossroads running in the middle of the park and rest of the park has been used as a lawn. If the area of the lawn is 2109 sq. m, then what is the width of the road?
- A. 2.91 m
- B. 3 m
- C. 5.82 m
- D. None of these
Answer: 3 m
Explanation: Area of the park = (60 × 40) m2 = 2400 m2. Area of the lawn = 2109 m2. ∴ Area of the crossroads = (2400 – 2109) m2 = 291 m2. Let the width of the road be x metres. Then, 60x + 40x – x2 = 291 ⇔ x2 – 100x + 291 = 0 ⇔ (x – 97) (x – 3) = 0 ⇔ x = 3 [∴ x ≠ 97].
Explanation: Area of the park = (60 × 40) m2 = 2400 m2. Area of the lawn = 2109 m2. ∴ Area of the crossroads = (2400 – 2109) m2 = 291 m2. Let the width of the road be x metres. Then, 60x + 40x – x2 = 291 ⇔ x2 – 100x + 291 = 0 ⇔ (x – 97) (x – 3) = 0 ⇔ x = 3 [∴ x ≠ 97].
990. A rectangular field has dimensions 25 m by 15 m. Two mutually perpendicular passages, 2 m wide have been left in its central part and grass has been grown in rest of the field. The area (in sq. metres) under the grass is
- A. 295
- B. 299
- C. 300
- D. 375
Answer: 299
Explanation: Area of the field = (25 × 15) m2 = 375 m2. Area of the passages = (25 × 2 + 15 × 2 – 2 × 2) m2 = 76 m2. Area under grass = (375 – 76) m2 = 299 m2.
Explanation: Area of the field = (25 × 15) m2 = 375 m2. Area of the passages = (25 × 2 + 15 × 2 – 2 × 2) m2 = 76 m2. Area under grass = (375 – 76) m2 = 299 m2.
991. A rectangular lawn 80 metres by 60 metres has two roads each 10 m wide running in the middle of it, one parallel to the length and the other parallel to the breadth. Find the cost of gravelling them at ₹ 30 per square metre.
- A. ₹ 3600
- B. ₹ 3900
- C. ₹ 36000
- D. ₹ 39000
Answer: ₹ 39000
Explanation: Area of the roads = (80 × 10 + 60 × 10 – 10 × 10) m2 = 1300 m2. ∴ Cost of gravelling = ₹ (1300 × 30) = ₹ 39000.
Explanation: Area of the roads = (80 × 10 + 60 × 10 – 10 × 10) m2 = 1300 m2. ∴ Cost of gravelling = ₹ (1300 × 30) = ₹ 39000.
992. A rectangular garden (60 m × 40 m) is surrounded by a road of width 2 m, the road is covered by tiles and the garden is fenced. If the total expenditure is ₹ 51600 and rate of fencing is ₹ 50 per metre, then the cost of covering 1 sq. m of road by tiles is
- A. ₹ 10
- B. ₹ 50
- C. ₹ 100
- D. ₹ 150
Answer: ₹ 100
Explanation: Length of the fence = 2(60 + 40) m = 200 m. Cost of fencing = ₹ (200 × 50) = ₹ 10000. Area of the road = [(64 × 44) – (60 × 40)] m2 = (2816 – 2400) m2 = 416 m2. Let the cost of tiling the rood be ₹ x per sq. m. ∴ 416x + 10000 = 51600 ⇒ 416x = 41600 ⇒ x = ₹ 100.
Explanation: Length of the fence = 2(60 + 40) m = 200 m. Cost of fencing = ₹ (200 × 50) = ₹ 10000. Area of the road = [(64 × 44) – (60 × 40)] m2 = (2816 – 2400) m2 = 416 m2. Let the cost of tiling the rood be ₹ x per sq. m. ∴ 416x + 10000 = 51600 ⇒ 416x = 41600 ⇒ x = ₹ 100.
993. A rectangular garden is 10 m wide and 20 m long. A path of uniform width is to be laid inside along its edges so that 96 m² is left for flowers. How wide must the path be?
- A. 1 m
- B. 2 m
- C. 2.1 m
- D. 2.5 m
Answer: 2 m
Explanation: Let the width of walk be x metres. Then, (20 – 2x) (10 – 2x) = 96 ⇔ 4x2 + 60x – 104 = 0 ⇔ x2 + 15x – 26 = 0 ⇔ (x – 13) (x – 2) = 0 ⇔ x = 2 [∴ x ≠ 13]
Explanation: Let the width of walk be x metres. Then, (20 – 2x) (10 – 2x) = 96 ⇔ 4x2 + 60x – 104 = 0 ⇔ x2 + 15x – 26 = 0 ⇔ (x – 13) (x – 2) = 0 ⇔ x = 2 [∴ x ≠ 13]
994. A path of uniform width runs round the inside of a rectangular field 38 m long and 32 m wide. If the path occupies 600 m2, then the width of the path is
- A. 5 m
- B. 10 m
- C. 18.75 m
- D. 30 m
Answer: 5 m
Explanation: Let the width of the path be x. Then, [(38 × 32) – {(38 – 2x)(32 – 2x)}] = 600 ⇒ [1216 – (1216 – 140x + 4x2)] ⇒ 4x2 – 140 x + 600 = 0 ⇒ x2 – 35x + 150 = 0 ⇒ x2 – 30x – 5x + 150 = 0 ⇒ (x – 30) (x – 5) = 0 ⇒ x = 5 m. [∴ x ≠ 30]
Explanation: Let the width of the path be x. Then, [(38 × 32) – {(38 – 2x)(32 – 2x)}] = 600 ⇒ [1216 – (1216 – 140x + 4x2)] ⇒ 4x2 – 140 x + 600 = 0 ⇒ x2 – 35x + 150 = 0 ⇒ x2 – 30x – 5x + 150 = 0 ⇒ (x – 30) (x – 5) = 0 ⇒ x = 5 m. [∴ x ≠ 30]
995. A 2 m wide path is built along the inner edge of a rectangular plot whose area is 96 sq. m, at ₹ 50 per square metre. What is the total construction cost?
- A. ₹ 2400
- B. ₹ 4000
- C. ₹ 4800
- D. Data inadequate
Answer: Data inadequate
Explanation: lb = 96 (Given) Area of pathway = [(l – 4)(b – 4) – lb] = 16 – 4(l + b), which cannot be determined. So, data is inadequate.
Explanation: lb = 96 (Given) Area of pathway = [(l – 4)(b – 4) – lb] = 16 – 4(l + b), which cannot be determined. So, data is inadequate.
996. What will be the cost of gardening 1 metre broad boundary around a rectangular plot having perimeter of 340 metres at the rate of ₹ 10 per square metre?
- A. ₹ 1700
- B. ₹ 3400
- C. ₹ 3440
- D. Cannot be determined
Answer: ₹ 3440
Explanation: 2(l + b) = 340 (Given). Area of the boundary = [(l + 2)(b + 2) – lb] = 2 (l + b) + 4 = 344. ∴ Cost of gardening = ₹ (344 × 10) = ₹ 3440.
Explanation: 2(l + b) = 340 (Given). Area of the boundary = [(l + 2)(b + 2) – lb] = 2 (l + b) + 4 = 344. ∴ Cost of gardening = ₹ (344 × 10) = ₹ 3440.
997. The breadth of a rectangular field is 3 4 of its length and its area is 300 sq. metres. What will be the area (in sq. metres) of the garden of breadth 1.5 metres developed around the field?
- A. 96
- B. 105
- C. 114
- D. Cannot be determined
Answer: 114
998. A lawn is in the shape of a rectangle of 80 m length and 50 m width. Outside the lawn there is a footpath of uniform 1 m width bordering the lawn. The area of the footpath is
- A. 264 m2
- B. 284 m2
- C. 4000 m2
- D. 4264 m2
Answer: 264 m2
Explanation: Area of the footpath = { } 2 [ ] (80 2) (50 2 (80 50) m +× + − × = [(82 × 52) – (80 × 50)] m2 = (4264 – 4000) m2 = 264 m2.
Explanation: Area of the footpath = { } 2 [ ] (80 2) (50 2 (80 50) m +× + − × = [(82 × 52) – (80 × 50)] m2 = (4264 – 4000) m2 = 264 m2.
999. A room 5 m × 8 m is to be carpeted leaving a margin of 10 cm from each wall. If the cost of the carpet is ₹ 18 per Sq. metre, the cost of carpeting the room will be
- A. ₹ 673.92
- B. ₹ 682.46
- C. ₹ 691.80
- D. ₹ 702.60
Answer: ₹ 673.92
Explanation: Area of the carpet = [(5 – 0.20) × (8 – 0.20)] m2 = (4.8 × 7.8) m2 = 37.44 m2. ∴ Cost of carpeting = ₹ (37.44 × 18) = ₹ 673.92.
Explanation: Area of the carpet = [(5 – 0.20) × (8 – 0.20)] m2 = (4.8 × 7.8) m2 = 37.44 m2. ∴ Cost of carpeting = ₹ (37.44 × 18) = ₹ 673.92.
1000. A typist uses a paper 30 cm by 15 cm. He leaves a margin of 2.5 cm at the top and bottom and 1.25 cm on either side. What percentage of paper area is approximately available for typing?
- A. 60%
- B. 65%
- C. 70%
- D. 80%
Answer: 70%
1001. The length of a rectangle is reduced by 20% and breadth is kept constant, and the new figure that is formed is a square. Consider the following statements: 1. The area of square is 25% less than the area of rectangle. 2. The perimeter of square is approximately 11% less than the perimeter of rectangle. 3. The diagonal of square is approximately 12% less than the diagonal of rectangle. Which of the statements given above is/are correct?
- A. 1 only
- B. 1 and 2
- C. 2 and 3
- D. 1, 2 and 3
Answer: 2 and 3
1002. If the length and breadth of a rectangular field are increased, the area increases by 50%. If the increase in length was 20%, by what percentage was the breadth increased?
- A. 20%
- B. 25%
- C. 30%
- D. Date inadequate
Answer: 25%
1003. If the breadth of a rectangle is decreased by 50%, then to double the area, its length is required to be increased by
- A. 150%
- B. 200%
- C. 300%
- D. 400%
Answer: 300%
1004. If the area of a rectangular plot increases by 30% while its breadth remains the same, what will be the ratio of the areas of new and old figures?
- A. 1 : 3 / 1: 3
- B. 3 : 1 / 3: 1
- C. 4 : 7 / 4: 7
- D. None of these
Answer: None of these
1005. The length of a rectangle is increased by 60%. By what percent would the width have to be decreased so as to maintain the same area?
- A. 37½%
- B. 60%
- C. 75%
- D. 120%
Answer: 37½%
1006. The length of a rectangle is decreased by r%, and the breadth is increased by (r + 5)%. Find r, if the area of the rectangle is unaltered.
- A. 5
- B. 8
- C. 10
- D. 20
Answer: 20
1007. The length of a rectangle is halved, while its breadth is tripled. What is the percentage change in area?
- A. 25% increase
- B. 50% increase
- C. 50% decrease
- D. 75% decrease
Answer: 50% increase
1008. A towel, when bleached, was found to have lost 20% of its length and 10% of its breadth. The percentage of decrease in area is
- A. 10%
- B. 10.08%
- C. 20%
- D. 28%
Answer: 28%
1009. If the length of a rectangle is increased by 50% and breadth is decreased by 25%, what is the percentage change in its area?
- A. 12.5% increase
- B. 10% increase
- C. 25% increase
- D. 20% decrease
Answer: 12.5% increase
1010. Two sides of a rectangle were measured. One of the sides (length) was measured 10% more than its actual length and the other side (width) was measured 5% less than its actual length. The percentage error in measure obtained for the area of the rectangle is
- A. 4.5%
- B. 5%
- C. 7.56%
- D. 15%
Answer: 4.5%
1011. If the length of a rectangle is increased by 10% and its breadth is decreased by 10%, the change in its area will be
- A. 1% increase
- B. 1% decrease
- C. 10% increase
- D. No change
Answer: 1% decrease
1012. An order was placed for supply of carpet of breadth 3 metre, the length of carpet was 1.44 times of breadth. Subsequently the breadth and length were increased by 25 and 40 percent respectively. At the rate of 45 per square metre, what would be the increase in the cost of the carpet?
- A. ₹ 398.80
- B. ₹ 437.40
- C. ₹ 583.20
- D. ₹ 1020.60
Answer: ₹ 437.40
1013. If each side of a rectangle is increased by 50%, its area will increase by
- A. 50%
- B. 125%
- C. 150%
- D. 200%
Answer: 125%
1014. The area of a rectangle gets reduced by 9 m2 if its length is reduced by 5 m and breadth is increased by 3 m. If we increase the length by 3 m and breadth by 2 m, the area is increased by 67 m2. The length of the rectangle is
- A. 9 m
- B. 15.6 m
- C. 17 m
- D. 18.5 m
Answer: 17 m
Explanation: Let the length and breadth of the rectangle be l and b metres respectively. Then, area of the rectangle = lb. lb – (l – 5) (b + 3) = 9 ⇒ lb – (lb + 3l – 5b – 15) = 9 ⇒ lb – lb – 3l + 5b + 15 = 9 ⇒ 3l – 5b = 6 ...(i) And, (l + 3)(b + 2) – lb = 67 ⇒ lb + 2l + 3b + 6 – lb = 67 ⇒ 2l + 3b = 61 ...(ii) Multiplying (i) by 2 and (ii) by 3 and subtracting, we get: – 19b = – 171 or b = 9. Putting b = 9 in (i) we get: 3l = 51 or l = 17 m.
Explanation: Let the length and breadth of the rectangle be l and b metres respectively. Then, area of the rectangle = lb. lb – (l – 5) (b + 3) = 9 ⇒ lb – (lb + 3l – 5b – 15) = 9 ⇒ lb – lb – 3l + 5b + 15 = 9 ⇒ 3l – 5b = 6 ...(i) And, (l + 3)(b + 2) – lb = 67 ⇒ lb + 2l + 3b + 6 – lb = 67 ⇒ 2l + 3b = 61 ...(ii) Multiplying (i) by 2 and (ii) by 3 and subtracting, we get: – 19b = – 171 or b = 9. Putting b = 9 in (i) we get: 3l = 51 or l = 17 m.
1015. The area of a grassy plot is 480 sq. m. If each side had been 5 m longer, the area would have been increased by 245 sq. m. Find the length of the fence to surround it.
- A. 87 m
- B. 88 m
- C. 90 m
- D. None of these
Answer: 88 m
Explanation: Let the length and breadth of the plot be l and b metres respectively. Then, lb = 480 And, (l + 5)(b + 5) = 725 ⇒ lb + 5(l + b) + 25 = 725 ⇒ 5(l + b) + 505 = 725 ⇒ (l + b) = 220 44. 5 = [∵ lb = 480] ∴ Length of the fence = 2(l + b) = (2 × 44) m = 88 m.
Explanation: Let the length and breadth of the plot be l and b metres respectively. Then, lb = 480 And, (l + 5)(b + 5) = 725 ⇒ lb + 5(l + b) + 25 = 725 ⇒ 5(l + b) + 505 = 725 ⇒ (l + b) = 220 44. 5 = [∵ lb = 480] ∴ Length of the fence = 2(l + b) = (2 × 44) m = 88 m.
1016. The length of a rectangular blackboard is 8 m more than its breadth. If its length is increased by 7m and its breath is decreased by 4 m, its area remains un-changed. The length and breadth of the rectangular blackboard is
- A. 24 m, 16 m
- B. 20 m, 24 m
- C. 28 m, 16 m
- D. 28 m, 20 m
Answer: 28 m, 20 m
1017. Carpeting a room costs ₹ 120. Had the room been 4 m narrower, the carpet would have cost ₹ 20 less. How wide is the room?
- A. 18.5 m
- B. 20 m
- C. 24 m
- D. 25 m
Answer: 24 m
Explanation: Let the length and breadth of the carpet be l and b metres respectively and let the rate of carpeting be ₹ x per metre. Then, lbx = 120 ...(i) And, l(b – 4)x = ⇒ lbx – 4lx = 100 ⇒ 4lx = 20 ...(ii) Dividing (i) by (ii), we get: 120 24 m. 4 20 lbx b lx = ⇒=
Explanation: Let the length and breadth of the carpet be l and b metres respectively and let the rate of carpeting be ₹ x per metre. Then, lbx = 120 ...(i) And, l(b – 4)x = ⇒ lbx – 4lx = 100 ⇒ 4lx = 20 ...(ii) Dividing (i) by (ii), we get: 120 24 m. 4 20 lbx b lx = ⇒=
1018. A rectangular field has length and breadth in the ratio 3 : 2. If the length alone is increased by 5 m, the area becomes 2600 sq. m. What is the breadth?
- A. 40 metres
- B. 60 metres
- C. 65 metres
- D. Cannot be determined
Answer: 40 metres
Explanation: Let length = (3x) metres and breadth = (2x) metres. Then, (3x + 5) × 2x = 2600 ⇔ 6x2 + 10x – 2600 = 0 ⇔ 3x2 + 5x – 1300 = 0 ⇔ (3x + 65) (x – 20) = 0 ⇔ x = 20. ∴ Breadth = 2x = 40 m.
Explanation: Let length = (3x) metres and breadth = (2x) metres. Then, (3x + 5) × 2x = 2600 ⇔ 6x2 + 10x – 2600 = 0 ⇔ 3x2 + 5x – 1300 = 0 ⇔ (3x + 65) (x – 20) = 0 ⇔ x = 20. ∴ Breadth = 2x = 40 m.
1019. A rectangular carpet has an area of 60 sq. m. If its diagonal and longer side together equal 5 times the shorter side, the length of the carpet is
- A. 5 m
- B. 12 m
- C. 13 m
- D. 14.5 m
Answer: 12 m
1020. A crossed a rectangular field along its diagonal in 15 seconds at 52 m/min, while B covered the sides in the same time at 68 m/min. What is the area of the field?
- A. 30 m2
- B. 40 m2
- C. 50 m2
- D. 60 m2
Answer: 60 m2
1021. If the diagonal and the area of a rectangle are 25 m and 168 m2, what is the length of the rectangle?
- A. 12 m
- B. 17 m
- C. 24 m
- D. 31 m
Answer: 24 m
1022. If the area of a rectangle is 2 3 , d where 2d is the length of its diagonal, then its perimeter is equal to
- A. 43 d
- B. 23 d
- C. 4( 3 1) + d / 4(3 1) + d
- D. 2( 3 1) + d / 2(3 1) + d
Answer: 2( 3 1) + d / 2(3 1) + d
1023. The diagonal of a rectangle is 41 cm and its area is 20 sq. cm. The perimeter of the rectangle must be
- A. 9 cm
- B. 18 cm
- C. 20 cm
- D. 41 cm
Answer: 18 cm
Explanation: 22 22 l b l b lb += += = 41 or 41. Also, 20. (l + b)2 = (l 2 + b2) + 2lb = 41 + 40 = 81 ⇒ (l + b) = 9. ∴ Perimeter = 2 (l + b) = 18 cm.
Explanation: 22 22 l b l b lb += += = 41 or 41. Also, 20. (l + b)2 = (l 2 + b2) + 2lb = 41 + 40 = 81 ⇒ (l + b) = 9. ∴ Perimeter = 2 (l + b) = 18 cm.
1024. A rectangular carpet has an area of 120 sq. metres and a perimeter of 46 metres. The length of its diagonal is
- A. 15 m
- B. 16 m
- C. 17 m
- D. 20 m
Answer: 17 m
1025. The diagonal of a rectangular field is 15 metres and the difference between its length and width is 3 metres. The area of the rectangular field is
- A. 9 m2
- B. 12 m2
- C. 21 m2
- D. 108 m2
Answer: 108 m2
Explanation: Let l and b be the length and breadth of the rectangle respectively. Then, 2 2 lb += 15 ⇒ (l 2 + b2) = (15)2 = 225. And, l + b = 3 ⇒ (l – b) 2 = 9 ⇒ l2 + b2 – 2 lb = 9 ⇒ 225 – 2lb = 9 ⇒ 2lb = 216 ⇒ lb = 108. Hence, area of the field = lb = 108 m2.
Explanation: Let l and b be the length and breadth of the rectangle respectively. Then, 2 2 lb += 15 ⇒ (l 2 + b2) = (15)2 = 225. And, l + b = 3 ⇒ (l – b) 2 = 9 ⇒ l2 + b2 – 2 lb = 9 ⇒ 225 – 2lb = 9 ⇒ 2lb = 216 ⇒ lb = 108. Hence, area of the field = lb = 108 m2.
1026. The diagonal of a rectangle is 10 cms and is twice the length of one of the sides. What is the area of the rectangle in sq. cm?
- A. 10 3
- B. 25
- C. 25 3
- D. 100
Answer: 25 3
1027. The length of a rectangle is three times of its width. If the length of the diagonal is 8 10 cm, then the perimeter of the rectangle is
- A. 15 10 cm
- B. 16 10 cm
- C. 24 10 cm
- D. 64 cm
Answer: 64 cm
1028. The diagonal of the floor of a rectangular closet is 1 7 2 feet. The shorter side of the closet is 1 4 feet. 2 What is the area of the closet in square feet?
- A. 1 5 4
- B. 1 132
- C. 27
- D. 37
Answer: 27
1029. Carpeting a room 18 m long with carpet 75 cm wide at ₹ 4.50 per metre costs ₹ 810. What is the breadth of the room?
- A. 7 m
- B. 7.5 m
- C. 8 m
- D. 8.5 m
Answer: 7.5 m
1030. How many metres of carpet 63 cm wide will be required to cover the floor of a room 14 m by 9 m?
- A. 185 m
- B. 200 m
- C. 210 m
- D. 220 m
Answer: 200 m
1031. A courtyard 25 m long and 16 m broad is to be paved with bricks of dimensions 20 cm by 10 cm. The total number of bricks required is
- A. 18000
- B. 20000
- C. 25000
- D. None of these
Answer: 20000
1032. An artist has completed one-fourth of a rectangular oil painting. When he will paint another 100 square centimetres of the painting, he would complete three-quarters of the painting. If the height of the oil painting is 10 cm, determine the length (in cm) of the oil painting.
- A. 10
- B. 15
- C. 20
- D. 25
Answer: 20
1033. A rectangular plot is half as long again as it is broad and its area is 2 3 hectares. Then, its length is
- A. 100 m
- B. 33.33 m
- C. 66.66 m
- D. 100 3 m 3
Answer: 100 m
1034. A rectangular paper, when folded into two congruent parts had a perimeter of 34 cm for each part folded along one set of sides and the same is 38 cm when folded along the other set of sides. What is the area of the paper?
- A. 140 cm2
- B. 240 cm2
- C. 560 cm2
- D. None of these
Answer: 140 cm2
1035. A 700-hectare field is split into two parts, and the difference between their areas is one-fifth of their average. How many hectares is the smaller part?
- A. 225
- B. 280
- C. 300
- D. 315
Answer: 315
1036. The ratio of length and breadth of a rectangle is 3 : 2 respectively. The respective ratio of its perimeter and area is 5 : 9. What is the breadth of the rectangle in metres?
- A. 6 m
- B. 8 m
- C. 9 m
- D. 13 m
Answer: 6 m
1037. A farmer wishes to start a 100 sq. m rectangular vegetable garden. Since he has only 30 m barbed wire, he fences three sides of the garden letting his house compound wall act as the fourth side fencing. The dimension of the garden is:
- A. 15 m × 6.67 m
- B. 20 m × 5 m
- C. 30 m × 3.33 m
- D. 40 m × 2.5 m
Answer: 20 m × 5 m
1038. A rectangular field has to be fenced on three sides, leaving one 20-foot side open. If its area is 680 sq. feet, how many feet of fencing are needed?
- A. 34
- B. 40
- C. 68
- D. 88
Answer: 88
Explanation: We have : l = 20 ft and lb = 680 sq. ft. So, b = 34 ft. ∴ Length of fencing = (l + 2b) = (20 + 68) ft = 88 ft.
Explanation: We have : l = 20 ft and lb = 680 sq. ft. So, b = 34 ft. ∴ Length of fencing = (l + 2b) = (20 + 68) ft = 88 ft.
1039. The area of a rectangular field is 52000 m2. This rectangular area has been drawn on a map to the scale 1 cm to 100 m. The length is shown as 3.25 cm on the map. The breadth of the rectangular field is
- A. 150 m
- B. 160 m
- C. 200.5 m
- D. 300.5 m
Answer: 160 m
1040. A rectangle has an area of 460 square metres and its length is 15% more than its breadth. Find the breadth.
- A. 15 metres
- B. 26 metres
- C. 34.5 metres
- D. None of these
Answer: None of these