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- Question
The percentage increase in the area of a rectangle, if each of its sides is increased by 20% is:
Options- A. 40%
- B. 42%
- C. 44%
- D. 46%
- Correct Answer
- 44%
ExplanationLet original length = x metres and original breadth = y metres.Original area = (xy) m2.
New length = ❨ 120 x ❩m = ❨ 6 x ❩m. 100 5 New breadth = ❨ 120 y ❩m = ❨ 6 y ❩m. 100 5 New Area = ❨ 6 x x 6 y ❩m2 = ❨ 36 xy ❩m2. 5 5 25 The difference between the original area = xy and new-area 36/25 xy is
= (36/25)xy - xy
= xy(36/25 - 1)
= xy(11/25) or (11/25)xy
∴ Increase % = ❨ 11 xy x 1 x 100 ❩% = 44%. 25 xy
Area problems
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- 1. A tank is 25 m long, 12 m wide and 6 m deep. The cost of plastering its walls and bottom at 75 paise per sq. m, is:
Options- A. Rs. 456
- B. Rs. 458
- C. Rs. 558
- D. Rs. 568 Discuss
Correct Answer: Rs. 558
Explanation:
Area to be plastered = [2(l + b) x h] + (l x b) = {[2(25 + 12) x 6] + (25 x 12)} m2 = (444 + 300) m2 = 744 m2. ∴ Cost of plastering = Rs. ❨ 744 x 75 ❩ = Rs. 558. 100 - 2. The difference between the length and breadth of a rectangle is 23 m. If its perimeter is 206 m, then its area is:
Options- A. 1520 m2
- B. 2420 m2
- C. 2480 m2
- D. 2520 m2 Discuss
Correct Answer: 2520 m2
Explanation:
We have: (l - b) = 23 and 2(l + b) = 206 or (l + b) = 103.Solving the two equations, we get: l = 63 and b = 40.
∴ Area = (l x b) = (63 x 40) m2 = 2520 m2.
- 3. The ratio between the perimeter and the breadth of a rectangle is 5 : 1. If the area of the rectangle is 216 sq. cm, what is the length of the rectangle?
Options- A. 16 cm
- B. 18 cm
- C. 24 cm
- D. Data inadequate
- E. None of these Discuss
Correct Answer: 18 cm
Explanation:
2(l + b) = 5 b 1 ⟹ 2l + 2b = 5b
⟹ 3b = 2l
b = 2 l 3 Then, Area = 216 cm2
⟹ l x b = 216
⟹l x 2 l = 216 3 ⟹ l2 = 324
⟹ l = 18 cm.
- 4. What is the least number of squares tiles required to pave the floor of a room 15 m 17 cm long and 9 m 2 cm broad?
Options- A. 814
- B. 820
- C. 840
- D. 844 Discuss
Correct Answer: 814
Explanation:
Length of largest tile = H.C.F. of 1517 cm and 902 cm = 41 cm.Area of each tile = (41 x 41) cm2.
∴ Required number of tiles = ❨ 1517 x 902 ❩ = 814. 41 x 41 - 5. 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?
Options- A. 2.91 m
- B. 3 m
- C. 5.82 m
- D. None of these Discuss
Correct Answer: 3 m
Explanation:
Area of the park = (60 x 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.
- 6. In how many different ways can the letters of the word 'OPTICAL' be arranged so that the vowels always come together?
Options- A. 120
- B. 720
- C. 4320
- D. 2160
- E. None of these Discuss
Correct Answer: 720
Explanation:
The word 'OPTICAL' contains 7 different letters.When the vowels OIA are always together, they can be supposed to form one letter.
Then, we have to arrange the letters PTCL (OIA).
Now, 5 letters can be arranged in 5! = 120 ways.
The vowels (OIA) can be arranged among themselves in 3! = 6 ways.
∴ Required number of ways = (120 x 6) = 720.
- 7. A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the box, if at least one black ball is to be included in the draw?
Options- A. 32
- B. 48
- C. 64
- D. 96
- E. None of these Discuss
Correct Answer: 64
Explanation:
We may have(1 black and 2 non-black) or (2 black and 1 non-black) or (3 black).∴ Required number of ways = (3C1 x 6C2) + (3C2 x 6C1) + (3C3) = ❨ 3 x 6 x 5 ❩ + ❨ 3 x 2 x 6 ❩ + 1 2 x 1 2 x 1 = (45 + 18 + 1) = 64. - 8. Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?
Options- A. 210
- B. 1050
- C. 25200
- D. 21400
- E. None of these Discuss
Correct Answer: 25200
Explanation:
Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4)= (7C3 x 4C2) = ❨ 7 x 6 x 5 x 4 x 3 ❩ 3 x 2 x 1 2 x 1 = 210. Number of groups, each having 3 consonants and 2 vowels = 210.
Each group contains 5 letters.
Number of ways of arranging
5 letters among themselves= 5! = 5 x 4 x 3 x 2 x 1 = 120. ∴ Required number of ways = (210 x 120) = 25200.
- 9. In how many ways can the letters of the word 'LEADER' be arranged?
Options- A. 72
- B. 144
- C. 360
- D. 720
- E. None of these Discuss
Correct Answer: 360
Explanation:
The word 'LEADER' contains 6 letters, namely 1L, 2E, 1A, 1D and 1R.∴ Required number of ways = 6! = 360. (1!)(2!)(1!)(1!)(1!) - 10. In how many different ways can the letters of the word 'DETAIL' be arranged in such a way that the vowels occupy only the odd positions?
Options- A. 32
- B. 48
- C. 36
- D. 60
- E. 120 Discuss
Correct Answer: 36
Explanation:
There are 6 letters in the given word, out of which there are 3 vowels and 3 consonants.Let us mark these positions as under:
(1) (2) (3) (4) (5) (6)
Now, 3 vowels can be placed at any of the three places out 4, marked 1, 3, 5.
Number of ways of arranging the vowels = 3P3 = 3! = 6.
Also, the 3 consonants can be arranged at the remaining 3 positions.
Number of ways of these arrangements = 3P3 = 3! = 6.
Total number of ways = (6 x 6) = 36.
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