Question
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 the 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?Solution
Let length of the rectangle = L, breadth of the rectangle = B Area= L*B Now, length reduced by 20%. So, now length – (4/5)L Breadth remains same So, new figure is square. So, (4/5)L = B So, New area = (4/5)L*B 1. Difference in area = [{L/B – (4/5) L*B}/ L*B]*100 = 20% (So, wrong as given in the statement 1 to be 25%) 2. Perimeter of square = 2(4/5) L + 2B = (8/5) L + 2B Perimeter of rectangle = 2(L + B) Difference = [{2L + 2B – (8/5)L – 2B}/ (2(L + B)]*100 = [(2/5)L/ 2(L + (4/5)L]*100 = 11.11% (So, right it is approx. 11%) 3. Diagonal of rectangle = √(L2 + B2) = √(L2 + (4/5 L)2 = (L/5)*√41 Diagonal of square =side*√2 = (4/5)L*√2 Require %age = [{(√41 - 4√2)*(L/5)}/(L/5)*√41]*100 = 11.65% (So, right it is approx. 12%) Hence option D.
Statements: M > L = K ≥ H, V > G > M, U < N = H
Conclusions:
I. V > U
II. H < G
III. L ≥ V
Statement: F ≥ G > I > E ≤ P, E = S ≥ P
Conclusion: I. F ≥ P II. G > P
Statement: P < Q; V < S > T; V < U > Q
Conclusion: I. T ≥ P II. P > T
Statements: B > A = D ≥ C = I ≥ H > E > F > G
Conclusions:
I. C ≥ E
II. E > C
III. A ≥ DStatements: G < O = H ≥ L > F < U≤ T ≥ Z
Conclusion:
I. L ≤ O
II. T > L
Statements:
A $ B * X © Y @ Z
Conclusions:
I. X @ Z
II. Z * A
III. Z % X
Statements: X < M = I > G < J ≤ S; Q ≥ M = J; S > T ≥ U
Conclusions:
I. S > M
II. Q > G
III. T ≥ I
IV....
Statements: E & F, H # I, G $ F, E % D, G @ H
Conclusions: I . D $ F II. F @ I ...
Statement: L > M = P < Q > R; S ≤ O < N; R > G > N
Conclusions:
I. Q > S
II. S < G
III. M < G
Statements: A % O & Z % O; O # C & E; E @ P # D
Conclusions : I. C @ P II. A % P ...