Quant Boosters  Kamal Lohia  Set 3

Maximum Area for a right triangular with hypotenuse 'L' is equal to = L²/4 (it can be easily derived by using AMGM inequality or could be understood by symmetry also that two legs of right triangle should be of equal length for maximum area with fixed hypotenuse length)
So answer is 100²/4 = 2,500 sq m.

Q9) If a, b and c are the sides of a triangle, and a² + b² + c² = bc + ca + ab, then the triangle is
a. equilateral
b. isosceles
c. rightangled
d. obtuseangled

a² + b² + c² = bc + ca + ab
i.e. 2(a² + b² + c²) = 2(bc + ca + ab)
i.e. (a²  2ab + b²) + (b²  2bc + c²) + (c²  2ca + a²) = 0
i.e. (a  b)² + (b  c)² + (c  a)² = 0
i.e. a = b = c
i.e. triangle is equilateral.Alternately, you can easily verify by putting a = b = c directly that whether triangle is equilateral or not and so on.

Q10) ABCDEFGH is a regular octagon. A and E are opposite vertices of the octagon. A frog starts jumping from vertex to vertex, beginning from A. From any vertex of the octagon except E, it may jump to either of the two adjacent vertices. When it reaches E, the frog stops and stays there. Let a(n) be the number of distinct paths of exactly n jumps ending in E. Then what is the value of a(2n – 1)?
a. 0
b. 4
c. 2n – 1
d. Cannot be determined

E is four steps ahead of A. So it is not possible to reach E from A in (2n  1) i.e. odd number of steps.

Q11) A farmer has decided to build a wire fence along one straight side of his property. For this, he planned to place several fenceposts at 6 m intervals, with posts fixed at both ends of the side. After he bought the posts and wire, he found that the number of posts he had bought was 5 less than required. However, he discovered that the number of posts he had bought would be just sufficient if he spaced them 8 m apart. What is the length of the side of his property and how many posts did he buy?
a. 100 m, 15
b. 100 m, 16
c. 120 m, 15
d. 120 m, 16

Let initially n + 1 posts were to be used i.e. the length of the fence must have been 6n as they were to be posted at a gap of 6 m from each other. So according to given scenario, we get
8(n  5) = 6n
i.e. n = 20
i.e. fence length = 6n = 120 m
and number of posts bought = (n + 1)  5 = 21  5 = 16.

Q12) A certain city has a circular wall around it, and this wall has four gates pointing north, south, east and west. A house stands outside the city, 3 km north of the north gate, and it can just be seen from a point 9 km east of the south gate. What is the diameter of the wall that surrounds the city?
a. 6 km
b. 9 km
c. 12 km
d. None of these

Ans: b. 9 km
Sol: This can be solved easily by using similarity of triangles.

Q13) Two sides of a quadrilateral plot measure 32 m and 24 m and the angle between them is a perfect right angle. The other two sides measure 25 m each and the other three angles are not right angles. What is the area of the plot?
a. 768 m²
b. 534 m²
c. 696.5 m²
d. 684 m²

This is a simple application of primitive Pythagorean triplets only.
Required area = (1/2)[8² (3 * 4) + 5² (3 * 4) + 5² (3 * 4)] = 6 * 114 = 684 m²

Q14) Euclid has a triangle in mind. Its longest side has length 20 and another of its sides has length 10. Its area is 80. What is the exact length of its third side?
a. rt(260)
b. rt(250)
c. rt(240)
d. rt(270)

Let θ be the angles between lengths 20 and 10.
So Area = 80 = (1/2) * 20 * 10 * Sinθ
i.e. Sinθ = 4/5, i.e. Cosθ = 3/5
Using cosine rule, we have third side = rt(20² + 10²  2 * 20 * 10 * Cosθ) = rt(260)

Q15) The area of the triangle whose vertices are (a, a), (a + 1, a + 1) and (a + 2, a) is
a. a³
b. 1
c. 2a
d. rt(2)

Just consider a = 0 and solve to get the base as 2 and height 1, so area = (1/2) * 2 * 1 = 1.

Q16) Neeraj has agreed to mow a lawn, which is a 20 m × 40 m rectangle. He mows it with 1 m wide strip. If Neeraj starts at one corner and mows around the lawn toward the centre, about how many times would he go round as he mowed complete the lawn?
a. 9.5
b. 10
c. 19.5
d. 20

Ans. Option b ( 10 ). It is simply half of smaller side. Draw the diagram and observe yourself.

Q17) Let A and B be two solid spheres such that the surface area of B is 300% higher than the surface area of A. The volume of A is found to be k% lower than the volume of B. The value of k must be
a. 85.5
b. 92.5
c. 90.5
d. 87.5

Equating surface areas we get that 2r(A) = r(B)
So, volume of A is 1/8 that of B i.e. 87.5% lesser.

Q18) How many integer solutions exist for x^2 – 3y^2 = 1376 ?