Quiz 5

Question 1

Time: 00:00:00

Four points A, B, C and D lie on a straight line in the X-Y plane, such that AB = BC = CD, and the length of AB is 1 metre. An ant at A wants to reach a sugar particle at D. But there are insect repellents kept at points B and C. The ant would not go within one metre of any insect repellent. The minimum distance in metres the ant must traverse to reach the sugar particle is

3/2

1 + pie

pie/2

1 + pie/2

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Question 2

Time: 00:00:00

Rectangular tiles each of size 70 cm by 30 cm must be laid horizontally on a rectangular floor of size 110 cm by 130 cm, such that the tiles do not overlap. A tile can be placed in any orientation so long as its edges are parallel to the edgesof the floor. No tile should overshoot any edge of the floor. The maximum number of tiles that can be accommodated on the floor is

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Hence, option 3.

Question 3

Time: 00:00:00

In the X-Y plane, the area of the region bounded by the graph of |x + y| + |x – y| = 4 is

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|x + y| + |x – y| = 4 ... (i)
Consider the case when x and y are both positive. This is the area of quadrant I
In this case two cases are possible.
Case I: x > y
In this case expression (i) becomes
x + y + x – y = 4
? 2x = 4
? x = 2
Case II: x < y
In this case expression (i) becomes
x + y + y – x = 4

? 2y = 4
? y = 2
? The area for the first quadrant is as shown in the figure.
Extending the same logic to other quadrants we get the area as shown in the
diagram.
? Its area = 4 × 4 = 16 sq. units
Hence, option 3.

Question 4

Time: 00:00:00

In the following figure, the diameter of the circle is 3 cm. AB and MN are two diameters such that MN is perpendicular to AB. In addition, CG is perpendicular to AB such that AE:EB = 1:2, and DF is perpendicular to MN such that NL:LM = 1:2. The length of DH in cm is

((4root2 - 1))/2

((2root2 - 1))/2

((4root2 - 1))/4

((2root2 - 1))/2

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AO = OD = 1.5 cm
AE + EB = 3cm and AE:EB = 1:2
∴ AE = 1 cm and EB = 2 cm
OE = AO – AE = 1.5 – 1 = 0.5 cm
Similarly, NL = 1 cm, M = 2 cm and OL = 0.5 cm
OEHL is a square as all its angles are right angles and OE = OL
∴ EH = HL = 0.5 cm
1.52 = 0.52 + (0.5 + DH)2
2.25 = 0.25 + 0.25 + DH + DH2
DH2 + DH – 1.75 = 0

Hence, option 2.

Question 5

Time: 00:00:00

Consider the triangle ABC shown in the following figure where BC = 12 cm, DB = 9 cm, CD = 6 cm and ?BCD = ?BAC. What is the ratio of the perimeter of the triangle ADC to that of the triangle BDC?

7/9

8/9

6/9

5/9

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m∠BCD = m∠BAC and B is common to triangles ABC and CBD.
∆ ABC is similar to ∆ CBD
AB/CB = BC/BD = AC/CD
AB/12 = 12/9 = AC/6
AB = 16 cm and AC = 8 cm
AD = AB – BD = 16 – 9 = 7 cm
∴ Perimeter of ∆ ABC = 7 + 6 + 8 = 21 cm
∴ Perimeter of ∆ BDC = 9 + 6 + 12 = 27 cm
∴ Required ratio = 21/27 = 7/9
Hence, option 1.

Question 6

Time: 00:00:00

P, Q, S, and R are points on the circumference of a circle of radius r, such that PQR is an equilateral triangle and PS is a diameter of the circle. What is the perimeter of the quadrilateral PQSR?

2r(1+root3)

4r(1+root8)

8r(1+root3)

4r(1+root3)

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∆ PQR is an equilateral triangle and PS is the diameter.
∴ m∠PQS = m∠PRS = 90° (angles subtended in a semi-circle)
PS bisects QPS as it is the median of ∆PQR
m∠PMQ = m∠PMR = 90°
∴ m∠QPS = m∠RPS = 30°
∴ m∠PSQ = m∠PSR = 60°
Radius = r
∴ PS = 2r
As ∆ PQS, ∆ PQM, ∆ MQS are 30°-60°-90° triangles,

Hence, option 1.

Question 7

Time: 00:00:00

100 sq units

114 sq units

50 sq units

200 sq units

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ans c

50 sq units

Question 8

Time: 00:00:00

ABCD is a rectangle. The points P and Q lie on AD and AB respectively. If the triangles PAQ, QBC and PCD all have the same areas and BQ = 2 then AQ = ?

1+ √5

1- √5

√7

-√7

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First draw the figure as described
let AB = CD = L, and AD =BC= B
now given QB = 2 we have AQ = L-2
if PD = K (let) we have PA = AP = (B- K)
now by given condition we have 3 tTriangleof equal areas
A ( PAQ) = A (QBC) = A (PCD)
all the three are right triangles where arae = 1/2 product of perpendicular sides
1/2 ( L-2) (B-K) = 1/2 (2B) = (1/2)(K)(L)
if we muliply by 2 we get
2B = (K)(L) gives K = 2B/L
substituting K = 2B/L in equality ( L-2) (B-K) = 2B = ( L-2) (B- 2B/L)
or multiply L on both sides gives 2L = $(L-2)^{2}$ or L^2 - 6L + 4 = 0
we solve for L we get L = $[ - (-6) ± √ (-6)^{2} - 4 (1) (4) ]/ 2$ = 3 ± √5
so we have two values (3 + √5 ) and (3 - √5 ) for L = AB

now length AQ = AB - 2 or we have (3 + √5 - 2) = 1+ √5 which is the positive value allowed for length
The other value is (3 - √5 - 2) = 1- √5 < o so not allowed

Hence we have AQ = 1+ √5 = 1 + 2.236 = 3.236 units

Question 9

Time: 00:00:00

On a plate in the shape of an equilateral triangle ABC with area 16 √3sq.cm , a rod GD, of height 8 cm, is fixed vertically. G is a point on the plate. If the areas of the triangles AGD and BGD are both equal to 4√19 sq.cm, find the area of the triangle CGD (in sq.cm).

3 sqrt 19

4 sqrt 19

12 sqrt 3

None of these

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ANS - 12 sqrt 3

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