Quiz-2

From statement I, we know that the height of the initial cone is 20cm. However, nothing is said about the small cone. Hence, we cannot answer the question using statement A. So, we can eliminate choices (C).
We are down to choices, (A), (B) or (D).

From Statement II,we know that the ratio of the volume of the small cone to that of the large cone is 1 : 15.
i.e. 13*π*r2*h : 13*π*R2*H is 1 : 15 (r is the base radius of the smaller cone and h is the height of the smaller cone)
or r2 * h : R2 * H is 1 : 15
From this information, we will not be able to find the answer to h. Hence, we can eliminate choice (A).

Combining the information in the two statements:
When a section is made the two cones are similar triangles. so, hH = rR
R = rHh
We know H = 20
h = 115 * rHh * rHh * Hr * r
i.e., h3 = 115 * H3. Substituting H = 20, we can get the value for h.
Choice (B) is therefore, the correct answer.

Hence, the answer is If the question cannot be answered with statement I alone or with statement II alone, but can be answered if both statements are used together.
Choice B is the correct answer.

Area = 4πr2
Cumulative area of the two pieces = 25%
more than that of sphere.
Area of 2 pieces = 1.25 × 4π2 = 5πr2
Extra area = πr2
Extra area = area of two new circles that are now created circles.
Area each new circle = πr22

Let radius of new circle be r1.
Now, πr12 = πr22
r1 = r√2 CAT : Geometry: Mensuration; Sphere
Now, r1 , h and r form a right angled triangle.
h2 + r12 = r2
h2 + (r√2)2 = r2
h = r√2
Choice (A) is therefore, the correct answer.

The question is " The cumulative surface area of these two pieces is 25% more than that of the sphere. Find h. "

Hence, the answer is r√2
Choice A is the correct answer.

As AB and CD are two chords that intersect at O, AP * PB = CP * PD
6 * 4 = CP * 3
CP = 8
From center O draw OM ⊥r AB and ON ⊥r CD.
From the center a line ⊥r to a chord bisects the chord.

So, we have AM = MB = 5 cm
MP = 1 cm, ON = 1 cm, CD = 11 cm, CN = 5.5 cm
ON2 + CN2 = OC2
12 + 5.52 + r2
1 + 30.25 = r2

Area = πr2
π * 31.25
31.25π = 125π/4 sq cms

The question is "What is the area of the circle?"

Hence, the answer is 125π/4 sq cms.
Choice A is the correct answer.

This question requires a good deal of visualization. Since, both the box and cans are hard solids, simply dividing the volume won’t work because the shape can’t be deformed.
Each cylindrical can has a diameter of 14 cm and while they are kept erect in the box will occupy height of 30 cm
Number of such cans that can be placed in a row = lDiameter = 7614 = 5 (Remaining space will be vacant)
Number of such rows that can be placed = WidthDiameter = 4614 = 3
Thus 5 * 3 = 15 cans can be placed in an erect position.
However, height of box = 45cm and only 30 cm has been utilized so far
Remaining height = 15 cm > 14 cm (Diameter of the can)
So, some cans can be placed horizontally on the base.
Number of cans in horizontal row = LengthofboxHeightofcan = 7630 = 2
Number of such rows = WidthofboxDiameterofcan = 4614 = 3
∴ 2 * 3 = 6 cans can be placed horizontally
∴ Maximum number of cans = 15+6 = 21

The question is "What is the maximum number of cans that can fit in the box?"

Hence, the answer is 21.
Choice D is the correct answer.

Notice, ∆PST ~ ∆SRT (all three sides equal). So, TS extended will meet PR at cross points of diagonal as shown:-

Now, A ∆ PST = A ∆ POT – A ∆ POS ---------- Equation(1)
A ∆POT = 1212 * PO * OT = 1212 * 2√222√22 (PO = 2√222√22 as diagonal of square = 2√a) * (OS + ST)
=√22√22 * (2√222√22 + 2)
= 1 * 2+√2√22+√2√2
= √2 + 1
A ∆ POS = 1414 * Area of Square
= 1414 * 2² = 1cm²
=> A ∆ POT = √2 +1 - 1 = √2 cm²

The question is " What is the area of ∆PST? "

Hence, the answer is √2 cm²

What is the relation between length of the string and a Circle ( p.s. it has something do with the circumference!!).

As, tangent from an exterior point makes right angle with the radius.
=> ADXB and ODX’E are squares,
∴ AO = BO = 40 cm
OD = DE = 30 cm
Length of string wound around circle x => 270^°/360^° ∗ 2 ∗ π ∗ 80
= 3/4 ∗ 80π
=60π
Similarly, Length of string wound around circle x’ => 270^°/360^° ∗ 2 ∗ π ∗ 30
= 3/4 ∗ 30π
= 45π/2
∴ Length of the string = 40 + 30 + 60π + 45π/2 ∗ π
= 70 + 165π/2

The question is "What is the total length of the string?"

Hence, the answer is 70 + 165π/2 cm.

Calculating the total surface area will be the key to the solution.

Total surface area of the box = 2(4 * 6 + 1 * 6 + 1 * 4)
= 2(24 + 6 + 4)
= 68 cm²
As the problem says, paper can’t be torn/cut a portion of paper will need to be fold, so, the area of paper required would be greater than 68 cm².
Only option b) gives the area greater 68 cm²

The question is " The dimension of the required paper could be"

Hence, the answer is 12 cm by 6 cm

Given,
Surface area of oil = 36 π = πr²
=> r = 6 cm
Now,
∆ABC ~ ∆AED
=> DEBCDEBC = AEACAEAC
=> 9696 = (3+h)h(3+h)h
=> h = 6 cm
∴ Volume of oil in the cone = πr²h
= π6² x 6
= 216π
=> Time taken = 216π1216π1 = 216π hours.

The question is " How long will it take the cone to be completely empty? "

Hence, the answer is 216π hours.

Let PQ, a side of equilateral triangle be b
By symmetry QT=ST=z (say)
A ∆ PQT = 1/2 * PQ * QT
= 1/2 * a * z
A ∆ TRU = 1/2 * RT * OR
= 1/2 * (a-z) * (a-z)
∴ AΔPQTAΔTRUAΔPQTAΔTRU = 1/2∗az1/2∗(a−z)21/2∗az1/2∗(a−z)2
= az(a−z)2az(a−z)2 --------------(p)
Since PQT and PTO are right angled triangles
PQ² + QT² = PT²
RT² + RU² = UT²
a² + z² = b² --------- (1)
And, (a-z)² + (a-z)² = b²--------- (2)
=> a² + z² = 2(a-z)²
=> a² + z² = 2a² + 2z² – 4az
=> a² + z²- 4az = 0
=> a² + z² – 2az = 2az (Please note how the solution is being managed here. You must always be aware of what you are looking for. Here, as equation -℗ we are looking for (a-z)2 in terms of az)
=> (a-z)² =2az
Putting in equation (p) = az/2(az) = 1/2

The question is "What is the ratio of A∆PQT to A∆TRU?"

Hence, the answer is 1 : 2

Distance = (Diagonalofcube–Diameterofsphere)2(Diagonalofcube–Diameterofsphere)2
= (5√3−5)2(5√3−5)2 (Since diagonal of the cube is given as √3 x l)
= 5(√3−1)25(√3−1)2
= 2.5(√3 – 1)

The question is "What is the shortest distance the fly must travel to reach the sweet? "

Hence, the answer is 2.5(√3 – 1) cm