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**Question 1: A right circular cone has height H and radius R. A small cone is cut off at the top by a plane parallel to the base. At what height above the base the section has been made?****Statement (I): H = 20 cm ****Statement (II): Volume of small cone: volume of large cone : 1:15**

(A) If the question can be answered with statement I alone but not statement II alone, or can be answered with statement II alone but not statement I alone.

(B) 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.

(C) If the question can be answered with either statement alone.

(D) If the question cannot be answered with the information provided.**Option BExplanation: **

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. 1/3*π*r^{2}*h : 1/3*π*R^{2}*H is 1 : 15 (r is the base radius of the smaller cone and h is the height of the smaller cone)

or r^{2} * h : R^{2} * 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, h/H = r/R

R = rH/h

We know H = 20

i.e., h^{3} = 1/15 * H^{3}. 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.**Question 2: A sphere of radius r is cut by a plane at a distance of h from its center, thereby breaking this sphere into two different pieces. The cumulative surface area of these two pieces is 25% more than that of the sphere. Find h.**

(A) r/√2

(B) r/√3

(C) r/√5

(D) r/√6**Option AExplanation: **Area = 4πr

Cumulative area of the two pieces = 25%

more than that of sphere.

Area of 2 pieces = 1.25 × 4π

Extra area = πr

Extra area = area of two new circles that are now created circles.

Area each new circle = πr

Let radius of new circle be r

Now, πr

r

Now, r_{1} , h and r form a right angled triangle.

h^{2} + r_{1}^{2} = r^{2}

h^{2} + (r/√2)^{2} = r^{2 }

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**Question 3: Two mutually perpendicular chords AB and CD meet at a point P inside the circle such that AP = 6 cms, PB = 4 units and DP = 3 units. What is the area of the circle?**

(A) 125π/4 sq cms

(B) 100π/7 sq cms

(C) 125π/8 sq cms

(D) 52π/3 sq cms**Option AExplanation:**

As AB and CD are two chords that intersect at O, AP * PB = CP * PD6 * 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

ON

1

1 + 30.25 = r

Area = πr

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.**Question 4: Cylindrical cans of cricket balls are to be packed in a box. Each can has a radius of 7 cm and height of 30 cm. Dimension of the box is l = 76 cm, b = 46 cm, h = 45 cm. What is the maximum number of cans that can fit in the box?**

(A) 15

(B) 17

(C) 22

(D) 21**Option D****Explanation:**

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 = l/Diameter = 76/14 = 5 (Remaining space will be vacant)

Number of such rows that can be placed = Width/Diameter = 46/14 = 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 = Lengthofbox/Heightofcan = 76/30 = 2

Number of such rows = Widthofbox/Diameterofcan = 46/14 = 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.**Question 5: PQRS is a square of sides 2 cm & ST = 2 cm. Also, PT=RT. What is the area of ∆PST?**

(A) 2 cm^{2}

(B) √3 cm^{2}

(C) √2 cm^{2}

(D) 1/√2cm^{2}**Option C****Explanation:**

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 = 1/2 * PO * OT = 1/2 * 2√2/2 (PO = 2√2/2 as diagonal of square = 2√a) * (OS + ST)

= √2 + 1

A ∆ POS = 1/4 * Area of Square

= 1/4 * 2^{2} = 1cm^{2}

=> A ∆ POT = √2 +1 - 1 = √2cm^{2}

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

Hence, the answer is √2cm^{2}**Question 6: A string is wound around two circular disk as shown. If the radius of the two disk are 40 cm and 30 cm respectively. What is the total length of the string?**

(A) 70 cm

(B) 70 + 165π cm

(C) 70 + 120π cm

(D) 70 + 165π/2 cm**Option C****Explanation:**

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.**Question 7: Figure below shows a box which has to be completely wrapped with paper. However, a single Sheet of paper need to be used without any tearing. The dimension of the required paper could be**** **

(A) 17 cm by 4 cm

(B) 12 cm by 6 cm

(C) 15 cm by 4 cm

(D) 13 cm by 4 cm**Option B****Explanation:**

Total surface area of the box = 2(4 * 6 + 1 * 6 + 1 * 4)

= 2(24 + 6 + 4)

= 68 cm^{2}

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 cm^{2}.

Only option b) gives the area greater 68 cm^{2}

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

Hence, the answer is 12 cm by 6 cm**Question 8: An inverted right circular cone has a radius of 9 cm. This cone is partly filled with oil which is dipping from a hole in the tip at a rate of 1cm2/hour. Currently the level of oil 3 cm from top and surface area is 36π cm2. How long will it take the cone to be completely empty?**

(A) 216π hours

(B) 1 hour

(C) 3 hours

(D) 36π hours**Option A****Explanation:**

Given,

Surface area of oil = 36 π = πr^{2}

=> r = 6 cm

Now,

∆ABC ~ ∆AED

=> DE/BC = AE/AC

=> 96 = (3+h)h

=> h = 6 cm

∴ Volume of oil in the cone = πr^{2}h

= π6^{2} x 6

= 216π

=> Time taken = 216π/1 = 216π hours.

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

Hence, the answer is 216π hours.**Question 9: A square PQRS has an equilateral triangle PTO inscribed as shown:**

**What is the ratio of A∆PQT to A∆TRU?**

(A) 1 : 3

(B) 1 : √3

(C) 1 : √2

(D) 1 : 2**Option D****Explanation: **

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)

Since PQT and PTO are right angled triangles

PQ^{2} + QT^{2} = PT^{2}

RT^{2} + RU^{2} = UT^{2}

a^{2} + z^{2} = b^{2} --------- (1)

And, (a-z)^{2} + (a-z)^{2} = b^{2}--------- (2)

=> a^{2} + z^{2} = 2(a-z)^{2}

=> a^{2} + z^{2} = 2a^{2} + 2z^{2} – 4az

=> a^{2} + z^{2}- 4az = 0

=> a^{2} + z^{2} – 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.**Question 10****: A spherical shaped sweet is placed inside a cube of side 5 cm such that the sweet just fits the cube. A fly is sitting on one of the vertices of the cube. What is the shortest distance the fly must travel to reach the sweet? **

(A) 2.5 cm

(B) 5(√3 – 1) cm

(C) 5(√2 – 1) cm

(D) 2.5(√3 – 1) cm**Option D****Explanation: **

Distance = (Diagonalofcube–Diameterofsphere)/2

(Since diagonal of the cube is given as √3 x l)

= 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**Directions for questions 11 & 12: Answer the questions based on the following information:**

A cow is tethered at point A by a rope. Neither the rope nor the cow is allowed to enter the triangle ABC Angle BAC = 30· Also AB =AC =10m.

**Question 11: What is the area that can be grazed by the cow if the length of the rope is 8 m?**

(A)

(B) 121π sq.m

(C) 132π sq.m

(D) **Option D****Explanation: **

The figure shows the prohibited area for the cow and the area the cow can graze. The length of the rope is 8 m whereas the length of the sides id 10 m. Hence the scenario is as explained in the figure above. The height of this triangle can be calculated and is approx 9.6m. So clearly greater than 8 m. So the area grazed by the cow = Area of the circle – Area of the sector of 30^{o }^{}**Question 12: What is the area that can be grazed by the cow if the length of the rope is 12 m? **

(A)

(B) 121π sq.m

(C) 132π sq.m

(D) **Option C****Explanation: **

Here the length of the rope is 12 m which is greater than the side of the triangle. Neither the cow nor the rope can enter the triangle. The area that can be grazed thereby is the area of the circle – the area of the sector of 30 degrees of the same circle which is equal to**Question 13: The figure below shows two concentric circles with centre O. PQRS is a square inscribed in the outer circle. It also circumscribes the inner circle, touching it at point B, C, D and A. What is the ratio of the perimeter of the outer circle to that of polygon ABCD?**

(A) π/4

(B) 3π/2

(C) π/2

(D) π**Option CExplanation: **

Let r be the radius of the inner circle. A, B, C and D are points on its circumference where the sides of the square PQRS touch it imply that the sides are tangent to the circle. The radius and tangent are perpendicular at the point of contact. Thus if we join the radius OB and OC then OB is perpendicular to PQ, OC is perpendicular to QR, OB = OC = r, Hence OBQC is a square of side r. OQ is the diagonal = √2r = R = radius of outer circle.

BC is also the diagonal of OQ = R

Clearly ABCD, is a square of side R and radius of outer circle is also R

Perimeter of ABCD = 4R

Perimeter of circle = 2πR

Hence ratio = 2πR/4R = π/2

(A) 6

(B) 5

(C) 4

(D) 3

An important property of the triangle is that the sum of its two sides should be greater than the third side

Let the sides of the triangle be a, b and c

Given a + b + c = 14 and by property of triangle: a + b > c, b + c >a, c + a > b

Let one side be 1, sum of rest will be 13. Possible combinations are (1,12),(2,11),(3,10),(4,9),(5,8),(6,7)

In all the combinations the property of triangle is violated. Hence 1 cannot be the side of the triangle

Then if one side is 2, only combination satisfied is (6,6).

Accordingly, following 4 combinations can only be formed –

(4,4,6), (5,5,4), (6,5,3) and (6,6,2).

(A) 5

(B) 4·5

(C) 3

(D) 3·5

Given, Area of the path = 516 m^{2}2

Area of the path = Area of the outer rectangle – Area of the pool

Area of pool = 60×20

Area of outer rectangle = (60 + 2x)(20+2x)

x cannot be negative and hence x = 3 meters.

So the path is 3 meters wide.

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