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Test: Perimeters, Areas and Volumes - 2 - UCAT MCQ


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10 Questions MCQ Test Quantitative Reasoning for UCAT - Test: Perimeters, Areas and Volumes - 2

Test: Perimeters, Areas and Volumes - 2 for UCAT 2024 is part of Quantitative Reasoning for UCAT preparation. The Test: Perimeters, Areas and Volumes - 2 questions and answers have been prepared according to the UCAT exam syllabus.The Test: Perimeters, Areas and Volumes - 2 MCQs are made for UCAT 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Perimeters, Areas and Volumes - 2 below.
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Test: Perimeters, Areas and Volumes - 2 - Question 1

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?

Detailed Solution for Test: Perimeters, Areas and Volumes - 2 - Question 1

As AB and CD are two chords that intersect at O, AP x PB = CP x PD
6 x 4 = CP x 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 x 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.

Test: Perimeters, Areas and Volumes - 2 - Question 2

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?

Detailed Solution for Test: Perimeters, Areas and Volumes - 2 - Question 2

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 x 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 = Length of box / Height of can = 76/30 = 2
Number of such rows = Width of box / Diameter of can  = 46/14  = 3
∴ 2 x 3 = 6 cans can be placed horizontally
∴ Maximum number of cans = 15 + 6 = 21

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Test: Perimeters, Areas and Volumes - 2 - Question 3

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 beCAT Mesuration: Cuboid

Detailed Solution for Test: Perimeters, Areas and Volumes - 2 - Question 3

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

Test: Perimeters, Areas and Volumes - 2 - Question 4

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?

Detailed Solution for Test: Perimeters, Areas and Volumes - 2 - Question 4

Given,
Surface area of oil = 36 π = πr2
⇒ r = 6 cm
Now,
∆ABC ~ ∆AED


h = 6 cm
∴ Volume of oil in the cone = (1/3)πr2h
= (1/3)π62 x 6
= 72π
⇒ Time taken = 72π/1
= 72π hours.

Test: Perimeters, Areas and Volumes - 2 - Question 5

An equilateral triangle ABC is inscribed in a circle with centre O. D is a point on the minor arc BC and ∠CBD = 40º. Find the measure of ∠BCD.

Detailed Solution for Test: Perimeters, Areas and Volumes - 2 - Question 5

∠ABC = ∠ACB = ∠BAC = 60°  [∵ ΔABC is an equilateral triangle]

Also, ∠BAC + ∠BDC = 180°

⇒ 60° + ∠BDC = 180°

⇒ ∠BDC = 180° - 60° = 120°

Also, ∠CBD + ∠BDC + ∠BCD = 180°

⇒ 40° + 120° + ∠BCD = 180°

⇒ ∠BCD = 180° - 40° - 120° = 20°

∴ The value of ∠BCD is 20°

Test: Perimeters, Areas and Volumes - 2 - Question 6

A sphere of radius 12 cm is melted and re-casted into a right circular cone of height 12 cm. The radius of the cone is.

Detailed Solution for Test: Perimeters, Areas and Volumes - 2 - Question 6

Let radius of the cone be r cm

According to the question

(1/3) × π × r2 × 12 = (4/3) × π × 12 × 12 × 12

⇒ r2 = 12 × 12 × 4

⇒ r = 12 × 2

∴ r = 24 cm

Test: Perimeters, Areas and Volumes - 2 - Question 7

AB is a diameter of a circle with center O. A tangent is drawn at point A. C is a point on the circle such that BC produced meets the tangent at P. If ∠APC = 62º, then find the measure of the minor arc AC(i.e.∠ ABC).

Detailed Solution for Test: Perimeters, Areas and Volumes - 2 - Question 7

Minor arc AC will create angle CBA

∠APC = 62º = ∠APB

∠BAP = 90° (diameter perpendicular to tangent)

In Δ APB,

∠APB + ∠BAP + ∠PBA = 180° 

⇒ ∠PBA = 180° - (90° + 62°)

⇒ ∠PBA = 28° 

∴ The measure of minor arc AC is 28° 

Test: Perimeters, Areas and Volumes - 2 - Question 8

To pack a set of books, Gautam got cartons of a certain height that were 48 inches long and 27 inches wide. If the volume of such a carton was 22.5 cubic feet, what was the height of each carton? [Use 1 foot = 12 inches.] 

Detailed Solution for Test: Perimeters, Areas and Volumes - 2 - Question 8

Volume of carton = volume of cuboid = Length × Breadth × Height 

⇒ volume of carton = 48 × 27 × Height

∵ 1 foot = 12 inches, then 22.5 cubic feet = 22.5 × 12 × 12 ×12

⇒ 22.5 × 12 × 12 × 12 = 48 × 27 × Height     

⇒ 38,880 = 1,296 × Height 

⇒ Height = 30 inches.

∴ The height of each cartoon is 30 inches.    

Test: Perimeters, Areas and Volumes - 2 - Question 9

There are 5 concentric circles that are spaced equally from each other by 1.25 cms. The innermost circle has a square of side √(32) cm inscribed in it. If a square needs to be inscribed in the outermost circle, what will be its area?

Detailed Solution for Test: Perimeters, Areas and Volumes - 2 - Question 9

From the figure, we can see that the diagonal of the square inscribed in innermost circle is the diameter of the innermost circle.
Side = √32
Or Diagonal = √32 * √2 = 8 cm
Since the circles are spaced at 1.25 cm, the distance between innermost and outermost circles = 1.25 * 4 = 5 cm
Therefore the diameter of the outermost circle = 5 cm + diagonal of inscribed square + 5cm = 18 cm
Now this 18 cm will be the diagonal of the square that needs to be inscribed in outermost circle.
Or a √2 = 18
a = 9 √2 cm
Area = 9 √2 x 9 √2 = 162 sq. cm

Test: Perimeters, Areas and Volumes - 2 - Question 10

Anil grows tomatoes in his backyard which is in the shape of a square. Each tomato takes 1 cm2 in his backyard. This year, he has been able to grow 131 more tomatoes than last year. The shape of the backyard remained a square. How many tomatoes did Anil produce this year?

Detailed Solution for Test: Perimeters, Areas and Volumes - 2 - Question 10

Let the area of backyard be x2 this year and y2 last year
∴ X2 - Y2 = 131
⇒ (X + Y)*(X - Y) = 131
Now, 131 is a prime number (a unique one too. Check out its properties on Google). Also, always identify the prime number given in a question. Might be helpful in cracking the solution.
⇒ (X + Y) x (X - Y) = 131 x 1
⇒ X + Y = 131
X - Y = 1
⇒ 2X = 132 ⇒ X = 66
and Y = 65
∴ Number of tomatoes produced this year = 662 = 4356

The question is "How many tomatoes did Anil produce this year?"

Hence, the answer is 4356

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