Quantitative Ability Test

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Quantitative Ability Test - 12

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Weekly Quiz Competition

Question 1
4 / -1

From a solid cube of side 8cm, a portion is cut out such that the cutting plane touches the midpoints of three adjacent sides of the cube. Find the surface area of the cut out portion.

A
10 + 5√2

B
24 + 8√3

C
32 + 4√3

D
16 + 5√2

Solution

The isometric view of the cutting plane is shown

The cut portion consists of 3 isosceles right triangles of sides $4,4,4 \sqrt{2}$ and one equilateral triangle of side $4 \sqrt{2}$
The surface area will be equal to the area of these 4 triangles.
Area of isosceles triangle $=0.5 * 4 * 4=8$
Area of 3 triangles $=24$
Area of equilateral triangle $=\frac{\sqrt{3}}{4} *(4 \sqrt{2})^{2}=8 \sqrt{3}$
Hence, surface area $=24+8 \sqrt{3}$
Question 2
4 / -1

The selling price of a cycle varies with the number of cycles sold. The selling price S, is given by S = 200 - 4x, where x is the number of cycles sold. If the shopkeeper wants to maximize his revenue, how many cycles should he sell?

A
24

B
25

C
28

D
27

Solution

Given,

The selling price, S = 200 - 4x

The revenue, will be given by R = (200 - 4x) x

Which can be written as R = 4(50 - x)x

Since the sum of 50 - x and xis constant, the product will be maximum when they are equal.

Hence, 50 - x = x

Thus, if he sells 25 cycles, the revenue will be maximum.
Question 3
4 / -1

Triangle PQR is an equilateral triangle with side 8 cm. The triangle is rotated about a line passing through Q and parallel to PR by 180°. What would be the volume of the figure generated?

A
256 π sq.cm.

B
112 π sq.cm.

C
128 π sq.cm.

D
224 π sq.cm.

Solution

The required volume would be half of the volume one would get by rotating the figure by $360^{\circ}$
Let $V^{\prime}$ be the volume of the figure when the triangle is rotated by $360^{\circ}$
So volume $V=$ Volume of the cylinder $-2^{\star}$ Volume of cone (base shown in dotted line
The cylinder will have a height, $h=8 \mathrm{cm}$ and radius, $r=$ altitude of the equilateral triangle $=4 \sqrt{3} \mathrm{cm}$
Volume of cylinder $=\pi * r^{2} * h=\pi * 48 * 8=384 \pi \mathrm{sq} . \mathrm{cm}$
The cone will have radius, $r=4 \sqrt{3} \mathrm{cm}$ and height, $\mathrm{h}^{\prime}=4 \mathrm{cm}$
Volume of cone $=\frac{1}{3} * \pi r^{2} * h=\frac{1}{3} * \pi 48 * 4=64 \pi \mathrm{sq} . \mathrm{cm}$
So $V=384 \pi-2^{*} 64 \pi=256 \pi$ sq. $\mathrm{cm}$
since the required volume is half of the volume we get by rotating the figure by 360 .
required volume $=0.5+256 \pi=128 \pi \mathrm{sq} \mathrm{cm}$
Hence, option $C$ is the right answer.

The required volume would be half of the volume one would get by rotating the figure by $360^{\circ}$
Let $V^{\prime}$ be the volume of the figure when the triangle is rotated by $360^{\circ}$
So volume $V=$ Volume of the cylinder $-2^{\star}$ Volume of cone (base shown in dotted line
The cylinder will have a height, $h=8 \mathrm{cm}$ and radius, $r=$ altitude of the equilateral triangle $=4 \sqrt{3} \mathrm{cm}$
Volume of cylinder $=\pi * r^{2} * h=\pi * 48 * 8=384 \pi \mathrm{sq} . \mathrm{cm}$
The cone will have radius, $r=4 \sqrt{3} \mathrm{cm}$ and height, $\mathrm{h}^{\prime}=4 \mathrm{cm}$
Volume of cone $=\frac{1}{3} * \pi r^{2} * h=\frac{1}{3} * \pi 48 * 4=64 \pi \mathrm{sq} . \mathrm{cm}$
So $V=384 \pi-2^{*} 64 \pi=256 \pi$ sq. $\mathrm{cm}$
since the required volume is half of the volume we get by rotating the figure by 360 .
required volume $=0.5+256 \pi=128 \pi \mathrm{sq} \mathrm{cm}$
Hence, option $C$ is the right answer.
Question 4
4 / -1

Raghu, Raghav and Raja ran on a racetrack at their respective constant speed. Raghu finished the race when Raghav was 200m behind and Raja was 350m behind. Raghav finished the race when Raja was 180 m behind. What is the length of the race track in meters?

A
1100

B
1200

C
1300

D
1400

Solution

Let the length of the race track be 'd' metres and speed of Raghu, Raghav and Raja be $x, y$ and z respectively.
We have, $\frac{z}{y}=\frac{d}{d-200}$ and $\frac{x}{z}=\frac{d}{d-350}$
So, $\frac{y}{x}=\frac{d-200}{d-350}$
Further, $\frac{y}{z}=\frac{d}{d-180}$
So we have, $\frac{d-200}{d-350}=\frac{d}{d-180}$
$d *(d-350)=(d-200) *(d-180)$
$d^{2}-350 d-d^{2}-180 d-200 d+200 * 180$
$30 d=200 * 180$
$d=1200 \mathrm{m}$
Thus the length of the track is $1200 \mathrm{m}$
Question 5
4 / -1

Two points are selected at random on the curve x² + y²= 49x. Find the probability that the length of the minor arc between the two points is not greater than 11 units.

A
0.25

B
0.17

C
0.34

D
0.50

Solution

The given curve is a circle of radius 7 units.

The circumference of the circle = 2 * π∗r = 44 units.

Now, let’s select any point x on the circumference of the circle.

The second point can be selected at any point which is at a distance of 11 units to the left of x or 11 units to the right of x.

Thus, out of 44 units 22 units is the required space.

44/ 22= 0.5
Question 6
4 / -1

A shopkeeper cheated a wholesaler, from whom he buys the goods, both in terms of quantity and price (by fraudulently paying less than market price for a given quantity) by 20% each. The same shopkeeper cheated the customer both in terms of quantity and price by 40% each. The shopkeeper claims to sell at no profit - no loss. What is the actual profit percentage for the shopkeeper?

A
250%

B
112.5%

C
125$

D
225%

Solution

Let the cost price for ‘y’ kg of goods be ‘x’ Rs.

We know that the shopkeeper cheated to wholesaler both in terms of quantity and price by 20% each.

So for the shopkeeper, 1.2y kg of goods cost 0.8x Rs.

Further, the shopkeeper claims to sell at no profit no loss but cheats the customer both in terms of quantity and price by 40% each.

Thus the shopkeeper sells 0.6 kg of goods at 1.4x Rs.

So the shopkeeper sells 1.2 kg of goods at 2.8x Rs.

Profit percentage for the shopkeeper = {2.8x - 0.8x} / {0.8x}*100

= 250% profit

Hence, option A is the right choice.
Question 7
4 / -1

Ram earned Rs. 20000 as an additional allowance. 25% of this income will have to be deducted as income tax. But if Ram deposit Rs. 12000 in the tax saving instrument which fetches 14% interest he would get a tax rebate of 11% on the original tax amount thus reducing his tax liability. What will be the approx. effective interest rate earned by Ram on the tax saving instrument?

A
19.45%

B
18.58%

C
17.63%

D
19.33%

Solution

Original tax to be paid on additional allowance = 0.25 * 20000 = Rs. 5000

Interest fetched on tax saving instrument = 0.14 * 12000 = Rs. 1680

Tax rebate = 0.11 * 5000 = Rs. 550

So total amount received = interest on tax saving instrument + tax rebate

= 1680 + 550 = Rs. 2230

Effective interest rate on tax saving instrument = 2230 / 12000 * 100 = 18.58%

Hence, option B is the right answer.
Question 8
4 / -1

What will be the remainder when 761376137613…..(578 digits) is divided by 625?

A
113

B
11

C
126

D
512

Solution

10,000 is divisible by 625.

The given number can be written as a multiple of 10,000 + last 4 digits.

The number of digits in the given number is 578

Had the number of digits been 576, the last 4 digits would have been 7613.

Since the number of digits is 578, the last 4 digits will be 1376.

As 10,000 is always divisible by 625, the last 4 digits will be 1376.

1376 on division by 625 leaves a remainder of 126.

Therefore, option C is the right answer.
Question 9
4 / -1

Find the area enclosed by the curves, given by (x - a)² + (y - b)² = 49(x−a) , where a² = 49

Assume that π =22/7

A
40

B
44

C
42

D
46

Solution

Given
$a^{2}=49$ and $b^{2}=49,$ Thus, $a=\pm 7$ and $b=\pm 7$
There will be 4 circles, each of radius 7 units. The area enclosed will be as shown in the given figure.

The enclosed area $=4 *[\operatorname{ar}(O B A C) \cdot \operatorname{ar}(\operatorname{sector}(A B C))]$
$=4 *\left[7^{2}-(0.25 * \pi * 49)\right]$
$=4 *[49-(0.25 * \pi * 49)]$
$=42$ units
Question 10
4 / -1

Four points are arranged in space such that distance between any two of the points is 6m. Find the volume of the solid formed by joining all the four points. (in cubic metres)

A
36√3

B
18√2

C
72√3

D
180

Solution

When four points are arranged such that they are equidistant, we get a equilateral triangle in a plane and point above the plane which is at a distance equal to the side of the triangle from all the three points. (This is a regular tetrahedron).

Now, just like how the volume of a cone is one-third that of a cylinder, the area of the tetrahedron will be one third the volume of a prism.

Volume of a prism = base area * height

Volume of tetrahedron = 1/3 * base area * height

Let a be the side of the equilateral triangle.

The distance from vertex to centroid $=\frac{2}{3} *$ altitude of triange $=\frac{2}{3} * \frac{\sqrt{3}}{2} * a=\frac{a}{\sqrt{3}}$

Let the height of the tetrahedron $=h$
Using Pythagoras theorem, $\frac{a^{2}}{3}+h^{2}=a^{2}$
$h=\sqrt{\frac{2}{3}} a$
Volume $=\frac{1}{3} * \frac{\sqrt{3}}{4} * a^{2} * \sqrt{\frac{2}{3}} a$
$=\frac{a^{3}}{6 \sqrt{2}}$
$a=6$
Hence, volume $=18 \sqrt{2}$

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Quantitative Ability Test - 12

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Quantitative Ability Test - 12

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SHARING IS CARING

Question 1

10 + 5√2

24 + 8√3

32 + 4√3

16 + 5√2

Solution

Question 2

24

25

28

27

Solution

Question 3

256 π sq.cm.

112 π sq.cm.

128 π sq.cm.

224 π sq.cm.

Solution

Question 4

1100

1200

1300

1400

Solution

Question 5

0.25

0.17

0.34

0.50

Solution

Question 6

250%

112.5%

125$

225%

Solution

Question 7

19.45%

18.58%

17.63%

19.33%

Solution

Question 8

113

11

126

512

Solution

Question 9

40

44

42

46

Solution

Question 10

36√3

18√2

72√3

180

Solution

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Question Analysis

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