Quantitative Ability Test

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Quantitative Ability Test - 11

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Weekly Quiz Competition

Question 1
4 / -1

If 2 log (x-1) = log 162log(x−1)=log16, then the sum of the possible values that xx can take is

A
2

B
8

C
-3

D
5

Solution

log(x−1)²=log16

=> (x-1)^2 = 16

=> (x-1) =±4

x=5 or -3

The actual function 2 log (x-1) 2log(x−1) is not defined for x = -3x=−3. Therefore, the only value that xx can take is 55. Therefore, option D is the right answer.
Question 2
4 / -1

Ben bought some pencils, erasers and pens. The cost of each pencil, eraser, and pen is Rs. 16, Rs. 20, and Rs. 22 respectively. The number of erasers bought is not less than a quarter of pencils and the number of pens is at least one-sixth of the pencils bought. Ben selected each item in such a way that the average cost of all the items he bought is minimized. Which of the following can be the number of total items bought?

A
45

B
52

C
68

D
71

Solution

The cost of a pencil, eraser, and pen is 16, 20, and 22 respectively.

Let the number of pencils bought be 12x.

The number of erasers bought is greater than or equal to 3x.

The number of pens bought is greater than or equal to 2x.

x need not necessarily be an integer.

But, if x is not an integer, the number of erasers or pens will be greater than 3x or 2x respectively. In such a case, the average cost will not be minimized.

Thus, to minimize the average cost, x has to be an integer.

Total number of items bought = 12x + 3x + 2x = 17x

Out of the given options, only 68 is a multiple of 17. Hence, option C is the correct answer.
Question 3
4 / -1

Karan sells rose milk packets. By the end of each day, 25% of the packets with him get spoiled and hence, he disposes them. He starts with 100 packets. He sells 20 packets of rose milk every day except the last day in which he sells 30 packets. By what percentage must he mark up the price of the packet, if he wants to end up with a profit of 26%?

A
50%

B
60%

C
70%

D
80%

Solution

In the first day, he would have sold 20 packets.

Of the 80 packets remaining, 25% (20 packets will get spoilt) and 60 packets will be remaining.

In the second day, he would have sold 20 packets.

40 pockets will remain by the end of the day of which 10 packets will get spoiled.

On the third day, he will sell all the 30 packets since it has been provided that he sells 30 packets on the last day.

Total number of packets sold = 70

Let the cost of 1 packet be x.

Total cost incurred = 100x.

Let the mark up percentage be y.

(1+y)*70x = 1.26*100x

1+y = 126/70

1+y = 1.8

=> y = 80%.

Therefore, the price of the packets must be marked up by 80% to get a profit of 26%. Therefore, option D is the right answer.
Question 4
4 / -1

Given, a =√5 + 1 . And ,a⁴−2a²−2a−24=na, find the value of n

A
18

B
17

C
15

D
14

Solution

\(a=\sqrt{5}+1\)
\((a-1)^{2}=5\)
\(a^{2}=4+2 a\)
\(a^{4}=16+4 a^{2}+16 a\)
\(a^{4}-2 a^{2}-2 a-24=2 a^{2}+14 a-8\)
\(a^{4}-2 a^{2}-2 a-24=8+4 a+14 a-8\)
\(a^{4}-2 a^{2}-2 a-24=18 a\)
Question 5
4 / -1

To go from point A to point B in a city public bus and private bus service are available. The speed of the public bus is 2.5 times that of the private bus. The public bus can carry 50 passengers while the private bus can carry 70 passengers. The average occupancy of the private bus service is twice that of the public bus service. The tickets for the private bus and the public bus cost Rs. 25 and Rs. 15 respectively. There is no wastage of time between trips throughout the day for both the bus services. What is the ratio of the average revenue of the private bus service to that of public bus service?

A
18 / 11

B
14 / 19

C
36 / 19

D
28 / 15

Solution

Since there is no wastage of time for both the buses, the ratio of the number of trips would be equal to the ratio of the speed of the buses.

The revenue of the bus service would be directly proportional to the speed, capacity, occupancy and ticket cost.

Thus the required ratio = speed ratio * capacity ratio * average occupancy ratio * cost ratio

= {70 * 2 * 25} / {2.5 * 50 * 15} = 28 / 15

Hence, option D is the right choice.
Question 6
4 / -1

Arun mixes coffee and chicory in the ratio 5:3. Then, he sells each unit of the mixture at a price 10% above the cost price of one unit of coffee powder. If Arun realizes a profit of 28% by doing so, by what percentage is the cost price of coffee costlier than the cost price of chicory?

A
50%

B
60%

C
75%

D
40%

Solution

Let the cost of coffee be "x' and chicory be \(y^{\prime}\)
cost price of the mixture \(=5 x+3 y\)
Selling price of the mixture \(=1.1+8 x=8.8 x\)
\(\Rightarrow\) Profit \(=8.8 x-5 x-3 y=3.8 x-3 y\)
Profit percentage \(=28 \%=0.28\)
\(\Rightarrow(3.8 x-3 y) /(5 x+3 y)=0.28\)
\(3.8 x-3 y=1.4 x+0.84 y\)
\(2.4 x=3.84 y\)
\(\Rightarrow x=1.6 y\)
Therefore, the CP of coffee powder is \(60 \%\) more than the CP of Chicory powder. Hence, option \(B\) is the right answer.
Question 7
4 / -1

A password for a security application consists of seven characters, where the first four characters are alphabets and the last three characters are digits. The digits are positive and repetition is allowed. The alphabets are not repeated. Find the number of passwords that can be formed.

A
261565200

B
261565300

C
261599200

D
26150000

Solution

The password is of the form

__ __ __ __ __ __ __

The first four alphabets can be arranged in ²⁶C_₄ * 4!= 358800

The last three characters can be written in 9 * 9 * 9 = 729 ways (9 positive digits and repetition is allowed)

Thus, the number of passwords = 358800 * 729= 261565200
Question 8
4 / -1

A milkman procures milk at some price. He mixes water with milk in the ratio 3:5 and then he announces that he will sell the milk only at a profit of 35%. If he uses a 900 ml cup instead of a 1000 ml one to sell the milk, then the overall profit percentage of the milkman is (Assume that water is available free of cost)

A
220%

B
210%

C
140%

D
175%

Solution

Let us assume that the milkman procures 125 litres of milk for ease of calculation. Let us assume the cost of 1 litre of milk to be Rs. 1.

He will add 75 litres of water to it (Since he mixes water to milk in the ratio 3:5).

He marks up the price of the milk by 35%.

Therefore, he will sell a litre of the mixture at Rs. 1.35.

He uses a 900 ml jar instead of a 1000 ml one.

Therefore, he will sell the 200 litres as 200/0.9 = 222.22 litres.

Total revenue = 222.22*1.35 = Rs. 300

Profit = Rs. 300 - Rs. 125 = 175

Cost = Rs. 125

Therefore, profit percentage = 175/125 = 140%. Hence, option C is the right answer.
Question 9
4 / -1

3 circles of equal size are tightly packed inside a larger circle such that the 3 circles touch each other externally and touch the larger circle internally. If the diameter of the smaller circle is ‘r’, then the area outside the smaller circles but inside the larger circle is

A
\(\pi r^{2}\left(\frac{2 \sqrt{3}-5}{3}\right)\)

B
\(\frac{\pi^{2}}{4}\left(\frac{4 \sqrt{3}-2}{3}\right)\)

C
\(\frac{\pi^{2}}{4}\left(\frac{2 \sqrt{3}-4}{3}\right)\)

D
\(\frac{\pi r^{1}}{4}\left(\frac{2 \sqrt{3}-13}{3}\right)\)

Solution

The radius of the smaller circle is \(\frac{T}{2}\) since it has been given that the diameter is \(r\)

On joining the radil of the 3 circles, we get an equilateral triangle with side \(r\)
The incentre of this equilateral triangle will also be the centre of the larger circle.
Inradius of the equilateral triangle \(=\frac{r}{2 \sqrt{3}}\) Radius of the outer circle \(=\) Radius of the smaller circle \(+\) Circumradius of the triangle.
\(\Rightarrow\) Radius of the outer circle \(=\frac{r}{2}+\frac{r}{\sqrt{3}}\) Area of the larger circle \(=\pi *\left[\frac{r}{2} *\left(1+\frac{2}{\sqrt{3}}\right)\right]^{2}\)
\(=\frac{\pi r r^{2}}{4} *\left(1+\frac{4}{3}+\frac{4}{\sqrt{3}}\right)\)
\(=\frac{\pi+r^{2}}{4} *\left(\frac{7+4 \sqrt{3}}{3}\right)\)
Area of the smaller circles \(=3+\frac{\pi^{2}}{4},\) since \(r\) is the diameter here. Difference between the areas \(=\frac{\pi r^{2}}{4}\left(\frac{7+4 \sqrt{3}}{3}-3\right)\) \(=\frac{\pi r^{2}}{4}\left(\frac{4 \sqrt{3}-2}{3}\right)\)
Therefore, option \(B\) is the right answer.
Question 10
4 / -1

Given, xx and yy are positive real numbers such that x + y = 4. The minimum value of the expression (x +1/x)² + (y +1/y)² is n. Find 4n.

A
50

B
54

C
56

D
51

Solution

We have to find the minimum value of the expresseion \(\left(x+\frac{1}{x}\right)^{2}+\left(y+\frac{1}{y}\right)^{2}\) \(x+y=4\)
\(\left(x+\frac{1}{x}\right)^{2}+\left(y+\frac{1}{y}\right)^{2}=x^{2}+y^{2}+\frac{1}{x^{2}}+\frac{1}{y^{2}}+4\)
We know.
since \(x\) and \(y\) are positive, squaring both sides,
\(x^{2}+y^{2} \geq 8\)
Also,
We know,
\(\frac{2}{t+\frac{1}{2}} \leq \frac{x+y}{2}\)
\(\frac{1}{x}+\frac{1}{y} \geq 1\)
Substituting in ii
\(\frac{\frac{1}{x^{2}}+\frac{1}{x^{2}}}{2} \geq 0.5\)
Using this in
\(\left(x+\frac{1}{z}\right)^{2}+\left(y+\frac{1}{y}\right)^{2}=x^{2}+y^{2}+\frac{1}{x^{2}}+\frac{1}{y^{2}}+4\)
\(\left(x+\frac{1}{x}\right)^{2}+\left(y+\frac{1}{y}\right)^{2} \geq 8+4+0.5\)
\(4 n=4 * 12.5=50\)

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Re-Attempt Weekly Quiz Competition

Quantitative Ability Test - 11

Result

Quantitative Ability Test - 11

Score

-

Rank

-

SHARING IS CARING

Question 1

2

8

-3

5

Solution

Question 2

45

52

68

71

Solution

Question 3

50%

60%

70%

80%

Solution

Question 4

18

17

15

14

Solution

Question 5

18 / 11

14 / 19

36 / 19

28 / 15

Solution

Question 6

50%

60%

75%

40%

Solution

Question 7

261565200

261565300

261599200

26150000

Solution

Question 8

220%

210%

140%

175%

Solution

Question 9

\(\pi r^{2}\left(\frac{2 \sqrt{3}-5}{3}\right)\)

\(\frac{\pi^{2}}{4}\left(\frac{4 \sqrt{3}-2}{3}\right)\)

\(\frac{\pi^{2}}{4}\left(\frac{2 \sqrt{3}-4}{3}\right)\)

\(\frac{\pi r^{1}}{4}\left(\frac{2 \sqrt{3}-13}{3}\right)\)

Solution

Question 10

50

54

56

51

Solution

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