Quantitative Ability Test

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Quantitative Ability Test - 10

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Weekly Quiz Competition

Question 1
4 / -1

Two circles are drawn such that the smaller circle touches the bigger circle internally at one point. Triangle PQR is drawn such that it circumscribes the smaller triangle. Side RQ is tangent to both the circle and sides PR and PQ intersect each other at the centre of the bigger circle. Side RP is extended to touch the bigger circle at point T such that RT = 20 cm. What is the radius of the smaller circle if the radius of the bigger circle is 7.5 cm?

A
10 /3 cm

B
3.5 cm

C
5 cm

D
17 / 6 cm

Solution

RT = 20 cm and DT = 2 * 7.5 = 15 cm
RD = RT - DT = 20 - 15 = 5 cm
By tangent secant theorem,
RB²=RD∗RT = 5 * 20 = 100
RB = 10 cm
Thus, RA = 10 cm
RP = RD + DP = 5 + 7.5 = 12.5
AP = RP - RA = 12.5 - 10 = 2.5 cm
For inner circle by tangent secant theorem,

AP²=PE∗PB
2.5²=PE∗7.5
PE = 5/6 cm
Diameter of the smaller circle = EB = PB - PE = 7.5 - 5/6 = 20/3 cm
Required radius = CB = 10/3 cm
Hence, option A is the right answer.
Question 2
4 / -1

Raghu bought 3 oranges, 2 apples and a banana and paid 20 rupees. If it is known that the prices of the fruits are positive integers, how many different combinations of prices is possible?

A
20

B
21

C
24

D
25

Solution

Let the price of an orange be \(^{4} x\), the price of an apple be 'a' and the price of a banana
be 'b".
We know that \(3 x+2 a+b=20\)
\(x\) can take 6 values -1,2,3,4,5 and \(6 .\) Let us check the number of possible cases
under each of the case.
When \(x=1\)
\(3+2 a+b=20\)
\(2 a+b=17\)
\(a=(17-b) / 2\)
As we can see, b can take all odd values between 1 and 15 (both inclusive). There are
8 such values.
When \(x=2\)
\(6+2 a+b=20\)
\(2 a+b=14\)
\(a=(14-b) / 2\)
b can take all even values between 2 and 12 (both inclusive). There are 6 such values.
When \(x=3\)
\(9+2 a+b=20\)
\(2 a+b=11\)
\(a=(11-b) / 2\)
b can take all odd values from 1 to 9 (both inclusive). There are 5 such values.
When \(x=4\)
\(12+2 a+b=20\)
\(2 a+b=8\)
\(a=(8-b) / 2\)
b can take all even values from 2 to 6 (both inclusive). There are 3 such values.
When \(x=5\)
\(15+2 a+b=20\)
\(2 a+b=5\)
\(a=(5-b) / 2\)
b can take 2 values \((1 \text { and } 3)\)
When \(x=6\)
\(18+2 a+b=20\)
\(2 a+b=2\)
\(a=(b-2) / 2\)
There are no values that \(b\) can take.
Therefore, the total number of solution sets is \(8+6+5+3+2=24 .\) Hence, option \(\mathrm{C}\)
is the right answer.
Question 3
4 / -1

If \(T=\frac{3}{5}+\frac{6}{5^{2}}+\frac{11}{d^{7}}+\frac{18}{5^{4}}+\frac{27}{5^{7}}+\frac{38}{5^{6}}+\ldots\)
What is the value of \(T ?\)

A
102 / 107

B
70 /77

C
155 / 172

D
31 / 32

Solution

\(\mathrm{T}=\frac{3}{5}+\frac{6}{5^{2}}+\frac{11}{5^{2}}+\frac{18}{5^{4}}+\frac{27}{5^{5}}+\frac{38}{5^{6}}+\ldots\)
Substitute \(\frac{1}{5}=x\)
\(T=3 x+6 x^{2}+11 x^{3}+18 x^{4}+27 x^{5}+38 x^{6}+\dots\)
\(T^{*} x=3 x^{2}+6 x^{3}+11 x^{4}+18 x^{5}+27 x^{6}+38 x^{7}+\ldots\)
Subtract equation (2) from equation (1)
\(T^{*}(1-x)=3 x+3 x^{2}+5 x^{3}+7 x^{4}+9 x^{5}+11 x^{6}+\dots\)
\(T^{*}(1-x) x=3 x^{2}+3 x^{3}+5 x^{4}+7 x^{5}+9 x^{6}+11 x^{7}+\dots\)
Subtract equation
(4) from equation (3)
\(T^{*}(1-x)^{*}(1-x)=3 x+0+2 x^{3}+2 x^{4}+2 x^{5}+2 x^{6}+\dots\)
\(T^{*}(1-x)^{*}(1-x)=3 x+2\left(x^{3}+x^{4}+x^{5}+x^{6}+\ldots\right)\)
\(T^{*}(1-x)^{*}(1-x)=3 x+2\left(\frac{x^{2}}{1-z}\right)\)
Resubstitute \(x=\frac{1}{5}\)
\(T^{*} \frac{4+4}{5 * 5}=\frac{3}{5}+2 * \frac{1 / 5^{4}}{4 / 5}\)
\(T=\frac{15}{16}+2 \cdot \frac{1}{61}\)
\(T=\frac{15}{16}+\frac{1}{32}\)
\(T=\frac{15}{16}+\frac{1}{32}\)
\(T=\frac{31}{32}\)
Hence, option \(D\) is the right choice.
Question 4
4 / -1

15 employees started working on a project and worked on it for 15 days. From the sixteenth day onwards, a new employee joined the existing team every day till the project was finished. It took 25 days to finish the entire project. The same project needed to be completed starting with the minimum possible even number of employees. Considering this how many days will it take to complete the project if two employees quit the project every day starting from the second day ?

A
21

B
16

C
42

D
32

Solution

Smallest value of n for which this condition satisfies is n = 21

So to complete the project 21*2 = 42 employees must work on the first day.

But we must note that 42 +40 +38 + … + 12 = 432

So the work gets completed on the day when 12 employees are working.

Thus the number of days required to complete the work = 21 - 5 = 16 days

Hence, option B is the right answer.
Question 5
4 / -1

Bourne is relaxing in the Caribbean Islands in a hammock. The hammock is tied to two trees which are 10 meters apart at height of 4m above the ground. Due to his weight, the hammock sags and it is in the shape of the parabola. The centre of the hammock touches the ground. At distance of 2m from the right tree, find the height of the hammock above the ground.

[The general equation of a parabola is y = ax² + bx + c, where a, b, and c are real numbers and a≠0]

(Assume hammock as a thin rope for calculations)

A
2.5 m

B
0.64 m

C
2 m

D
1.44 m

Solution

Let's represent the given information in the Cartesian plane.

The general equation of a parabola is \(y=a x^{2}+b x+c,\) where \(a, b,\) and \(c\) are real numbers and \(a \neq 0\)
It passes through (0,0) . Hence, \(c=0\)
The minimum of a quadratic equation occurs at \(x=-\frac{b}{2 a}\). Here, it occurs at \(x=0 .\) Hence, \(b=0\)
The curve is of the form \(y=k x^{2}\)
Substituting the points,
\(4=25 k\)
\(k=0.16\)
\(y=0.16 x^{2}\)
\(2 \mathrm{m}\) from the right end corresponds to \(x=3\)
\(y=0.16 * 9=1.44\)
Question 6
4 / -1

If \(f(x)=\frac{1}{x}, g(x)=\log x, h(x)=x^{2}\), then which of the following has the highest value at \(x=1734 ?\)

A
f ( g ( h ( x ) ) )

B
g ( f ( h ( x ) ) )

C
h ( f ( g ( x ) ) )

D
g ( h ( f ( x ) ) )

Solution
Question 7
4 / -1

A set X consists of 507 elements which are squares of natural numbers selected at random. List Y consists of elements that are remainders when the elements of set X are divided by 11. The least number of times the element that gets repeated the most can appear in the list Y is _______ .

A
46

B
81

C
85

D
47

Solution

The squares of natural numbers leave the remainders - 0, 1, 4, 9, 5 and 3 when divided by 11.
There are six remainders which are possible.
Even if we assume equal distribution of remainders, we get (84, 84, 84, 85, 85, 85)
Thus, we can definitely find at least 85 numbers which leave the same remainder when divided by 11.
Question 8
4 / -1

If \(b=\left[\log _{5} a\right]^{3}+12 *\left[\log _{5} a\right]-6 *\left[\log _{5} a\right]^{2}-8\) and \(a^{b}=125 .\) If \(a\) and \(b\) are positive numbers, what is the value of \(\log _{5} b ?\)

A
0

B
1

C
2

D
3

Solution

\(a^{b}=125\)
Taking log to the base 5 on both the sides, we get.
\(b=\log _{5} a=\log _{5} 125=3\)
Let \(\log _{5} a=r\)
So, \(b * r=3\)
We have, \(b=\left[\log _{5} a\right]^{3}+12 *\left[\log _{5} a\right]-6 *\left[\log _{5} a\right]^{2}-8\)
Using equation (1) and (2).
\(\frac{3}{r}=r^{3}+12 r-6 r^{2}-8\)
\(3=r *(r-2)^{3}\)
since \(r\) is an positive integer,
\(r=3\)
Thus, \(b=1\) and \(a=125\)
Hence, \(\log _{5} b=\log _{5} 1=0\)
Question 9
4 / -1

Amar and Balu run around a circular track. If they decide to run in the opposite directions, the distance covered by Amar will be 90 m more than the distance covered by Balu when they meet for the first time. The length of the track is 60 m more than the distance covered by Amar. What will be the difference between the distances covered by Amar and Balu when they meet for the first time had they decided to run in the same direction?

A
70 m

B
140 m

C
150 m

D
210 m

Solution

Let the distance covered by Balu be ‘b’, Amar be ‘a’ and the circumference of the ground be ‘d’.
When they meet for the first time, total distance covered = d.
=> a + b = d
Also, it has been given that a-b = 90
Length of track,d = a + 60
We know that a+b = d
=> b = 60 m and a = 150 m and length of the track = 210 m.
If 2 persons run in the same direction, then the difference between the distances run by them when they meet for the first time will be equal to the length of the track. Therefore, option D is the right answer.
Question 10
4 / -1

In a factory, 40% of the bulbs produced are blue in colour, 20% of the bulbs are green in colour and the rest are red in colour. 30% of the red bulbs are defective while the total percentage of defective bulbs is 24%. 80% of the blue bulbs are not defective. If the number of defective blue bulbs is 120, what is the number of non-defective green bulbs?

A
60

B
120

C
240

D
360

Solution

Let the total number of bulbs be ‘x’.
Number of blue bulbs = 0.4x
Number of green bulbs = 0.2x
Number of red bulbs = 1 - (0.4x + 0.2x) = 0.4x
Number of defective blue bulbs = (1-0.8)*0.4x = 0.2*0.4x = 0.08x
Number of defective red bulbs = 0.3*0.4x = 0.12x
Total number of defective bulbs = 0.24x
=> Number of defective green bulbs = Total number of defective bulbs - (Number of defective blue bulbs +Number of defective red bulbs)
Number of defective green bulbs = 0.24x - (0.12x + 0.08x) = 0.04x
=> Number of non-defective green bulbs = 0.2x - 0.04x = 0.16x
We know that number of defective blue bulbs = 0.08x = 120.
=> Number of defective green bulbs = 0.16x = 2*0.08x = 2*120 = 240.
Therefore, option C is the right answer.

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Quantitative Ability Test - 10

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Quantitative Ability Test - 10

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SHARING IS CARING

Question 1

10 /3 cm

3.5 cm

5 cm

17 / 6 cm

Solution

Question 2

20

21

24

25

Solution

Question 3

102 / 107

70 /77

155 / 172

31 / 32

Solution

Question 4

21

16

42

32

Solution

Question 5

2.5 m

0.64 m

2 m

1.44 m

Solution

Question 6

f ( g ( h ( x ) ) )

g ( f ( h ( x ) ) )

h ( f ( g ( x ) ) )

g ( h ( f ( x ) ) )

Solution

Question 7

46

81

85

47

Solution

Question 8

0

1

2

3

Solution

Question 9

70 m

140 m

150 m

210 m

Solution

Question 10

60

120

240

360

Solution

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