Quantitative Ability Test

Result

Quantitative Ability Test - 14

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
4 / -1

Eighth term of an AP is zero. What is the ratio of the 38th term and the 18th term of that AP?

A
2

B
3

C
1

D
0

Solution

8th term is 0
\(\Rightarrow a+7 d=0\)
\(\frac{38 \text { th term }}{18 \text { th term }}=\frac{a+37 d}{a+17 d}\)
\(=\frac{a+7 d+30 d}{a+7 d+10 d}\)
Putting \(a+7 d=0\)
\(=\frac{30 d}{10 d}=3\)
Hence, option \(\mathrm{B}\) is the correct answer.
Question 2
4 / -1

Ram painted fours walls of a cubic-hall using Red, Blue and Yellow colour in such a manner that he painted one wall each by Red, Blue, Yellow and Orange color. The orange colour is made by mixing Red and Yellow in a fixed ratio. If Ram had 150 litres of Red, 170 litres of Yellow and 115 litres of Blue paint before painting and He’s left with the same quantity of each type of color after the process then find out the total quantity(in litres) of paint left with him?

A
75

B
79

C
73

D
69

Solution

Ram will be left with the same quantity of each type only when an excess quantity of Red Paint and Yellow Paint as compare to Blue paint is used to make Orange colour.

So total quantity of orange colour consumed = Part that came from Red colour + Part that came from the Yellow colour = (150 - 115) + (170 - 115) = 90 litres

Since the area of each wall is same hence the same quantity of each colour of paint must have consumed.

Quantity of Red paint left = Quantity of Blue paint left = Quantity of Yellow paint left = 115 - 90 = 25 litres

Hence total quantity of paint left with Ram = 3*25 = 75 litres.
Question 3
4 / -1

A vessel is in the form of a cylinder mounted on a hemisphere. The diameter of the hemisphere is \(18\) cm and the ratio of the height of cylinder and radius of the hemisphere is \(2:1\). Find the volume of the vessel.

A
\(1944 \pi\)

B
\(1844 \pi\)

C
\(1954 \pi\)

D
\(1664 \pi\)

Solution

Given, diameter of hemishphere \(=18 {~cm}\)

Radius of vessel \(=9 {~cm}\)

Radius of hemisphere \(=\) Radius of cylinder \(=9 {~cm}\)

Ratio of total height of cylinder: Radius of hemispherre \(=2: 1\)

Height of cylinder \(=2 \times 9=18 {~cm}\)

Total volume of vessel = Volume of cylinder + volume of hemisphere \(=\pi r^{2} h+\frac{2}{3} \pi r^{3}\)

\(=\pi \times 9 \times 9 \times 18+2 \pi \times 3 \times 9 \times 9\)

\(=\pi \times 9 \times 9 \times 18+\pi \times 6 \times 9 \times 9\)

\(=\pi \times 9 \times 9 \times 24\)

\(=1944 \pi\)
Question 4
4 / -1

A, B and C started working together on a project, but A left the work 9 days before the completion and B left the work 7 days before completion. Had A worked for one more day, the work would have been completed two days before the actual time. Had B worked for two days more, the work would have been completed three days before the actual time. In how many days can A, B and C together do a different work which can be done by C alone in 18 days?

A
4

B
-4

C
5

D
-7

Solution

If A works for one more day, the total number of days decreases by two. This means that C’s working days reduce by two. Thus, A is twice as efficient as C.

Similarly, if B works for two more days, C’s working days reduce by three.

Thus, B is 1.5 times more efficient than C.

The efficiencies of A, B and C are in the ratio 4 : 3 : 2.

New work is done by C in 18 days

=> New work = 2 * 18 = 36 units

(A + B + C)’s 1 day’s work = (4 + 3 + 2) = 9 units

Number of days required by all of them to complete the work = 36/9 = 4 days

Hence, the correct answer is 4 days.
Question 5
4 / -1

Function \(f(x)\) is defined as \(f(x)=x^{5}-3\). If \(f(a)=0\) then find the value of \((a-1)\left(a^{15}+a^{16}+a^{17}+\ldots+a^{44}\right) ?\)

A
21600

B
2160

C
19656

D
6534

Solution

Given that \(f(x)=x^{5}-3\)
\(f(a)=0\) i.e. a is a root of polynomial \(x^{5}-3=0\) therefore we can say that \(a^{5}-3=0\) or \(a^{5}=3\)
For finding out the value of
\(\Rightarrow(a-1)\left(a^{15}+a^{16}+a^{17}+\ldots+a^{44}\right)\)
\(\Rightarrow(a-1) \frac{a^{13}\left(a^{30}-1\right)}{(a-1)}\)
\(\Rightarrow a^{15}\left(a^{30}-1\right)\)
\(\Rightarrow\left(a^{5}\right)^{3}\left(\left(a^{5}\right)^{6}-1\right)\)
\(\Rightarrow(3)^{3}\left((3)^{6}-1\right)\)
\(\Rightarrow 27(729-1)\)
\(\Rightarrow 19656\) Hence option \(\mathrm{C}\) is correct answer.
Question 6
4 / -1

Two circles with radii r₁ and r₂are drawn such that a triangle with sides 35 cm, 37 cm and 12 cm is inscribed and circumscribed by the two circles. What is the ratio of the areas of the two circles (assume that the ratio is greater than 1)

A
11.56

B
13.69

C
7.29

D
10.24

Solution

We can see that \(37^{2}=12^{2}+35^{2}\)

Hence given triangle is an right-angled triangle with side \(37 \mathrm{cm}\) as hypotenuse

We are given that \(r_{2}>r_{1}\) therefore we can say that \(r_{1}\) is in-radius and \(r_{2}\) is
circumradius.
Therefore \(r_{1}=\frac{\text { Area of trinagle }}{4} \quad\) (Area of the given triangle \(=\frac{1}{2} \times 35 * 12=210\)
\(\mathrm{sq}-\mathrm{cm}\) and \(\mathrm{s}=\frac{35+12+37}{2}=42 \mathrm{cm}\)
\(\Rightarrow r_{1}=\frac{210}{42}=5 \mathrm{cm}\)
Circumradius of the triangle \(r_{2}=\frac{a+b+c}{4 \times A r e a}\) of triangle
\(\Rightarrow r_{2}=\frac{12+35 \cdot 37}{4 \times 210}=18.5 \mathrm{cm}\)
Therefore the ratio of areas of circles with radius \(r_{2}\) to radius \(r_{1}\)
\(\Rightarrow \frac{\pi r_{0}^{2}}{\pi r_{1}^{2}}\)
\(\Rightarrow\left(\frac{18.5}{5}\right)^{2}\)
\(\Rightarrow 13.69\)
Therefore option B is the correct answer.
Question 7
4 / -1

Suppose that \(a=\frac{2 b}{3}\) and \(a^{b}=b^{a} .\) The sum of \(a\) and \(b\) can be expressed as a rational number \(\frac{P}{q},\) where \(p\) and \(q\) are co-prime positive integers. Find the value of \(p+q ?\)

A
35

B
91

C
47

D
53

Solution

Let us assume that \(a=k b\) where \(k\) can take any positive value.
We are given that \(a^{b}=b^{a}\)
\(\Rightarrow(k b)^{b}=b^{k b}\)
\(\Rightarrow\left(k^{b}\right) \times\left(b^{b}\right)=\left(b^{b}\right)^{k}\)
\(\Rightarrow(k)^{b}=\left(b^{b}\right)^{k-1}\)
Taking \(b^{t h}\) root each side
\(\Rightarrow k=(b)^{k-1}\)
\(\Rightarrow b=(k)^{\frac{1}{k-1}}\)
Bu substituting \(k=\frac{2}{3}\)
\(\Rightarrow b=\left(\frac{2}{3}\right)^{-3}\)
\(\Rightarrow b=\left(\frac{3}{2}\right)^{3}\)
\(\Rightarrow b=\frac{27}{8}\)

Hence \(a=\frac{2}{3} \times \frac{27}{8}\)
\(\Rightarrow a=\frac{9}{4}\)
We can calculate \(a+b=\frac{9}{4}+\frac{27}{8}\)
\(\Rightarrow a+b=\frac{45}{8}=\frac{p}{q}\)
Hence \(p+q=45+8=53,\) Therefore option \(\mathrm{D}\) is the correct answer.
Question 8
4 / -1

A recently conducted study suggests that 6% of mobile users suffer from migraine and 60% of migraine patients are mobile users. If 90% of the population uses mobile phones, what is the percentage of population suffering from migraine?

A
8

B
9

C
10

D
5

Solution

Let us assume that total population = 100x

Then number of mobile phone users = 90 / 100 *100x = 90x

Number of mobile phone users suffering from migraine = 6 / 100×90x=5.4x

It is given that 80% of migraine patients are mobile users then we can say that

Total patients suffering from migraine 5.4x / 0.6= 9x

Hence we can say that 9% population is suffering from migraine. Hence B is the correct answer.
Question 9
4 / -1

n the given diagram O is the centre of the circle and angle∠ABO+angle∠ACO = 55°. All the points A, B and C lie on the circumference of the circle. Find the value of angle∠OCB?

A
37.5

B
45

C
35

D
42.5

Solution

We know that angle made by a chord on centre is twice the angle made by same chord on circumference.Therefore

angle∠BOC = 2angle∠BAC

Let us assume that angle∠BAC = x° then angle∠BOC = 2x°

Then external \(\angle \mathrm{BOC}=360-2 \mathrm{x}^{\circ}\)
In quadrilateral ABOC,
\(\Rightarrow \angle A B O+\angle B O C+\angle O C A+\angle C A B=360^{\circ}\)
\(\left.\Rightarrow \angle A B O+360-2 x^{2}+\angle O C A+x^{*}=360^{\circ} \quad \text { (It is given that } \angle A B O+\angle A C O=55^{\circ}\right)\)
\(\Rightarrow 55^{\circ}+360-2 x^{\circ}+x^{\circ}=360^{\circ}\)
\(\Rightarrow x^{\circ}=55^{\circ}\)
Therefore \(\angle \mathrm{BOC}=2^{*} 55^{\circ}=110^{\circ}\)
\(\ln \triangle \mathrm{BOC}\)
\(\angle \mathrm{OBC}+\angle \mathrm{BCO}+\angle \mathrm{COB}=180^{\circ}\)
We know that \(\mathrm{OB}\) and \(\mathrm{OC}\) are the radius of circle therefore \(\angle \mathrm{OBC}=\angle \mathrm{BCO}\)
\(\Rightarrow 2 \angle B C O+110^{\circ}=180^{\circ}\)
\(\Rightarrow \angle B C O=35^{\circ}\)
Hence option \(C\) is the correct answer.
Question 10
4 / -1

Ram drives a car at fixed speed and stops for a five-minute break at the end of every 2 km. Shyam drives a motorcycle at three-fourths the speed of Ram and stops for a five-minute break at the end of every 4 km. If both started at the same time and reach 100 km at the exact same time then find out the time(in minutes) taken by each of them?

A
620

B
612

C
630

D
600

Solution

Let us assume that time taken to reach \(100 \mathrm{km}\) is \(^{\prime} t^{\prime}\) minutes.
Ram stops once after every \(2 \mathrm{km}\). Therefore he will stop 49 times before completing \(100 \mathrm{km}\)
Total time consumed by Ram in breaks \(=49^{*} 5=245\) minutes.
Hence we can say that Ram actually drove for \(^{\prime} t-245^{\prime}\) minutes.
Hence distance covered by Ram in one minute \(=\frac{100}{t-245} \mathrm{km}\)

Similarly, Shyam will stop 24 times before reaching 100 km mark.
Total time consumed by Shyam in breaks \(=24^{*} 5=120\) minutes.
Hence we can say that Shyam actually drove for \(^{\prime} t-120^{\prime}\) minutes.
Hence distance covered by Shyam in one minute \(=\frac{100}{t-120} \mathrm{km}\)
It is given that Speed of Shyam is \(\frac{3}{4}\) speed of Ram. Therefore,
\(\Rightarrow \frac{100}{t-120}=\frac{3}{4} \times \frac{100}{t-245}\)
\(\Rightarrow 4 t-980=3 t-360\)
\(\Rightarrow t=620\) minutes
Hence we can say that Both Ram and shyam will reach \(100 \mathrm{km}\) mark in 620 minutes.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Quantitative Ability Test - 14

Result

Quantitative Ability Test - 14

Score

-

Rank

-

SHARING IS CARING

Question 1

2

3

1

0

Solution

Question 2

75

79

73

69

Solution

Question 3

\(1944 \pi\)

\(1844 \pi\)

\(1954 \pi\)

\(1664 \pi\)

Solution

Question 4

4

-4

5

-7

Solution

Question 5

21600

2160

19656

6534

Solution

Question 6

11.56

13.69

7.29

10.24

Solution

Question 7

35

91

47

53

Solution

Question 8

8

9

10

5

Solution

Question 9

37.5

45

35

42.5

Solution

Question 10

620

612

630

600

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.