Quantitative Ability Test

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Quantitative Ability Test - 6

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Weekly Quiz Competition

Question 1
4 / -1

If 50% of a certain number is equal to 3/4th of another number, what is the ratio between the numbers?

A
3 : 2

B
2 : 5

C
5 : 2

D
3 : 4

Solution

First number = x Second number $=y$
$\therefore x \times \frac{50}{100}-y \times \frac{3}{4}$
$\Rightarrow \frac{x}{2}=y \times \frac{3}{4}$
$\Rightarrow \frac{x}{y}=\frac{3}{4} \times 2=\frac{3}{2}$
Question 2
4 / -1

There is a rectangular Garden whose length and width is 120m X 40m.There is a walkway of uniform width around garden. Area of walkway is 1032m². Find width of walkway

A
1

B
2

C
3

D
4

Solution

Let the width of rectangle be x so the length & breath is increased by 2x

So new total area along with walkway is (120+2x)*(40+2x)

So (120+2x)*(40+2x)-120*40=1032

=> (120+2x)*(40+2x) =5832

=> (60+x)*(20+x) = 1458 =63*23

=>x=3
Question 3
4 / -1

Directions For Questions

Percentagewise distribution of employees in six different professions Total number of employees $=26800$

...view full instructions

What is the difference between the total number of employees in teaching and medical profession together and the number of employees in management profession ?

A
6770

B
7700

C
6700

D
7770

Solution

Total number of employees (in percent) in teaching and medical profession = $(15 + 27) %$

= $42 %$

Total number of employees (in percent) in management = $17 %$

Difference (in percent) = $(42 - 17) %$

$25 %$

Actual required difference = $25 % o f 26800$

= $\frac{25}{100} \times 26800$

= $6700$
Question 4
4 / -1

Directions For Questions

Percentagewise distribution of employees in six different professions Total number of employees $=26800$

...view full instructions

In management profession three-fourth of the number of employees are female. What isthe number of male employees in management profession?

A
1239

B
1143

C
1156

D
1139

Solution

In management profession three fourth of employees
are female i.e. $\frac{3}{4} \times 100=75 \%$ In management profession $25 \%$ demployees are male Now total number of employees in management
protession $=\frac{26800 \times 17}{100}=4556$
Number of male employees in management profession
$=\frac{4556 \times 25}{100}=1139$
Question 5
4 / -1

Directions For Questions

Percentagewise distribution of employees in six different professions Total number of employees $=26800$

...view full instructions

25% of employees from film production profession went on a strike. What is the number of employees from film production who have not participated in the strike?

A
3271

B
3819

C
3948

D
1273

Solution

According to the question, $25 \%$ of the employees trom fim production profession went on strike $75 \%$ of the employees of film production have not participated in strike.

Now total employoos of flim production $=\frac{26800 \times 19}{100}$
$=5092$
Number of employees of flim production who have not
participated in strike $=\frac{5092 \times 75}{100}=3819$
Question 6
4 / -1

In the battle site, there are 1800 soldiers. If each soldier consumes 6 kg per day, the provisions available in the fort will last for 30 days. If some more soldiers join, the provisions available will last for 20 days given each soldier consumes 5 kg per day. Find the number of soldiers joining the battle site in that second case?

A
1400

B
1440

C
1500

D
1660

Solution

Assume x soldiers join the battle site. 1800 soldiers have provision for 30 (days for which provisions last them)(rate of consumption of each soldier)

= (1800)(30)(6) kg = 324000

Also provisions available for (1800 + x) soldiers is (1800 + x)(20)(5) kg

As the same provisions are available

=> 324000 = (1800 + x)(20)(5)

1800+x = 324000/100 => x = 3240-1800 = 1440.
Question 7
4 / -1

10 kg of sugar costing of Rs. 25 per kg is mixed with 15 kg of sugar costing of Rs. 16 per kg. Find the price per kg of the mixture.

A
19.6 kg

B
19.7 kg

C
15.5 kg

D
15 kg

Solution

Total cost of $10 \mathrm{kg}=(10)(25)=\mathrm{Rs} .250$
Total cost of $15 \mathrm{kg}=(15)(16)=\mathrm{Rs} .240$
Average cost of the mixture $=\frac{\text { Total } \operatorname{mat}}{\text { Total quantity }}=\frac{(250+240)}{25}=\frac{490}{25}=$ Rs.19.6 kg.
Question 8
4 / -1

The m men working m hours per day for m days produce m units of work , then the unit of work produced by k men working k hours a day for k day is ?

A
m²/ k²

B
m³ / k³

C
k^{2 /} m²

D
k³ / m³

Solution

Therefore m men working for m hours/ days for m days produce m units of work.

Therefore 1 man working 1 hour/days for 1 day produce m/m³= 1/m² units of work

Therefore k men working for k hours a day for k days produce k²/m² units of work.
Question 9
4 / -1

When an odd number of two digits is divided by an even number of two digits, then quotient is 0.625. If the odd number is less than the even number by 5, then what is the ratio between odd number and even number?

A
9 : 6

B
8 : 5

C
6 : 9

D
5 : 8

Solution

Let the odd number be $x$ and the even number be $y$
Given, $\frac{x}{y}=0,625=\frac{5}{8}$
Question 10
4 / -1

In the given figure, $A B$ is the tangent to the circle and $B C D$ is the diameter of the circle. If $\angle B A C=25^{\circ}$, then find $\angle {BCA}$.

A
$75^{\circ}$

B
$80^{\circ}$

C
$90^{\circ}$

D
$65^{\circ}$

Solution

Given:

$A B$ is tangent to the circle.

$BCD$ is diameter.

We know that,

Angles of a triangle add up to $180^{\circ}$.

Since tangent to the circle is perpendicular to the diameter at that point, $\angle A B C=90^{\circ} $

Sum of angles of $\triangle {ABC}=180^{\circ}$

$\angle {BCA}+90^{\circ}+25^{\circ}=180^{\circ}$

$\Rightarrow \angle {BCA}=180^{\circ}-115^{\circ}$

$\Rightarrow \angle {BCA}=65^{\circ}$

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Re-Attempt Weekly Quiz Competition

Quantitative Ability Test - 6

Result

Quantitative Ability Test - 6

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SHARING IS CARING

Question 1

3 : 2

2 : 5

5 : 2

3 : 4

Solution

Question 2

1

2

3

4

Solution

Question 3

6770

7700

6700

7770

Solution

Question 4

1239

1143

1156

1139

Solution

Question 5

3271

3819

3948

1273

Solution

Question 6

1400

1440

1500

1660

Solution

Question 7

19.6 kg

19.7 kg

15.5 kg

15 kg

Solution

Question 8

m2 / k2

m3 / k3

k2 / m2

k3 / m3

Solution

Question 9

9 : 6

8 : 5

6 : 9

5 : 8

Solution

Question 10

\(75^{\circ}\)

\(80^{\circ}\)

\(90^{\circ}\)

\(65^{\circ}\)

Solution

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m²/ k²

m³ / k³

k^{2 /} m²

k³ / m³