Ratio and Proportion Test

Result

Ratio and Proportion Test - 16

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
1 / -0

A local television station sells time slots for programs in $$30$$-minute intervals. If the station operates $$24$$ hours per day, every day of the week, what is the total number of $$30$$-minute time slots the station can sell for Tuesday and Wednesday?

A
$$96$$

B
$$24$$

C
$$48$$

D
$$12$$

Solution

$$\text{Numbers of slots in 1 hour}=2$$
$$\text{Each day has 24 hours.}$$
$$\text{Numbers of slot in 2 days = }24\times 2\times2=96$$
Question 2
1 / -0

The total cost of 30 identical erasers is y dollars. At this rate, find the total cost in dollars of 70 of these erasers in terms of y.

A
$$\dfrac {3y}{7}$$

B
$$\dfrac {4y}{7}$$

C
$$\dfrac {7y}{4}$$

D
$$\dfrac {7y}{3}$$

E
$$70y$$

Solution

Given that,
The cost of $$30$$ erasers is $$y$$ dollars.
To find out,
The cost of $$70$$ erasers.

The cost of $$30$$ erasers $$=y$$ dollars

Hence, the cost of one eraser $$=\dfrac{y}{30}$$ dollars [unitary method]

$$\therefore\ $$ The cost of $$70$$ erasers $$=\dfrac{y}{30}\times70$$

$$=\dfrac{7y}{3}$$ dollars.

Hence, the cost of $$70$$ erasers is $$\dfrac{7y}{3}$$ dollars.
Question 3
1 / -0

Sixteen ounces is equal to one pound, and one tonne is equal to $$2,000$$ pounds. How many ounces equal two tonnes?

A
$$4,000$$

B
$$16,000$$

C
$$32,000$$

D
$$32,016$$

E
$$64,000$$

Solution

Given, $$16$$ ounces $$=1$$ pound
$$1$$ tonne $$=2000$$ pound
$$2$$ tonne will be $$=4000$$ pound
As we know,
$$1$$ pound $$=16$$ ounces
$$4000$$ pound will be $$=$$ $$16\times 4000=64000$$ ounces
So, $$2$$ tonne will be $$64000$$ ounces.
Question 4
1 / -0

On a house blueprint, $$2$$ feet is represented by $$1$$ inch. If a room on the blueprint measures $$5$$ inches by $$6$$ inches, what is the area of the actual room?

A
$$30$$ square feet

B
$$60$$ square feet

C
$$90$$ square feet

D
$$120$$ square feet

Solution

If $$1$$ inch is $$2$$ feet, then $$5$$ inches is $$10$$ feet, while $$6$$ inches is $$12$$ feet.
Hence, the actual dimensions of the room is $$10$$ feet by $$12$$ feet.
Thus the area is $$10\times 12=120$$ sq feet.
Question 5
1 / -0

In the ratio between $$12 m$$ and $$13\ kg$$, the antecedent is _____

A
$$12$$

B
$$13$$

C
$$12\ m$$

D
Not possible

Solution

The ratio of two unlike quantities is not defined. Hence, the ratio between $$12\ m$$ and $$13\ kg$$ cannot be found. Thus, the antecedent is not possible.
Question 6
1 / -0

The ratio of length of two cars, M and N whose lengths are $$3\ m$$ and $$4\ m$$ is $$3 : 4$$. Here, $$4$$ is called ________

A
Consequent

B
Antecedent

C
Dividend

D
Constant

Solution

The ratio of lengths of car M and N is $$3 : 4$$. Here, $$4$$ being the second term is called consequent.
Question 7
1 / -0

If the antecedent is $$16$$ and the consequent is $$12$$, then the ratio is _________

A
$$12 : 16$$

B
$$16 : 12$$

C
$$6 : 2$$

D
Cannot be determined

Solution

$$Antecedent:Consequent=16:12$$
Question 8
1 / -0

If the consequent is $$13$$ and the antecedent is $$11$$, then the ratio is _______

A
$$11 : 13$$

B
$$13 : 11$$

C
$$3 : 1$$

D
None of these

Solution

First term $$:$$ Second Term
$$\downarrow \quad \quad \quad \quad \quad \downarrow$$
Antecedent Consequent
Since antecedent $$=11$$ and consequent $$=13$$
$$\therefore$$ antecedent $$:$$ consequent $$=11:13$$
Question 9
1 / -0

The area of the circle and the area of an regular polygon of sides of perimeter equal to that of the circle are in the ratio of

A
$$\tan { \left( \cfrac { \pi }{ n } \right) } :\cfrac { \pi }{ n } $$

B
$$\cot { \left( \cfrac { \pi }{ n } \right) } :\cfrac { \pi }{ n } $$

C
$$\cos{ \left( \cfrac { \pi }{ n } \right) } :\cfrac { \pi }{ n } $$

D
$$\sin { \left( \cfrac { \pi }{ n } \right) } :\cfrac { \pi }{ n } $$

Solution

Let $$r$$ be the radius of the circle and a be the side of polygon.
The perimeter of circle $$= 2\pi r$$
The perimeter of a polygon with n sides $$= na$$
So $$2\pi r = na$$
So $$a = \dfrac {2\pi r}{n}$$
Area of the circle, $$A_1= \pi r^2$$
Area of a regular polygon, $$A_2=\bigg( \dfrac{1}{4} \bigg) na^2 \cot \bigg( \dfrac{\pi}{n}\bigg)$$

$$=\dfrac14\ n\ \bigg(\dfrac {4\pi ^2 r^2}{n^2}\bigg) \cot \bigg(\dfrac{\pi}{n}\bigg)$$

$$=\bigg( \dfrac{\pi ^2 r^2}{n} \bigg) \cot \bigg( \dfrac{\pi}{n} \bigg)$$

$$\Rightarrow \dfrac{A_1}{A_2}=\dfrac{\pi r^2}{\dfrac{\pi^2 r^2}{n} \cot \bigg( \dfrac{\pi}{n} \bigg)}$$

$$=\dfrac{\tan \bigg( \dfrac{\pi}{n}\bigg)}{\dfrac{\pi}{n}}$$

So the ratio is $$\tan \bigg( \dfrac{\pi}{n}\bigg) : \dfrac{\pi}{n}$$
Question 10
1 / -0

If $$14$$ milliliters of a certain liquid has a mass of $$16$$ grams. What is the mass (in grams) of $$28$$ liters of this liquid? ($$1$$ liter = $$1,000$$ milliliters)

A
$$8$$

B
$$32$$

C
$$3,200$$

D
$$8,000$$

E
$$32,000$$

Solution

Given that $$14$$ milliliters have $$16$$ grams.
Therefore $$14$$ liters will have $$16000$$ grams.
So, $$28$$ liters will have $$16000 \times 2 = 32000$$ grams.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Ratio and Proportion Test - 16

Result

Ratio and Proportion Test - 16

Score

-

Rank

-

SHARING IS CARING

Question 1

$$96$$

$$24$$

$$48$$

$$12$$

Solution

Question 2

$$\dfrac {3y}{7}$$

$$\dfrac {4y}{7}$$

$$\dfrac {7y}{4}$$

$$\dfrac {7y}{3}$$

$$70y$$

Solution

Question 3

$$4,000$$

$$16,000$$

$$32,000$$

$$32,016$$

$$64,000$$

Solution

Question 4

$$30$$ square feet

$$60$$ square feet

$$90$$ square feet

$$120$$ square feet

Solution

Question 5

$$12$$

$$13$$

$$12\ m$$

Not possible

Solution

Question 6

Consequent

Antecedent

Dividend

Constant

Solution

Question 7

$$12 : 16$$

$$16 : 12$$

$$6 : 2$$

Cannot be determined

Solution

Question 8

$$11 : 13$$

$$13 : 11$$

$$3 : 1$$

None of these

Solution

Question 9

$$\tan { \left( \cfrac { \pi }{ n } \right) } :\cfrac { \pi }{ n } $$

$$\cot { \left( \cfrac { \pi }{ n } \right) } :\cfrac { \pi }{ n } $$

$$\cos{ \left( \cfrac { \pi }{ n } \right) } :\cfrac { \pi }{ n } $$

$$\sin { \left( \cfrac { \pi }{ n } \right) } :\cfrac { \pi }{ n } $$

Solution

Question 10

$$8$$

$$32$$

$$3,200$$

$$8,000$$

$$32,000$$

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.