Numerical Applications Test 37

Result

Numerical Applications Test 37

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

Vasu can run $$4$$ Km in $$48$$ minutes. If Vikas can run twice as fast as Vasu, how many minutes does it take Vikas to run $$6$$ Km?

A
$$24$$

B
$$30$$

C
$$36$$

D
$$48$$

Solution

Vasu takes $$=\dfrac{48}{4}=12$$ minutes to run $$1$$ km.

Vikas can run twice as fast as Vasu, means Vikas takes $$6$$ min to run $$1$$ km.

Hence, he takes, $$= 6\times 6 = 36$$ min. to run $$6$$ km.
Question 2
1 / -0

A single frame of $$35$$ mm film is about three-quarters of an inch long. A film reel holds up to $$1000$$ feet of film. Find the number of reels required for a $$2:47:00$$ ($$2$$ hr $$47$$ min) film shot at $$24$$ frames per second.

A
13

B
14

C
15

D
16

Solution

There are $$12$$ inches in one foot, so a reel is $$12 \times 1,000 = 12,000$$ inches long. Set up a proportion to determine how many frames per reel: $$=$$ $$\dfrac{1 frame }{\dfrac{3}{4}}=12000$$
$$\dfrac{3}{4}x=12000$$
Cross-multiply to get $$\dfrac{3}{4}x=12000$$.
Divide both sides by $$\dfrac{3}{4}$$ to get $$x = 16,000$$ frames per reel. Next, find the number of frames the film requires.
Convert the time to seconds.
There are $$60$$ minutes in an hour, so $$2$$ hours and $$47$$ minutes is equal to $$(2\times 60)+47=167$$ minutes.
There are $$60$$ seconds in a minute, so there are $$60\times 167=10020$$ seconds in this film.
If each second consists of $$24$$ frames, then there are $$24\times 10020=240480$$ frames in this film.
To determine the number of reels, divide by the number of frames per reel: $$= 15.03$$ reels. Because $$15$$ reels does not hold quite enough frames, the film requires $$16$$ reels, which is (D).
Question 3
1 / -0

A water tank has two pipes to fill it. Pipe A can fill the tank in $$2$$ hours , only if tank is empty. Pipe B can empty the tank in $$3$$ hours, only if tank is full. If both the pipes start at the same time when the task is empty, calculate the number of hours will it take for the tank to be filled to $$60\%$$ of its capacity?

A
$$2.4$$

B
$$3.6$$

C
$$6$$

D
$$60$$

Solution

Work efficiency of pipe A in filling tank = $$\dfrac{1}{2}$$ per hour

Work efficiency of pipe B in emptying tank = $$-\dfrac{1}{3}$$ per hour

$$\therefore$$ work done by A and B together = $$\dfrac{1}{2} - \dfrac{1}{3} = \dfrac{1}{6}$$per hour

Hence time taken to fill the tank $$100$$% = $$6$$ hours

$$\therefore$$ time taken to fill $$60$$% = $$\dfrac{6\times 60}{100}$$ = $$3.6$$hours

Answer is option B
Question 4
1 / -0

The size of song downloaded from internet is $$4$$ megabytes. Edwird downloads music at a rate of $$256$$ kilo bytes per second.how many songs can Edwird download in $$2$$ hours? (1 Megabytes =1024 kilobytes).

A
$$128$$

B
$$450$$

C
$$1,800$$

D
$$1,920$$

Solution

Given: The size of song downloaded from internet $$= 4$$ megabytes.
Per second $$= 256$$ kilobytes
$$2$$ hours $$= 7200$$ seconds.
For $$7200$$ seconds to download $$= 7200 \times 256 = 1843200 $$ kilobytes
Convert kilobytes to megabytes,
So, $$\dfrac{843200}{1024}=1800$$
Edwin download in $$2$$ hours $$=$$ $$\dfrac{1800}{4} = 450$$
Question 5
1 / -0

A coastal geologist estimates that a certain countrys beaches are eroding at a rate of $$1.5$$ feet per year.
According to the geologists estimate, how long will it take, in years, for the countrys beaches to erode
by $$21$$ feet?

A
$$14$$ years

B
$$12$$ years

C
$$8$$ years

D
$$15$$ years

Solution

If the beaches are eroding at the rate of 1.5 feet per year, then for eroding by 21 feet, it will take $$ \frac {21}{1.5} = 14 $$ years
Question 6
1 / -0

Working for $$10$$ months as a teacher in High School, David earned $$\$3,200$$ per month from September to June. He then undertook a job as a barista at a cafe, where he earned $$\$2,000$$ per month during July and August. Find the average monthly pay of David for the $$12$$ months.

A
$$\$2400$$

B
$$\$2500$$

C
$$\$2600$$

D
$$\$2800$$

E
$$\$3000$$

Solution

David earned $$\$3200$$ per month in High school
So in $$10$$ months he earned $$=$$ $$10\times 3200=$32000$$
David earned $$\$2000$$ per month in cafe
So in $$2$$ month (july,august) he earned $$=$$ $$2\times 2000=$4000$$
His total earning $$=$$ $$32000+4000=36000$$
$$\therefore$$ Average monthly pay $$=\dfrac{36000}{12}=$3000$$
Question 7
1 / -0

Jennilina's hobby is to paint the rooms at her college. At top speed, she could paint $$5$$ identical rooms during one $$6$$-hour shift. Find the time it took her to paint each room.

A
$$50$$ minutes

B
$$1$$ hour and $$10$$ minutes

C
$$1$$ hour and $$12$$ minutes

D
$$1$$ hour and $$15$$ minutes

E
$$1$$ hour and $$20$$ minutes

Solution

There are $$60$$ minuet in $$1$$ hour
So in $$6$$ hours there are $$=$$ $$6\times 60=360$$ minutes
So Jennilina takes $$360$$ minutes to paint $$5$$ rooms.
So she paint one room in $$=$$ $$\dfrac{360}{5}=72$$ minutes.
$$72$$ minutes is equal to $$1$$ hour and $$12$$ minutes.
Question 8
1 / -0

The value of $$\dfrac {(n + 2)! - (n + 1)!}{n!} $$ is:

A
$$(n + 2)!$$

B
$$(n + 1)!$$

C
$$(n + 2)^{2}$$

D
$$(n + 1)^{2}$$

E
$$n$$

Solution

The value of $$\dfrac { (n+2)!-(n+1)! }{ n! } $$
$$=\dfrac { (n+2)(n+1)n!-(n+1)n! }{ n! } $$
$$=\dfrac { (n+1)(n!)(n+2-1) }{ n! } $$
$$=\dfrac { (n+1)(n+1) }{ 1 } $$
$$={ (n+1) }^{ 2 }$$
Question 9
1 / -0

The diagram shows a right pyramid that has an isosceles triangular base. If the volume of the pyramid is $$330$$ cm$$^3$$, calculate its height, h.

A
$$5.5cm$$

B
$$6.5cm$$

C
$$7.5cm$$

D
None of these

Solution
Question 10
1 / -0

The pyramid in Figure 3 is composed of a square base of area $$16$$ and four isosceles triangles, in which each base angle measures $$60^o$$. Calculate the volume of the pyramid.

A
$$7.39$$

B
$$9.24$$

C
$$15.08$$

D
$$21.33$$

E
$$24.05$$

Solution

Given, area of square base $$16$$ and four isosceles triangles in which base angle $$60^o$$
Then side of base of triangle $$=$$ $$\sqrt{16}=4$$
The Length from top to base of pyramid $$=$$ $$(4)^{2}-\left ( \dfrac{4}{2} \right )^{2}=2\sqrt{3}$$
Height of pyramid $$=$$ $$h=(2\sqrt{3})^{2}-\left ( \dfrac{4}{2} \right )^{2}\Rightarrow h=2\sqrt{2}$$
Then, volume of pyramid $$=$$ $$\dfrac{4\times 4\times 2\sqrt{2}}{3}=15.08$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Numerical Applications Test 37

Result

Numerical Applications Test 37

Score

-

Rank

-

SHARING IS CARING

Question 1

$$24$$

$$30$$

$$36$$

$$48$$

Solution

Question 2

13

14

15

16

Solution

Question 3

$$2.4$$

$$3.6$$

$$6$$

$$60$$

Solution

Question 4

$$128$$

$$450$$

$$1,800$$

$$1,920$$

Solution

Question 5

$$14$$ years

$$12$$ years

$$8$$ years

$$15$$ years

Solution

Question 6

$$\$2400$$

$$\$2500$$

$$\$2600$$

$$\$2800$$

$$\$3000$$

Solution

Question 7

$$50$$ minutes

$$1$$ hour and $$10$$ minutes

$$1$$ hour and $$12$$ minutes

$$1$$ hour and $$15$$ minutes

$$1$$ hour and $$20$$ minutes

Solution

Question 8

$$(n + 2)!$$

$$(n + 1)!$$

$$(n + 2)^{2}$$

$$(n + 1)^{2}$$

$$n$$

Solution

Question 9

$$5.5cm$$

$$6.5cm$$

$$7.5cm$$

None of these

Solution

Question 10

$$7.39$$

$$9.24$$

$$15.08$$

$$21.33$$

$$24.05$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday