Knowing our Numbers Test

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Knowing our Numbers Test - 23

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Weekly Quiz Competition

Question 1
1 / -0

Find an even $$4$$ digit number using $$2, 7, 5,1$$.

A
$$7512$$

B
$$5217$$

C
$$7125$$

D
$$1275$$

Solution

An even number is an integer which is "evenly divisible" by two. This means that if the integer is divided by $$2$$, it yields no remainder.

(a) $$7512$$ is divisible by $$2$$ as $$\dfrac {7512}{2}=3756$$, so it is a four digit even number.

(b) $$5217$$ is not divisible by $$2$$, so it is not a four digit even number.

(c) $$7125$$ is not divisible by $$2$$, so it is not a four digit even number.

(d) $$1275$$ is not divisible by $$2$$, so it is not a four digit even number.

Hence, $$7512$$ is a four digit even number using the digits $$2,7,5,1$$.
Question 2
1 / -0

Write 'Ninety- Six' in Roman numerals.

A
$$XCVI$$

B
$4XVII$$

C
$$XCV$$

D
$$XCVII$$

Solution

In Roman numeral form,
$$90 = XC$$
$$6 = VI$$

$$\Rightarrow$$ Ninety-six $$= XCVI$$
Question 3
1 / -0

Roman numeral for $$498$$ is ______.

A
$$CDCXVIII$$

B
$$CDCXIV$$

C
$$CDXCVIII$$

D
$$CDXCVII$$

Solution
Question 4
1 / -0

Round off to the nearest $$10, 100,$$ and $$1000$$:
$$1289$$

A
$$1290,1389,1009$$

B
$$1290,1300,1000$$

C
$$1280, 1300, 1200$$

D
$$1280,1200,1000$$

Solution

$$\Rightarrow$$ Round off to 10 for 1289 is 1280 or 1290 but nearest to 1289 is 1290.
$$\Rightarrow$$ So, round off to the nearest $$10$$ is $$1290$$.
$$\Rightarrow$$ Round off to 100 for 1289 is 1200 or 1300 but nearest to 1289 is 1300.
$$\Rightarrow$$ So, round off to the nearest $$100$$ is $$1300$$.
$$\Rightarrow$$ Round off to 1000 for 1289 is 1000 or 2000 but nearest to 1289 is 1000.
$$\Rightarrow$$ So, round off to the nearest $$1000$$ is $$1000$$.
$$\therefore$$ The answer is $$1290,\,1300,\,1000.$$

Question 5

1 / -0

Roman number representation of $$655$$ is

DCLV

CLV

XXI

ML

Solution

Converting to Roman

Iteration	Decimal number	Highest decimal	Highest roman	Tempoaray result
$$1$$	$$655$$	$$500$$	$$D$$	$$D$$
$$2$$	$$155$$	$$100$$	$$C$$	$$DC$$
$$3$$	$$55$$	$$50$$	$$L$$	$$DCL$$
$$4$$	$$5$$	$$5$$	$$V$$	$$DCLV$$

$$655=DCLV$$

So option $$A$$ is correct.

Question 6
1 / -0

There is a rod of length $$4\ cm$$ and another rod of length $$500\ mm$$ has been joined to the first rod. Then the length of the rod(in cm) formed by joining these $$2$$ is :

A
$$450\ cm$$

B
$$54\ cm$$

C
$$9\ cm$$

D
$$4.5\ cm$$

Solution

Length of first Rod is $$4cm$$

Length of second Rod is $$500mm$$

Total Length of the Rod by joining these two rods is $$4cm$$ $$500mm$$

We know that, $$1$$ $$centimeter =10$$ $$mm$$, $$1$$ $$mm =\dfrac{1}{10}$$ $$cm$$

We need to convert $$4$$ $$cm$$ $$500$$ $$mm$$ to $$centimeter$$
$$4cm$$ $$500$$ $$mm$$ $$ = 4cm + 500mm$$
Now, $$500$$ $$mm =\dfrac{1}{10} \times 500$$ $$cm$$ $$= 50$$ $$cm$$
$$\therefore 4cm\ 500\ mm=4cm + 500mm=(4+50)cm$$ $$=54\ cm$$

So, Option $$B$$ is correct
Question 7
1 / -0

Round-off $$13$$ to nearest $$10$$.

A
$$10$$

B
$$20$$

C
$$13$$

D
$$1$$

Solution

The given number is $$13$$
Its unit digit is $$3$$ which is less than 5.
So, we replace unit digit by $$0$$
$$\therefore$$ Rounding off 13 to nearest 10 gives $$10$$.
Question 8
1 / -0

The correct expanded form for $$2.06$$ is _______ .

A
$$\left( 2\times 10 \right) +\left( 6\times \cfrac { 1 }{ 10 } \right) $$

B
$$\left( 2\times 1 \right) +\left( 6\times \cfrac { 1 }{ 10 } \right) $$

C
$$\left( 2\times 1 \right) +\left( 6\times \cfrac { 1 }{ 100 } \right) $$

D
$$\left( 2\times 1 \right) +\left( 6+\cfrac { 1 }{ 10 } \right) $$

Solution

The number given is $$2.06$$.

We observe that the whole number part of the above number is $$2$$ which essentially is $$2 \times 1$$
The decimal part of the above number is $$.06$$ which is $$6 \times \dfrac{1}{100}$$

Thus, $$2.06 = (2 \times 1) + (6 \times \dfrac{1}{100})$$
Question 9
1 / -0

Estimate the sum $$579+7994$$ by estimating the numbers to its nearest hundreds.

A
$$8600$$

B
$$8573$$

C
$$7500$$

D
$$1200$$

Solution

$$\Rightarrow$$ $$579+7994=8573$$
$$\Rightarrow$$ In number $$8573$$, its tens digit is 7, which is greater than 5.
$$\Rightarrow$$ So, we increase the digits at hundreds place by 1 and replace each one of the digits at tens and ones place by 0 to round off the given number to nearest to 100.
$$\therefore$$ The number nearest to 100 is $$8600.$$

Question 10

1 / -0

Roman number of $$2050$$ is?

CX

LXI

MML

XXII

Solution

Conversion to Roman

Iteration	Decimal number	Highest decimal value	Highest roman numeral	Temporary result
$$1$$	$$2050$$	$$1000$$	$$M$$	$$M$$
$$2$$	$$1050$$	$$1000$$	$$M$$	$$MM$$
$$3$$	$$50$$	$$50$$	$$L$$	$$MML$$

So $$2050$$ in Roman is $$MML$$

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Re-Attempt Weekly Quiz Competition

Knowing our Numbers Test - 23

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Knowing our Numbers Test - 23

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SHARING IS CARING

Question 1

$$7512$$

$$5217$$

$$7125$$

$$1275$$

Solution

Question 2

$$XCVI$$

$4XVII$$

$$XCV$$

$$XCVII$$

Solution

Question 3

$$CDCXVIII$$

$$CDCXIV$$

$$CDXCVIII$$

$$CDXCVII$$

Solution

Question 4

$$1290,1389,1009$$

$$1290,1300,1000$$

$$1280, 1300, 1200$$

$$1280,1200,1000$$

Solution

Question 5

DCLV

CLV

XXI

ML

Solution

Question 6

$$450\ cm$$

$$54\ cm$$

$$9\ cm$$

$$4.5\ cm$$

Solution

Question 7

$$10$$

$$20$$

$$13$$

$$1$$

Solution

Question 8

$$\left( 2\times 10 \right) +\left( 6\times \cfrac { 1 }{ 10 } \right) $$

$$\left( 2\times 1 \right) +\left( 6\times \cfrac { 1 }{ 10 } \right) $$

$$\left( 2\times 1 \right) +\left( 6\times \cfrac { 1 }{ 100 } \right) $$

$$\left( 2\times 1 \right) +\left( 6+\cfrac { 1 }{ 10 } \right) $$

Solution

Question 9

$$8600$$

$$8573$$

$$7500$$

$$1200$$

Solution

Question 10

CX

LXI

MML

XXII

Solution

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