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Byju's Answer

Standard VII

Mathematics

Percentages

Divide 900 ...

Question

Divide $900$ into two parts such that $60 %$ of one part is equal to $30 %$ of the other.

200

and

800.

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500

and

400.

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700

and

800.

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300

and

600.

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Solution

The correct option is D $300$ and $600.$
Let one part of $900$ be $x$
$∴$ other part is $900 - x$
Given,
$60 %$ of $x = 30 %$ of $(900 - x)$
$\Rightarrow \frac{60 x}{100} = \frac{30 (900 - x)}{100}$
$\Rightarrow 2 x = 900 - x$
$\Rightarrow 3 x = 900$
$\Rightarrow x = 300$
One part is $300$ and
other part is $900 - 300 = 600$

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Similar questions

Q. Find the mean of the following data .

Class Interval	Frequency
$0 - 100$	$2$
$100 - 200$	$5$
$200 - 300$	$4$
$300 - 400$	$12$
$400 - 500$	$17$
$500 - 600$	$20$
$600 - 700$	$20$
$700 - 800$	$9$
$800 - 900$	$7$
$900 - 1000$	$4$

Q. The median of the following data is

525

. Find the values of

x

and

y

, if the total frequency is

100

Class interval	Frequency
$0 - 100$	$2$
$100 - 200$	$5$
$200 - 300$	$x$
$300 - 400$	$12$
$400 - 500$	$17$
$500 - 600$	$20$
$600 - 700$	$y$
$700 - 800$	$9$
$800 - 900$	$7$
$900 - 1000$	$4$

Q. The median of the following data in

525

. Find the missing frequency, if it is given that there are

100

observations in the data:

Class interval	Frequency	Class interval	Frequency
0-100	$2$	500-600	$20$
100-200	$5$	600-700	$f_{2}$
200-300	$f_{1}$	700-800	$9$
300-400	$12$	800-900	$7$
400-500	$17$	900-1000	$4$

Q. Divide

900

into two parts such that

60 %

of one part exceeds

30 %

of the other by

270

The following table gives the life time of 400 neon lamps:

$\begin{matrix} Life time (in hours) & 300 - 400 & 400 - 500 & 500 - 600 & 600 - 700 & 700 - 800 & 800 - 900 & 900 - 1000 Number of lamps & 14 & 56 & 60 & 86 & 74 & 62 & 48 \end{matrix}$

A bulb is selected at random. Find the probability that the life time of the selected bulb is:

(i) less than 400 (ii) between 300 to 800 hours (iii) at least 700 hours.

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Divide 900 into two parts such that 60% of one part is equal to 30% of the other.

The correct option is D 300 and 600.Let one part of 900 be x∴ other part is 900−xGiven,60% of x=30% of (900−x)⇒60x100=30(900−x)100⇒2x=900−x⇒3x=900⇒x=300One part is 300 and other part is 900−300=600

Divide $900$ into two parts such that $60 %$ of one part is equal to $30 %$ of the other.