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2 years ago

Arnav was 1.5 m tall. In the last couple of years, his height has increased by 20%.

Mathematics

2 answers:

adell [148]2 years ago

4 0

Answer:

He/she grew 0.3 meters so 1.8m

Step-by-step explanation:

I just did the math

Sedbober [7]2 years ago

3 0

Answer: 1.8 m

Step-by-step explanation:

The original height is taken as 100 % , since the height increased by 20 % , this means that the new height is 120% of the original height. That

120% of 1.5

which is

$\frac{120}{100}$ x 1.5

1.5 could be written as 15/ 10 in fraction , so we have

$\frac{120}{100}$ x $\frac{15}{10}$

= $\frac{6}{5}$ x $\frac{15}{10}$

= $\frac{18}{10}$

= 1.8

Therefore Arnav is 1.8m today

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Stels [109]

Answer: The answer is 4 and 32.

Step-by-step explanation: Let "A", "B" and "C" represents the set of students who were taking Arabic, Bulgarian and Chinese respectively.

The, according to the given information, we have

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Let 'p' represents the number of students who take all the three languages, then

$n(A\cap B\cap C)=p.$

Also,

$n(A)-n(A\cap B)-n(A\cap C)+n(A\cap B\cap C)=9\\\\\Rightarrow n(A\cap B)+n(A\cap C)=24+p~~~~~~~~~~~~~~(a),\\\\n(A\cap B)+n(B\cap C)=20+p~~~~~~~~~~~~~~(b),\\\\n(B\cap C)+n(A\cap C)=20+p~~~~~~~~~~~~~~~(c).$

From here, we get after subtracting equation(c) from (b) that

$n(A\cap B)=n(A\cap C)=14.$

Therefore,

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$n(A\cap B\cap C)=4$ and

$n(A\cup B\cup C)=n(A)+n(B)+n(C)-n(A\cap B)-n(B\cap C)-n(A\cap C)+n(A\cap B\cap C)\\\\\Rightarrow n(A\cap B\cap C)=33+32+40-9-12-20+4\\\\\Rightarrow n(A\cap B\cap C)=68.$

Thus, the number of students who take all the three languages is 4 and the number of students who take none of the languages is 100-68 = 32.

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Answer:

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