Topic: Simultaneous equations Class 9 ICSE

Mathematics

1 answer:

BARSIC [14]3 years ago

8 0

Denote the number by 10a + b, where both a and b are integers picked from {1, 2, 3, …, 9}.

"the difference of whose digits is 3" ==> |a - b| = 3

"4 times the number is equal to 7 times the number obtained by reversing the digits" ==> 4 (10a + b) = 7 (10b + a)

Simplifying the second equation, you get

40a + 4b = 70b + 7a

33a - 66b = 0

a - 2b = 0

Assume a > b, in which case we would have |a - b| = a - b, so

a - b = 3

Eliminate a to solve for b, then for a :

(a - b) - (a - 2b) = 3 - 0

b = 3 ==> a = 6

Then the original number is 63.

If instead we had assumed a < b, we would have had |a - b| = b - a = 3. Then

2 (b - a) + (a - 2b) = 2×3 + 0

-a = 6

But neither a nor b can be negative, so this case is moot.

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irina1246 [14]

Using the it's concept, the correct calculation for the probability of choosing a red grape and then without putting the red grape back into the bowl, choosing a green grape is:

$p = \frac{5}{2} \times \frac{1}{23} = \frac{5}{46}$

<h3>What is a probability?</h3>

A probability is given by the number of desired outcomes divided by the number of total outcomes.

In this problem, we have that there is a total of 6 + 10 + 8 = 24 grapes. For the first grape, 10 are red, hence the probability that the first grape is red is:

$pR = \frac{10}{24} = \frac{5}[12}$

For the second grape, considering that the first was red and was removed, there will be 23 grapes, out of which 6 will be green, hence the probability that the second grape is green is:

$pG = \frac{6}{23}$