What number should be subtracted from each of the numbers 23 30 57 and 78 so that the differences are in proportion?

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What number should be subtracted from each of the numbers 23, 30, 57 and 78 so that the remainders are in proportion?

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 What number should be subtracted from each of the numbers 23,30,57 and 78 so that the remainders are in proportion?

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Consider x be subtracted from each term

So, 23–x,30–x,57–x and 78–x are proportional

It can be written as

(23–x):(30–x)::(57–x):(78–x)

(30–x) (23–x) ​ = (78–x) (57–x) ​

By cross multiplication

(23–x)(78–x)=(30–x)(57–x)

By further calculation

1794–23x–78x+x 2 =1710–30x–57x+x 2 ⇒x 2 –101x+1794–x 2 +87x–1710=0 So we get –14x+84=0 ⇒14x=84 ⇒x= 14 84 ​ =6

Therefore, 6 is the number to be subtracted from each of the numbers.

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What number should be subtracted from each of the numbers 4, 10 and 28 so that the remainder may be in continued proportion?

Answer (1 of 2): Answer is 1. Let x be the number to subtract from each number for continued proportion. We have (10-x)/(4-x) = (28-x)/(10-x) => (10-x)^2 = (28-x)(4-x) => 100–20x+x^2 = 112–32x+x^2 => 32x-20x = 112–100 => 12x = 12 or x = 1. The continued proportion is 3, 9, 27.

What number should be subtracted from each of the numbers 4, 10 and 28 so that the remainder may be in continued proportion?

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Suruti Ranjan Behera

M.Tech in Mechanical Engineering & Mechanical Machine Design, Kalinga Institute of Industrial Technology (KIIT) (Graduated 2019)Author has 179 answers and 647.5K answer views3y

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What should be subtracted from each of the numbers 54, 71, 75 , 99 so that the remainders may be proportional?

Let assume x is the number, if you subtracte from each of the numbers mentioned above in the question, so that the remainders may be proportion.

As per question,

54 - x , 71 - x , 75 - x , and 99 - x are in proportion.

(54 - x) / (71 - x) = (75 - x) / ( 99 - x)

(54 - x)( 99 - x) = (75 - x) (71 - x)

x^2 - 99x - 54x + 5346 = x^2 - 75x - 71x + 5325

(multiplying both side)

- 153x + 5346 = -146x + 5325(x^2 cancelled both side)

153x- 146x = 5346 - 5325

7x = 21 ∴ x = 3.

So, 3 should be subtracted from each of the numbers54,71,75,99 so that the remainders may be proportional.

Hope it's will help you.

Thanks for re Related questions

What number should be subtracted from each of the numbers 23,30,57 and 78 so that the remainders are in proportion? (A) 4 (B) 3 (C) 6 (D) 7?

What number must be subtracted from each of the numbers 18, 78, 19 and 83, so that the remainder will be in proportion?

In a class of 20 students in an examination in mathematics, 2 students scored 100 marks each, 3 got zero each, and the average of the rest was 40. What is the average for the whole class?

80 is subtracted from 30% of a number, the result is 50. What is the number?

How many terms are there in Arithmetic Progression 7, 10, 13,…43?

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What number should be subtracted from each of the numbers 23,30,57 and 78 so that the remainders are in proportion? (A) 4 (B) 3 (C) 6 (D) 7?

The answer is (C) 6

Let the required number be ‘x’. after subtracting x from the given numbers, we have 23-x, 30-x, 57-x, 78-x.

At this point, you can choose to be lazy and see which option fits, or solve it the “good” way.

(23−x)/(30−x)=(57−x)/(78−x)

(23−x)/(30−x)=(57−x)/(78−x)

Transposing, we get x 2 −101x+1794= x 2 −87x+1710

x2−101x+1794=x2−87x+1710

14x=84 14x=84 x=6 x=6 S. S. Agarwal

Math Competition Problems, Most Viewed Writer - QuoraAuthor has 442 answers and 1.4M answer views3y

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When 9 is subtracted from a number and then divided by 2, the answer is 4. What is the number?

Ans. Let's assume the required number be X and X-9 be Y.

So equation should be,

First equation, X - 9 =Y

Second equation, Y / 2 = 4

By solving the second equation we get,

Y = 4×2 Y = 8

Now putting the value of Y in first equation, we get

X - 9 = Y

X = 8 +9 [putting Y = 8]

X = 17.

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Mahi Sharma

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When 12 is subtracted from 3times a certain number and the result is divided by 4. The answer is 18. What is the number?

I am not very good at maths, but I think this is the correct answer.

Answer is 28. Explanation:

So, let us assume the number to be x.

For forming the equation:

Now, according to the question, if twelve is subtracted from three times the number, ie.

Three times the number will be 3x.

Subtract 12 from it, we get 3x-12.

Now, when we divide it from 4, ie.,

3x-12/4, we get the answer as 18.

So the equation is 3x-12/4=18.

Now, solving the equation:

We cross multiply.

So, the equation formed will be 3x-12=18*4

=) 3x-12=72 =) 3x=72+12 =) 3x=84 =) x= 84/3 =) x=28.

Therefore, the number is 28.

Hope it helps you. Upvote if

Rebecca Harvey

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What two numbers make 10 adding and 3 subtracting?

Presuming you don't need whole numbers, 3.5 and 6.5.

3.5 + 6.5 = 10 6.5 - 3.5 = 3

Do you need the method/workings?

I'm a little rusty, but the way I'd do this would e with algebra.

X + y = 10 X - y = 3

Add them together = x + x + y - y = 13

= 2x = 13

Therefore x = 13 ÷ 2 = 6.5

To calculate y, substitute 6.5 with x in a previous equation:

X + y = 10 6.5 + y = 10

Therefore y = 10 - 6.5 = 3.5

Dorota Zarzycka

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Which number should be subtracted from each of the numbers 11, 23 and 53 so that the resulting numbers would be in continued proportion?

3.

(11–3)/(23–3)=(23–3)/(53–3)

8/20=20/50

The way you arrive at it is as follows:

(11-x)/(23-x)=(23-x)/(53-x)

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What number should be subtracted from each of the numbers 23 30 57 and 78 so that the remainders are...

Free solutions for M L Aggarwal - Understanding ICSE Mathematics - Class 10 Chapter 7 - Ratio and Proportion Ratio and Proportion Exercise 7.2 question 7. These explanations are written by Lido teacher so that you easily understand even the most difficult concepts

ML Aggarwal Solutions Class 10 Mathematics Solutions for Ratio and Proportion Exercise 7.2 in Chapter 7 - Ratio and Proportion

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What number should be subtracted from each of the numbers 23, 30, 57, and 78 so that the remainders are in proportion?

Answer:

Consider x be subtracted from each term

23 – x, 30 – x, 57 – x and 78 – x are proportional

It can be written as

23 – x: 30 – x :: 57 – x: 78 – x

(23 – x)/ (30 – x) = (57 – x)/ (78 – x)

By cross multiplication

(23 – x) (78 – x) = (30 – x) (57 – x)

By further calculation

\begin{array}{l} 1794-23 x-78 x+x^{2}=1710-30 x-57 x+x^{2} \\ x^{2}-101 x+1794-x^{2}+87 x-1710=0 \end{array}

1794−23x−78x+x 2 =1710−30x−57x+x 2 x 2 −101x+1794−x 2 +87x−1710=0 ​ So we get -14x + 84 = 0 14x = 84 x = 84/14 = 6

Therefore, 6 is the number to be subtracted from each of the numbers.

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