The sum of three numbers is 98. The ratio of the first to the second is 2/3, and the ratio of the second to the third is 5/8. The second number is:
(a) 15, (b) 20, (c) 30, (d) 32, (e) 33
Answers
Answered by
4
Let The Numbers Be x , y and z
Then,
x + y + z = 98 ----[Given]
x : y = 2 / 3
x = 2y / 3
y : z = 5 / 8
Using Invertendo --[It means Making Numerator Denominator and
z : y = 8 / 5 making Denominator Numerator Its There in
z = 8y / 5 10th std ICSE syllabus]
Now,
x + y + z = 98
Substituting The Value of x and z in the given equation
2y + y + 8y = 98
---- ----
3 5
LCM = 15
10y + 15y + 24y = 98
----------------------
15
49y = 98 * 15
y = 98 * 15
----------
49
y = 2 * 15
y = 30
So,
The Second Number = 30
ANSWER = (C) = 30
Hope U Understood
Mark This Brainliest If This Helped You
Thankx
Then,
x + y + z = 98 ----[Given]
x : y = 2 / 3
x = 2y / 3
y : z = 5 / 8
Using Invertendo --[It means Making Numerator Denominator and
z : y = 8 / 5 making Denominator Numerator Its There in
z = 8y / 5 10th std ICSE syllabus]
Now,
x + y + z = 98
Substituting The Value of x and z in the given equation
2y + y + 8y = 98
---- ----
3 5
LCM = 15
10y + 15y + 24y = 98
----------------------
15
49y = 98 * 15
y = 98 * 15
----------
49
y = 2 * 15
y = 30
So,
The Second Number = 30
ANSWER = (C) = 30
Hope U Understood
Mark This Brainliest If This Helped You
Thankx
Raj12345:
Amazing!!!!!!!!
Answered by
0
Let the three numbers be x, y and z.
Now,
Its given that their sum is 98.
So, x + y + z = 98 --Eq.1
Again, given ratios –
x : y = 2 : 3 --Eq.2
y : z = 5 : 8 --Eq.3
So, now we have to find the second number i.e. y
Finding the value of x in terms of y.
Using Eq.2, we can write
x/y = 2/3
x = 2y/3 --Eq.4
Now, again finding the value of z in terms of y.
Using Eq.3, we can write
y/z = 5/8
z = 8y/5 --Eq.5
Putting the values of x and z according to Eq.4 and Eq.5 in Eq.1
x + y + z = 98
2y/3 + y + 8y/5 = 98
Making the denominators common
10y/15 + 15y/15 + 24y/15 = 98
49y/15 = 98
Transposing 15 and 49 to RHS
y = 98×15/49
y = 2 × 15
y = 30
So, the answer is (C)30
Now,
Its given that their sum is 98.
So, x + y + z = 98 --Eq.1
Again, given ratios –
x : y = 2 : 3 --Eq.2
y : z = 5 : 8 --Eq.3
So, now we have to find the second number i.e. y
Finding the value of x in terms of y.
Using Eq.2, we can write
x/y = 2/3
x = 2y/3 --Eq.4
Now, again finding the value of z in terms of y.
Using Eq.3, we can write
y/z = 5/8
z = 8y/5 --Eq.5
Putting the values of x and z according to Eq.4 and Eq.5 in Eq.1
x + y + z = 98
2y/3 + y + 8y/5 = 98
Making the denominators common
10y/15 + 15y/15 + 24y/15 = 98
49y/15 = 98
Transposing 15 and 49 to RHS
y = 98×15/49
y = 2 × 15
y = 30
So, the answer is (C)30
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