The annual income of Victor and Angela are in the ratio 8 : 3 and their annual expenditures are in the ratio 4 : 1. If each save Rs. 2000 per annum. What is the annual expense of Angela?a)2000b)4000c)5000d)2500Correct answer is option 'D'. Can you explain this answer?

Question

Answer

Answer

Answer: B
From the given ratios,

incomes be 8x, 3x
expenses be 4y, y
then 8x-4y= 2000 (1)
3x- y = 2000 (2)
so, 5x= 3y
by substituting, y= 2500, x= 1500

Angela's expense: 2500

Answer

Answer

Victor annual income =8a
Angela = 3a
both savings =2000
Victor annual expenditure=4b
Angela=b

savings +expenditure=income

Victor, 2000+4b=8a (i)
Angela, 2000+b=3a(ii)

4×(ii)-(i)

4b+8000=12a
- 4b+2000=8a

0 + 6000=4a
a=6000/4=1500

put a=1500 in eq (i)
2000+4b=8a
4b= 8×1500-2000
4b= 10000
b=10000/4
b=2500

here, annual income of Angela is (b) 2500

Answer

: y : z will still be the same as the original ratio.

Let the original ratio be a:b:c, so we can write:

x = a/(a+b+c) * 5000
y = b/(a+b+c) * 5000
z = c/(a+b+c) * 5000

Now, let's apply the given reductions:

x' = (a-50)/(a+b+c) * 5000
y' = (b-100)/(a+b+c) * 5000
z' = (c-150)/(a+b+c) * 5000

We want to show that:

x'/y' = x/y and y'/z' = y/z

Simplifying x'/y':

x'/y' = [(a-50)/5000]/[(b-100)/5000]
= (a-50)/(b-100)

Simplifying x/y:

x/y = [(a/(a+b+c)*5000)]/[(b/(a+b+c)*5000)]
= a/b

We need to show that (a-50)/(b-100) = a/b:

a(b-100) = b(a-50)
ab - 100a = ab - 50b
50b = 100a
b = 2a

Substituting b = 2a in the original equation x+y+z = 5000, we get:

x + 2x + z = 5000
3x + z = 5000

Now, let's simplify y'/z':

y'/z' = [(b-100)/5000]/[(c-150)/5000]
= (b-100)/(c-150)

Simplifying y/z:

y/z = [(b/(a+b+c)*5000)]/[(c/(a+b+c)*5000)]
= b/c

We need to show that (b-100)/(c-150) = b/c:

b(c-150) = c(b-100)
bc - 150b = bc - 100c
50c = 150b
c = 3b

Substituting c = 3b in the equation 3x + z = 5000, we get:

3x + 3by = 5000 - z

Since we know that x+y+z = 5000, we can substitute z = 5000 - x - y in the above equation to get:

3x + 3by = 5000 - x - y

Simplifying:

4x + 4y = 3by

4x/b + 4y/b = 3y/b

4x/b + 4 = 3y/b

4x/b = 3y/b - 4

Substituting b = 2a, we get:

4x/2a = 3y/2a - 4

2x/a = (3/2)(y/a) - 4

2x/a + 4 = (3/2)(y/a)

4a/(2x+a) = 2/(y/a)

4a(y/a) = 2(2x+a)/(2x+a)

4y = 4x + 2a

2y = 2x + a

Substituting

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