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

► If new values of x, y, z are x′, y′ and z′, and respectively then x′ : y′ = 4 : 5, y′ : z′ = 3 : 4

⇒ x′ : y′ : z′ = 12 : 15 : 20
⇒ x + y + z = 5000
⇒ x′ + 50 + y′ + 100 + z′ + 150 = 5000 x′ + y′ + z′ = 4700
⇒ 12k + 15k + 20k = 4700 k = 100

► x = 1200 + 50 = 1250
► y = 1500 + 100 = 1600 z = 2000 + 150 = 2150
► x + y = 1250 + 1600 = 2850

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

Answer

Income ratio - 8:3 = 8x : 3x
expenditure ratio - 4:1 = 4y : y

ATQ,
saving of victor = 8x - 4y = 2000
angela = 3x - y = 2000
so from above two equations,
x = 1500
y = 2500
Annual expense of Angela = ₹ 2500

Answer

Let the number of total employees in the company be 100x, and the total salary of all the employees be 100y.

It is given that 20% of the employees work in the manufacturing department, and the total salary obtained by all the manufacturing employees is one-sixth of the total salary obtained by all the employees in the company.

Hence, the total number of employees in the manufacturing department is 20x, and the total salary received by them is (100y/6)

Average salary in the manufacturing department = (100y/6*20x) = 5y/6x

Similarly, the total number of employees in the nonmanufacturing department is 80x, and the total salary received by them is (500y/6)

Hence, the average salary in the nonmanufacturing department = (500y/6*80x) = 25y/24x

Hence, the ratio is:- (5y/6x): (25y/24x)

=> 120: 150 = 4:5

The correct option is D

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