Question

Read the following information carefully to answer the question given below:
The sum of the incomes of A and B is more than that of C and D taken together. The sum of the incomes of A and C is the same as that of B and D taken together. Moreover, A earns half as much as the sum of the incomes of B and D.
If A's income be Rs. 80,000 per annum and the difference between the incomes of B and D be the same as A's income, then B's income is:
A. Rs. 40,000
B. Rs. 60,000
C. Rs. 80,000
D. Rs. 1,20,000

Answer 1

Hint: Form equations / inequations using the given information. If we add / subtract the same quantity or multiply / divide by a positive quantity, the order of the quantities in relation to each other does not change. If we multiply / divide by a negative quantity, the order of the quantities gets reversed.

Complete step by step answer:
According to the question, the following set of relations can be identified:
 $ A+B>C+D $ ... (1)
 $ A+C=B+D $ ... (2)
 $ A=\dfrac{1}{2}(B+D) $ ... (3)
 $ A=80,000 $ ... (4)
 $ B-D=A $ ... (5)
From equations (3) and (4), we get:
 $ 80,000=\dfrac{1}{2}(B+D) $
⇒ $ B+D=1,60,000 $ ... (6)
From equations (4) and (5), we get:
 $ B-D=80,000 $ ... (7)
Adding equations (6) and (7), gives us:
 $ (B+D)+(B-D)=1,60,000+80,000 $
⇒ $ 2B=2,40,000 $
On dividing by 2, we get:
⇒ $ B=1,20,000 $

Therefore, the correct answer is D. Rs. 1,20,000.

Note: Some Rules of Equality:
If $ a=b $, then $ a\pm x>b\pm y $, $ a{{(x)}^{\pm 1}}=b{{(x)}^{\pm 1}} $ and $ \dfrac{x}{a}=\dfrac{x}{b} $ for all x.
If $ a=b $ and $ x>y $, then $ a+x>b+y $, but we cannot say anything definite about $ a-x,b-x $ or $ ax,bx $.

Some Rules of Inequalities:
If $ a>b $ , then $ a\pm x>b\pm x $ .
If $ a>b $ , then $ a{{(x)}^{\pm 1}}>b{{(x)}^{\pm 1}} $ if x>0 and $a{(x)^{\pm 1}}$ < $b{(x)^{\pm}}$ if $ x<0 $ .
If $ a>b $ , then $ \dfrac{x}{a}$ > $\dfrac{x}{b}$ if a,b>0 OR a,b<0.
If a>b and x>y, then a+x>b+y, but we cannot say anything definite about a-x and b-x or ax and bx.