Price increased by M% from 1991 to 1992 and by N% from 1992

Let price on 1991 be P
in 1992=P(1+m/100)
1993=P(1+m/100)(1+n/100)
%increase= change(1993)-original(1991)/original(1991)
=P(1+m/100)(1+n/100)-P/P
P cancels out.
(1+m/100)(1+n/100)-1
(10^4+100m+100n+mn-10^4)/10^4
(100m+100n+mn)/10^4------(a)
Given in (2) 100m+100n+mn=4300-------->m+n+mn/100=43----(b)
taking 100 out common from (a)
(m+n+mn/100)/10^2
substituting from(b) we get
43/100
43%
Ans B

For Statement 2:It seems like a lengthy process.Any shortcuts ?

It took me 4 mins to solve.

There is a formula for cumulative percentage changes:

a + b + ab/100 = total percentage change
a = first period change
b = second period change.

In statement (2) we are simply given this formula, multiplied by 100.

100M + 100N + MN = 4300

equals

M + N + MN/100 = 43

The cumulative increase is 43 %.

Price increased by \(m\%\) from 1991 to 1992 and by \(n\%\) from 1992 to 1993. What was the percentage increase in price from 1991 to 1993?

Increasing some value by \(x\%\) is the same as multiplying by \(1 + \frac{x}{100}\). For example, if you increase something by 10% you multiply by \(1 + \frac{10}{100}= 1.1\). Hence, after the increase by \(m\%\) and then by \(n\%\), the price would become \((1 + \frac{m}{100})(1 + \frac{n}{100}) = 1 + \frac{m}{100} + \frac{n}{100} + \frac{mn}{10,000}\) times the original price.

(1) \(mn = 300\).

From the above, we cannot deduce the value of \((1 + \frac{m}{100})(1 + \frac{n}{100}) \). Not sufficient.

(2) \(100m + 100n + mn = 4300\).

Dividing the above by 10,000 gives \(\frac{m}{100} + \frac{n}{100} + \frac{mn}{10,000}=0.43\). Adding 1 to both sides gives \(1 + \frac{m}{100} + \frac{n}{100} + \frac{mn}{10,000}=1.43\). Therefore, the price increased by 43%. Sufficient.


Answer: B