Coin Problems

The key to these problems is to keep the quantity of coins distinct from the value of the coins. An example will illustrate.

Example : Laura has 20 coins consisting of quarters and dimes. If she has a total of $3.05, how many dimes does she have?

(A) 3 (B) 7 (C) 10 (D) 13 (E) 16

Let D stand for the number of dimes, and let Q stand for the number of quarters. Since the total number of coins in 20, we get D + Q = 20, or Q = 20 - D. Now, each dime is worth 10 cents, so the value of the dimes is 10D. Similarly, the value of the quarters is 25Q = 25(20 - D). Summarizing this information in a table yields

Notice that the total value entry in the table was converted from $3.05 to 305 cents. Adding up the value of the dimes and the quarters yields the following equation:

10D + 25(20 - D) = 305
10D + 500 - 25D = 305
-15D = -195
D = 13

Hence, there are 13 dimes, and the answer is (D).

Maximum Value of an expression

(A^x ) (B^y) (C^z) will be maximum when A/x = B/y = C/z

Ex (a+x)^3 (a-y)^4 will have the maximum value when

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

i.e 4a+4x=3a-3y i.e a=3y+4x

Age Problems

Typically, in these problems, we start by letting x be a person's current age and then the person's age a years ago will be x - a and the person's age a years in future will be x + a. An example will illustrate.

Example : John is 20 years older than Steve. In 10 years, Steve's age will be half that of John's. What is Steve's age?

(A) 2 (B) 8 (C) 10 (D) 20 (E) 25

Steve's age is the most unknown quantity. So we let x = Steve's age and then x + 20 is John's age. Ten years from now, Steve and John's ages will be x + 10 and x + 30, respectively. Summarizing this information in a table yields

Since "in 10 years, Steve's age will be half that of John's," we get

(x + 30)/2 = x + 10
x + 30 = 2(x + 10)
x + 30 = 2x + 20
x = 10

Hence, Steve is 10 years old, and the answer is (C).

Prime Numbers

How many divisors does 1200193 have and what is the sum of all those divisors?

Ans: (From the post http://www.cat4mba.com/node/3533#comment-1255 - user rajorshi)

The big pain being to fing the prime factors of 1200193 which i wont take, who knows the number could be prime and iwould have wasted half  a day.

let me give the formula generally :

if p = a^m  * bⁿ  * c^o  * ..... where a,b,c are prime.

thrn the number of divisors = (m+1)(n+1)(o+1).....

the sum of divisors = [(a^m+1  - 1)(bⁿ⁺¹  - 1)(c^o+1  - 1)......] / [(a-1)(b-1)(c-1).....]