PSA_PRACTICE_QNR23. Kay spends $134 on 11 tickets of two different classes. Class A tickets are $3 more expensive than class B. How many class A tickets and at what price does she buy?
Ans: Let, she buys ‘a’ tickets of Class A and ‘b’ tickets of Class B.
a + b = 11
Also, let us assume price of Class B ticket is ‘p’ and that of Class A is ‘p+3’
Total amount spent on tickets= 134.
a*(p+3) + b*(p)= 134
=> p*a+ p*b + 3a= 134
=> p(a+b) + 3a =134
=> 11p + 3a =134 —– (Z)
Now we have two variables but one equation; let us re-arrange the equation in the below fashion
3a = 134 – 11p
=> 3a = (132 +2) -9p -2p
=> 3a= (132-9p) +2(1-p)
The purpose of the last 3 lines is to try to express LHS as multiple of 3. Similarly, on the RHS, we expressed 134 as (132+2) where 132 is a multiple of 3. Also 11p is expressed as 9p+2p where 9p is a multiple of 3.
Since, LHS is a multiple of 3, RHS should also be. In RHS, (132-9p) is a multiple of 3, thus 2(1-p) should also be a multiple of 3.
p=1 is the first value; from equation (Z), a=41.
Other values of (p,a) are (4,30) ,(7,19) , (10,8) …
The only possible set is (10,8) as a+b is given as 11 and other values of a are violating this condition. Kay buys 8 class A tickets at $10 each.