Permutation-Combination
Aptitude

 Back to Questions
Q.

Jay wants to buy a total of 100 plants using exactly a sum of Rs 1000. He can buy Rose plants at Rs 20 per plant or marigold or Sun flower plants at Rs 5 and Re 1 per plant respectively.

If he has to buy at least one of each plant and cannot buy any other type of plants, then in how many distinct ways can Jay make his purchase?

 A.

2

 B.

3

 C.

4

 D.

5

 Hide Ans

Solution:
Option(B) is correct

Let the number of Rose plants be $a$.

Let number of marigold plants be $b$.

Let the number of Sunflower plants be $c$.

$⇒ 20a+5b+1c=1000$; 

and $a+b+c=100$

Solving the above two equations by eliminating $c$,

$19a+4b=900$

$b= \left(\dfrac{900-19a}{4}\right) = 225 - \dfrac{19a}{4}$ ---------- (1)

$b$ being the number of plants, is a positive integer, and is less than 99, as each of the other two types have at least one plant in the combination i.e .:

$0 < b < 99$ ---------- (2)

Substituting (1) in (2),

$0 < 225 - \dfrac{19a}{4} < 99$

⇒ $225 < -\dfrac{19a}{4} < (99-225)$

⇒ $4×225 >19a > 126×4$

⇒$\dfrac{900}{19} > a >504$

a is the integer between 47 and 27 ---------- (3)

From (1), it is clear, a should be multiple of 4.

Hence, possible values of a are $(28,32,36,40,44)$

For $a=28$ and $32$, $a+b > 100$

For all other values of $a$, we get the desired solution:

$a=36, b=54, c=10$

$a=40, b=35, c=25$

$a=44, b=16, c=40$

Three solutions are possible.


(1) Comment(s)


Prashanth
 ()

is this not correct ??? if he buy likes this

a=37,b=47,c=5