Dutch-Bangla Bank Ltd || PO (15-11-2019) || 2019

All

Here, to solve the problem, we need to use some laws from the primary concept of sets.

Let, n(F) = 60; n(M) = 45 n(Om)=45; n(T) = 100

All courses taken, n(F  M  Om )=10

None taken, n(F  M  Om )^ =10

We know, z(F  M  Om )=n(T)-n(F  M  Om)^

= n(F  M  Om)= 100 - 10 = 90

= So, we can write, n (F  Om)

= n(F)+n(M)+n(Om)-n(FM)-n(MOm)-n(OmF)+n(FMOm)

= 90=60+45+45-{n(FM)+n(MOm)+n(OmF}+10

=90=160-ln(F  M)+n(M  Om )+n(Om  F)

n(F  M)+n(M  Om )+n(Om  F)=70

The number of students who took 2 of these 3 courses is

=70-3×(F  M  Om)= 70 - (3 ×10) = 70 - 30 = 40

Given, length = 90 feet and Breadth = 66 feet

Stimulated length of tree plantation = 90-(5+5)=90-10 = 80 feet

Stimulated breadth of tree plantation = 66-(5+5)=66-10=56 feet

The number of columns bc = 804+1 [leaving 4 feet in rows and columns]

=20+1=21

And the number of rows be = 564+1 [leaving 4 feet in rows and columns]

== 14+1=15

Maximum number of trees that can be planted = 21 × 15 = 315