A.
A senior citizen invest Tk. 50 lac in a fixed deposit scheme at 11.5 % annual interest for six months. In every six months he withdraws Tk. 2 Lac from his principal plus interest earned. What will be his principal amount to invest after two years?
Ans: 42 Lac
B.
After traveling 108 km, a cyclist observed that he would have required 3 hrs less if he could have traveled at a speed 3 km/hr more. At what speed did he travel?
S*T = 108
Again, (S+3)(T-3) = 108
=> ST – 3S + 3T – 9 = ST
=> 3T – 3S = 9
=> T – S = 3
=> T = S+3
Now, S*(S+3) = 108
=> S² + 3S – 108 = 0
=> S² + 12S – 9S – 108 = 0
=> S(S+12) – 9(S+12) = 0
=> (S+12)(S-9) = 0
S = 9 (Ans.)
SHORTCUT:
108 = 9*12 = 12*9
D = S*T
So, S = 9 and T = 12
Ans: 9
C. Three numbers are in A.P. and their sum is 30. Also, the sum of their squares is 308. Find the numbers.
Solution:
Let,
Second Term = a
Common deference = d
So, first term will be = a – d
Third term will be = a + d
A.T.Q.,
a- d + a + a+d = 30
or, 3a = 30
so, a = 10
So, first term = 10 – d
Second term = 10
Third term = 10 + d
Again A.T.Q.,
(10 – d)2 + (10)2 + (10+d)2 = 308
Or, 100 – 20d + d2 + 100 + 100 + 20d + d2 = 308
Or, 2d2 + 300 = 308
Or, 2d2 = 8
Or, d2 = 4
So, d = 2
So first term = 8, second term = 10 , third term = 12
Ans: 8,10,12
OR,
Let the Three numbers: a,a+x,a+2x
Atq, 3a+3x=30
A+X=10
X=10-A
And,
A.A+10.10+(20-A)(20-A)=308
500+2A.A-40A-308=0
A.A -20A+192=0
A=8,12
A=8
8,10,12
Ans: 8,10,12
D.
A square is inscribed inside a circle. What is the area of the square, if the radius of the circle is 10 cm?
Here,
Radius of circle = 10 cm
So, Diameter of circle = 10*2 = 20 cm
Then, Diagonal of square = 20 cm
Therefore, Area of square = d^2/2 = 20^2/2 = 400/2 = 200 cm^2
Ans: 200 CM2
A:
Ans:
B.
Ans: 10 days
C.
Ans: 3/8, 3/8, 3/8
D.
