Then, number of members owning only 2 cars = 20.
Number of members owning 3 cars = 40% of 80 = 32.
Number of members owning only 1 car = 100 - (20 + 32) = 48.
Thus, 48% of the total members own one car each.
Then, 3x + x = 4x = total number of students.
Thus, to find exact value of x, the total number of students must be divisible by 4.
There are 11 gapes between the two corner trees (11 x 2) metres and 1 metre on each side is left.
Therefore Length = (22 + 2) m = 24 m.
B-3 = E ...(i)
B + 3 = D ...(ii)
A+B = D + E+10 ...(iii)
B = C + 2 ...(iv)
A+B + C + D + E= 133 ...(v)
From (i) and (ii), we have : 2 B = D + E ...(vi)
From (iii) and (vi), we have : A = B + 10 ...(vii)
Using (iv), (vi) and (vii) in (v), we get:
(B + 10) + B + (B - 2) + 2B = 133 ⟺ 5B = 125 ⟺ B = 25.
= (No. of digits in 1- digit page nos. + No. of digits in 2-digit page nos. + No. of digits in 3- digit page nos.)
= (1 x 9 + 2 x 90 + 3 x 267) = (9 + 180 + 801) = 990.
So, the bells will toll together after every 420 seconds i.e. 7 minutes.
Now, 7 x 8 = 56 and 7 x 9 = 63.
Thus, in 1-hour (or 60 minutes), the bells will toll together 8 times, excluding the one at the start.
Let the runs scored by E be x.
Then, runs scored by D = x + 5; runs scored by A = x + 8;
runs scored by B = x + x + 5 = 2x + 5;
runs scored by C = (107 - B) = 107 - (2x + 5) = 102 - 2x.
So, total runs = (x + 8) + (2x + 5) + (102 - 2x) + (x + 5) + x = 3x + 120.
Therefore 3x + 120 =180 ⟺ 3X = 60 ⟺ x = 20.
Then, we have :
d - 1 = s and 2 (s - 1) = d.
Solving these two equations, we get: d = 4, s = 3.
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