If Rs. 10 be allowed as true discount on a bill of Rs. 110 due at the end of a certain time, then the discount allowed on the same sum due at the end of double the time is?

Total events = n (s) = 2⁵ = 32
n(E) of getting heads = 1
p(E) = 1/32
? n(E) = 1 - p(E) = 1 - 1/32 = 31/32

Number of trees that can be planted on one side of road = (1760 / 20) + 1
= 88 + 1
=89

? Trees on the both side = 2 x 89=178

1307 x 1307 = (1307)²
= (1300 + 7 )²
= (1300)² + (7)² + 2 x 1300 x 7
= 1690000 + 49 +18200
= 1708249

? 2?R - 2?r = (176-132)
? 2?(R-r) = 44
? R-r = ( 44 x 7 )/ (2 x 22)
= 7 m

Given Exp. = (1/2 x 1/4 + 20) / (2 + 20) = (161/8) x (1/22)
= 161 / 176

NA

If the required fraction be P
According to the question
(P x P) / (1/P) = 18²⁶/₂₇
? P³ = 512/27
? P = 8/3 = 2²/₃

(0.04)³ = 0.04 x 0.04 x 0.04
= 0.000064

We have p³ + q³ + r³ ? 3pqr = (p + q + r) (p² + q² + r² ?pq ? qr - rp)
Here p = a ? 4, q = b ? 3, r = c ?1
So, given expression is (p + q + r) (p² + q² + r² ? pq ? qr ? rp)
= (a ? 4 + b ? 3 + c ? 1) (p² + q² + r² ? pq ? qr ? rp)
= (a + b + c ? 8) (p² + q² + r² ? pq ? qr ? rp)
= (8 ? 8) (p² + q² + r² ? pq ? qr ? rp)
? (a ? 4)³ + (b ? 3)³ + (c ? 1)³ ? 3 (a ? 4) (b ? 3) (c ? 1) = 0

Let 3571 + x - 6086 = 115
Then, x = 6086 +115 -3571
= 6201-3571
= 2630