Let us assume the digits of the original number are unit's digit a and ten's digit b.
The Original Number will be 10a + b.
After interchanging the digits the new number will be 10b + a.
According to question,
The number obtained by interchanging the two digits of a two-digit number is lesser than the original number by 54.
New Number = Original Number - 54
10b + a = 10a + b - 54
? 10b + a - 10a - b = -54
? 9b - 9a = -54
? a - b = 6....................................(1)
Again according to question,
Sum of the digits of original number = 10
a + b = 10..................................................(2)
Add the equation (1) and (2), we will get
a - b + a + b = 10 + 6
2a = 16
a = 8
Put the value of a in Equation (2) , we will get
8 + b = 10
b = 10 - 8
b = 2
Put the value of a and b for original number, we will get
10a + b = 10 x 8 + 2 = 80 + 2 = 82
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
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