let us assume the ten's digit number be x and unit's place digit be y.
Then two digit Original Number = 10x + y
According to question ? y = 2x - 1 ................(1)
When digits are interchanged then new number = 10y + x
If the digits at unit's and ten's place are interchanged, the difference between the new and the original number is less than the original number by 20,
New Number - Original Number = Original number - 20
? 10y + x - (10x + y) = 10x + y - 20
? 10y + x - 10x - y - 10x - y = - 20
? 8y - 19x = - 20
? 19x - 8y = 20 ...................(2)
Put the value of y from equation (1) in above equation (2).
? 19x - 8(2x - 1) = 20
? 19x - 16x - 8 = 20
? 7x = 20 + 8
? 7x = 28
? x = 28/7
? x = 4
Put the value of y in equation (1)
From (i) y = 2 x 4 -1 = y = 7
? original number = 10x + y =10 x 4 + 7 = 47.
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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