Let the speed of rowing in still water be u km/h.
Distance in downstream motion = 21 km
and speed downstream = (u + 4) km/h
? Time taken = 21/ (u + 4)
Distance upstream motion = 21 km and speed upstream = (u +4) km
? Time taken = 21/(u - 4)
According to the question.
21/(u + 4) + 2 = 21/(u - 4)
? (21 + 2u + 8)/u + 4 = 21/(u - 4)
? (2u + 29)/(u + 4) = 21/(u - 4)
? (u - 4) (2u + 29) = 21(u + 4)
? 2u2 - 8u + 29u - 116 = 21u + 84
? 2u2 + 21u - 116 = 21u + 84
? 2u2 = 84 + 116
? u2 = 200/2
? u2 = 100
? u = 10 km/h
Area of rhombus =(d1 x d2)/2
=(1 x 1.5)/2 m2
= 0.75 m2
Required number of arrangements = 6 ! / 3 ! [? S has come thrice ]
= [ 6 x 5 x 4 x 3! ] /3 !
= 120
Required LCM = a x 2 x 5 x 7 = 70a
Straight forward answer 12, 24, 36
B.D. for 3⁄2 years | = Rs. 558. | |||||||
B.D. for 2 years |
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= Rs. 744 |
T.D. for 2 years = Rs. 600.
∴ Sum = | B.D. x T.D. | = Rs. | ❨ | 744 x 600 | ❩ | = Rs. 3100. |
B.D. - T.D | 144 |
Thus, Rs. 744 is S.I. on Rs. 3100 for 2 years.
∴ Rate = | ❨ | 100 x 744 | ❩% | = 12% |
3100 x 2 |
Internal radius = 3 cm.
Volume of iron |
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= 462 cm3. |
∴ Weight of iron = (462 x 8) gm = 3696 gm = 3.696 kg.
Let E = Event of getting same number on both the occasions
= (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)
? n(E) = 6
? Required probability = 6/62 = 1/6
= √(1 - 2a)2 + 3a
= (1 - 2a) + 3a
= (1 + a)
= (1 + 0.1039)
= 1.1039
Given expression
= [(0.5)3 + (0.3)3] / [(0.5)2 - 0.5 x 0.3 + (0.3)2]
= [a3 + b3] / [a2 - ab + b2]
= [(a + b)(a2 - ab + b2)] / [a2 -ab + b2]
= a+b
= 0.5 + 0.3
= 0.8
Speed of boat in still water = 15 - 1.5 = 13.5 km/hr
Speed of boat in upstream = 13.5 - 1.5 = 12 km/hr
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