Given, xyz = 1, ax = b, by = c
Now, b = ax
Apply the law of algebra .
Take the power y of both side equation.
? by= axy
Take the power z of both side equation.
? byz = axyz
? ( by )z = axyz
Replace the by with c and xyz with 1 as per given equation,
? cz = a
∴ The watch gains | ❨ | 2 + 4 | 4 | ❩min. | or | 34 | min. in 170 hrs. |
5 | 5 |
Now, | 34 | min. are gained in 170 hrs. |
5 |
∴ 2 min. are gained in | ❨ | 170 x | 5 | x 2 | ❩hrs | = 50 hrs. |
34 |
∴ Watch is correct 2 days 2 hrs. after 12 p.m. on Monday i.e., it will be correct at 2 p.m. on Wednesday.
Perimeter = Distance covered in 8 min. = | ❨ | 12000 | x 8 | ❩m = 1600 m. |
60 |
Let length = 3x metres and breadth = 2x metres.
Then, 2(3x + 2x) = 1600 or x = 160.
∴ Length = 480 m and Breadth = 320 m.
∴ Area = (480 x 320) m2 = 153600 m2.
Also, lb = 20.
(l + b)2 = (l2 + b2) + 2lb = 41 + 40 = 81
⟹ (l + b) = 9.
∴ Perimeter = 2(l + b) = 18 cm.
Let 40% of A = | 2 | B |
3 |
Then, | 40A | = | 2B |
100 | 3 |
⟹ | 2A | = | 2B |
5 | 3 |
⟹ | A | = | ❨ | 2 | x | 5 | ❩ | = | 5 |
B | 3 | 2 | 3 |
∴ A : B = 5 : 3.
So, in a day, the hands point in the opposite directions 22 times.
∴ Last day of 1st century is Friday.
200 years contain (5 x 2) ≡ 3 odd days.
∴ Last day of 2nd century is Wednesday.
300 years contain (5 x 3) = 15 ≡ 1 odd day.
∴ Last day of 3rd century is Monday.
400 years contain 0 odd day.
∴ Last day of 4th century is Sunday.
This cycle is repeated.
∴ Last day of a century cannot be Tuesday or Thursday or Saturday.
∴ B covers 200 m in | ❨ | 7 | x 200 | ❩ | = 40 sec. |
35 |
B's time over the course = 40 sec.
∴ A's time over the course (40 - 7) sec = 33 sec.
∴ In 24 hours, they are at right angles 44 times.
T.D. |
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= Rs. 400. |
So, the day on 8th Dec, 2007 will be 1 day beyond the day on 8th Dec, 2006.
But, 8th Dec, 2007 is Saturday.
∴ 8th Dec, 2006 is Friday.
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