Let the number be 81a and 81b where a and b are two numbers prime to each other.
? 81a + 81b = 1215
? a + b = 1215/81 = 15
Now, find two numbers, whose sum is 15, the possible pairs are (14, 1) , (13, 2), (12, 3), (11, 4), (10, 5), (9, 6 ), (8, 7) of these the only pairs of numbers that are prime to each other are (14, 1) , (13, 2), (11, 4), and (8, 7).
Hence, the required number are
(14 x 81, 1 x 81), (13 x 81, 2 x 81), (11 x 81, 4 x 81), (8 x 81, 7 x 81)
or (1134, 81), (1053, 162), (891, 324), (648, 567)
So, there are four such pairs .
Then, | n | [2a + (n - 1)d] = 1800 |
2 |
⟹ | n | [2 x 6 + (n - 1) x 6] = 1800 |
2 |
⟹ 3n (n + 1) = 1800
⟹ n(n + 1) = 600
⟹ n2 + n - 600 = 0
⟹ n2 + 25n - 24n - 600 = 0
⟹ n(n + 25) - 24(n + 25) = 0
⟹ (n + 25)(n - 24) = 0
⟹ n = 24
Number of terms = 24.
Given expression = [(0.05)2 + (0.41)2 +(0.073)2] / [(0.005)2 + (0.041)2 + (0.0073)2]
Which is in form of [(a2 + b2 + c2)] /[ (a/10)2 + (b/10)2 + (c/10)2] = 100 x (a2 + b2 + c2) / (a2+b2+c2)
= 100
So, A, B and C will again meet at the starting point in 2772 sec. i.e., 46 min. 12 sec.
Given expression = [(.356)2 -2 x .356 x .106 + (.106)2] / [(.632)2+2 x .632 x .368 +(.368)2]
= (a2 -2ab + b2) / (c2+2cd+d2)
= (a - b)2 / (c + d)2
= (.356 -.106)2 / (.632 + .368)2
= (.25)2
= .0625
(469 + 174)2 - (469 - 174)2 | =? |
(469 x 174) |
Given exp. = | (a + b)2 - (a - b)2 |
ab |
= | 4ab |
ab |
= 4 (where a = 469, b = 174.)
Given expression
=[(3.65)2 + (2.35)2 -2 x 3.65 x 2.35 ] / 1.69
=a2 + b2 - 2ab / 1.69, Where a = 3.65 and b = 2.35
=(a - b)2 / 1.69
= (3.65 - 2.35)2 / 1.69
= (1.3)2 / 1.69
= 1.69 / 1.69
= 1
❨ | xb | ❩ | (b + c - a) | . | ❨ | xc | ❩ | (c + a - b) | . | ❨ | xa | ❩ | (a + b - c) | =? |
xc | xa | xb |
Given Exp. |
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Given expression = (0.01 - 0.0001) / (0.00001 + 1)
= (.0099 / .0001) + 1
= 99 + 1
= 100
(243)n/5 x 32n + 1 | =? |
9n x 3n - 1 |
Given Expression |
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The Given Number are 1.08, .36 and 0.9
G. C. D of 108, 36 and 90 is 18
? Required G. C. D. = 0.18
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