A' s 1 day's work = 1/8
B' s 1 day's work = 1/10
? A' s share : B' s share
= 1/8 : 1/10 = 5/40 : 4/40 = 5 : 4
Let A' s share = 5k
and B' s share = 4k.
According to the question,
5k + 4k = 900
? 9k = 900
? k = 100
? A' s share = 5k = 5 x 100 = ? 500
and B' s share = 4k = 4 x 100 = ? 400.
0.53 = 53 / 99
(x x 5) = (0.75 x 8) ⟹ x = | ❨ | 6 | ❩ | = 1.20 |
5 |
(1 - 2) = -1
(3 - 4) = -1
(5 - 6) = -1
and so on.
? Total terms of -1 will be 50.
? Answer = -1 x 50 = -50
Let us draw a figure below as per given question.
Let AB = h meter be the height of the tower and two ships are situated at D and C respectively; such that, CD = 200 m; ?ADC = 30° ?ACB = 45° and BC = x meter (say)
Now from right triangle ABC,
tan 45° = h/x ? 1 = h/x
? x = h
Again from right triangle ABD,
tan 30° = h/( 200 + x )
? 1/?3 = h/( 200 + x )
Since x = h , we will get.
? 1/?3 = h/(200 + h )
? (200 + h ) = h X ?3
? h X ?3 - h = 200
? h x 1.732 - h = 200
? h(1.732 - 1) = 200
? h = 200/0.732 = 273.2 m = 273 m
? Probability in each trial (shooting) = 0.3
? Reqd. probability = (0.3)10
As per the given question ,
Food of 150 men for 45 days = 150 × 45 = 6750 unit
After 10 days , Food of 150 men for 150 × 10 = 1500 unit
And after 10 days remaining men = 150 - 25 = 125
and remaining food = 6750 - 1500 = 5250 unit
Let D be the number of days for which the remaining food .
So, 125 × D = 5250
? D = 5250 /125 = 42 days
Let the required number of units of work = k
According to the formula,
W1= m and W2 = k
M1T1D1W2 = M2T2D2W1
? m x m x m x k = n x n x n x m
? x= (m x n3) / m3 = n3/m2
Let the ages of mother, father and boys be M, F, B1 and B2, respectively.
The total age of four member
= 19 x 4
= 76 yr.
Given (B1 + B1)/2 = 11/2
? B1 + B2 = 11
M + F + B1 + B2 = 76 - 11
? M + F = 65 .......(i)
According to the question,
(B1 + B2 + F) / 3 = (B1 + B2 + M) / 3 + 3
? B1 + B2 + F) = B1 + B2 + M + 9
? F = M + 9
? F - M = 9 ............. (ii)
From Eqs (i) and (ii), we get
F = 37 yr and M = 28 yr
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