5 men + 3 boys can reap 23 hectares in 4 days ......(i)
3 men + 2 boys can reap 7 hectares in 2 days ......(ii)
? From (i),
14( 5 men + 3 boys) can reap 23 x 14 hectares in 4 days
Now, from (ii)
? 23 (3 men + 2 boys) can reap 7 x 2 x 23 hectares in 4 days
? 14(5 men + 3 boys) = 23 (3 men + 2 boys)
? 70 men + 42 boys = 69 men + 46 boys
? 1 men = 4 boys
Now, 5 men + 3 boys = 23 boys
? 23 boys can reap 23 hectare 4 days
? 1 boy can reap 1 hectares in 4 days
? 4 boys can reap 1 hectare in 1 day
? 4 x 45 boys can reap 45 hectaes in 1 day
? 4 x 45 / 6 boys can reap 45 hectares in 6 days
? 30 boys can reap 45 hectares in 6 days
But 30 boys = 28 boys + 2 boys = 7 men + 2 boys
Hence, 2 boys can assist 7 men for the work.
The number of pieces of chocolate left with Manju
= 1- [(1/4) + (1/3) + (1/6)
= 1 - (3 + 4 + 2)/12
= 1 - 9/12 = (12 - 9) / 12
= 3/12
Third power of 4 = 43 = 4 × 4 × 4 = 64
Total number of cards n(S) = 52
Number of red cards n(E) = 26
? P(E) = n(E)/n(S) = 26/52 = 1/2
Given Exp. = [(4 x 5 x 6) + (5 x 6) + (2 x 6) + (2 x 3)] / [(2 x 3 x 4 x 5 x 6)]
= 168 / (24 x 30)
= 7 / 30
Speeds (in m/sec) of A and B are
9 / 2 * 5 / 18 = 5 / 4 and 6 * 5 / 18 = 5 / 3, respectively.
A has a start of 190 m.
So, A has to run 1000 - 190 = 810 m, while B 1000 m.
Time taken by B to cover 1000 m = 3 5 × 1000 = 600 seconds.
In this time, A covers 5 × 600 / 4 = 750 m.
So, B reaches the winning post while A remains 810 - 750 = 60 m behind.
? B wins by 60 m.
Let 30% of 80/y =24
? 24y = (30 x 80)/100
? y = (30 x 80)/ (24 x 100) =1
Let the radius of the park be r, then
?r + 2r = 288
(? + 2)r = 288
? [(22/7) + 2]r = 288
? r = (288 x 7)/36 = 56
? Area of the park = (1/2)?r2
= (1/2) x (22/7) x 56 x 56
= 4928
1178.999 x 25.0001 - ?? x 16.0011 = 29075
? 1179 x 25 - ?? x 16 = 29075
? 29475 - ?? x 16 = 29075
? ?? x 16 = 29475 - 29075 = 400
? ?? = 25
? ? = 25 x 25 = 625
So, 96 is wrong.
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