? 1/5A : 1/8B = 3 : 4
? 8A/5B = 120/160
? A/B = 120/160 x 5/8 = 15/32
? First part = Rs. (94 x 15/47) = Rs. 30
? 5A = 3B = 2C = x
? A = x/5, B=x/3 and C =x/2
? A : B : C = x/5 : x/3 : x/2 = 6 : 10 : 15
Gold in C =(7/9 + 7/18) = 21/18 = 7/6
Copper in C = (2/9 + 11/18) = 15/18 = 5/6
? Gold : Copper = 7/6 : 5/6 = 7 : 5
Milk = (729 x 7/9) = 567 ml
Water = (729 x 2/9) = 162 ml
Let water to be added is w
? 567 / (162 + w) = 7/3
? 3 x 567 - 7 x 162 = 7 x w
? 1701 - 1134 = 7 x w
? 7w = 1701 - 1134
? w = 567/7 = 81 ml
? 2A/5 + 40 = 2B/7 + 20 = 9C/17 + 10 = k
? A = 5(k - 40)/2, B =7(k - 20)/2 and C = 17(k - 10)/9
? 5(k - 40)/2 +7(k - 20)/2 +17/(k - 10 )/9 = 600
? 45k - 1800 + 63k - 1260 + 34k - 340 = 10800
? 142k = 14200
? k = 14200 / 142 = 100
Hence A's share = (5/2) x ( 100 - 40) = Rs. 150
Let, 1 gm of gold be mixed with y gm of copper to give (1 +x) gm of mixture.
Now, 1G =19W and 1C = 9W and mixture = 15W
Now, 1 gm gold + y gm copper = (1 + y) gm mixture
? 19 W + 9W x y = (1 + y ) x 15W
Thus 4W = 6W x y
? y = 4W/6W = 4/6 = 2/3
So the required ratio is 1 : 2/3 i.e. 3 : 2
Suppose C gets Re.1. Then B gets Re. (1/4)
? A gets = Re. (2/3 x 1/4) = Re. 1/6
? A : B : C = 1/6 : 1/4 : 1 = 2 : 3 : 12
Hence B's share = Rs. (680 x 3/17) = Rs. 120
? Remainder = Rs. [ 2430 - (5 + 10 + 15)] = Rs. 2400
&there4 A's share = Rs. [ (2400 x 3/12) + 5 ] = Rs. 605
Ratio of sides 1/3 : 1/4 : 1/5 = 20 : 15 : 12
Length of smallest side = 94 x (12/47) cm. = 24 cm.
Suppose C gets Rs. x
Then, B gets Rs. (x + 8) and A gets Rs. (x + 15)
? x + x + 8 + x + 15 = 53
? x =10
so, A : B : C = 25 : 18 : 10
A : B : C = 100 : 65 : 35 = 20 : 13 : 7
If C's share is Rs. 7, the sum is Rs. 40.
If C's share is Rs. 28, the sum is Rs.(40/7 x 28) = Rs. 160.
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