Speed of truck 640/10 = 64 km/h
Speed of car = 640/8 = 80 km/h
So, the required ratio = 64/80 = 4/5 = 4 : 5
Total number of student = 8670
Total number of boys = 4545
? Total number of girls = 8670 - 4545 = 4125
? Required ratio = 4545 : 4125 = 303 : 275
Let original number of boys = 4N and number of girls = 5N
According to the question
(4N + 10)/5N = 6/5
? 30N = 20N + 50
? 10N = 50
? N = 50/10 = 5
? Number of girls = 5N = 5 x 5 = 25
Let number of boys = 4N and number of girls = 5N
Accounting to the question
4N / (5N - 100) = 6/7
? 28N = 6 ( 5N - 100 )
? 28N = 30N - 600
? 2N = 600
? N = 600/2 = 300
? Number of boys = 4N = 4 x 300 = 1200
Given that
Sanjay + Suresh = 120 ...(i)
Sanjay - Suresh = 30 ...(ii)
By adding Eqs (I) and (ii) we get
2 x sanjay = 150
? sanjay = 150/2 = 75 kg
? Weight of Suresh = 75 - 35 = 45 kg
? Suresh : Sanjay = 45 : 75 = 3 : 5 = 3/5 = 0.6
Given that, x/2y = 6/7
? x/y = 12/7
By componenod dividendo,
(x - y) / (x + y) = (12 - 7) / (12 + 7) = 5/19
? (x - y) / (x + y) + 14 /19 = 5/19 + 14/19 = 19/19 = 1
Case I Speed = Distance/ Time
Case II Speed = (Distance/2) / (2 x Time) = Distance / (4 x Time)
? Required ratio = 1/4 = 1:4
The ratio among A, B, C and D = 1/3 : 1 / 4 : 1/5 : 1/6
On rearranging the ratio = 60/3 : 60/4: 60/5 : 60/6 = 20 : 15 : 12 : 10
So minimum number of pens can be when the common ration is 1.
So minimum number of pen = 20 + 15 + 12 + 10 = 57
If three parts be P, Q and R,
then, P/2 = Q/3 = R/5
? P : Q : R = 2 : 3 : 5
? P = (2/10) x 990 = 198
Q = (3/10) x 990 = 297 and
R = (5/10) x 990 = 495
Let the quantity be k.
Then, (a + k) / (b + k) = c/d
? (a + k)d = (b + k)c
? ad + dk = bc + ck
? ad - bc = ck - dk
? ad - bc = k(c - d)
? k = (ad - bc) / (c - d)
Given, A : B = 2 : 3
and B : C = 4 : 5
Then, A : B : C = 8 : 12 : 15
7there4; Share of B = [12/(8 + 12 + 15)] x 7000
= (12 x 7000)/35
= 2400
= ? 2400
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