If p : q : r = 2 : 3 : 5, then what will be the ratio of (p + q + r) : (3p + q – r)?

 

Let us assume S as number of shirts and T as number of trousers
Given that each trouser cost = Rs.70 and that of shirt = Rs.30
Therefore, 70 T + 30 S = 810
=> 7T + 3S = 81......(1)
T = ( 81 - 3S )/7

We need to find the least value of S which will make (81 - 3S) divisible by 7 to get maximum value of T
Simplifying by taking 3 as common factor i.e, 3(27-S) / 7
In the above equation least value of S as 6 so that 27- 3S becomes divisible by 7

Hence T = (81-3xS)/7 = (81-3x6)/7 = 63/7 = 9

Hence for S, put T in eq(1), we get
S = 81-7(9)/3 = 81-63 / 3 = 18/3 = 6.
The ratio of T:S = 9:6 = 3:2.

Santhosh's investment = 6000 x 70/100 = Rs.4200
Rajitha's investment = 4200 x 5/4 = Rs.5250
Therefore, total amount invested = 6000 + 4200 + 5250 = Rs.15450.
Required ratio = 5250 : 15450 = 35 : 103.

Given total rupees = 20 Rs
No. of one rupee stamps = 3
Now, remaining money = Rs. 17
With that he buys only 2 and 5 rupee stamps
Let number of Rs. 5 stamps = K
Let number of Rs. 2 stamps = L
5K + 2L = 17
K = 3, L = 1 (possible)
L = 6, K = 1 (possible)
=> But given that they are different in number so, K is not equal to 3
one rupee stamps = 3
2 two stamps = 6
5 rupee stamps = 1
Total number of stamps = 10.

6,400 gents, 2400 ladies in that company.
So total 8,800employees.
for 8,800 employees, we want 2400 ladies.
for 12,100employees we want how many ladies ?
=> (12,100/8,800) x 2400 = 3300.
So we want 3300 - 2400 = 900 more ladies.

Ratio of the share of a lady, a gents and a girl is 7 : 4 : 3
No of ladies, gents and girls are 5, 3, 3
Thus effective ratio of ladies, gents and girls is 7 x 5 : 4 x 3 : 3 x 3 = 35 : 12 : 9
so part of gents = (12/56) x 7444 = Rs. 1595.142
so part of 1 gents = 1595.142/3 = Rs. 531.71.

1st supporter recieve 440 calls
2nd supporter recieve 360 calls
3rd supporter recieve 300 calls

So total calls = 1100 calls ;
Calls this month= 1500
So remaining calls to be distributed is 400

So Now Ratio 1st:2nd:3rd ==> 440:360:300
=> 22:18:15

Now No. of More Calls 2nd supporter will get => [18/(22+18+15)] x 400
=> (18/55) x 400
=> 131 Calls
So 131 more Calls than last month.

If there are total 100 candidates,
40 candidates are really capable and
60 candidates are incapable.
Of the candidates who are really capable = 80%
40*(80/100)=32 pass the test and
of the incapable= 30%
60*(30/100)=18 pass the test.
Total candidates who passed the test = 32+18 = 50
The proportion of the really capable students who can pass the test to the total students who can pass = 32/50 = 16/25.

Seats for Mathematics, Physics and Biology in a school are in the ratio 5 : 7 : 8.

Seats for Mathematics, Physics and Biology in a school are 5x, 7x and 8x.

There is a proposal to increase these seats by 40%, 50% and 75% respectively.

Then Seats for Mathematics, Physics and Biology in a school will be

1.4 X 5x, 1.5 X 7x and 1.75 X 8x and

ratio will be 7 : 10.5 : 14  => 2:3:4.