Ratio And Proportions - Selected Problems in Ratio and Proportion

Ratio And Proportions

Well if you are planning to go for any interview, sitting for any placement, or any competitive exam for any job application, pin pointing to especially Bank PO exams / Bank Clerical Exams or any exam for recruitment to govt posts, you cannot miss out this topic of Mathematics. Then better have a look at it (Ratio and Proportion). Atelast 3 to 4 Questions in mathematics under algebra will surely be framed from this section because this is neither tough nor lengthy but good and concise questions can be framed from here to test your calculations and understanding. SO DONT MISS OUT THE EASY MARKS IN MATHEMATICS....Just take pain of 10 minutes to digest this capsule on Ratio & Proportion and make yourself little aware / refresh.



Problem in Ratio and Proportion (Mathematics) :
Three positive numbers are in the ratio 2:3:4 and the sum of their square is 1421. Find the numbers?

Solution:
Let the numbers in the ratio 2:3:4 be 2x,3x and 4x
Sum of their squares = (2x)2 + (3x)2 + (4x)2 =4x2 + 9x2 + 16x2 = 29x2 = 1421
x2 = 1421/29 = 49
x = 7

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Continued Proportion –
a, b and c are said to be in continued proportion if a/b =b/c or means b2=ac
Similarly we can also have a/b=b/c=c/d=d/e.......... for more numbers

Problem in Ratio and Proportion (Mathematics):
If a, b, c, d are in continued proportion, the prove that a. b. d. e = c4

Solution :
Since they are in continued proportion, we can have a/b =b/c =c/d =d/e =say k
so we have
a=bk
b=ck
c=dk or d=c/k
d=ek or e=d/k

We have to prove a. b. c. e = (bk).(ck). (c/k) . (d/k) = (ckk).(ck).(c/k).(c/kk) = c4 k3 / k3 = c4

Problem in Ratio and Proportion (Mathematics):
If a/b=c/d=e/f
prove that, (a+3c-5e)/(b+3d-5f) = a/b = c/d = e/f

Solution :
Let a/b=c/d=e/f = k(say)
(bk+3dk – 5fk) / (b+3d – 5f) = k (b+3d – 5f)/(b+3d – 5f) = k
Hence Proved

Problem in Ratio and Proportion (Mathematics) :
If p/(b-c) = m/(c-a) = n/(a-b) then prove that ap+bm+cn = 0

Solution:
Take each ratio = k
p=(b-c)k
m=(c-a)k
n=(a-b)k

Then ap+bm+cn = a(b-c)k + b(c-a)k + c(a-b)k = abk -ack + bck – abk +ack -cbk = 0



Problem in Ratio and Proportion (Mathematics) :
Ram, Sham and Suresh start business investing in the ratio 1/2 : 1/3: 1/4. The time for which each of them invested their money was in the ratio 8:6:12 respectively. If they get profit of Rs 18000 from the business. Then how much share of profit will each one get ?

Solution:
ratio of Investment = 1/2:1/3:1/4=6:4:3 and time for which investment is done = 8:6:12
So in actual their amount was in business in the ratio 6*8 : 4*6 : 3*12 =48:24:36 =4:2:3
So they should also share the profit in this ratio 4:2:3
So Ram gets = 4/9 *18000 =8000
Sham gets = 2/9*18000 = 4000
Suresh gets = 3/9*18000 = 6000

{This is actually based on the concept that a small sum put in for large time and large sum put in for small time, so actually we see for how much period we can have money on same scale. You can compare by money * time }

Problem in Ratio and Proportion (Mathematics) :
p,q and r are three positive numbers and Q=(p+q+r)/2
If (Q-p):(Q-q):(Q-r) = 2:5:7, then find the ratio of p,q and r ?

Solution
:
Substitute the value of Q =(p+q+r)/2 everywhere
you will get
(p+q+r-2p)/2 : (p+q+r-2q)/2 : (p+q+r -2r)/2 = 2:5:7
(p+q+r-2p) : (p+q+r-2q) : (p+q+r -2r) = 2:5:7
(q+r-p) : (p-q+r) : (p+q -r) = 2:5:7
or (q+r-p) : (p-q+r) : (p+q -r) = 2k:5k:7k
To solve this you can assume the following equations
q+r-p =2k --------( 1 )
p-q+r =5k --------( 2 )
p+q-r =7k --------( 3 )

Adding (1) & (2)
q+r-p + p-q+r = 7k
2r = 7k
r = 7k/2

Adding (2) & (3)
2p = 12k
p=6k

Adding (1) & (3)
2q = 9k
q=9k/2

So p:q:r = 6k: 9k/2 : 7k/2 = 6 : 9/2 : 7/2 =12:9:7

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