Ratio And Proportion Questions For CAT PDF Set-2

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Ratio And Proportion Questions For CAT PDF Set-2
Ratio And Proportion Questions For CAT PDF Set-2

Ratio And Proportion Questions For CAT PDF Set-2:

Download Ratio And Proportion Questions For CAT PDF Set-2. Practice important problems on Ratio And Proportion with detailed explanations and solutions.

Download Ratio And Proportion Questions For CAT PDF Set-2

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Question 1: Given that p:q = 2:3, q:r = 1:2, r:s = 3:5 and s:t = 2:5, find the ratio of p+q and s+t.

a) 1:4
b) 1:5
c) 1:6
d) 1:7

Question 2: Eight years ago, the ratio of ages of Akhil and Akash was 1:5. 12 years from now, the ratio changes to 7 : 15. Find the sum of the antecedent and the consequent of the ratio of their present ages, when the ratio is in its lowest form.

a) 3
b) 4
c) 6
d) 1

Question 3: The electricity bill consists of a fixed component and a variable component. The average electricity bill in Ramesh’s house amounts to Rs 1200 if he uses 150 units of current. It amounts to Rs 2480 if he uses 400 units of current. Find the amount(in rupees) that Ramesh has to pay if he uses 750 units of current.

a) 2078
b) 3070
c) 4272
d) 5383

Question 4: Ravi has three alloys A, B and C with copper percentages as 36%, 50% and 60% respectively in them. In which of the following ratios could he melt the three alloys and mix them so as to get a percentage of 45% in the mixture?

a) 3:2:1
b) 4:3:2
c) 5:3:2
d) 3:4:6

Question 5: Given that a:b = 2:3, c:d = 4:5, d:e = 3:1 and a:d = 2:5, find the value of $\frac{ab+cd}{bc+de}$

a) 52:31
b) 61:78
c) 31:52
d) 78:61

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Answers & Solutions:

1) Answer (D)

p:q = 2:3, q:r = 1:2, r:s = 3:5 and s:t = 2:5
Let p = 2k and q = 3k
r = 2*3k = 6k
s = $\frac{5}{3}*6k$ = 10k
t = $\frac{5}{2}*10k$ = 25k
p + q = 5k
s + t = 35k
Ratio of p+q to s+t = 5k : 35k = 1 : 7

2) Answer (B)

Let the present ages of Akhil and Akash be x and y respectively.
$\frac{x-8}{y-8}$ = $\frac{1}{5}$
=> 5x – 40 = y – 8
=> 5x – y = 32
=> y = 5x – 32
$\frac{x+12}{y+12}$ = $\frac{7}{15}$
=> 15x + 180 = 7y + 84
=> 15x + 180 = 7(5x – 32) + 84
=> 15x + 96 = 35x – 224
=> 20x = 320
=> x = 16
=> y = 48
x:y = 16:48 = 1:3
Sum of antecedent and consequent = 1 + 3 = 4

3) Answer (C)

Let the fixed component be Rs x and the variable component be Rs y per unit.
x + 150y = 1200
x + 400y = 2480
=> 250y = 1280
y = Rs 5.12
x + 150*5.12 = 1200
=> x = 1200 – 768 = Rs 432
x + 750y = 432 + 750*5.12 = 432 + 3840 = Rs 4272

4) Answer (C)

Let the ratio of the alloys A, B and C melted be x : y : z.
$\frac{36x+50y+60z}{x+y+z}$ = 45
=> 36x + 50y + 60z = 45x + 45y + 45z
=> 9x = 5y + 15z

From the single equation, we can’t find the exact ratio, but we can identify whether a given ratio satisfies the equation.
In this case, 9x = 5y+15z is satisfied only by (5,3,2) and is not satisfied by the other three (3,2,1); (4,3,2) and (3,4,6)

Hence, option (c) is the correct answer.

5) Answer (D)

a:b = 2:3 => a = 2x and b = 3x
a:d = 2:5 => d = 5x
c:d = 4:5 => c = 4x
e:d = 1:3 => e = 5x/3
ab+cd = $26x^2$
bc+de = $12x^2+25x^2/3$ = $61x^2/3$
ab+cd : bc+de = $26x^2$ : $61x^2/3$ = 78 : 61

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