Practise math problems to ace competitive exams

This article is in continuation to the last article focusing on the ratio and proportion topic. Here are some practice questions along with solutions that will help you in your preparation for the State government recruitment tests. In a cricket match, there are three types of tickets say A, B and C each costing Rs […]

This article is in continuation to the last article focusing on the ratio and proportion topic. Here are some practice questions along with solutions that will help you in your preparation for the State government recruitment tests.

In a cricket match, there are three types of tickets say A, B and C each costing Rs 1,000, Rs 500 and Rs 200 respectively. The ratio of the ticket sold of category A, B and C is 3 : 2 : 5. If the total collection from selling the tickets is Rs 2.5 crore, find the total number of tickets sold.
a) 50000 b) 40000 c) 45000 d) 60000
Ans: a
Solution:
Sold tickets of A,B,C categories are 3x, 2x, and 5x respectively.
Total Collection = (3x × 1000) (2x × 500) (5x × 200) = 25000000
5000x = 25000000
x = 5000
Total sold tickets = 3x 2x 5x = 10x = 50000

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In a competitive exam, the number of passed students was four times the number of failed students. If there had been 35 fewer appeared students and 9 more had failed, the ratio of passed and failed students would have been 2 : 1, then the total number of students appeared for the exam are?
a) 150 b) 145 c) 155 d) 160
Ans: c
Solution:
Let the number of failed students be x
=>; Number of passed students = 4x
So, total number of students was 5x
From the given data,

If total number of students be 5x – 35

=>; 4x – 35 – 9/x 9 = 2/1

=>; 4x – 44 = 2(x 9)
=>; 4x – 2x = 18 44
=>; x = 31
Total number = 31×5
= 155

For any two numbers m, n ; (m n) : (m-n) : mn = 7: 1: 60 . Find the value of 1/m : 1/n
a) 4:3 b) 8:7 c) 3:4 d) 7:8
Ans: c
Solution:
m nm − n = 7×x ⇒ mn = 4×3 xm nm-n = 7×x ⇒ mn = 4 × 3x
Again mn=12×2mn = 12×2
and mn = 60x
so, 60x = 12×2 ⇒x=560x=12×2 ⇒x=5
=>; m = 20 and n = 15
Hence, 1m:1n = 120 : 115 = 3:4

There are two containers — the first one contains 1 litre pure water and the second one contains 1 litre pure milk. Now 5 cups of water from the first container is taken out and is mixed well in the second container. Then, 5 cups of this mixture is taken out and is mixed in the first container. Let A denote the proportion of milk in the first container and B denote the proportion of water in the second container, then
a) A < B b) A = B c) A >; B d) Can’t be determined
Ans: b
Solution:
Here the ratio of mixtures (i.e., milk, water) does not matter. But the important point is that whether the total amount ( either pure or mixture ) being transferred is equal or not. Since the total amount ( i.e., 5 cups) being transferred from each one to another is similar, hence A = B.

If y/(x – z) = (y x)/z = x/y then find z : y : x ?
a) 4:3 b) 8:7 c) 3:4 d) 7:8
Ans: b
Solution:
Given, y/(x – z) = (y x)/z = x/y
yz = xy x2 − yz − xz ….(1) yz = xy x2 – yz – xz ….(1)
Also xy = yx − zxy = yx-z ⇒ x2 − xz = y2 ….(2) ⇒ x2 – xz = y2 ….(2)
Using (1) and (2), we get yz = xy – yz y2y2
2yz = xy y2y2
2z = x y
Only option (B) satisfies the equation.

At a casino in Mumbai, there are three tables A, B and C. The payoffs at A is 10:1, at B is 20:1 and at C is 30:1. If a man bets Rs 200 at each table and wins at two of the tables, what is the maximum and minimum difference between his earnings can be?
a) 2500 b) 2000 c) 4000 d) none of these
Ans: c
Solution:
Maximum earning will be only when he will win on the maximum yielding table.
A —->; 10:1
B —->; 20:1
C —->; 30:1
i.e, he won on B and C but lost on A
20 × 200 30 × 200 – 1 × 200 = 9800
minimum earning will be when he won on table A and B and lost on that table 3.
Therefore, 10 x 200 20 × 200 – 1 × 200 = 6000-200 = 5800