Below are three problems based on probability / random events.
Question 1
A boy has several coloured balls in his bag - 12 white, 15 orange and 25 pink. The lights are out and it is totally dark. Now, how many balls he must take out to make sure that he has a pair of each color?
a)40 b)42 c)6 d)12
Answer : b)42


Solution:

In this problem, the worst case scenario will be as follows.

Step 1 : He picks all of same colored balls of the largest group. Largest group is pink. Therefore, he will be picking all 25 pink balls first.
Step 2 : Then he picks all of same colored balls of the next largest group. Second largest group is Orange. Therefore, he will be picking 15 orange balls.
Step 3 : Now, if he starts picking the remaining balls which are white, he could very well stop at 2 balls as he would be having a pair of white ball anyways.

Therefore total picks = 25 + 15 + 2 = 42.

 Shortcut :

If confused, for these kinds of problems you have to add higher values first + 2, leaving the least group. Here least group is white.
Therefore our answer will be 25 pink + 15 orange + 2 = 42

Hence the answer is 42.

 Question 2

A girl has some bangles and bracelets in her cupboard - 15 red, 20 yellow, 17 white, 25 green and 22 black. The lights are out and it is totally dark. In spite of darkness, she can make out the difference between a bangle and bracelet. she takes out an item out of the cupboard only if she is sure that it is a bracelet. How many bracelets must she take out to make sure that she has a pair of green, black, red?
a)85 b)45 c)69 d)59

Answer : a)85

 Solution:

It is not given that how many of those items are bangles and how many are bracelets.
If we consider that all the items are bracelets, as per worst case scenario then attempts 20 + 17 + 25 + 22 + 2 = 86.(as discussed in first question)
The answer got to be 86, provided all the items are bracelets but which might not be the case. Therefore, you have to find an option that is lesser than but closest to 86 which is 85.

Hence the answer is 85.

Question 3

A box contains keys of different colors - 15 blue, 20 green and 25 black. In the case of darkness, how many more attempts are needed,so that we get a pair of each color than a single key of each color?
a)2 b)3 c)1 d)4

Answer : c)1

Solution:

Part:1 To find the number of attempts needed to take out the pair of each color

consider the worst case (similar to first two problems),
Picking all black is first(first biggest group) and all green(second biggest group) followed by 2 keys which are blue.
so total 25 + 20 + 2 = 47 attempts are needed. ...(1)

Part:2 To find the number of attempts needed to take out a key of each color

Similarly, pick all the black's first and all the green's. Now since a singe key of each color is sufficient, you could stop with just one more pick, which will ensure there is at least one blue colored ball.

So totally 25 + 20 + 1 = 46 attempts are needed. ...(2)

Value of (2) - (1) = 47- 46 = 1 attempt is needed more to make the pair of each color than a single key of each color.

Hence the answer is 1.

 Arithmetic: 

  you can expect one or two super simple questions like those below.

Question 1

One-fourth of sum of a number and 5 is 1/3rd of the same number. What is the number ?
a)15 b) 7 c) 34 d) 55

Answer : a)15

Solution:

Based on the question statement, we can form the below equation.

1/4 x(X+5) = 1/3 x X
3 x(X+5) = 4X
3X + 15 = 4X
4X - 3X = 15
X = 15

Question 2

When 8 is subtracted out from two third of X, the result is 25 more than one third of X. Find the largest prime number which is less than X.

a) 113 b) 97 c) 89 d) 131

Answer : b) 97

Solution:

Part-1 : To find X.

Below equation can be formed based on data in question.

(2/3)x X - 8 = 25 + 1/3 x X
(2/3)x X - (1/3)x X = 25+8
X/3 = 33
X=99.

Part-2 : To find the required prime

97 is the largest prime number which is lesser than 99. Hence 97 is the answer.

Question 3

6 reduced from 1/5 of a number is 7 reduced from half of the same number.

a)10/3 b)22/3 c)7/3 d)none of these

Answer: a) 10/3

Solution:

X/5 - 6 = X/2 - 7
(X-30)/5 = (X-14)/2
2X-60 = 5X-70
5X-2X = 70-60
3X = 10
X = 10/3

Easy Areas: 

These problems can be solved by knowing basic area formulas.

Question 1

Messrs. Siva Constructions, leading agents in Chennai prepared models of their lands in the shape of a rectangle and triangle. They made models having same area. The length and width of rectangle model are 24 inches and 8 inches respectively. The base of the triangle model is 16 inches. What is the altitude of triangle model from the base to the top?

a) 24 inches b) 8 inches c) 20 inches d) 32 inches

Answer : a) 24 inches

Solution :

Area of rectangle model -- length x breadth = 24 x 8 = 192 sq. inches.
Area of triangle model is also 192 sq.inches.
Its base - 16 inches
Area of a triangle - 1/2 x base x height
1/2 x 16 x height = 192
Height = (192 x 2) / 16 = 24 inches.

Question 2

Fisher-Price, leading toy manufacturers made a rectangle toy and triangle toy having same area. The length of the rectangle was 30 inches. The triangle toy’s height was 36 inches. How will you express the base y of the triangle as a function of the breadth x of the rectangle ?

a. 20x/3 b. 10x/3 c. 16x d. 22x

Answer : b. 10x/3

Solution :

Base of the triangle = y.
Breadth of the rectangle = x.
Area of the rectangle = length x breadth = 30x
Area of the triangle = 1/2 x base x height = 36y/2 = 18y
Equating the areas of triangle and rectangle, we get, 30x = 18y or y = 30x/18 = 10x/3

Question 3

Ideal Toy company, New York brought out two models – one rectangle and another hexagon in shape. The area of the two are same. The base and height of the triangle are 48” and √3” respectively. Find the length of each of the sides of the hexagon.

a) 2” b) 4” c) 24” d) 8”

Answer : b) 4”

Solution :

Area of triangle = 1/2 x base x height = 1/2 x 48 x √3 = 24√3 sq. inches
Since, the areas of triangle and hexagon are equal , Area of hexagon = 24√3 sq. inches ....(1)
If the side of hexagon is x inches, then its area = (3√3/2 )r2 ...(2)
Since equations 1 and 2 equal,
24√3 = (3√3/2 )r2
16 = r2
Or r = 4 inch
Problems based on Dice 

Below are three probability questions for practice.

Question 1

Four friends namely Rahul, Ravi, Rajesh and Rohan contested for a dairy milk chocolate. To decide which friend will get the chocolate they decided to throw two dice. Every friend was asked to choose a number and if the sum of the numbers on two dice equals that number, the concerned person will get the chocolate. Rahul's choice was7, Ravi's choice was 9, Rajesh's choice was 10 and Rohan's choice was 11. Who has the maximum probability of winning the amount.

a) Rahul b) Ravi c) Rajesh d) Rohan

Answer : a) Rahul

Solution :

Number 7 will appear more often –(1,6), (2,5), (3,4), (4,3), (6,1), (5,2) --- 6 cases
Number 9 -- (3,6),(6,3), (4,5) (5,4) ---4 cases
For number 10 -- (4,6) ,(6,4) (5,5) ---3 cases
For number 11 -- (5,6),(6,5)...2 cases.
Since number 7 has the maximum chance of appearing, it will have the maximum probability as well. Hence, Rahul will most probably be the winner.

Question 2

Yegaraj, Zimbo, Kodanda, Ramiah and Kashi were given undescribed measure of gold coins, silver bars etc. Differences arose among them as to how these are to be distributed. Different strategies/options/methodologies were discussed. Finally it was decided to throw two dice. Every friend was asked to choose a number. Assume Yegaraj chooses 9 and Zimbo chooses 10. By how much percentage, probability of Yegaraj winning is more than that of Zimbo.

a) 30% b)66.67% c)33.33% d) 25%

Answer : c) 33.33%

Solution :

Possible cases for the sum of the numbers to be 9 -- (3,6),(6,3), (4,5) (5,4) ---4 cases
Possible cases for the sum to be 10 -- (4,6) ,(6,4) (5,5) ---3 cases
Total number of combinations when two dice are thrown = 36 (... 6 numbers on first die X 6 numbers on the second die)
P1 = probability of Yegaraj winning = Possible cases for the sum to be 9 / Total combinations = 4/36
P2 = probability of Zimbo winning = Possible cases for the sum to be 10 / Total combinations = 3/36
Percentage by which P1 is more than P2 = (P1 - P2) / P2 x 100% = (4/36 - 3/36) / 3/36 x 100% = 1/3 x 100% = 33.33%

Question 3

A professor of Statistics posed the following question to his students:
“Three dices (six sided) are thrown one after the other. What is the probability of getting a different number on each dice?”

a) 1/12 b) 1/3 c) 5/9 d) 4/9

Answer : c) 5/9

Solution :

Dice rolled first can show any of the six numbers – so there are six possibilities. Second dice shall not have the number rolled by first dice. So there are 5 possibilities. Similarly for third dice has four possibilities—i.e. it should not show the number shown by first dice and/or second dice.
Based on the above argument, Number of possibilities for three dice to show different numbers = 6 x 5 x 4 = 120.
Also we know that the total possibilities when three dice are thrown = 6 x 6 x 6 = 216
Probability of getting a different number on each dice = Number of Possibilities for three dice to show different numbers / Total Possibilities = 120/216 = 5/9

Tallying Costs
Below are three problems based on costs calculation and tallying.
Question 1
Santosh Iyer had Rs.35 with him and he went for shopping. Vendor was selling one apple for Rs.5, two bananas for Rs.5 and three guavas for Rs. 5. If Santosh Iyer wants to give equal number of fruits to three of his friends what is the combination (for each of his friends) in which he can buy and give fruits?
a) Two apples, two bananas and two guavas
b) One apple, two bananas and one guava
c) Three apples, two bananas, two guavas
d) None of these.
Answer : b) One apple, two bananas and one guava
Solution :
Cost Observations :
Cost of 1 apple = Rs. 5
Cost of 2 bananas = Rs. 5. Therefore cost of 1 banana = Rs. 5/2
Cost of 3 guavas = Rs. 5. Therefore cost of 1 guava = Rs. 5/3
Now its time to inspect option by option, if the total money spent tallies with the amount originally had.
i) Consider Option a) 2 Apples, 2 Bananas, 2 Guavas
If a combination 2 apples, 2 bananas and 2 guavas are given to each friend (totally 3 friends), he would have to buy 2 x 3 = 6 apples, 2 x 3 = 6 bananas and 2 x 3 = 6 guavas
Total Cost = 6 apples x Cost of 1 apple + 6 bananas x Cost of 1 banana + 6 guavas x Cost of 1 guava
= 6 x 5 + 6 x 5/2 + 6 x 5/3 = 30 + 15 + 10 = Rs .55
But Rs. 55 is not equal to Rs. 35 which Santosh had initially. Therefore, option a cannot be the right answer.
ii) Consider Option b) 1 Apple, 2 Bananas and 1 Guava
If option b combination is given to each of his friends, in total he would have to buy 3 apples, 6 bananas and 3 guavas.
Total Cost = 3 apples x Cost of 1 apple + 6 bananas x Cost of 1 banana + 3 guavas x Cost of 1 guava
= 3 x 5 + 6 x 5/2 + 3 x 5/3 = 15 + 15 + 5 = Rs. 35
Here, Total cost Rs. 35 tallies perfectly with the amount of Rs. 35 that Santosh had initially. Therefore, option b is correct.
Question 2
Srinisha went to a leading stationery shop for buying gifts for her four friends—Two pens cost Rs.40, each eraser costs Rs.15 and two sharpeners are sold for Rs.12.50. Srinisha had Rs.300 with her. In which combination she can buy these items so that she can present her friends equal number of pens, sharpeners and erasers?
a) Two pens, one eraser and one sharpener
b) One pen, two erasers and four sharpeners
c) One pen, four erasers and two sharpeners
d) Two pens, Two erasers and two sharpeners
Answer : b) One pen, two erasers and four sharpeners
Solution :
This question is very similar to question number 1.
Cost Observations :
Cost of 1 pen = Rs. 40/2 = Rs. 20
Cost of 1 eraser = Rs. 15
Cost of 1 sharpener = Rs. 12.5/2
Consider option a) 2 pens, 1 eraser, 1 sharpener
If option a) combination is to be given to each of her 4 friends, she should buy 8 pens, 4 erasers and 4 sharpeners
Total Cost = 8 pens x Cost of 1 pen + 4 erasers x Cost of 1 eraser + 4 sharpeners x Cost of 1 sharpener
= 8 x 20 + 4 x 15 + 4 x 12.5/2 = Rs. 465
But Rs. 465 is not equal to Rs. 300 which Srinisha had initially. Therefore, option a is not correct.
Consider option b) 1 pen, 2 erasers and 4 sharpeners
For 4 friends, she should buy 4 pens, 8 erasers and 16 sharpeners.
Total Cost = 4 pens x Cost of 1 pen + 8 erasers x Cost of 1 eraser + 16 sharpeners x Cost of 1 sharpener
= 4 x 20 + 8 x 15 + 16 x 12.5/2 = Rs. 300
Total cost of Rs. 300 matches perfectly with Rs. 300 that Srinisha had initially. Hence, option b is correct.
Question 3
Pragathi won Super Singer Contest and wanted to give fruits to her five friends equal number of each fruits. In what combination can she purchase and give if she has Rs.1175 with her.? -- 4 Papayas cost Rs. 100, One avocado costs Rs. 62.50, ten apples cost Rs.300.
a) One papaya, two avocados, 2 apples
b) Two papayas, two avocados, 1 apple
c) Two papayas, 2 avocados, 2 apples
d) Three apples, two avocados, 2 apples.
Answer : c) Two papayas, 2 avocados, 2 apples
Solution :
Cost Observations :
One papaya costs Rs. 25
One avocado costs Rs. 62.50
One apple costs Rs. 30
Consider option a) 1 papaya, 2 avocado, 2 apples
To give the above combination to each of her 5 friends, she should buy 5 papayas, 10 avocados and 10 apples
Total Cost = 5 papayas x Cost of 1 papaya + 10 avocados x Cost of 1 avocados + 10 apples x Cost of 1 apple
= 5 x 25 + 10 x 62.5 + 10 x 30 = Rs. 1050
But Rs. 1050 is not equal to Rs. 1175, which she had initially. Hence, option a is not correct.
Consider option b) 2 papayas, 2 avocados, 1 apple
To give the above combination to each of her 5 friends, she should buy 10 papayas, 10 avocados and 5 apples
Total Cost = 10 papayas x Cost of 1 papaya + 10 avocados x Cost of 1 avocados + 5 apples x Cost of 1 apple
= 10 x 25 + 10 x 62.5 + 5 x 30 = Rs. 1025
But Rs. 1025 is not equal to Rs. 1175, which she had initially. Hence, option b is not correct.
Consider option c) 2 papayas, 2 avocados, 2 apples
To give the above combination to each of her 5 friends, she should buy 10 papayas, 10 avocados and 10 apples
Total Cost = 10 papayas x Cost of 1 papaya + 10 avocados x Cost of 1 avocados + 10 apples x Cost of 1 apple
= 10 x 25 + 10 x 62.5 + 10 x 30 = Rs. 1175
Total cost is exactly equal to Rs. 1175, which she had initially. Hence, option c is correct.

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