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Test: Problem Solving- 2 - UPSC MCQ


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15 Questions MCQ Test CSAT Preparation - Test: Problem Solving- 2

Test: Problem Solving- 2 for UPSC 2024 is part of CSAT Preparation preparation. The Test: Problem Solving- 2 questions and answers have been prepared according to the UPSC exam syllabus.The Test: Problem Solving- 2 MCQs are made for UPSC 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Problem Solving- 2 below.
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Test: Problem Solving- 2 - Question 1

Six points lie on a circle. How many quadrilaterals can be drawn joining these points? 

Detailed Solution for Test: Problem Solving- 2 - Question 1

The number of quadrilaterals that can be formed is 6C4 = (6 x 5 x 4 x 3)/(4 x 3 x 2) = 15.

Test: Problem Solving- 2 - Question 2

There are 3 children of a lady. In how many ways is it possible to dress them for a party if the first child likes 3 dresses, second likes 4 and the third likes 5 but the third child has out grown one of them? Each child has a different set of clothes. 

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Test: Problem Solving- 2 - Question 3

How many three-digit odd numbers can be formed from the digits 1, 3, 5, 0 and 8?  [repetition is allowed ]

Detailed Solution for Test: Problem Solving- 2 - Question 3

To find out how many three-digit odd numbers can be formed from the digits 1, 3, 5, 0, and 8, we need to consider the constraints:

  1. The number must be three digits long.
  2. It must be odd.
  3. We can use the digits 1, 3, 5, 0, and 8.

For a number to be odd, its last digit must be odd. Given the digits, we have three choices for the last digit (1, 3, 5) to ensure the number is odd.

For the first digit, we can choose any of the four digits except 0 (because a three-digit number cannot start with 0). This gives us 4 options (1, 3, 5, 8).

For the middle digit, we can choose any of the remaining four digits (after choosing the first and the last, but remembering digits can be reused because it's not specified that digits cannot repeat). This gives us 5 options since all five digits are available for use again.

Therefore, the total number of possible three-digit odd numbers is given by the product of the number of options for each position, which is 4×5×34×5×3.

Let's calculate this:

The total number of three-digit odd numbers that can be formed from the digits 1, 3, 5, 0, and 8 is 60. Therefore, the correct answer is option 2: 60.

Test: Problem Solving- 2 - Question 4

Find the number of words formed by permuting all the letters of the word INDEPENDENCE. 

Detailed Solution for Test: Problem Solving- 2 - Question 4

Test: Problem Solving- 2 - Question 5

There are 12 children in a party. For a game they have to be paired up. How many different pairs can be made for the game? 

Test: Problem Solving- 2 - Question 6

How many different differences can be obtained by taking only 2 numbers at a time from 3, 5,2,10 and 15? 

Test: Problem Solving- 2 - Question 7

In a fruit market, watermelon costs Rs 25 per fruit, mango costs Rs 20 per fruit, an apple costs Rs 15 per fruit, and orange costs Rs 10 per fruit. If Indu bought 2 watermelons, 5 mangoes, 3 apples, and some oranges and paid a bill of Rs 285, what was the number of oranges purchased by her?

Detailed Solution for Test: Problem Solving- 2 - Question 7

Test: Problem Solving- 2 - Question 8

ow many five digit numbers can be formed using the digits 0, 2, 3,4and 5, when repetition is allowed such that the number formed is divisible by 2 and 5? 

Test: Problem Solving- 2 - Question 9

In how many ways can five rings be worn in 3 fingers? 

Detailed Solution for Test: Problem Solving- 2 - Question 9

When arranging the rings on the fingers, we have three choices for each ring (since each ring can be worn on any of the three fingers). Therefore, the total number of arrangements is 35, which is calculated as follows:

35=3×3×3×3×3=243

So, there are 243 ways to wear five rings on three fingers.

Test: Problem Solving- 2 - Question 10

How many pentagons can be drawn by joining the vertices of a polygon with 10 sides? 

Detailed Solution for Test: Problem Solving- 2 - Question 10

      = 252

Test: Problem Solving- 2 - Question 11

Find the number of words formed by permuting all the letters of the word INDEPENDENCE such that the E???s do not come together. 

Test: Problem Solving- 2 - Question 12

Ten different letters of an alphabet are given. Words with 6 letters are formed with these alphabets. How many such words can be formed when repetition is not allowed in any word? 

Test: Problem Solving- 2 - Question 13

 

If P(2n+1,n-1):P(2n-1,n) = 3:5, find n.

Test: Problem Solving- 2 - Question 14

A polygon has 20 diagonals. How many sides does it have? 

Test: Problem Solving- 2 - Question 15

A box contains 7 red, 6 white and 4 blue balls. In how many ways can a selection of 3 balls be made,if all the balls are red ?

Detailed Solution for Test: Problem Solving- 2 - Question 15

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