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UPSC Exam  >  UPSC Test  >  CSAT Preparation  >  Test: Problem Solving- 2 - UPSC MCQ

Problem Solving- 2 - Free MCQ Practice Test with solutions, UPSC


MCQ Practice Test & Solutions: Test: Problem Solving- 2 (15 Questions)

You can prepare effectively for UPSC CSAT Preparation with this dedicated MCQ Practice Test (available with solutions) on the important topic of "Test: Problem Solving- 2". These 15 questions have been designed by the experts with the latest curriculum of UPSC 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 30 minutes
  • - Number of Questions: 15

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Test: Problem Solving- 2 - Question 1
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Six points lie on a circle. How many quadrilaterals can be drawn joining these points? 

Detailed Solution: 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
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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. 

Detailed Solution: Question 2

The first child has 3 choices, the second has 4, and the third, having outgrown one dress, has 4 options. The total number of ways is 3 * 4 * 4 = 48.

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Test: Problem Solving- 2 - Question 3
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How many three-digit odd numbers can be formed from the digits 1, 3, 5, 0 and 8?  [repetition is allowed ]

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

Detailed Solution: Question 4

Test: Problem Solving- 2 - Question 5
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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? 

Detailed Solution: Question 5

To determine the number of unique individual pairs that can be formed from 12 children, we use combinations. The formula for combinations is C(n, k) = n! / (k!(n - k)!), where n=12 and k=2. C(12,2) = 12! / (2! * 10!) = (12×11)/2 = 66. Thus, the number of different individual pairs is 66.

Test: Problem Solving- 2 - Question 6
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How many different differences can be obtained by taking only 2 numbers at a time from 3, 5,2,10 and 15? 

Detailed Solution: Question 6

The unique differences calculated from the pairs are: 1, 2, 3, 5, 7, 8, 10, 12, and 13. This results in a total of 9 different absolute differences.

Test: Problem Solving- 2 - Question 7
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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: Question 7

Test: Problem Solving- 2 - Question 8
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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? 

Detailed Solution: Question 8

The last digit must be 0 for divisibility by 2 and 5, allowing the first digit to have 4 choices (excluding 0) and each of the next three digits having 5 choices each. Thus, total numbers = 4 * 53 = 500.

Test: Problem Solving- 2 - Question 9
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In how many ways can five rings be worn in 4 fingers? 

Detailed Solution: Question 9

Test: Problem Solving- 2 - Question 10
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How many pentagons can be drawn by joining the vertices of a polygon with 10 sides? 

Detailed Solution: Question 10

      = 252

Test: Problem Solving- 2 - Question 11
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Find the number of words formed by permuting all the letters of the word INDEPENDENCE such that the E???s do not come together. 

Detailed Solution: Question 11

1. Total Permutations Without Restrictions: The word has 12 letters with repetitions: I(1), N(3), D(2), E(4), P(1), C(1). Total permutations = 12! / (3! * 2!) = 1,663,200. 2. Permutations With All 'E's Together: Treat the four 'E's as a single entity, reducing the problem to arranging 9 entities: I(1), N(3), D(2), P(1), C(1), [EEEE](1). Permutations = 9! / (3! * 2!) = 30,240. 3. Desired Permutations: Subtract the permutations with all 'E's together from the total: 1,663,200 - 30,240 = 1,632,960. Thus, the number of ways to arrange the letters so that no two 'E's are consecutive is 1,632,960.

Test: Problem Solving- 2 - Question 12
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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? 

Detailed Solution: Question 12

To determine the number of six-letter words that can be formed from ten different letters without repetition, we use permutations since the order matters and repetition is not allowed. The formula for permutations without repetition is: P(n, r) = n! / (n - r)!. Here, n = 10 (total letters) and r = 6 (letters to choose). Plugging in the values: P(10, 6) = 10! / (10 - 6)! = 10! / 4!. Calculating this: 10 × 9 × 8 × 7 × 6 × 5 = 151200. Thus, the number of possible words is 151200.

Test: Problem Solving- 2 - Question 13
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If P(2n+1,n-1):P(2n-1,n) = 3:5, find n.

Detailed Solution: Question 13

[1. Express the Permutations: P(2n + 1, n - 1) = (2n + 1)! / (n + 2)! and P(2n - 1, n) = (2n - 1)! / (n - 1)!. 2. Form the Ratio: (P(2n + 1, n - 1) / P(2n - 1, n)) = ((2n + 1)(2n) * (n - 1)!)/((n + 2)(n + 1)n!). 3. Set the Ratio Equal to (3/5): ((2n + 1)(2n)/(n(n + 1)(n + 2)) = (3/5). 4. Cross-Multiply and Simplify: 5(2n + 1)(2n) = 3n(n + 1)(n + 2) leads to 17n2 - n - 6 = 0. 5. Solve the Quadratic Equation: Using the quadratic formula, we find that n = 4 is the valid solution.]

Test: Problem Solving- 2 - Question 14
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A polygon has 20 diagonals. How many sides does it have? 

Detailed Solution: Question 14

To solve for the number of sides (n) in a polygon with 20 diagonals, we use the formula: n(n - 3)/2 = Number of diagonals. Plugging in the given value: n(n - 3)/2 = 20 n² - 3n - 40 = 0. Solving this quadratic equation using the quadratic formula yields n = 8 as the valid solution. Thus, the polygon has 8 sides.

Test: Problem Solving- 2 - Question 15
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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: Question 15

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About this UPSC Practice Test

This test contains 15 MCQs on Problem Solving- 2 with detailed solutions for each question. It is part of the CSAT Preparation course on EduRev, designed to align with the current UPSC syllabus. Each solution explains not just the correct answer but also gives you an understanding of the topic — not just to attempt the question but also help you prepare for the exam with practice.

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