UPSC Insta–DART (Daily Aptitude and Reasoning Test) 11 Feb 2025
Kartavya Desk Staff
Considering the alarming importance of CSAT in UPSC CSE Prelims exam and with enormous requests we received recently, InsightsIAS has started Daily CSAT Test to ensure students practice CSAT Questions on a daily basis. Regular Practice would help one overcome the fear of CSAT too.We are naming this initiative as Insta– DART – Daily Aptitude and Reasoning Test. We hope you will be able to use DART to hit bull’s eye in CSAT paper and comfortably score 100+ even in the most difficult question paper that UPSC can give you in CSP-2021. Your peace of mind after every step of this exam is very important for us.
Looking forward to your enthusiastic participation (both in sending us questions and solving them on daily basis on this portal).
#### Quiz-summary
0 of 5 questions completed
Questions:
#### Information
Best of Luck! 🙂
You have already completed the quiz before. Hence you can not start it again.
Quiz is loading...
You must sign in or sign up to start the quiz.
You have to finish following quiz, to start this quiz:
0 of 5 questions answered correctly
Your time:
Time has elapsed
You have reached 0 of 0 points, (0)
#### Categories
• Not categorized 0%
• Question 1 of 5 1. Question Mr. Sanjay Patil has a few children, such that each daughter of Mr. Sanjay Patil has the same number of brothers as she has sisters. Also, each son has twice as many sisters as he has brothers. What is the number of sons and daughters of Mr. Sanjay Patil? (a) 9 daughters and 8 sons (b) 5 daughters and 3 sons (c) 7 daughters and 4 sons (d) 4 daughters and 3 sons Correct Answer: D Explanation: Consider d = number of daughters, and s = number of sons It is given that each daughter has the same number of brothers as she has sisters. So, d – 1 = s Now, each son has twice as many sisters as he has brothers. So, 2(s – 1) = d or 2(d – 1 – 1) = d (putting the value of s) or 2(d – 2) = d or 2d – 4 = d d = 4 Since, d – 1 = s or 4 – 1 = s or s = 3 Therefore, the number of sons and daughters that Mr. Sanjay Patil has is 3 and 4 respectively. Incorrect Answer: D Explanation: Consider d = number of daughters, and s = number of sons It is given that each daughter has the same number of brothers as she has sisters. So, d – 1 = s Now, each son has twice as many sisters as he has brothers. So, 2(s – 1) = d or 2(d – 1 – 1) = d (putting the value of s) or 2(d – 2) = d or 2d – 4 = d d = 4 Since, d – 1 = s or 4 – 1 = s or s = 3 Therefore, the number of sons and daughters that Mr. Sanjay Patil has is 3 and 4 respectively.
#### 1. Question
Mr. Sanjay Patil has a few children, such that each daughter of Mr. Sanjay Patil has the same number of brothers as she has sisters. Also, each son has twice as many sisters as he has brothers. What is the number of sons and daughters of Mr. Sanjay Patil?
• (a) 9 daughters and 8 sons
• (b) 5 daughters and 3 sons
• (c) 7 daughters and 4 sons
• (d) 4 daughters and 3 sons
Explanation:
Consider d = number of daughters, and s = number of sons
It is given that each daughter has the same number of brothers as she has sisters.
So, d – 1 = s
Now, each son has twice as many sisters as he has brothers.
So, 2(s – 1) = d
or 2(d – 1 – 1) = d (putting the value of s)
or 2(d – 2) = d
or 2d – 4 = d
Since, d – 1 = s
or 4 – 1 = s
or s = 3 Therefore, the number of sons and daughters that Mr. Sanjay Patil has is 3 and 4 respectively.
Explanation:
Consider d = number of daughters, and s = number of sons
It is given that each daughter has the same number of brothers as she has sisters.
So, d – 1 = s
Now, each son has twice as many sisters as he has brothers.
So, 2(s – 1) = d
or 2(d – 1 – 1) = d (putting the value of s)
or 2(d – 2) = d
or 2d – 4 = d
Since, d – 1 = s
or 4 – 1 = s
or s = 3 Therefore, the number of sons and daughters that Mr. Sanjay Patil has is 3 and 4 respectively.
• Question 2 of 5 2. Question Shankar has 5 jackets, 4 track pants and 3 pair of goggles. In how many ways can he wear these items (choosing one each)? (a) 30 (b) 40 (c) 60 (d) 12 Correct Answer: C Explanation: There are 5 jackets, 4 track pants and 3 pair of goggles. A jacket can be chosen in 5 different ways. After a jackets is chosen, a track pants can be chosen in 4 different ways. Then a pair of goggles can be chosen in 3 different ways. Required number of ways to wear these items (choosing one each) = 5×4×3 = 60 Incorrect Answer: C Explanation: There are 5 jackets, 4 track pants and 3 pair of goggles. A jacket can be chosen in 5 different ways. After a jackets is chosen, a track pants can be chosen in 4 different ways. Then a pair of goggles can be chosen in 3 different ways. Required number of ways to wear these items (choosing one each) = 5×4×3 = 60
#### 2. Question
Shankar has 5 jackets, 4 track pants and 3 pair of goggles. In how many ways can he wear these items (choosing one each)?
Explanation:
There are 5 jackets, 4 track pants and 3 pair of goggles.
A jacket can be chosen in 5 different ways. After a jackets is chosen, a track pants can be chosen in 4 different ways. Then a pair of goggles can be chosen in 3 different ways.
Required number of ways to wear these items (choosing one each) = 5×4×3 = 60
Explanation:
There are 5 jackets, 4 track pants and 3 pair of goggles.
A jacket can be chosen in 5 different ways. After a jackets is chosen, a track pants can be chosen in 4 different ways. Then a pair of goggles can be chosen in 3 different ways.
Required number of ways to wear these items (choosing one each) = 5×4×3 = 60
• Question 3 of 5 3. Question How many numbers greater than 100000 can be formed by using the digits 0, 1, 2, 2, 3, 4? (a) 300 (b) 360 (c) 600 (d) 120 Correct Answer: A Explanation: Since, 100000 is a 6-digit number and the number of digits to be used are also 6, so we need to use all these digits. Also, the numbers have to be greater than 100000, so they can begin with 1, 2, 3 or 4. When 1 is fixed at the extreme left position, the remaining digits are 3, 0, 2, 4 and 2. So, the number of numbers beginning with 1 = 5!/2! = (5×4×3×2)/2 = 60 When 2 is fixed at the extreme left position, the remaining digits are 1, 3, 0, 2 and 4. So, the number of numbers beginning with 2 = 5! = 5×4×3×2 = 120 When 3 is fixed at the extreme left position, the remaining digits are 1, 0, 2, 4 and 2. So, the number of numbers beginning with 3 = 5!/2! = (5×4×3×2)/2 = 60 When 4 is fixed at the extreme left position, the remaining digits are 1, 3, 0, 2 and 2. So, the number of numbers beginning with 4 = 5!/2! = (5×4×3×2)/2 = 60 Therefore, the required number of numbers = 60 + 120 + 60 + 60 = 300 Incorrect Answer: A Explanation: Since, 100000 is a 6-digit number and the number of digits to be used are also 6, so we need to use all these digits. Also, the numbers have to be greater than 100000, so they can begin with 1, 2, 3 or 4. When 1 is fixed at the extreme left position, the remaining digits are 3, 0, 2, 4 and 2. So, the number of numbers beginning with 1 = 5!/2! = (5×4×3×2)/2 = 60 When 2 is fixed at the extreme left position, the remaining digits are 1, 3, 0, 2 and 4. So, the number of numbers beginning with 2 = 5! = 5×4×3×2 = 120 When 3 is fixed at the extreme left position, the remaining digits are 1, 0, 2, 4 and 2. So, the number of numbers beginning with 3 = 5!/2! = (5×4×3×2)/2 = 60 When 4 is fixed at the extreme left position, the remaining digits are 1, 3, 0, 2 and 2. So, the number of numbers beginning with 4 = 5!/2! = (5×4×3×2)/2 = 60 Therefore, the required number of numbers = 60 + 120 + 60 + 60 = 300
#### 3. Question
How many numbers greater than 100000 can be formed by using the digits 0, 1, 2, 2, 3, 4?
Explanation:
Since, 100000 is a 6-digit number and the number of digits to be used are also 6, so we need to use all these digits.
Also, the numbers have to be greater than 100000, so they can begin with 1, 2, 3 or 4.
When 1 is fixed at the extreme left position, the remaining digits are 3, 0, 2, 4 and 2.
So, the number of numbers beginning with 1 = 5!/2! = (5×4×3×2)/2 = 60
When 2 is fixed at the extreme left position, the remaining digits are 1, 3, 0, 2 and 4.
So, the number of numbers beginning with 2 = 5! = 5×4×3×2 = 120
When 3 is fixed at the extreme left position, the remaining digits are 1, 0, 2, 4 and 2.
So, the number of numbers beginning with 3 = 5!/2! = (5×4×3×2)/2 = 60
When 4 is fixed at the extreme left position, the remaining digits are 1, 3, 0, 2 and 2.
So, the number of numbers beginning with 4 = 5!/2! = (5×4×3×2)/2 = 60
Therefore, the required number of numbers = 60 + 120 + 60 + 60 = 300
Explanation:
Since, 100000 is a 6-digit number and the number of digits to be used are also 6, so we need to use all these digits.
Also, the numbers have to be greater than 100000, so they can begin with 1, 2, 3 or 4.
When 1 is fixed at the extreme left position, the remaining digits are 3, 0, 2, 4 and 2.
So, the number of numbers beginning with 1 = 5!/2! = (5×4×3×2)/2 = 60
When 2 is fixed at the extreme left position, the remaining digits are 1, 3, 0, 2 and 4.
So, the number of numbers beginning with 2 = 5! = 5×4×3×2 = 120
When 3 is fixed at the extreme left position, the remaining digits are 1, 0, 2, 4 and 2.
So, the number of numbers beginning with 3 = 5!/2! = (5×4×3×2)/2 = 60
When 4 is fixed at the extreme left position, the remaining digits are 1, 3, 0, 2 and 2.
So, the number of numbers beginning with 4 = 5!/2! = (5×4×3×2)/2 = 60
Therefore, the required number of numbers = 60 + 120 + 60 + 60 = 300
• Question 4 of 5 4. Question Each of the 3 persons is to be given some identical items such that the product of the numbers of items received by each of the three persons is equal to 30. In how many maximum different ways can this distribution be done? (a) 21 (b) 24 (c) 27 (d) 33 Correct Answer: C Explanation: Suppose three people have been given a, b and c number of items. Then, a × b × c = 30 Now, there can be 5 cases: Case I: When one of them is given 30 items and rest two 1 item each. So, the number of ways for (30 × 1 × 1) = 3! / 2! = 3 (As two of them have same number of items) Case II: Similarly, number of ways for (10 × 3 × 1) = 3! = 6 Case III: Number of ways for (15 × 2 × 1) = 3! = 6 Case IV: Number of ways for (6 × 5 × 1) = 3! = 6 Case V: Number of ways for (5 × 3 × 2) = 3! = 6 Here, either of these 5 cases are possible. Hence, total number of ways = 3 + 6 + 6 + 6 + 6 = 27 Incorrect Answer: C Explanation: Suppose three people have been given a, b and c number of items. Then, a × b × c = 30 Now, there can be 5 cases: Case I: When one of them is given 30 items and rest two 1 item each. So, the number of ways for (30 × 1 × 1) = 3! / 2! = 3 (As two of them have same number of items) Case II: Similarly, number of ways for (10 × 3 × 1) = 3! = 6 Case III: Number of ways for (15 × 2 × 1) = 3! = 6 Case IV: Number of ways for (6 × 5 × 1) = 3! = 6 Case V: Number of ways for (5 × 3 × 2) = 3! = 6 Here, either of these 5 cases are possible. Hence, total number of ways = 3 + 6 + 6 + 6 + 6 = 27
#### 4. Question
Each of the 3 persons is to be given some identical items such that the product of the numbers of items received by each of the three persons is equal to 30. In how many maximum different ways can this distribution be done?
Explanation:
Suppose three people have been given a, b and c number of items.
Then, a × b × c = 30
Now, there can be 5 cases:
Case I: When one of them is given 30 items and rest two 1 item each.
So, the number of ways for (30 × 1 × 1) = 3! / 2! = 3
(As two of them have same number of items)
Case II: Similarly, number of ways for (10 × 3 × 1) = 3! = 6
Case III: Number of ways for (15 × 2 × 1) = 3! = 6
Case IV: Number of ways for (6 × 5 × 1) = 3! = 6
Case V: Number of ways for (5 × 3 × 2) = 3! = 6
Here, either of these 5 cases are possible.
Hence, total number of ways = 3 + 6 + 6 + 6 + 6 = 27
Explanation:
Suppose three people have been given a, b and c number of items.
Then, a × b × c = 30
Now, there can be 5 cases:
Case I: When one of them is given 30 items and rest two 1 item each.
So, the number of ways for (30 × 1 × 1) = 3! / 2! = 3
(As two of them have same number of items)
Case II: Similarly, number of ways for (10 × 3 × 1) = 3! = 6
Case III: Number of ways for (15 × 2 × 1) = 3! = 6
Case IV: Number of ways for (6 × 5 × 1) = 3! = 6
Case V: Number of ways for (5 × 3 × 2) = 3! = 6
Here, either of these 5 cases are possible.
Hence, total number of ways = 3 + 6 + 6 + 6 + 6 = 27
• Question 5 of 5 5. Question In two schools A and B, boys and girls are in the ratio of 4 : 5 and 3 : 1 respectively. A group of students has to be made for cleaning the beach. In what ratio should the students be selected from the two schools to get equal number of boys and girls in the group? (a) 7 : 2 (b) 8 : 1 (c) 9 : 2 (d) 10 : 1 Correct Answer: C Explanation: Fraction of boys in school A = 4/(4 + 5) = 4/9 Fraction of boys in school B = 3/(3 + 1) = 3/4 Fraction of boys in group = 1/2 Applying Allegation Method, we get: So, required ratio = (1/4) : (1/18) = 9 : 2 Incorrect Answer: C Explanation: Fraction of boys in school A = 4/(4 + 5) = 4/9 Fraction of boys in school B = 3/(3 + 1) = 3/4 Fraction of boys in group = 1/2 Applying Allegation Method, we get: So, required ratio = (1/4) : (1/18) = 9 : 2
#### 5. Question
In two schools A and B, boys and girls are in the ratio of 4 : 5 and 3 : 1 respectively. A group of students has to be made for cleaning the beach. In what ratio should the students be selected from the two schools to get equal number of boys and girls in the group?
• (d) 10 : 1
Explanation:
Fraction of boys in school A = 4/(4 + 5) = 4/9
Fraction of boys in school B = 3/(3 + 1) = 3/4
Fraction of boys in group = 1/2
Applying Allegation Method, we get:
So, required ratio = (1/4) : (1/18) = 9 : 2
Explanation:
Fraction of boys in school A = 4/(4 + 5) = 4/9
Fraction of boys in school B = 3/(3 + 1) = 3/4
Fraction of boys in group = 1/2
Applying Allegation Method, we get:
So, required ratio = (1/4) : (1/18) = 9 : 2
• Official Facebook Page HERE
• Follow our Twitter Account HERE