UPSC Insta–DART (Daily Aptitude and Reasoning Test) 6 Aug 2024

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).

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• Question 1 of 5 1. Question Find the greatest 4 digit number which when divided by 12,15,20 and 35 leaves no remainder. a) 9999 b) 9660 c) 8249 d) 9960 Correct Solution:- LCM of 12, 15, 20 and 35 is 420. Divide 9999 greatest 4 digit number by 420, and subtract the remainder. Answer would be 9660. Incorrect Solution:- LCM of 12, 15, 20 and 35 is 420. Divide 9999 greatest 4 digit number by 420, and subtract the remainder. Answer would be 9660.

#### 1. Question

Find the greatest 4 digit number which when divided by 12,15,20 and 35 leaves no remainder.

Solution:-

LCM of 12, 15, 20 and 35 is 420. Divide 9999 greatest 4 digit number by 420, and subtract the remainder. Answer would be 9660.

Solution:-

LCM of 12, 15, 20 and 35 is 420. Divide 9999 greatest 4 digit number by 420, and subtract the remainder. Answer would be 9660.

• Question 2 of 5 2. Question A, B, and C start jogging around a circular field and complete a single round in 18 seconds, 22seconds and 30 seconds respectively. In how much time will they meet again at the starting point? a) 16min 30sec b) 990 min c) 20min d) 99min Correct Solution: LCM of 18, 22 and 30 is 990. Hence they meet in 990 sec which is 16min 30sec. Incorrect Solution: LCM of 18, 22 and 30 is 990. Hence they meet in 990 sec which is 16min 30sec.

#### 2. Question

A, B, and C start jogging around a circular field and complete a single round in 18 seconds, 22seconds and 30 seconds respectively. In how much time will they meet again at the starting point?

• a) 16min 30sec

• b) 990 min

Solution: LCM of 18, 22 and 30 is 990. Hence they meet in 990 sec which is 16min 30sec.

• Question 3 of 5 3. Question Both the HCF and the difference of two numbers is 6. If the LCM of the two numbers is a 4 digit number, then what is the maximum possible value of the bigger number? a) 290 b) 41 c) 246 d) 260 Correct Answer: Let the numbers be 6k and 6 k(k + 1). For the HCF to be 6, k and k + 1 are always co-prime. LCM of 6k and 6(k + 1) is 6 k (k + 1). 6 k (k + 1) is a 5-digit number for k ≥ 41. Hence, the maximum possible value of k is 40 for which the LCM is a 4 digit number. Hence, the required number = 6(k + 1) = 6 × 41 = 246. Incorrect Answer: Let the numbers be 6k and 6 k(k + 1). For the HCF to be 6, k and k + 1 are always co-prime. LCM of 6k and 6(k + 1) is 6 k (k + 1). 6 k (k + 1) is a 5-digit number for k ≥ 41. Hence, the maximum possible value of k is 40 for which the LCM is a 4 digit number. Hence, the required number = 6(k + 1) = 6 × 41 = 246.

#### 3. Question

Both the HCF and the difference of two numbers is 6. If the LCM of the two numbers is a 4 digit number, then what is the maximum possible value of the bigger number?

Let the numbers be 6k and 6 k(k + 1).

For the HCF to be 6, k and k + 1 are always co-prime.

LCM of 6k and 6(k + 1) is 6 k (k + 1).

6 k (k + 1) is a 5-digit number for k ≥ 41.

Hence, the maximum possible value of k is 40 for which the LCM is a 4 digit number.

Hence, the required number = 6(k + 1) = 6 × 41 = 246.

Let the numbers be 6k and 6 k(k + 1).

For the HCF to be 6, k and k + 1 are always co-prime.

LCM of 6k and 6(k + 1) is 6 k (k + 1).

6 k (k + 1) is a 5-digit number for k ≥ 41.

Hence, the maximum possible value of k is 40 for which the LCM is a 4 digit number.

Hence, the required number = 6(k + 1) = 6 × 41 = 246.

• Question 4 of 5 4. Question Three planets revolve round the Sun once in 200, 250 and 300 days, respectively in their own orbits. When do they all come relatively to the same position as at a certain point of time in their orbits? A) After 3000 days B)After 2000 days C)After 1500 days D)After 1200 days Correct Answer: (A) After 3000 days. Solution: Given that, three planets revolves the Sun once in 200, 250 and 300 days, respectively in their own orbits. Required time = LCM of (200, 250 and 300) = 3000 days Hence, after 3000 days they all come relatively to the same position as at a certain point of time in their orbits. Hence, option A is correct. Incorrect Answer: (A) After 3000 days. Solution: Given that, three planets revolves the Sun once in 200, 250 and 300 days, respectively in their own orbits. Required time = LCM of (200, 250 and 300) = 3000 days Hence, after 3000 days they all come relatively to the same position as at a certain point of time in their orbits. Hence, option A is correct.

#### 4. Question

Three planets revolve round the Sun once in 200, 250 and 300 days, respectively in their own orbits. When do they all come relatively to the same position as at a certain point of time in their orbits?

• A) After 3000 days

• B)After 2000 days

• C)After 1500 days

• D)After 1200 days

Answer: (A) After 3000 days.

Given that, three planets revolves the Sun once in 200, 250 and 300 days, respectively in their own orbits.

Required time = LCM of (200, 250 and 300) = 3000 days

Hence, after 3000 days they all come relatively to the same position as at a certain point of time in their orbits.

Hence, option A is correct.

Answer: (A) After 3000 days.

Given that, three planets revolves the Sun once in 200, 250 and 300 days, respectively in their own orbits.

Required time = LCM of (200, 250 and 300) = 3000 days

Hence, after 3000 days they all come relatively to the same position as at a certain point of time in their orbits.

Hence, option A is correct.

• Question 5 of 5 5. Question The greatest five digit number which is exactly divisible by each one of the number 12, 18, 21 and 28 is A)98286 D)92888 C)99284 D)99792 Correct Answer: (D) 99792 Explain it LCM of 12, 18, 21 and 28 = 252 When we divide greatest five digit no. 99,999 by 252, we get 207 remainder Now, subtracting 207 by 99,999, we get 99999-207 = 99,792. 99,792 is largest five digit number divisible by 12, 18, 21 and 28. Incorrect Answer: (D) 99792 Explain it LCM of 12, 18, 21 and 28 = 252 When we divide greatest five digit no. 99,999 by 252, we get 207 remainder Now, subtracting 207 by 99,999, we get 99999-207 = 99,792. 99,792 is largest five digit number divisible by 12, 18, 21 and 28.

#### 5. Question

The greatest five digit number which is exactly divisible by each one of the number 12, 18, 21 and 28 is

Answer: (D) 99792

Explain it

LCM of 12, 18, 21 and 28 = 252

When we divide greatest five digit no. 99,999 by 252,

we get 207 remainder

Now, subtracting 207 by 99,999,

we get 99999-207 = 99,792.

99,792 is largest five digit number divisible by 12, 18, 21 and 28.