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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).
Wish you all the best ! 🙂
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Question 1 of 5
1. Question
Compound interest on rupees 8000 for 1 year at 10% per annum compounded half yearly is
Correct
Answer -D
Explanation – 8000 * (209/200 )∗ (209/200) = 8736.20 So, interest = 8736.20 − 8000 = 736.20
Incorrect
Answer -D
Explanation – 8000 * (209/200 )∗ (209/200) = 8736.20 So, interest = 8736.20 − 8000 = 736.20
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Question 2 of 5
2. Question
In 12 years, M will be thrice as old as N was 12 years ago. If M is now 8 years older than N, the present age of N is (in years):
Correct
Answer :A
Let N’s present age = x years. Then, M’s present age = (x + 8) years. ∴ (x + 8) + 12 = 3(x – 12) ⇒ x + 20= 3x – 36 ⇒ 2x =56 ⇒ x=28
Incorrect
Answer :A
Let N’s present age = x years. Then, M’s present age = (x + 8) years. ∴ (x + 8) + 12 = 3(x – 12) ⇒ x + 20= 3x – 36 ⇒ 2x =56 ⇒ x=28
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Question 3 of 5
3. Question
What is the least value of K. so that 123K578 is divisible by 11.
Correct
Answer : B
Answer Justification : Given: The number is 123K578 The number is divisible by 11 when the difference between the sum of the digit at odd places and the sum of the digit at even places is either 0 or divisible by 11.
Difference = the sum of odd the digit at the odd place – the sum of the digit at even place Calculation: Number = 123K578
Odd place digits are 8, 5, 3 and 1.
Even place digits are 7, K, and 2 Difference = the sum of odd the digit at the odd place – the sum of the digit at even place ⇒ 0 = (8 + 5 + 3 + 1) – (7 + K + 2) ⇒ 0 = 17 – 9 – K ⇒ K = 8 ∴ The least value of K is 8.
Incorrect
Answer : B
Answer Justification : Given: The number is 123K578 The number is divisible by 11 when the difference between the sum of the digit at odd places and the sum of the digit at even places is either 0 or divisible by 11.
Difference = the sum of odd the digit at the odd place – the sum of the digit at even place Calculation: Number = 123K578
Odd place digits are 8, 5, 3 and 1.
Even place digits are 7, K, and 2 Difference = the sum of odd the digit at the odd place – the sum of the digit at even place ⇒ 0 = (8 + 5 + 3 + 1) – (7 + K + 2) ⇒ 0 = 17 – 9 – K ⇒ K = 8 ∴ The least value of K is 8.
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Question 4 of 5
4. Question
The mean marks of 30 students in a class is 58.5. Later on it was found that 75 was wrongly recorded as 57. Find the correct them.
Correct
Answer : D
Answer Justification : Explanation: Correct mean = 30x 58.5 – 57 + (75/30) (1755+18)/30 = 1773/30 = 59.1
Incorrect
Answer : D
Answer Justification : Explanation: Correct mean = 30x 58.5 – 57 + (75/30) (1755+18)/30 = 1773/30 = 59.1
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Question 5 of 5
5. Question
How many arrangements can be made out of the letters of the word ‘ENGINEERING’ ?
Correct
Answer : B
Answer Justification : Explanation:
The word ‘ENGINEERING’ has 11 letters. But in these 11 letters, ‘E’ occurs 3 times, ‘N’ occurs 3 times, ‘G’ occurs 2 times, ‘I’ occurs 2 times and rest of the letters are different. Hence, number of ways to arrange these letters =11!/(3!)(3!)(2!)(2!) =11×10×9×8×7×6×5×4×3×2/(3×2)(3×2)(2)(2) =277200
Incorrect
Answer : B
Answer Justification : Explanation:
The word ‘ENGINEERING’ has 11 letters. But in these 11 letters, ‘E’ occurs 3 times, ‘N’ occurs 3 times, ‘G’ occurs 2 times, ‘I’ occurs 2 times and rest of the letters are different. Hence, number of ways to arrange these letters =11!/(3!)(3!)(2!)(2!) =11×10×9×8×7×6×5×4×3×2/(3×2)(3×2)(2)(2) =277200
