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
If a number P is a two-digit natural number which when multiplied by 3 gives a three-digit number ending in digit Q, then consider the following statements and a question:
Statement I: Q is half of P.
Statement II: Q is an odd number.Question: What is the value of P?
Which one of the following is correct in respect of the above Statements and the Question?
Correct
Answer – D
Explanation:
Incorrect
Answer – D
Explanation:
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Question 2 of 5
2. Question
If the sum of two numbers X and Y is 50, then consider the following statements and a question:
Statement I: X is 10 more than twice Y.
Statement II: Y is an even number.
Question: What is the value of X?Which one of the following is correct with respect to above statements
Correct
Answer – A
Solution –
Statement I alone:
X = 2Y + 10 and X + Y = 50 ⇒ (2Y + 10) + Y = 50 ⇒ 3Y + 10 = 50 ⇒ 3Y = 40 ⇒ Y = 40/3 ⇒ X = 2 × (40/3) + 10 = 80/3 + 10 ⇒ Unique ⇒ Sufficient.Statement II alone:
Y even and X + Y = 50 ⇒ many possibilities ⇒ Not sufficient.Conclusion: (a) Statement I alone is sufficient.
Incorrect
Answer – A
Solution –
Statement I alone:
X = 2Y + 10 and X + Y = 50 ⇒ (2Y + 10) + Y = 50 ⇒ 3Y + 10 = 50 ⇒ 3Y = 40 ⇒ Y = 40/3 ⇒ X = 2 × (40/3) + 10 = 80/3 + 10 ⇒ Unique ⇒ Sufficient.Statement II alone:
Y even and X + Y = 50 ⇒ many possibilities ⇒ Not sufficient.Conclusion: (a) Statement I alone is sufficient.
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Question 3 of 5
3. Question
From the first 200 natural numbers, how many are not divisible by any of 2, 3, 5, or 7?
Correct
Answer: (b)
Explanation:
Using Inclusion–Exclusion for 1–200:- By 2,3,5,7: 100+66+40+28=234
- Subtract pairs: 33+20+14+13+9+5=94
- Add triples: 6+4+2+1=13
- Quadruple: 000
Count divisible by at least one = 234−94+13=153
Not divisible by any = 200−153= 47
Incorrect
Answer: (b)
Explanation:
Using Inclusion–Exclusion for 1–200:- By 2,3,5,7: 100+66+40+28=234
- Subtract pairs: 33+20+14+13+9+5=94
- Add triples: 6+4+2+1=13
- Quadruple: 000
Count divisible by at least one = 234−94+13=153
Not divisible by any = 200−153= 47
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Question 4 of 5
4. Question
Consider the sequence
AB_CC_A_BCCC_BBC_C that follows a certain pattern. Which one of the following completes the sequence?
Correct
Answer:(c)
Given: AB_CC_A_BCCC_BBC_C
Since, there are 18 letters in the series, we can divide them in groups of 6 or 3. Dividing them in groups of 6, we get:
[ A B _ C C _ ] ; [ A _ B C C C ] ; [ _ B B C _ C ]
The pattern being repeated is ABBCCC. So, we can fill in the blanks as follows:
[ A B B C C C ] ; [ A B B C C C ] ; [ A B B C C C ]
Therefore, the required letters to complete the sequence are: B, C, B, A, C
Incorrect
Answer:(c)
Given: AB_CC_A_BCCC_BBC_C
Since, there are 18 letters in the series, we can divide them in groups of 6 or 3. Dividing them in groups of 6, we get:
[ A B _ C C _ ] ; [ A _ B C C C ] ; [ _ B B C _ C ]
The pattern being repeated is ABBCCC. So, we can fill in the blanks as follows:
[ A B B C C C ] ; [ A B B C C C ] ; [ A B B C C C ]
Therefore, the required letters to complete the sequence are: B, C, B, A, C
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Question 5 of 5
5. Question
If 6∗8=106 * 8 = 106∗8=10 and 11∗60=6111 * 60 = 6111∗60=61, then what is 33∗5633 * 5633∗56 equal to?
Correct
Answer: (c)
The given numbers are Pythagorean triplets, wherein:
a2+b2=c2So, (62+82)=36+64=100=102
And, (112+602)=121+3600=3721=612Similarly, (332+562)=1089+3136=4225=652
Hence, 33∗56=6533 * 56 = 6533∗56=65.
Incorrect
Answer: (c)
The given numbers are Pythagorean triplets, wherein:
a2+b2=c2So, (62+82)=36+64=100=102
And, (112+602)=121+3600=3721=612Similarly, (332+562)=1089+3136=4225=652
Hence, 33∗56=6533 * 56 = 6533∗56=65.
