[Mission 2022] Insta–DART (Daily Aptitude and Reasoning Test) 29 September 2021

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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

For any integer n, what is HCF (22n + 7,33n + 10) equal to?

Correct

Sol.(b)

HCF of (22n + 7, 33n + 10) is always 1.

Illustration: For n = 1, HCF (29, 43) = 1

For n = 2, HCF (51, 76) = 1

For n= 3, HCF (73, 109) = 1

Incorrect

Sol.(b)

HCF of (22n + 7, 33n + 10) is always 1.

Illustration: For n = 1, HCF (29, 43) = 1

For n = 2, HCF (51, 76) = 1

For n= 3, HCF (73, 109) = 1

Question 2 of 5

2. Question

A, B and C can do a piece of work in 11 days, 20 days and 55 days respectively, working alone. How soon can the work be done if A is assisted by B on odd days and C on even days?

Correct

Ans- Solution: B

o (A+B)’s 1 day’s work = (1/11 + 1/20 ) = 31/220 o (A+C)’s 1 day’s work = ( 1/11 + 1/55) = 6/55 o Work done in 2 days by them= ( 31/220 + 6/55) = 55/220 = ¼. o Therefore, Whole work will be done in 8 days

Incorrect

Ans- Solution: B

o (A+B)’s 1 day’s work = (1/11 + 1/20 ) = 31/220 o (A+C)’s 1 day’s work = ( 1/11 + 1/55) = 6/55 o Work done in 2 days by them= ( 31/220 + 6/55) = 55/220 = ¼. o Therefore, Whole work will be done in 8 days

Question 3 of 5

3. Question

Amit covers one-fourth of the total distance at 20 kmph, one-fourth at 10 kmph and rest of his journey at 80 kmph. Find Amit’s average speed for the whole distance?

Correct

Solution: A

Average speed = Total Distance/Total Time

Let the total distance = D Km Therefore, Time taken to complete first D/4 distance = t1 =(D/4)km/20kmph Time taken to complete second D/4 distance = t2 = (D/4)/10kmph Time taken to complete remaining distance (that is D/2) = t3 = (D/2)/80kmph Therefore, Average speed = Total Distance/Total Time = (D)/(t1+t2+3) = D/(D/80 + D/40 + D/160) = D/(2D + 4D + D)/160 = 160 D / 7D Therefore, Average Speed = 22. 85kmph

Incorrect

Solution: A

Average speed = Total Distance/Total Time

Let the total distance = D Km Therefore, Time taken to complete first D/4 distance = t1 =(D/4)km/20kmph Time taken to complete second D/4 distance = t2 = (D/4)/10kmph Time taken to complete remaining distance (that is D/2) = t3 = (D/2)/80kmph Therefore, Average speed = Total Distance/Total Time = (D)/(t1+t2+3) = D/(D/80 + D/40 + D/160) = D/(2D + 4D + D)/160 = 160 D / 7D Therefore, Average Speed = 22. 85kmph

Question 4 of 5

4. Question

A zoo consists of only Tigers and Lions. The number of tigers in a zoo is three times the number of lions. Which one of the following numbers cannot represent the total number of animals in the zoo?

Correct

Ans-. Solution: B

• Let the number of lions = x • We know that the number of tigers in a zoo is three times the number of lions. Hence, the number of tigers = 3x. • Then, x + 3x = 4x = total number of animals. • Thus the total number of animals is a number that must be divisible by 4. • Here, 42 is the only number which is not divisible by 4

Incorrect

Ans-. Solution: B

• Let the number of lions = x • We know that the number of tigers in a zoo is three times the number of lions. Hence, the number of tigers = 3x. • Then, x + 3x = 4x = total number of animals. • Thus the total number of animals is a number that must be divisible by 4. • Here, 42 is the only number which is not divisible by 4

Question 5 of 5

5. Question

There are two containers P and Q. P contains 56 kg of salt and Q contains 36 kg of sugar. From P 24 kg of salt is taken out and poured into Q. Then 20kg of the mixture from Q is taken out and poured into P. Find the ratio of final quantity of salt to sugar in container P

Correct

Ans-Solution: D •

Initially, the Amount of sugar in Q = 36kg • Now, 24kg of salt is poured in Q, • Total quantity in Q becomes = 36kg (sugar) + 24kg (salt) = 60kg (mixture) • The ratio of salt to sugar in Q becomes = 24 : 36 = 2 : 3 • Now, the Amount of salt in P = 56 – 24 = 32 kg • Again, 20kg of the mixture is taken out from Q and poured into P Therefore, quantity of salt and sugar in P becomes = [32 + 20 * (2/5)] kg of salt + 20 * (3/5) kg of sugar = (32 + 8) kg of salt + 12 kg of sugar = 40 kg of salt + 12 kg of sugar Required ratio = (40/12) = (10/3)

Incorrect

Ans-Solution: D •

Initially, the Amount of sugar in Q = 36kg • Now, 24kg of salt is poured in Q, • Total quantity in Q becomes = 36kg (sugar) + 24kg (salt) = 60kg (mixture) • The ratio of salt to sugar in Q becomes = 24 : 36 = 2 : 3 • Now, the Amount of salt in P = 56 – 24 = 32 kg • Again, 20kg of the mixture is taken out from Q and poured into P Therefore, quantity of salt and sugar in P becomes = [32 + 20 * (2/5)] kg of salt + 20 * (3/5) kg of sugar = (32 + 8) kg of salt + 12 kg of sugar = 40 kg of salt + 12 kg of sugar Required ratio = (40/12) = (10/3)

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