#### Quiz-summary

0 of 5 questions completed

Questions:

- 1
- 2
- 3
- 4
- 5

#### Information

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 ! 🙂

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:

#### Results

0 of 5 questions answered correctly

Your time:

Time has elapsed

You have reached 0 of 0 points, (0)

#### Categories

- Not categorized 0%

- 1
- 2
- 3
- 4
- 5

- Answered
- Review

- Question 1 of 5
##### 1. Question

33 men can do a job in 30 days. If 44 men started the job together and after every day of the work, one person leaves. What is the minimum number of days required to complete the whole work?

CorrectYour Answer :C

If 33 men can do a job in 30 days, then there are 33 X 30 = 990 pieces of job.

If 44 pieces of work is done on the first day, then 43 pieces will be done on second day. Cumulatively, 44+43+42+………. = 990 days. This is the problem of sum of series with starting number as a = 44 and difference d = -1.

From the formula for Sum of the series S = n/2( 2a + (n-1) d) 990 = n/2 (2 × 44 + (n-1) × -1

1980 = n (89 – n) 1980 = 89n – n2 n^2 – 89n + 1980=0 Solving quadratic equation, n = 44 or 45. Therefore, minimum number of days required is 44 days.

IncorrectYour Answer :C

If 33 men can do a job in 30 days, then there are 33 X 30 = 990 pieces of job.

If 44 pieces of work is done on the first day, then 43 pieces will be done on second day. Cumulatively, 44+43+42+………. = 990 days. This is the problem of sum of series with starting number as a = 44 and difference d = -1.

From the formula for Sum of the series S = n/2( 2a + (n-1) d) 990 = n/2 (2 × 44 + (n-1) × -1

1980 = n (89 – n) 1980 = 89n – n2 n^2 – 89n + 1980=0 Solving quadratic equation, n = 44 or 45. Therefore, minimum number of days required is 44 days.

- Question 2 of 5
##### 2. Question

If a leap year begins with Sunday, which of the following days in that year will come for 53 times?

- Sunday
- Monday
- Tuesday
- Saturday

CorrectYour Answer : A

In a leap year, there are 366 days. There will be 52 complete weeks (52*7=364 days) and 2 days are left. So, in a cycle, 1st and 2nd day of that year will repeat as 365th and 366th days of that year.

IncorrectYour Answer : A

In a leap year, there are 366 days. There will be 52 complete weeks (52*7=364 days) and 2 days are left. So, in a cycle, 1st and 2nd day of that year will repeat as 365th and 366th days of that year.

- Question 3 of 5
##### 3. Question

Anil and Bharati entered into partnership with Rs.700 and Rs.600 respectively. After 3 months Anil withdrew 2/7 0f his stock but 3 months later, he puts back 3/5 of what he had withdrawn. The profit at the end of the year is rs.726. How much of this should Anil receive?

CorrectAnswer : B

Anil’s investment for first 3 months = Rs.700 Anil’s investment for next 3 months = 700 – (2/7) of 700 = 700 – 200 = Rs.500 Anil’s investment for last 6 months = 500 + (3/5) of withdrawn amount = 500 + (3/5) of (2/7) of 700 = 500 + 120 = Rs.620 The ratio of shares in the profit of Anil and Bharati = {(700 × 3) + (500 × 3) + (620×6)} : 600 × 12 = {2100 + 1500 + 3720} : {7200} = 7320 : 7200 = 61:60 The share of Anil = (61/121) of Rs.726 = Rs.366

IncorrectAnswer : B

Anil’s investment for first 3 months = Rs.700 Anil’s investment for next 3 months = 700 – (2/7) of 700 = 700 – 200 = Rs.500 Anil’s investment for last 6 months = 500 + (3/5) of withdrawn amount = 500 + (3/5) of (2/7) of 700 = 500 + 120 = Rs.620 The ratio of shares in the profit of Anil and Bharati = {(700 × 3) + (500 × 3) + (620×6)} : 600 × 12 = {2100 + 1500 + 3720} : {7200} = 7320 : 7200 = 61:60 The share of Anil = (61/121) of Rs.726 = Rs.366

- Question 4 of 5
##### 4. Question

There are eight friends P, Q, R, S, T, U, V and W who invested in some business in the ratio 1:2:3:4:5:6:7:8 and the duration for which they invested the money is in the ratio 8:7:6:5:4:3:2:1 respectively. Who will get the maximum profit at the end of the year?

CorrectYour Answer : C

The share in the profit will be proportional to the product of investment and the time duration. Therefore, the ratio of shares in the profit of P, Q, R, S, T, U, V and W = 1×8 : 2×7 : 3×6 : 4×5 : 5×4 : 6×3 : 7×2 : 8×1 = 8 : 14 : 18 : 20 : 20 : 18 : 14 : 8 So, both S and T will get the maximum share.

IncorrectYour Answer : C

The share in the profit will be proportional to the product of investment and the time duration. Therefore, the ratio of shares in the profit of P, Q, R, S, T, U, V and W = 1×8 : 2×7 : 3×6 : 4×5 : 5×4 : 6×3 : 7×2 : 8×1 = 8 : 14 : 18 : 20 : 20 : 18 : 14 : 8 So, both S and T will get the maximum share.

- Question 5 of 5
##### 5. Question

A sum of Rs.8400 was taken as a loan. This is to be paid in two equal instalments. If the rate of interest is 10% per annum, compounded annually, then the value of each instalment is:

CorrectYour Answer – C

Let the instalment be Rs.’x’ Liability at the end of the first year = 8400 + 10% of 8400 = 8400 + 840 = Rs.9240 Principle for the beginning of second year = Rs.(9240 – x) Liability at the end of second year = (9240-x) + 10% of (9240-x) = 1.1(9240-x) But, liability at the end of second year = instalment 1.1(9240-x) = x 10164 – 1.1x = x 2.1x = 10164 x = 10164/2.1 = Rs.4840 Therefore, the value of each instalment = Rs.4840

IncorrectYour Answer – C

Let the instalment be Rs.’x’ Liability at the end of the first year = 8400 + 10% of 8400 = 8400 + 840 = Rs.9240 Principle for the beginning of second year = Rs.(9240 – x) Liability at the end of second year = (9240-x) + 10% of (9240-x) = 1.1(9240-x) But, liability at the end of second year = instalment 1.1(9240-x) = x 10164 – 1.1x = x 2.1x = 10164 x = 10164/2.1 = Rs.4840 Therefore, the value of each instalment = Rs.4840