csat-2013 questions€¦ · 57/11, 1st floor, old rajinder nagar (opp. syndicate bank), new...

57/11, 1ST Floor, Old Rajinder Nagar (opp. Syndicate Bank), New Delhi-60. #9560774800, 011-45562800

CSAT-2013 QUESTIONS Q-1) In a rare coin collection, there is one gold coin for every three non-gold coins. 10 more gold coins are added to the collection and the ratio of gold coins to non-gold coins would be 1:2. Based on the information, the total number of coins in the collection now becomes

a) 90 b) 80 c) 60 d) 50

Usual method:

Let the number of gold coins = x Then number of gold coins = 3x After adding 10 gold coins, the number of gold coins = x + 10 Gold coins/non-gold coins = 1/2 (x+10)/3x = 1/2 2x + 20 = 3x X = 20 i.e. originally number of gold coins = 20, non-gold coins = 3 x 20 = 60 hence, number of coins in the collection now becomes (20+10)+60=90 *Estimated time to find an answer: 100 secs

Using technique:

As new ratio of gold coins & non-gold coins is 1:2.hence,total must be divisible by 3. This eliminates option b) and d). Before adding 10 gold coins to collection, ratio of gold to non-gold coins was 1 : 3. hence,earlier total must be divisible by 4. Option c) 60 – 10 =50 not divisible by 4 Option a) 90 -10 = 80 divisible by 4 Answer : a) *Estimated time to find an answer: 20 secs

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Q-2) A gardener has 1000plants. He wants to plant them in such a way that the number of rows and the number of columns remain the same. What is the minimum number of plants that he needs for this purpose?

a) 14 b) 24 c) 32 d) 34

USUAL METHOD:

Let the no. Of rows and columns be=x

Total no of plants= X.X

(Which is nothing but a perfect square)

As 1000 is not a perfect square we need to check for a perfect square above and nearest to 1000

Now, by observation, 31x31=961, 32x32=1024,

So he needs to add 24 more trees to get 1024. i.e. ANS (b)

*Estimated time 90secs

Using technique:

In perfect square tens digit must be even always except those ending with 6. Eg: 246, 1354, 225 can never be perfect square

Now in question option (a),(c),(d) can be easily eliminated.

Answer (b)

*Estimated time to find an answer: 10 secs



Q-3)A train travels at certain average speed for a distance of 63 km and then travels a distance of 72 km at an average speed of 6km/hr more than its original speed. If it takes 3 hrs to complete the total journey, what is the original speed of the train in km/hr?

a) 24 b) 33 c) 42 d) 66

Usual method:

let the original speed be S.

Total time taken = T1 + T2 = 3hrs

= 63/S + 72/(S + 6) =3

Try putting options, option c) satisfies equation.

63/42 +72/48 =3

Answer: c


Using technique: Overall avg. speed = total distance/total time = 135 / 3 =45 kmph The average speed must be in between original speed and speed in second spell. The only possible option amongst given is 42 kmph. Check: if original speed= 42 ,speed in 2nd spell = 42+6=48 Hence, avg. speed = (42+48)/2= 45 kmph *Estimated time to find an answer: 30 secs



Q-4) Four friends A,B,C & D distribute some money among themselves in such a manner that A gets one less than B, C gets 5 more than D, D gets 3 more than B. Who gets the smallest amount ?

a) A b) B c) C d) D

Usual method:

Let the amount got by A = x, Then the amount of B = x +1 The amount of D= (x+1) + 3 = x+4 The amount of C = (x+4)+5 = x+9 It is clear that everyone except A has got something more than some other (+ indicates more than).So answer cannot be B,C or D. Ans : a) *Estimated time to find an answer: 70 secs

Using technique: Take assumption Let B has Rs.2,then A will have Rs.1 and D will get Rs.5 and C will get Rs.10.so A has least. Answer: a) *Estimated time to find an answer: 20 secs



Q-5) In a class of 45 students, a boy is ranked 20th. When to buys joined, his rank was dropped by one what is his new rank from the end?

a) 25th b) 26th c) 27th d) 28th

Using technique:

After 2 boys joined,total strength of class = 47 New rank of boy = 21st Rank from end = 47-21+1 = 27 Answer: c) *Estimated time to find an answer: 20 secs



Q-6) A thief running at 8km/hr is chased by a policeman whose speed is 10km/hr. If the thief is 100m ahead of the policeman, then the time required for the policeman to catch the thief will be

a) 2mins b) 3mins c) 4mins d) 6mins

Usual method:

Relative speed of police & thief = 10 – 8 =2 kmph

(as they are in opposite direction)

Distance between police and thief = 100m =0.1km

Time taken by police to catch thief = distance/relative speed

= 0.1/2 =0.05hrs

= 0.05 x 60min = 3 min

Answer : b)


Using technique:

Try to do mentally Relative speed is 2kmph i.e. 2000 m in 60 min. Hence,1000m in 30 min and 100m in 3 min Answer: b) *Estimated time to find an answer: 20 secs



Q-7) A train travels at certain average speed for a distance of 63 km and then travels a distance of 72 km at an average speed of 6km/hr more than its original speed. If it takes 3 hrs to complete the total journey, what is the original speed of the train in km/hr?

a) 24 b) 33 c) 42 d) 66

Usual method:

let the original speed be S.

Total time taken = T1 + T2 = 3hrs

= 63/S + 72/(S + 6) =3

Try putting options, option c) satisfies equation.

63/42 +72/48 =3

Answer: c


Using technique:

Overall avg. speed = total distance/total time = 135 / 3 =45 kmph The average speed must be in between original speed and speed in second spell. The only possible option amongst given is 42 kmph. Check: if original speed= 42 ,speed in 2nd spell = 42+6=48 Hence, avg. speed = (42+48)/2= 45 kmph *Estimated time to find an answer: 30 secs



Q-8) A sum of rupees 700 has to be used to give seven cash prizes to the students of a school for their overall academic performance. If each prize is 20 rupees less than its preceding prize, what is the least value of prize?

a) 30 b) 40 c) 60 d) 80

USUAL METHOD:

Let the least value of prize be x Then the next value is x+20, and the next value is x+40, so on upto 7 values Hence it is an arithmetic progression with sum= 700 x+x+20+x+40+....+x+120=700 7x+ (20+40+....+120)=700 7x+20(21)=700 7x= 280 X=40 Answer(b) *Estimated time with the usual method is 120sec

Using technique:

For an arithmetic progression (A.P), Average of an A.P= Middle Term Sum =Average X No. Of Terms Average = Sum / Number = 700/70=100 Hence, student at the middle i.e (at position 4) will receive rupees 100 So, 3rd student =80 2nd student = 60 1st student = 40= least value Answer (b) *Estimated time to find an answer: 30 secs



Q-9) Out of 120 applications for a post, 70 are male and 80 have a driver’s license. What is the ratio between the minimum to maximum no of males having driver’s license?

a) 1 to 2 b) 2 to 3 c) 3 to 7 d) 5 to 7

Using technique:

Maximum males with drivers license:

All 70 males can have drivers license, and rest females

Hence, maximum = 70

Minimum males with drivers license

For minimum males, females should be maximum .

Number of females = 120 -70 = 50

Hence, 50 females with license

Male with driver license = 80-50 = 30= minimum

Required ratio = 30 : 70 = 3:7

Answer : c




Q-10) In a garrison there was food for 1000 soldiers for one month. After 10 days, 1000 more soldiers joined the garrison. How long would the soldiers be able to carry with the remaining food?

a) 25 days b) 20days c) 15days d) 10days

Usual method: Total provision is equivalent to 1000 x 30 = 30,000 man days

Of this total, 1000 x 10 = 10,000 man days consumed

Balance provision = 30,000 -10,000 = 20,000 man days

Total number of men now = 2000 men

Number of days = 20000/2000 = 10 days

Answer: d

Using technique: After 10 days, Time remaining = 20 days As number of people got doubled,time would be half as consumption is constant. Hence, 10 days *Estimated time to find an answer: 20 secs



Q-11) The tank full petrol in Arun’s motorcycle lasts for 10 days. If he starts 25% more every day, how many days will the tank-full petrol last?

a) 5 b) 6 c) 7 d) 8

Usual method:

Let arun’s daily consumption of petrol initially = 100 units Capacity of tank = 10 x 100 = 1000 units Now arun’s daily consumption = 100+25=125 Number of days tank full = 1000/125 = 8 Answer: d *Estimated time to find answer : 60 sec

Using technique:

Using multiplying factor, Old x (5/4) = 10 Old = 8 Answer: d) *Estimated time to find answer : 20 sec



Q-12) A person can walk a certain distance and drive back in six hours. He can also walk both ways in 10 hours. How much time will he take to drive both ways?

a) two hrs b) two & a half hrs c) five & a half hrs d) four hrs

Using technique: 2 way walk = 10 hrs Hence, 1 way walk = 5 hrs 1 way walk + 1 way drive = 6 hrs 1 way drive = 6-5 = 1 hr 2 way drive = 2 hrs Answer: a)

*Estimated time to find answer : 20 sec



Q-13) A contract on construction job specifies a penalty for delay in completion of a work beyond a certain date is as follows:200/- for 1st day, 300/- for 3rd day , etc. The penalty for each succeeding day being 50/- more than that of preceding day. How much penalty should the contractor pay if he delays the work by 10 days?

a) 4950 b) 4250 c) 3600 d) 650

Usual method

Penalty on first day = Rs.200 Subsequent increment = Rs.50 Total penalty for 10 days = 200 x 10 + 50 (1+2+…..+9) =2000 + 50x45 = Rs.4250

Using technique: For an AP, Sum = middle term X 10 Middle term = (number of terms + 1)/2 =1+10/2= 5.5 i.e; middle term lies between 5 & 6. 5th term will be 400 and by adding half of common difference we will get 5.5th term = 425 Sum = 425 x 10 = 4250 Answer (b) *Estimated time to find an answer: 30 secs


csat-2013 questions€¦ · 57/11, 1st floor, old rajinder nagar (opp. syndicate bank), new...

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