# Quest-Global placement paper set 2

Posted on :01-04-2016

Q1) Walking 5/6 of its usual speed, a train is 10min late. Find the usual time to cover the journey?

ANS: 50 min

Solution:

New speed = 5/6 of usual speed
New time = 6/5 of usual time
Therefore, (6/5 of usual time)  usual time = 10min
Therefore Usual time = 50min

Q2) A train running at 54 kmph takes 20 seconds to pass a platform. Next it takes 12 seconds to pass a man walking at 6 kmph in the same direction in which the train is going. Find the length of the train and the length of the platform.

ANS: length of the train=160m
length of the platform=140 m.

Solution:

Let the length of the train be x meters and length of the platform be y meters.
Speed of the train relative to man=(54-6) kmph =48 kmph.
=(48*5/18) m/sec =40/3 m/sec.
In passing a man, the train covers its own length with relative speed.
Therefore, length of the train=(Relative speed *Time)
=(40/3 * 12) m =160 m.
Also, speed of the train=(54 * 5/18) m/sec=15 m/sec.
Therefore, x+y/2xy=20 or x+y=300 or y=(300-160 m=140 m.
Therefore, Length of the platform=140 m.

Q3) A man is standing on a railway bridge which is 180m long. He finds that a train crosses the bridge in 20seconds but himself in 8 seconds. Find the length of the train and its speed.

ANS: length of train=120m
Speed of train=54kmph

Solution:

Let the length of the train be x meters
Then, the train covers x meters in 8 seconds and (x + 180) meters in 20 seconds.
Therefore x/8 = (x+180)/20 ó 20x = 8(x+180) ó x = 120
Therefore Length of the train = 120m
Speed of the train = 120/8 m/sec = 15 m/sec =15 * 18/5 kmph = 54kmph

Q4)A man sells an article at a profit of 25%. If he had bought it at 20 % less and sold it for Rs.10.50 less, he would have gained 30%. Find the cost price of the article?

ANS: Rs. 50

Solution:

Let the C.P be Rs.x.
1st S.P =125% of Rs.x.= 125*x/100= 5x/4.
2nd C.P=80% of x. = 80x/100 =4x/5.
2nd S.P =130% of 4x/5. = (130/100* 4x/5) = 26x/25.
Therefore, 5x/4-26x/25 = 10.50 or x = 10.50*100/21=50.
Hence, C.P = Rs. 50.

Q5)A grosser purchased 80 kg of rice at Rs.13.50 per kg and mixed it with 120 kg rice at Rs. 16 per kg. At what rate per kg should he sell the mixture to gain 16%?

ANS: Rs.17.40 per kg.

Solution:

C.P of 200 kg of mix. = Rs (80*13.50+120*16) = Rs.3000.
S.P = 116% of Rs 3000= Rs (116*3000/100) = Rs.3480.
Rate of S.P of the mixture = Rs.3480/200.per kg. = Rs.17.40 per kg.

Q6)Two persons A and B working together can dig a trench in 8 hrs while A alone can dig it in 12 hrs. In how many hours B alone can dig such a trench?

ANS: 24hours.

Solution:

(A+B)s one hours work =1/8, As one hours work =1/12
Therefore, Bs one hours work =( 1/8-1/12) =1/24.
Hence, B alone can dig the trench in 24 hours.

Q7)A and B can do a piece of work in 12 days ; B and C can do it in 20 days. In how many days will A, B and C finishes it working all together? Also, find the number of days taken by each to finish it working alone?

ANS: 60 days

Solution:

(A+B)s one days work=1/12; (B+C)s one days work=1/15 and (A+C)s one days work=1/20.
Adding, we get: 2(A+B+C)s one days work = (1/12+1/15+1/20)=1/5
Therefore, (A+B+C)s one days work=1/10
Thus, A, B and C together can finish the work in 10 days.
Now, As one days work
= [A+B+C)s one days work] - [(B+C)s one days work]
= 1/10-1/15)
= 1/30.
Therefore, A alone can finish the work in 30 days.
Similarly, Bs 1 days work = (1/10 -1/20) = 1/20.
Therefore, B alone can finish the work in 20 days.
And, Cs 1 days work= (1/10-1/12) = 1/60.
Therefore, C alone can finish the work in 60 days.

Q8)A is twice as good a workman as B and together they finish a piece of work in 18 days. In how many days will A alone finish the work?

ANS: 27 days.

Solution:

(As 1 days work): (Bs 1 days work) = 2:1
(A + B)s 1 days work = 1/18
Divide 1/18 in the ratio 2:1.
Therefore As 1 days work = 1/18* 2/3 = 1/27
Hence, A alone can finish the work in 27 days.

Q9)2 men and 3 boys can do a piece of work in 10 days while 3 men and 2 boys can do the same work in 8 days. In how many days can 2 men and 1 boy do the work?

ANS: 12 ½ days.

Solution:

Let 1 mans 1 days work = x and 1 boys 1 days work =y
Then, 2x+3y=1/10 and 3x+2y=1/8.
Solving, we get: x=7/200 and y=1/100.
Therefore (2 men+ 1boy)s 1 days work =(2*7/200+1*1/100)=16/200=2/25
So, 2 men and 1 boy together can finish the work in 25/2 =12 1/2 days.

Q10)What was the day of the week on 12th January, 1979?

ANS: Friday

Solution:

Number of odd days in (1600 + 300) years = (0 + 1) = 1 odd day.
78 years = (19 leap years + 59 ordinary years) = (38 + 59) odd days = 6 odd days 12 days of January have 5 odd days.
Therefore, total number of odd days= (1 + 6 + 5) = 5 odd days.
Therefore, the desired day was Friday.

Q11) Find the day of the week on 16th july, 1776.

ANS: Tuesday

Solution:

16th july, 1776 means = 1775 years + period from 1st january to 16th july
Now, 1600 years have 0 odd days.
100 years have 5 odd days.
75 years = 18 leap years + 57 ordinary years
= (36 + 57) odd days = 93 odd days
= 13 weeks + 2 odd days = 2 odd days
Therefore, 1775 years have (0 + 5 + 2) odd days = 0 odd days.
Now, days from 1st Jan to 16th july; 1776
Jan Feb March April May June July
31 + 29 + 31 + 30 + 31 + 30 + 16 = 198 days
= (28 weeks + 2 days) odd days
Therefore, total number of odd days = 2
Therefore, the day of the week was Tuesday

Q12) Find the angle between the minute hand and hour hand of a click when the time is 7.20?

ANS: 100deg

Solution:

Angle traced by the hour hand in 12 hours = 360 degrees.
Angle traced by it in 7 hrs 20 min i.e. 22/3 hrs = [(360/12) * (22/3)] = 220 deg.
Angle traced by minute hand in 60 min = 360 deg.
Angle traced by it in 20 min = [(360/20) * 60] = 120 deg.
Therefore, required angle = (220 - 120) = 100deg.

Q13)The minute hand of a clock overtakes the hours hand at intervals of 65 min of the correct time. How much of the day does the clock gain or lose?

ANS: the clock gains 10 10/43 minutes

Solution:

In a correct clock, the minute hand gains 55 min. spaces over the hour hand in 60 minutes.To be together again, the minute hand must gain 60 minutes over the hour hand.55 minutes are gained in 60 min.
60 min. are gained in [(60/55) * 60] min == 65 5/11 min.
But they are together after 65 min.
Therefore, gain in 65 minutes = (65 5/11 - 65) = 5/11 min.
Gain in 24 hours = [(5/11) * (60*24)/65] = 10 10/43 min.
Therefore, the clock gains 10 10/43 minutes in 24 hours.

Q14)A clock is set right at 8 a.m. The clock gains 10 minutes in 24 hours. What will be the true time when the clock indicates 1 p.m. on the following day?

ANS:. 48 min. past 12.

Solution:

Time from 8 a.m. on a day to 1 p.m. on the following day = 29 hours.
24 hours 10 min. of this clock = 24 hours of the correct clock.
145/6 hrs of this clock = 24 hours of the correct clock.
29 hours of this clock = [24 * (6/145) * 29] hrs of the correct clock
= 28 hrs 48 min of the correct clock.
Therefore, the correct time is 28 hrs 48 min. after 8 a.m.
This is 48 min. past 12.

Q15) At what time between 2 and 3 o clock will the hands 0a a clock together?

ANS: 10 10/11 min. past 2.

Solution:

At 2 o clock, the hour hand is at 2 and the minute hand is at 12, i.e. they are 10 min space apart. To be together, the minute hand must gain 10 minutes over the other hand. Now, 55 minutes are gained by it in 60 min.
Therefore, 10 min will be gained in [(60/55) * 10] min = 10 10/11 min.
Therefore, the hands will coincide at 10 10/11 min. past 2.

Q16) A sum of money amounts to Rs.6690 after 3 years and to Rs.10035 after 6 years on compound interest. Find the sum.

ANS: Rs. 4460

Solution:

Let the Sum be Rs. P. Then
P [1 + (R/100)]^3 = 6690..(i)
P [1 + (R/100)]^6 = 10035..(ii)

On dividing, we get [1 + (R/100)]^3 = 10035/6690 = 3/2.
P * (3/2) = 6690 or P = 4460.
Hence, the sum is Rs. 4460.

Q17) Simple interest on a certain sum is 16/25 of the sum. Find the rate percent and time, if both are numerically equal.

ANS: Rate = 8% and Time = 8 years

Solution:

Let sum = X. Then S.I. = 16x/25

Q18) Let rate = R% and Time = R years. Therefore, x * R * R/100 = 16x/25 or R^2 = 1600/25, R = 40/5 = 8 Therefore, Rate = 8% and Time = 8 years. Find

i. S.I. on RS 68000 at 16 2/3% per annum for 9 months.
ii. S.I. on RS 6250 at 14% per annum for 146 days.
iii. S.I. on RS 3000 at 18% per annum for the period from 4th Feb 1995 to 18th April 1995.

ANS:

i. RS 8500.
ii. RS 350.
iii. RS 108.

Solution:

i. P = 68000, R = 50/3% p.a. and T = 9/12 year = ¾ years
Therefore, S.I. = (P * Q * R/100)
= RS (68000 * 50/3 * ¾ * 1/100) = RS 8500.
ii. P = RS 6265, R = 14% p.a. and T = (146/365) year = 2/5 years.
Therefore, S.I. = RS (6265 * 14 * 2/5 *1/100) = RS 350.
iii. Time = (24 + 31 + 18) days = 73 days = 1/5 year
P = RS 3000 and R = 18% p.a.
Therefore, S.I. = RS (3000 * 18 * 1/5 * 1/100) = RS 108

Q19) A sum at simple interest at 13 ½% per annum amounts to RS 2502.50 after 4 years. Find the sum.

ANS: sum = RS 1625

Solution:

Let sum be x. Then,
S.I. = (x * 27/2 * 4 * 1/100) = 27x/50
Therefore, amount = (x + 27x/50) = 77x/50
Therefore, 77x/50 = 2502.50 or x = 2502.50 * 50 / 77 = 1625
Hence, sum = RS 1625

Q20) A sum of money doubles itself at C.I. in 15 years. In how many years will it become eight times?

ANS: 45 years.

Solution:

P [1 + (R/100)]^15 = 2P [1+(R/100)]^15 = 2..............(i)
Let P [1 + (R/100)]^n = 8P P [1+(R/100)]^n = 8 = 2^3
= [{1 + (R/100)}^15]^3.
è [1 + (R/100)]^n = [1 + (R/100)]^45.
è n = 45.
Thus, the required time = 45 years.

Q21) A certain sum amounts to Rs. 7350 in 2 years and to Rs. 8575 in 3 years. Find the sum and rate percent.

ANS: Sum = Rs. 5400,Rate=16 2/3 %.

Solution:

S.I. on Rs. 7350 for 1 year = Rs. (8575-7350) = Rs. 1225.
Therefore, Rate = (100*1225 / 7350*1) % = 16 2/3 %.
Let the sum be Rs. X. then, x[1 + (50/3*100)]^2 = 7350.
è x * 7/6 * 7/6 = 7350.
è x = [7350 * 36/49] = 5400.
Therefore, Sum = Rs. 5400.

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