Find the day of the week on
16th july, 1776.
Ans: Tuesday
Sol: 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
Find the angle between the
minute hand and hour hand of a click when the time is
7.20?
Ans: 100deg
Sol: 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.
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
Sol: 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.
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.
Sol: 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.
At what time between 2 and 3
o’ clock will the hands 0a a clock together?
Ans: 10 10/11 min. past 2.
Sol: 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.
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
Sol: 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.
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
Sol: Let sum = X. Then S.I. = 16x/25
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.
Sol:
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
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
Sol: 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
A sum of money doubles
itself at C.I. in 15 years. In how many years will it become eight times?
Ans.45 years.
Sol: P [1 + (R/100)]^15 = 2P è [1 + (R/100)]^15 = 2……….(i)
Let P [1 + (R/100)]^n = 8P è 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.
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 %.
Sol: 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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