# Quant Boosters - Vikas Saini - Set 5

• How to find the day of any century
In a non leap year 365 days.
365 = 7 x 52 + 1.
It means one odd day in a non leap year.
In a year every single day of week repeats by 52 times at least, whereas 1 odd day will be remain in a year itself.
For a leap year 2 odd days will be remain.
In a century we find 76 non leap years and 24 leap years.
So total no of odd days in a century = 76 x 1 + 24 x 2 = 124 days.
124 mod 7 = 5 days.

In a century there is 5 odd days.
100 years = 5 odd days.
200 years = 5 x 2 mod 7 = 3 odd days
300 years = 5 x 3 mod 7 = 1odd day.
400 years = (5 x 4 +1) mod 7 = 0 odd.

It means after 400 years no odd days.

For month odd days code
January - 0
February - 3
March - 3
April - 6
May - 1
June - 4
July - 6
August - 2
September - 5
October - 0
November - 3
December - 5

And code for century
100 years = +4.
200 years = +2.
300 years = 0.
400 years = -1.

An awesome formula to find the day if you know the date :
Day = ( Date + Month code + year + [year/4]+ century code ) mod 7

Given question,
Date = 15
Month code = 2
Year = 47
century code = 0(1900 = +300)
= (15 + 2 + 47 + [47/4] + 0 ) mod 7
= (15 + 2 + 47 + 11) mod 7
= 75 mod 7
= 5.
5th day of week is Friday.

Some more examples

What was the day 26th January 1950.
Day = 26
Month code = 0
Year = 50
century code = 0.
Day = (26+0+50+[50/4]+0) mod 7
= (26+50+12+0) mod 7
= 88 mod 7
= 4.
4th day of week = Thursday.

What was the day on 12 December 1729.
Date = 12.
Month code = 5
Year = 29.
Century code = +4 (+100 year)
Day = (12+5+29+[29/4]+4) mod 7
= (12+5+29+7+4) mod 7
= 57 mod 7
= 1
1st day of week is Monday

What will be the day on 30th September 2016.
Date = 30.
Month code = 5.
Year = 16.
Century code = -1 ( +400)
Day = (30 + 5 + 16 + [16/4] -1) mod 7
= (30+5+16+4-1) mod 7
= 5
5th day of week is Friday.

• Q19) What is the angle between Hour hand and minute hand at 4:45 PM ?

• Finding angle between Hands
12 h = 360°
1 h = 30°
60 m = 30°
1 m = 0.5°
Angle = | 5.5 M - 30 H |
M = minute
H = Hour

Given question,
5 hours = 5 x 30° = 150°
45 min = 45 x 0.5° = 22.5°
Total = 150° - 22.5° = 127.5°

using Shortcut :-
Angle = | 5.5 M - 30 H |
= | 5.5 x 45 - 30 x 4 |
= 127.5 °

One more example -

To find the angle between hour hand and minute hand at 3:30
3 Hours = 30° x 3 = 90°
30 min = 0.5° x 30 = 15°
Total = 90° - 15° = 75°.
Shortcut :-
Angle = | 5.5 M - 30 H |
= | 5.5 x 30 - 30 x 3 |
= 75°

• Q20) A watch gains 5 sec in every 3 minutes and was set right at 8 AM. So what time will it show at 10 PM on the same day ?

• Some concepts before the solution

Relative speed of 2 clock hands
12 H = 60 div
1 H = 5 div.
SH = 60 M = 50 div.
SH = 5/60 div/min.
SH = 1/12 div/min.
SM = 60 div / 60 min = 1div/1min
Relative speed = 1 - 1/12 = 11/12 div/min.
T = 60 / (11/12)
T = 65 + (5/11)
T = 1:05:5/11
Both hand will coincide after every 1:05:5/11.

Examples

By what time both hands will coincide in between 5-6 PM ?
(1:05:5/11) x 5
= 5:25:25/11
= 5:27:3/11

By what time both hand will coincide in between 9 - 10 PM ?
(1:05:5/11) x 9
= 9:45:45/11
= 9:49:1/11.

By what time in between 2-3 PM both hand will make angle of 30°.
angle = | 5.5 M - 30 H |
30° = | 5.5 x M - 30 x 2 |
30° = | 5.5 M - 60° |
90° = 5.5 M
M = 900 / 55 = 180/11 = 16 (4/11).
Time = 2:16:4/11.

By what time in between 5-6 PM both hands will make angle of 60°.
angle = | 5.5 M - 30 H |
60° = | 5.5 M - 30 x 5 |
60° = 5.5 M - 150
210° = 5.5 M
M = 2100/55 = 420/11 = 38(2/11)
Time = 5:38:2/11.

Given question

3 Min = 5 sec
1 min = 5/3 sec.
60 min = 100 sec.
1 Hour = 100 sec.
14 Hour = 1400 sec.
1400 sec = 23 min 20 seconds.
Time = 10:23:20 sec.

• Q21) If all the roots of the equation (x – m)^2 * (x – 10) + 4 = 0 are integers, find the number of distinct values that ‘m’ can have ?

• (x – m)^2 * (x-10) = -4.
-4 = 4 x (-1) , 1 x (-4),

(x – m)^2 (x -10) = 4 x (-1)
x – 10 = -1.
x = 9.
(9 – m)^2 = 4.
(9 – m) = 2, -2.
m = 7, 11.

(x – m)^2 (x – 10) = 1 x (-4)
x – 10 = -4.
x = 6.
( 6 – m)^2 = 1.
6 – m =1, -1.
m = 5, 7.

Hence m can take three values.

• Q22) Find the number of integer solutions of the equation x^2/y = 4x – 3, where x and y are non zero real numbers

• Suppose x^2 / y = t.
t = 4x – 3.
x = 1, t = 1, y = 1.
x = 3, t = 9, y = 1.
Only two integer solutions.

• Q23) If a, b are integers then how many ordered pairs (a, b) satisfy the equation a^2 + ab + b^2 = 1 ?

• a^2 + ab + ab + b^2 – ab = 1
(a+b)^2 = 1 + ab.
a = 0, b = 1.
a = 1, b = 0.
a = 0, b = -1.
a = -1, b = 0.
a = 1, b = -1.
a = -1, b=1.
6 ordered pairs.

• Q24) x^2 * y^3 = 8, where x, y > 0. What is the minimum value of 4x + 3y ?

• 4x + 3y = P.
4x = 2k
x = k/2.
3y = 3k.
y = k.
(k/2)^2 k^3 = 8.
K^5/4 = 8.
K = 2.
4x + 3y = 2k + 3k = 5k = 10.

• Q25) [ log (1) + log (1+3) + log (1+3+5) ... log(1+3+5…….19) - 2 (log 1 + log 2 ... log 7) = m + nx + ay. If log 2 = x and log 3 = y, then find the value of m, n & a ?

• (log 1 + log 4 + log 9……..log 100) – 2(log 1 + log 2…….log 7)
= 2(log 1 + log 2 + log 3……..log 10) – 2( log 1 + log 2…………log 7)
= 2 (log 8 +log 9 + log 10)
= 2(3log2 + 2log3 + 1)
= 2 + 6log 2 + 4log 3
= 2 + 6x + 4y
m = 2, x =6, y =4.

• Q26) All three roots of the cubic equation x^3 - 10x^2 + 31x - k = 0 are prime numbers. What is the value of K ?

• Let’s three roots l,m and n.
l + m + n = 10.
lm + mn + nl = 31.
lmn = k.
all roots are prime.
Hence l = 2, m = 3, n = 5.
lmn = 30.

• {x} = x – [x]

(a) 0
(b) 1
(c) 2
(d) 3

• Hence 0.

• Q28) When ‘2’ is added to each of the three roots of x^3 - Ax^2 + Bx - C = 0, we get the roots of x^3 + Px^2 + Qx - 18 = 0. If A, B, C, P and Q are all non zero real numbers then what is the value of (4A + 2B + C).

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