Quant Boosters  Vikas Saini  Set 5

Here values of x = 2,3,2.
Firstly we take extreme value of x = 3.
At x = 3
2x3+3x+2+5x2 = 20.
At x = 4,
it will increase by 10,because 2+3+5 = 10.
243+34+2+542 = 30.
At x = 5,
it will give value 40.
253+35+2+552 = 40.
So max value of x = 5.
For min value of x = 2.
at x = 2,
2x3+3x+2+5x2 = 30.
at x = 3
It will give 40.
So sum of values of x = 5  3 = 2.

Q15) Find the range of x if x+2 + x3 + x5 < 15.

First we take extreme value of x = 5.
At x = 5,
x+2+x3+x5 = 9.
at x = 6,
x+2+x3+x5 = 12.
at x = 7,
x+2+x3+x5 = 15.
For minimum take x = 2(min)
at x = 2,
x+2+x3+x5 = 12.
at x = 3,
x+2+x3+x5 = 15.
x € (3,7).

Q16) Find minimum value of x1 + 2x1 + 3x1 + 4x1 + 5x1

x1+2x1/2+3x1/3+4x1/4+5x1/5
Total terms = 1+2+3+4+5 = 15.
median = 8th term.
we will get 8th term at x1/4 = 0
= > x = 1/4.
at x = 1/4
x1+2x1+3x1+4x1+5x1
= 3/4 + 2/4 + 1/4 + 1/4
= 7/4.

Q17) Find the range of x if x1 + x5 + x9 + x13 ≤ 42

first we take extreme value at x13 = 0
= > x = 13
at x = 13.
x1+x5+x9+x13 = 24.
if we inverse value of x by 1 then total value will increase by 4.
we need to increase value of 18,we will have to increase value by 4.5.
To increase by 4 = +1.
To increase by 1 = +1/4.
To increase by 18 = 18/4 = 4.5.
at x = 13+4.5 = 17.5
x1+x5+x9+x13
= 16.5 + 12.5 + 8.5 + 4.5
= 42.
max value of x = 17.5
for minimum value of x1 = 0
= > x = 1.
at x = 1
x1+x5+x9+x13
= 24.
again to increase by 18,we need to reduce value of x by 4.5.
x = 1  4.5 = 3.5
x1+x5+x9+x13
= 4.5 + 8.5 + 12.5 + 16.5
= 42.
x € [3.5,17.5]

Q18) What was the day on 15th August 1947 ?

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  5And 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 7Given 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 MondayWhat 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+41) 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 = HourGiven 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 56 PM ?
(1:05:5/11) x 5
= 5:25:25/11
= 5:27:3/11By 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 23 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 56 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 * (x10) = 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 ?