# Quant Boosters - Vikas Saini - Set 5

• x = 8,10,12 ... 100.
total terms = (100-8 )/2 + 1 = 47.
value of x for which total product should be negative.
so for 8-10,if we put x = 9 then one value will be positive,rest 46 values will be negative. Product of negative nos(even no of negative terms) give a positive value.
For x = 11, 2 positive & 45 negative
hence product negative.
so negative value of x = {(100-8 )/2}/2 = 23.
Hence 23 values of x for which product < 0.

• Q11) Find the number of integer values of x for which (x+1)(x-3)(x+5)(x-7) ... (x+97)(x-99) < 0

• Here x = -1,-5,-9 ... -97.
x = 3,7 ... 99
for x € (-1,-99) product will come negative.
total terms = (-1+97)/4 + 1
= 25.
but in between two terms there will be few values for which product will become negative.
Let's take any two term -1 & -5.
in between -1 & -5, 3 values x=-2,-3,-4 product will become negative. It means only 3 values in between any two terms.
Hence total values = 3 x 25 = 75.

• Q12) Find the number of integer values of x for which (x-1)(x+5)(x-9) ... (x+61) < 0

• x = 1,9....57.
x = -5,-13.....-61.
here terms of x which will give product in negative.
total term = (-5+61)/8 + 1
= 7 + 1
= 8.
total values in between -5 & -13 which will give negative values are :-
-6,-7,-8,-9,-10,-11,-12
total 7 values.
Total = 8 x 7 = 56.

• Q13) Find the number of integer values of x for which (x+1)(x-2)(x+3) ... (x-98) < 0

• x = -1,-3 ... -97
x = 2,4 .... 98
so total terms which will give negative value = (-1+97)/2 + 1
= 49.
The value in between two -1 & -3 which will give negative value is only one -2.
so total values = 49 x 1 = 49.

• Q14) Find the sum of all values of x which satisfy the equation 2|x-3| + 3|x+2| + 5|x-2| = 40.

• Here values of x = 2,3,-2.
Firstly we take extreme value of x = 3.
At x = 3
2|x-3|+3|x+2|+5|x-2| = 20.
At x = 4,
it will increase by 10,because 2+3+5 = 10.
2|4-3|+3|4+2|+5|4-2| = 30.
At x = 5,
it will give value 40.
2|5-3|+3|5+2|+5|5-2| = 40.
So max value of x = 5.
For min value of x = -2.
at x = -2,
2|x-3|+3|x+2|+5|x-2| = 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| + |x-3| + |x-5| < 15.

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

• Q16) Find minimum value of |x-1| + |2x-1| + |3x-1| + |4x-1| + |5x-1|

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

• Q17) Find the range of x if |x-1| + |x-5| + |x-9| + |x-13| ≤ 42

• first we take extreme value at x-13 = 0
= > x = 13
at x = 13.
|x-1|+|x-5|+|x-9|+|x-13| = 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
|x-1|+|x-5|+|x-9|+|x-13|
= 16.5 + 12.5 + 8.5 + 4.5
= 42.
max value of x = 17.5
for minimum value of x-1 = 0
= > x = 1.
at x = 1
|x-1|+|x-5|+|x-9|+|x-13|
= 24.
again to increase by 18,we need to reduce value of x by 4.5.
x = 1 - 4.5 = -3.5
|x-1|+|x-5|+|x-9|+|x-13|
= 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 - 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 ?

61

61

61

64

46

61

61

61