Quant Boosters  Vikas Saini  Set 5

x = 8,10,12 ... 100.
total terms = (1008 )/2 + 1 = 47.
value of x for which total product should be negative.
so for 810,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 = {(1008 )/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)(x3)(x+5)(x7) ... (x+97)(x99) < 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 (x1)(x+5)(x9) ... (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)(x2)(x+3) ... (x98) < 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 2x3 + 3x+2 + 5x2 = 40.

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 ?