Quant Boosters  Vikas Saini  Set 5

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 ?

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).

Let three roots of first equation are l, m, n respectively.
18 = (l+2)(m+2)(n+2)
18 = (lm+2l+2m+4)(n+2)
18 = lmn + 2lm + 2ln + 4l +2mn + 4m + 4n + 8.
10 = lmn + 2(lm+ln+mn) + 4(l+n+m)
10 = C+2B+4A.

Q29) If a, b and c are root of the equation 3x^3 + 42x + 93 = 0, then what is value of a^3 + b^3 + c^3 ?