## Answers by Dr Gulshan S.

Hi Shawn Let C = Number of child tickets and A = Number of adult tickets  C+ A = 159 Gives C = 159 - A also C*6 + A*9.30 = 1257.60  (159 -A)*6  + A*9.30 =1257.60 A = 303.60/3.30 = 92

f(x) = 4x+8 g(x) = 5 Using   f(x) + g(x) =( f+g) x = 4x+8 +5 =4x +13   so (f+g) (2) = 4*2+13 = 8+13 = 21

As 7≥ -3k   7 ≤ 3k   7/3 ≤ K   k≥ 7/3 k≥ 2.333 None of the options is correct as no value of k is ≥ 2.3333

G= Number of girls T= total No. of kids B = No. of boys B = T - G Given G/T = 5/8 = G/ ( B + G) = G/ 24 or G = 5*24/8 =15 No. of girls = 15 No. of Boys = 9

Let W = wind speed and P = speed of plane    D = (P -W ) *t1 against the wind  and  D = (P + W) * t2 with the wind t2 = Distance / (P +W)   W = (D/2)* ( t1 -t2) /( t1*t2) Given   D = 175 , t1 = 70 min and t2 = 50 min   Now...

Let the number be X   (2X -7) < (X + 3)   X < 10

Speed = Distance /time Distance covered in 2 hours = 20 miles Remaining distance = 60 - 20 = 40 miles Additional time taken = Distance / speed = 40/ 30 hours   and the total time = 2 hours + 4/3 hours

Hi Alana Let a = Number of adult tickets and c=Number of child tickets sold  Given a=4 c Total Sales = 933.60 = 5.30*c + 8.40*a                                = 5.30*c  + 8.40 *4*c Find...

Number of adult tickets bought in \$24 =24/6 = 4 Number of student tickets bought for \$24 = 24/4 = 6 Difference = 6-4 =2

Hi Ally   V = derivative of r wrt time   V = dr/dt V= d/dt(-7tsint) = -7t. cos t -7sint and acceleration  is derivative of velocity  wrt time   dV/dt =d/dt ( -7t.cost -7sint) or second derivative of position    = 7t sint...

Hi  Let a= Number of adult tickets sold Then number of child tickets sold = 147 + a The ratio =( a + 147 )/ a= 8/1 a+147 = 8a   7a = 147 a= 147/7 = 21   Number of child ticket = 147 +21 = 168 So total tickets sold  = 168...

Hi  Yvette Let there be M men in the group of 250 members W= Number of women in the group of 250 So M + W = 250  Given W/M = 19/81 ratio of Number of women to number of men and W = (19/81) M M + (19/81M = 250  M =(250 * 81) / 100 202.5 and...

Hi Victoria L= 5W Given Perimeter P = 2L +2W = 10 W + 2W = 12W =110 cm 12 W = 110 cm   W = 110/12 cm     L = 5 W     = 5*110/12 cm

Hi Julia No value of X in the interval (0,2) will be valid for this equation

Hi Jackson The answer is Vf   If we substitute  Vf = Vi +at in the given equation , the equation does not have Vf in it any more

Hi James as the points are collinear  As P, Q, R, S, T  Given PT = 20 PQ=QR = RS  With QS = 6 So QR = RS = 3 PQ = 3 ST = 11 RT =14 SP =9 RP =6  Hope all covered

Hi Julia  Here is the solution From the above equation acceleration due to gravity at Mars is 0.98 m/s (a) Using V = u-gt =10 - 0.98*1 V =9.02 m/s (b) At  t= a,  V = 10- 1.86 *a/2  m/s (c) Time to reach the highest point t= 10*2/1.86 sec So total...

Hi Kahlil   A = 345 km/h B = 100 km/hr Angle between the two = 90 +35 = 125 degree   R 2 = A 2 + B 2 - 2 A*B* Cos ( angle between the two) Find R

Hi Chris   Speed = Distance / time Time of first leg =Distance/speed  = 120/100 = 1.2 hr = 1hr 12 min Time for Second leg = 80/75 = 1.07 hr = 1hr 4 min  Total time of travel = 2.27 hr approx. = 2hr 16 min Average speed = total distance / total time = 200/2...

Hi Zainab Time taken by first packet to hit the ground is t H = 1/2 gt 2 as initial velocity is zero Using g = 10 m/s 2  800 = 5t 2 Gives t= 12.65 sec Now to know where is second packet after    time available to second packet is 12.65- 1.0 sec = 11...