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Q170. ANS-8

Humraj!!!!!!!!!!!

Humraj

Sol.. 171

2/3 +5/9 - 4/5 = 19/45

My solutions

161) I am not very sure bat the solution..

the possible numbers for length, breadth and height are  1 2 3, 2 3 4, 3 4 5. each value can be either length, breadth or height.  for example, in solution 1 2 3, ht = 1/2/3 , bdth = 1/2/3 and ht = 1/2/3.(3 cuboids)

Hence total no.of cuboids = 9

162) Easy to prove.. 8n + 6n = 14n. Hence divisible by 7

163) 200 solutions.

Y can take any values between (-1 to -50) and (1 to 50). Based on this x valu will change. Ex, if Y = 49 / -49 then X = 2 / -2.. Hence 4 pairs of x and y. Hence for 50 valyes of y we have 200 pairs of (x,y)

164)2.5 hours

Abi

Answers

169) 12 minutes

170) sure had 8 apples

Abi

Solution

M161. How many cuboids exist for which the volume is less than 100 units3 and the integer side lengths are in an arithmetic sequence?
Ans.   Let the side be xyz<=100
          The sides are in A.P. then y-x = z-y =d>=1
          Considet the cuboid of d=1,
          (x)(x+1)(x+2)
          Possible Pairs are (1,2,3),(2,3,4),(3.4.5)
          for d=2 (x)(x+2)(x+4) = (1,3,5),(2,4,6),(3,5,7)
          for d=3 (x)(x+3)(x+6) = (1,4,7),(2,5,8)
          for d=4 (x)(x+4)(x+8) = (1,5,9)
          for d=5 (x)(x+5)(x+10) =(1,6,11)
          for d=6 (x)(x+6)(x+12) =(1,7,13)
          Thus total cuboids are =11

M162. Prove that 6^n + 8^n is divisible by 7 iff n is odd.
Ans  If n is odd then =(6+8)xP thus divisible by 7

M164. A truck driving east at 50 mph passes a certain mile marker. A motorcyclist also driving east passes that same mile marker 45 minutes later. If the motorcyclist is driving 65 mph, how long will it take for the motorcyclist to pass the truck?
Ans. The distance of travelled by teuck in 45 mts = 3/4 hr = 50x3/4=150/4 miles and relative spped = 15 mph so time required for overtake = 150/4x15 =10/4=2hr30mts

M166. The velocity v of a body moving in a straight line t seconds after starting from rest is v = 4t3 - 12 t2 meters/second
a. How many seconds after starting does its acceleration become zero?
b. Mary began walking home from school, heading south at a rate of 4 mph. Sharon left school at the same time heading north at 6 mph.
c. How long will it take for them to be 3 miles apart?
Ans.a) v' = 12t^2-24t =0 or t=0 or t=2 sec. Thus after 2 sec.
       b) The relative speed = 10 mph they will be 10 miles apart in 60 min
           Thus they will be 3 miles apart in 18 min.

M167. The sum of the infinite series 1- (1/2) + (1/4) - (1/8) + ... is
Ans. Here r =-1/2 and a =1 then a/(1-r) =1/(1+1/2)=2/3 Ans

M168. Two airplanes leave an airport simultaneously, one heading east; the other, west. The eastbound plane travels at 140 mph and the westbound plane travels at 160 mph. How long will it take for the planes to be 750 miles apart?
Ans. Here relative velocity = 160+140 =300 mph thus in 1hr 300 miles and in 2hr30 min it is 750 miles apart.

M169. Sachin can paint a bat in 45 minutes; Rahul, in 30 minutes. If Sachin
begins painting his bat and Rahul joins him to help after 15 minutes, how
long will it take both to finish the job?
Ans.   The work completed in 1/3 and left over is 2/3 work and together they can finish left over work in (45x30/75)x2/3 min = 12 min

M170.Farmer Brown told Bob and Sue that they could pick apples from his tree, but that neither of them could take more than 20. They worked for a while, and then Bob asked Sue, "Have you picked your limit yet?"
Sue replied, "Not yet. But if I had twice as many as I have now, plus half as many as I have now, I would have my limit." How many did Sue have?
Ans.  Let it be x and y and x<=20 and y<=20 and  2x+y/2 =20 then x+y/4=10 (9,4),(8,8),(7,12),(6,16) Possible solutions


M171.Q. The probability that Nike will get Cricket team contract is 2/3 and the probability that Nike will get Hockey team contract is 5/9. If the probability of getting at least one contract is 4/5, what is the probability of getting both the contract?
a. 19/45
b. 13/45
c. 12/35
d. 11/23
Ans. Nike prababilty to P(C) = 2/3 and ~P(C) = 1/3 and Nike P(H)=5/9 and ~P(H)=4/9
       Probabilty of getting at least 1 contract  = 4/5
       Then P(C or H) =4/5=2/3 + 5/9 -P(C and H) => P(C and H) = 19/45 Ans.