The cost of fuel for running the engine of an army tank is proportional to the square of the speed and Rs. 64 hour for a speed of 16 kmph. Other costs amount to Rs. 400 hour. the tank has to make a journey of 400 km at a constant speed.
Answer: D Cost of fuel is proportional to square of the speed E = KS2 =>E= 64 and S= 16 => K =1/4 Total cost = 1/4*S2t +400t Most economical speed, checking options we get most economical speed at 40km/hr.
Q. No. 2:
The total cost for the journey at this most economical speed is
Answer: B Cost of fuel is proportional to square of the speed E = KS2 =>E= 64 and S= 16 => K =1/4 Total cost = 1/4*S2t +400t Most economical speed, checking options we get most economical speed at 40km/hr. Total cost at 40km/hr = 1/4*402 +400(400/40) => Rs 8,000
Q. No. 14:
The currencies in countries X and Y are denoted by Xs. and Ys. respectively. The exchange rate in 1990 was 1 Xs. = 0.6 Ys. The price level in 2006 in X and Y are 150 and 400 respectively with 1990 as a base of 100. The exchange rate in 2006, based solely on this purchasing power parity consideration, is 1 Xs.=
Answer: C In 1990, 1 Xs = 0.6 Ys Price levels in 2006 change by 150 and 400 for X and Y respectively with 1990 as base 100. In 2006, 150 × 1Xs = 400 × 0.6 Ys Xs = 240/150 Ys => Xs = 1.6 Ys
Q. No. 15:
In a family of husband, wife and a daughter, the sum of the husband’s age, twice the wife’s age, and thrice the daughters age is 85; while t he sum of twice the husband’s age, four times the wife’s age, and six times the daughter’s age is 170. It is also given that the sum of five times the husband’s age, ten times the wife’s age and fifteen time the daughter’s age equals 450. The number of possible solutions, in terms of the ages of the husband, wife and the daughter, to this problem is
Answer: A Let the age of husband wife and daughter be denoted by h, w and d respectively h + 2w + 3d = 85 ...Eq. (i) 2h + 4w + 6d = 170 ...Eq. (ii) 5h + 10w + 15d = 450 ...Eq. (iii) Multiplying the first equation by 5 we get 5h + 10w + 15d = 425 but Eq (iii) gives 5h + 10w + 15d = 450 No solution possible.
Q. No. 16:
Ramesh has two examinations on Wednesday - Engineering Mathematics in he morning Engineering drawing in t he afternoon. He has a fixed amount of time to read the textbooks of both these subjects on Tuesday. During this time he can read 80 pages of Engineering Mathematics and 100 pages of Engineering Drawing. Alternatively, he can also read 50 pages of Engineering Mathematics and 250 pages of Engineering Drawing. Assume that the amount of time it takes to read one page of the textbook of either subject is constant. Ramesh is condident about Engineering Drawing and wants to devote full time ti reading Engineering Mahtematics. The number of Engineering Mathematics text book pages he can fead on Tuesday is
Answer: C In the given time Ramesh can read 80 pages of Engineering maths and 100 pages of Engineering drawing. Or He can read 50 pages of Engineering Maths and 250 pages of Engineering drawing. => 30 pages of Engineering Maths ≈ 150 pages of Engineering Drawing. =>10 pages of Engineering Maths ≈ 50 pages of Engineering Drawing. So in the given time Ramesh can read 80 + (100/50 * 10 ) = 80 + 20 = 100pages of Engineering Maths.
Q. No. 17:
On January 1, 2004 two new societies s1 and s2 are formed, each n numbers. On the first day of each subsequent month, s1 adds b members while s2 multiples its current numbers by a constant factor r. Both the societies have the same number of members on July 2, 2004. If b = 10.5n, what is the value of r?
Answer: A There will be an increase of 6 times. No. of members s1 will be in A.P. On July 2nd , 2004, s1 will have n + 6 b members = n + 6 × 10.5 n = 64n No. of members in s2 will be in G.P On July 2nd, 2004 Number of members in s2= nr6 => 64n = nr6 => r=2
Q. No. 18:
N persons stand on the circumference of a circle at distinct points. Each possible pair of persons, not standing next to each other, sings a two-minute song one pair after the other. If the total time taken for singing is 28 minutes, what is N?
Answer: B Each person will form a pair with all other persons except the two beside him. Hence he will form (n – 3) pairs. If we consider each person, total pairs = n (n – 3) but here each pair is counted twice. Hence actual number of pairs = n(n-3)/2 They will sing for = n(n-3)/2 * 2= n(n-3) =28 => Hence, n= 7 by discarding -ve value of n .