Aptitiude Questions Y5

41. A sum of money doubles itself at C.I. in 15 years. In how many years will it become eight times?



Ans.45 years.



Sol: P [1 + (R/100)]^15 = 2P è [1 + (R/100)]^15 = 2……….(i)

Let P [1 + (R/100)]^n = 8P è P [1 + (R/100)]^n = 8 = 2^3

= [{1 + (R/100)}^15]^3.

è [1 + (R/100)]^n = [1 + (R/100)]^45.

è n = 45.

Thus, the required time = 45 years.



42. A certain sum amounts to Rs. 7350 in 2 years and to Rs. 8575 in 3 years. Find the sum and rate percent.



Ans: Sum = Rs. 5400,Rate=16 2/3 %.



Sol: S.I. on Rs. 7350 for 1 year = Rs. (8575-7350) = Rs. 1225.

Therefore, Rate = (100*1225 / 7350*1) % = 16 2/3 %.

Let the sum be Rs. X. then, x[1 + (50/3*100)]^2 = 7350.

è x * 7/6 * 7/6 = 7350.

è x = [7350 * 36/49] = 5400.

Therefore, Sum = Rs. 5400.



43. A, B and C start a business each investing Rs. 20000. After 5 months A withdrew Rs. 5000, B withdrew Rs. 4000 and C invests Rs. 6000 more. At the end of the year, a total profit of Rs. 69,900 was recorded. Find the share of each.



Ans. A’s share = Rs. 20,500

B’s share = Rs. 21200

C’s share = Rs. 28200



Sol: Ratio of the capitals of A, B and C

= (20000*5+ 15000*7) : (20000*5+16000*7): (20000*5+26000*7)

=205000: 212000 : 282000 = 205:212:282

Therefore, A’s share = Rs. ( 69900*205/699) = Rs. 20,500

B’s share = Rs. (69900*212/699) = Rs. 21200,

C’s share = Rs. (69900*282/699) = Rs. 28200.



44. Sanjiv started a business by investing Rs. 36000. After 3 months Rajiv joined him by investing Rs. 36000. Out an annual profit of Rs. 37100, find the share of each?



Sol: Ratio of their capitals= 36000*12:36000*9 = 4:3

Sanjiv’s share= Rs. ( 37100*4/7) = Rs. 21200.

Rajiv’s share = Rs. ( 37100*3/7) = Rs.15900.



45. If 20 men can build a wall 56m long in 6 days, what length of a similar wall can be built by 35 men in 3 days?



Ans. Length=49m.



Sol: Since the length is to be found out, we compare each item with the length as shown below.

More men, more length built (Direct).

Less days, less length built (Direct).

Men 20:35 :: 56: x

Similarly, days 6:3 :: 56: x.

Therefore, 20*6*x= 35*3*56 or x= 49.

Hence, the required length= 49m.



46.If 9 engines consume 24 metric tonnes of coal, when each is working 8 hours a day; how much coal will be required for 8 engines, each running 13 hours a day, it being given that 3 engines of the former type consume as much as 4 engines of latter type.



Ans:26metric tonnes.



Sol: We shall compare each item with the quantity of coal.

Less engines, less coal consumed (direct)

More working hours, more coal consumed (direct)

If 3 engines of former type consume 1 unit, then 1 engine will consume 1/3 unit.

If 4 engines of latter type consume 1 unit, then 1 engine will consume 1/4 unit.

Less rate of consumption, less coal consumed (direct).

Therefore, number of engines 9:8 :: 24:x

Working hours 8:13 :: 24:x

Rate of consumption 1/3:1/4 :: 24:x.

9*8*1/3*x= 8*13*1/4*24 or x= 26.

Therefore, required consumption of coal 26 metric tonnes.



47. A contract is to be completed in 46 days and 117 men were set to work, each working 8 hours a day. After 33 days 4/7 of the work is completed. How many additional men may be employed so that the work may be completed in time, each man now working 9 hours a day?



Ans.81



Sol: Remaining work = 1-4/7 =3/7.

Remaining period = (46-33) days =13 days.

Less work, less men (direct)

Less days, more men (indirect).

More hours per day, less men (indirect)

Therefore, work 4/7:3/7 ::117/x

Days 13:33 :: 117/x

Hrs/day 9:8:: 117/x

Therefore, 4/7*13*9*x= 3/7*33*8*117 or x= 198.

Therefore, additional men to be employed =(198-117) =81.



48. A garrison of 3300 men had provisions for 32 days, when given at the rate of 850gms per head. At the end of 7 days, reinforcement arrives and it was found that the provisions will last 17 days more, when given at the rate of 825gms per head. What is the strength of the reinforcement?



Ans: 1700



Sol: The problem becomes:

3300 men taking 850gms per head have provisions for (32-7) or 25 days. How many

men taking 825gms each have provisions for 17 days?

Less ration per head, more men (indirect).

Less days, more men (indirect)

Ration 825:850::3300:x

Days 17:25::3300:x

Therefore, 825*17*x= 850*25*3300 or x= 5000.

Therefore, strength of reinforcement = 5000-3300 = 1700.



49. Find the slant height, volume, curved surface area and the whole surface area of a cone of radius 21 cm and height 28 cm.



Sol: Slant Height, l = v(r^2 + h^2) =v(21^2 + 28^2) = v1225 = 35 cm

Volume = 1/3?r^2h = (1/3 * 22/7 * 21 * 21 * 28) cm^3 = 12936 cm^3

Curved surface area = ?rl = 22/7 * 21 *35 cm^3 = 2310 cm^2

Total Surface Area = (?rl + ?r^2) = (2310 + 22/7 * 21 * 21) cm^2 = 3696 cm^2



50. If the radius of the sphere is increased by 50%, find the increase percent in volume and the increase percent in the surface area.



Sol: Let the original radius = R. Then, new radius = 150/100 R = 3R/2

Original Volume = 4/3?R^3, New volume = 4/3?(3 R/2)^3 = 9?R^3/2

Original surface area = 4?R^2 , New surface area = 4?(3R/2)^2 = 9?R^2

Increase % in surface area = (5?R^2/4?R^2 * 100)% = 125%