Average = (x_{1}+x_{2}+x_{3}+...+x_{n}) / N. |
W . A = (n1*a1 + n2 * a2) / (n1 + n2) n1= strength of group 1 n2 = strength of group 2 a1 = average of group 1 a2 = average of group 2 |
Method to calculate average speed taken
Step 1:Average speed = (2S1 * S2)/(S1 + S2) [S1 and S2 are speeds]
Step 2:
The ratio of speeds is 60 : 100 = 3 : 5 (say r_{1}: r_{2}) Then, divide the difference of speeds (40 in this case) by r_{1} + r_{2} (3 + 5 = 8, in this case) to get one part. (40/8 = 5, in this case)
The required answer will be three parts away (i.e. r_{1} parts away) from the lower speed
Example : Ratio of speeds = 20 : 40 = 1 : 2
Divide difference of 20 into 3 parts (r_{1} + r_{2}) = 20/3 = 6.66. Required average speed = 20 + 1 × 6.66
Quantity of cheaper / Quantity of dearer = |
Ans .
75
Average speed = (total distance)/(total time). If we assume distance between 2 points to be "d" Then Average speed = 2d/[(d/60) + (d/100)] = (2 × 60 × 100)/ (60 + 100) = (2 × 60 × 100)/160 = 75
Ans .
x+n
If the average age of a group of persons is x years today then after n years their average age will be (x + n).
Ans .
108
Since the average increases by 2 runs per innings it is equivalent to 2 runs being added to each score in the first 25 innings. Now, since these runs can only be added by the runs scored in the 26th inning, the score in the 26th inning must be 25 × 2 = 50 runs higher than the average after 26 innings (i.e. new average = 58). Hence, runs scored in 26th inning = New Average + Old innings × Change in average = 58 + 25 × 2 = 108
Ans .
29
The teacher after fulfilling the average of 14 (for the group to which he belonged) is also able to give 0.5 years to the age of each of the 30 students. Hence, he has 30 × 0.5 = 15 years to give over and above maintaining his own average age of 14 years. Age of teacher = 14 + 30 × 0.5 = 29 years
Ans .
80
The replacement has the effect of reducing the average marks for each of the 20 students by 4. Hence, the replacement must be 20 × 4 = 80 marks below the original. Hence, answer = 10 marks.
Ans .
The first sentence gives you a total of 144 for A, B and C’s marks. Second sentence: When D joins the group, the total becomes 44 × 4 = 176. Hence D must get 32 marks. Alternatively, you can reach this point by considering the first 2 statements together as: D’s joining the group reduces the average from 48 to 44 marks (i.e. 4 marks). This means that to maintain the average of 44 marks, D has to take 4 marks from A, 4 from B and 4 from C Æ A total of 12 marks. Hence, he must have got 32 marks. From here: The first part of the third sentence gives us information about E getting 3 marks more than 32 Æ Hence, E gets 35 marks. Now, it is further stated that when A is replaced by E, the average marks of the students reduces by 1 to 43. Mathematically this can be shown as A + B + C + D = 44 × 4 = 176 while, B + C + D + E = 43 × 4 = 172 Subtracting the two equations, we get A – E = 4 marks. Hence, A would have got 39 marks. Alternatively, you can think of this as: The replacement of A with E results in the reduction of 1 mark from each of the 4 people who belong to the group. Hence, the difference is 4 marks. Hence, A would get 4 marks more than E i.e. A gets 39 marks.
Ans .
18
we get that the total from Monday to Wednesday was 81 while from Tuesday to Thursday was 72. The difference is arising out of the replacement of Monday by Thursday. This can be mathematically written as Mon + Tue + Wed = 81 Tue + Wed + Thu = 72 Hence, Mon – Thu = 9 Thu = (2/3) × Mon; thus we get Thursday = 18 °C
Ans .
20
using weighted average : \( \frac{10 * 0.2 + 20*0.3 + 30*0.4}{10+20+30} = 20/60 = 33.33%\)
Ans .
39
P = 25 × 15 – 24 × 14 = 375 – 336 = 39
Ans .
79
D’s weight = 4 × 80 – 3 × 84 = 320 – 252 = 68. E’s weight = 68 + 3 = 71. Now, we know that A + B + C + D = 4 × 80 = 320 and B + C + D + E = 78 × 4 = 312. Hence, A’s weight is 8 kg more than E ’s weight. A = 71 + 8 = 79.
Ans .
40
Today’s total age of A, B and C = 30 × 3 = 90. Today’s total age for B and C = 25 × 2 = 50. C ’s age = 90 – 50 = 40
Ans .
39.8
there are five prime numbers between 30 and 50. They are 31,37,41,43 and 47. Therefore the required average=(31+37+41+43+47)/5 So 199/5 = 39.8
Ans .
20.5
sum of first n natural numbers=n(n+1)/2; So,sum of 40 natural numbers=(40*41)/2 which is 820. Therefore the required average=(820/40) which is 20.5.
Ans .
73.5
Required average =7(1+2+3+…….+20)/20 which is (7*20*21)/(20*2) which is (147/2)= 73.5.
Ans .
30
let the numbers be x,x+2,x+4 and x+6. then, (x+(x+2)+(x+4)+(x+6))/4) = 27 i.e. (4x+12)/4 = 27 i.e. x+3=27 i.e. x=24. Therefore the largest number=(x+6)=24+6=30
Ans .
37.25
total weight of(36+44) students=(36*40+44*35)kg =2980kg. Therefore weight of the total class=(2980/80)kg = 37.25kg
Ans .
117
Let the average expenditure of all nine be Rs.x Then 12*8+(x+8)=9x or 8x=104 or x=13. Total money spent = 9x=Rs.(9*13)=Rs.117
Ans .
72
Let the third number be x. Then second number = 3x. First number=3x/2. Therefore x+3x+(3x/2)=(44*3) or x=24 So largest number= 2nd number = 3x = 72
Ans .
78
Clearly 13th result = (sum of 25 results)-(sum of 24 results) =(18*25)-(14*12)+(17*12) =450-(168+204) =450-372 =78
Ans .
: 6th result = (58*6 + 63*6 - 60*11) = 66
Ans .
31
Let A,B,c represent their individual wgts. Then, A+B+C=(45*3)Kg=135Kg A+B=(40*2)Kg=80Kg & B+C=(43*2)Kg=86Kg B=(A+B)+(B+C)-(A+B+C) =(80+86-135)Kg =31Kg
Ans .
Total age of 39 persons = (39 x 15) years = 585 years. Average age of 40 persons = 15 yrs 3 months = 61/4 years. Total age of 40 persons = (_(61/4 )x 40) years= 610 years. :. Age of the teacher = (610 - 585) years = 25 years.
Ans .
71
Total weight increased =(1.8 x 10) kg =18 kg. :. Weight of the new man =(53 + 18) kg = 71 kg
Ans .
420
Let the original average expenditure be Rs. x. Then, 42 (x - 1) - 35x=42 -> 7x= 84 -> x =12. Original expenditure = Rs. (35 x 12) = Rs. 420.
Ans .
39
Let the average after 17th inning = x. Then, average after 16th inning = (x - 3). :. 16 (x - 3) + 87 = 17x or x = (87 - 48) = 39
Ans .
67.2
Required average speed = ((2xy)/(x+y)) km / hr =(2 x 84 x 56)/(84+56)km/hr = (2*84*56)/140 km/hr = 67.2 km/hr
Q = V * ( 1 - (k/V) ) ^ ^{n } |
Q. If average age of 10 people is 32 what must be the age of the new joinee if the average age increases by 5.
A. Old average = 10*32=320, New average = 11 * 37 = 407 so new joinee must have age = 407 - 320 = 87.
Q. there are two sections A and B of a class consisting of 36 and 44 students respectively. If the average weight of section A is 40kg and that of section B is 35kg, find the average weight of the whole class?
A. n1 =36, n2 = 44, a1 = 40 , a2 = 35 now apply weighted average formula (see above).
Q. Nine persons went to a hotel for taking their meals 8 of them spent Rs. 12 each on their meals and the ninth spent Rs. 8 more than the average expenditure of all the nine. What was the total money spent by them?
A. Let 'x' be the average. The equation is 12*8 + (x+8) = 9*x as total sum spent by all 9 is addition of Rs. 12 each spent by first 8 i.e. 12*8 and the Rs. 8 more than average spent by last one i.e. x+8.
Q. The average age of a class of 39 students is 15 years. If the age of the teacher be included, then the average increases by 3 months. Find the age of the teacher.
A. The teachers age = (Group average) + (increase in average age of entire group)
= 15 yrs + ( 40 * 3 months)
= 15 yrs + (120 months / 10 yrs)
= 25 years
Q. The average weight of 10 oarsmen in a boat is increased by 1.8 kg when one of the crew, who weighs 53 kg is replaced by a new man. Find the weight of the new man.
A. Weight of new man = 53 + (1.8*10) = 53 + 18 = 71 kg.
Q. There were 35 students in a hotel. Due to the admission of 7 new students, ; the expenses of the hotel were increased by Rs. 42 per day while the average expenditure per student diminished by Rs 1. What was the original expenditure of the hotel?
A. Let original average expenditure per student be 'x'.
So we get 42(x-1) - 35x = 42. As for 42 students the average expense decreased by 1 and so total was 42*(x-1) and for 35 students it was 35x.
Q. Distance between two stations A and B is 200 km. A train covers the journey from A to B at 100 km per hour and returns back to A with a uniform speed of 50 km per hour. Find the average speed of the train during the whole journey.
A. time taken to go from A to B = 2 hrs and from B to A is 4 hrs.
Average speed = total distance / total time = 400 / (2+4) km/hr
OR
Directly use formula 2xy / (x+y) where x = speed from A to B, y = speed from B to A.
average speed = 2*100*50 / (100+50) = 400 / 6 km/hr
Q. In what ratio must rice at Rs. 9.30 per kg be mixed with rice at Rs. 10.80 per kg so that the mixture be worth Rs. 10 per kg ?
A. d = 10.8, c = 9.3 and m = 10
Quantity of cheaper / Quantity of dearer = (10.8 - 10)/(10-9.3) = 8/7
Q. How much water must be added to 60 litres of milk at 1 ½ litres for Rs. 20 So as to have a mixture worth Rs.10 2/3 a litre ?
A. 1 liter milk = 40/3 rupees
Quantity of milk / quantity of water = (32/3 - 0 ) / (40/3 - 32/3) = (32/3:8/3)
So 4:1
Q. How many kgs. of wheat costing Rs. 8 per kg must be mixed with 36 kg of rice costing Rs. 5.40 per kg so that 20% gain may be obtained by selling the mixture at Rs. 7.20 per kg ?
A. If selling price is 7.2 then cost
price is 6 as 20% gain is made.
quantity of rice / wheat = (6 - 5.4) / (8-6) = 0.6 / 2 = 3/10.
So for 3 kg of rice 10kg of wheat is needed so we need x kg of wheat for 36 kg of rice. 3:10 :: 36:x
x = 36*10 / 3 = 120 kg
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