Question 1

A man divides his total route of journey into three equal parts and decide to travel the three parts with speed of 15 km/h. 10 km/h and 5 km/ h respectively. Find his average speed of during journey.

Accepted Answer

If three equal distances are covered at different spees of A, B and C km/h, respectively, then average speed during whole journey = 3ABC/(AB + BC + CA) km/h ∴ Required average speed = (3 x 15 x 10 x 5) / (15 x 10 + 10 x 5 + 15 x 5) = (3 x 15 x 10 x 5)/(150 + 50 + 75) = (3 x 15 x 10 x 5)/275 = 8.18 km/h

Question 2

The average temperature of Delhi for four days in a particular month is 48°C. If the average temperature of second and third days 34°C and the ratio of the temperature of first and fourth days is 9 : 11 then find out the temperature of first and fou…

Accepted Answer

Total temperature of first and fourth days= (4 x 48 - 2 x 34)°C = (192 - 68)°C= 124°C Now, according to the question Temperature of the first day = (9/20) x 124 = 55.8°C Temperature of fourth day = (11/20) x 124 = 68.2°C

Question 3

A man spends average of ₹ 1694. 70 per month for the first 7 month and ₹ 1810.50 per month for the next 5 months. His monthly salary if he saves ₹ 3084.60. during the whole year is?

Accepted Answer

Let the monthly salary of a man = ₹ N ∴ The annual salary = ₹ 12N According to the question, Annually spends of a man = 7 x 1694.70 + 5 x 1810.50 = ₹ 20915.40 ∴ His monthly spends = 20915.40/12 = ₹ 1742.95 And monthly saving = 3084.60/12 = ₹ 257.05 ∴ His monthly salary = 1742.95 + 257.05 = ₹ 2000

Question 4

In the afternoon, a student read 100 pages at the rate of 60 pages/h. In the evening, when she was tired, she read 100 more pages at the rate of 40 pages/h . What was her average rate of reading in pages per hour?

Accepted Answer

Required average rate of reading = 2AB / (A + B) = 2 x 60 x 40 / (60 + 40) = (2 x 60 x 40)/100 = 48 page/h

Question 5

Mr. Bundda travels in this car and covers 1/4 part of his journey with 8 km/h, 3/5 part with 6 km/h and remaining 3/20 part with a speed of 10 km/h. Find out his average speed during the whole journey.

Accepted Answer

According to the average speed formula, P = 1/4, Q =3/5, R = 3/20 x = 8 km/h, y = 6 km/h, z = 10km/h ∴ Required average speed = 1 / [(P/x) + (Q/y) + (R/z)] =1 / [(1/32) + (1/10) + (3/200)] = 1/[(25 + 80 + 12)/800] = 800/117 = 6.83 km/h

Question 6

A person covers 9 km with a speed of 3 km/h, 25 km with a speed of 5 km/h and 30 km with a speed of 10 km/h. Find out the average speed of person.

Accepted Answer

Here, P = 9 km/h, Q = 25 km/h, R = 30 km x = 3 km/h, y = 5 km/h and z = 10 km/h ∴ Required average speed = [P + Q + R] / [(P/x) + (Q/y) + (R/z)] = [9+25 + 30] / [(9/3) + (25/5) + (30/10)] = 64/(3 + 5 + 3) = 64/11 = 59/11 km/h

Question 7

The daily minimum temperature of a place in a certain week. begining from Monday was 20°, 18.6°, 17°, 21.4°, 19° and 19.8°. It was known that the average of whole week was 19.5°. The minimum temperature on Sunday was?

Accepted Answer

Let temperature on sunday be x° C. (20° + 18.6° + 17° + 21.4° + 19° + 19.8° + x) / 7 = 19.5° ⇒ 115.8 + x = 19.5 x 7 ⇒ x = 136.5 - 11.58 ⇒ x = 20.7°C

Question 8

Average weight of three boys P, T and R is 54 1/ 3 kg, while the average weight of three boys T, F and G is 53 kg. What is the average weight of P, T, R, F and G?

Accepted Answer

According to the question (P + T + R) / 3 = 541/3 ⇒ P + T + R = (163/3) x 3 ⇒ P + T + R = 163 ............(i) Also (T + F + G)/3 = 53 ⇒ T + G + F = 159 ..........(ii) Clearly, from the two Eqs. (i) and (ii), the problem cannot be solved.

Question 9

The average runs scored by a cricketer is 42 innings, is 30. The difference between his maximum and minimum scores in an inning is 100. If these two innings are not taken into consideration, then the average score of remaining 40 inning is 28. Calcu…

Accepted Answer

Let the minimum score = x ∴ Maximum score = x + 100 ∴ x + (x + 100) = 30 x 42 - 40 x 28 ⇒ 2x + 100 = 1260 - 1120 = 140 ⇒ 2x = 140 - 100 = 40 ⇒ x = 20 Hence, the maximum score = x + 100 = 20 + 100 = 120

Question 10

A car reached Raipur from Somgarh in 35 min With an average speed of 69 km/h. If the average speed is increased by 36 km/h, how much time will it take to cover the same distance?

Accepted Answer

Distance between Raipur and Somgarh = Average speed x Time = 69 x 35/60= 161/4 km New speed = (69 + 36) km/h = 105 km/h ∴ Time = Distance/ Speed = 161/4 x 105 h = 161 x 60/4 x 105min = 23 min

Question 11

A cricketer has a certain average for 10 inning. In the eleventh innings. he scored 216 runs, there by increasing his average by 12 runs. Find out his new average?

Accepted Answer

Let average after 10th innings = x. According to the question. 10x + 216 = 11(x +12) ⇒ x = 216 - 132 = 84 ∴ New average = 84 + 12 = 96

Question 12

A cricketer whose bowling average is 12.4 runs per wicket, takes 5 wickets for 26 runs and there by decreases his average by 0.4. Find out the number of wickets taken by him till the last match?

Accepted Answer

Let x be the number of wickets taken till the last match According to the question, (12.4 x + 26) = 12(x + 5) ⇒ 12.4 x + 26 = 12x + 60 ⇒ 0.4 x = 34 ∴ x = 34/0.4 = 85

Question 13

The average age of two boys and their father is greater then average age of those two boys and their mother by 3 yr. The average age of the four is 19 yr. If the average age of the two boys be 5 1/ 2, then find the age of the father and mother.

Accepted Answer

Let the ages of mother, father and boys be M, F, B1 and B2, respectively. The total age of four member = 19 x 4 = 76 yr. Given (B1 + B1)/2 = 11/2 ⇒ B1 + B2 = 11 M + F + B1 + B2 = 76 - 11 ⇒ M + F = 65 .......(i) According to the question, (B1 + B2 + F) / 3 = (B1 + B2 + M) / 3 + 3 ⇒ B1 + B2 + F) = B1 + B2 + M + 9 ⇒ F = M + 9 ⇒ F - M = 9 ............. (ii) From Eqs (i) and (ii), we get F = 37 yr and M = 28 yr

Question 14

A has double the money of B and B has 50% more money then C. If average money of all the three persons is ₹ 12000, how money does A have?

Accepted Answer

Let money of C be ₹ N. According to the question Total money of B = N + N x 50% = N + 50N/100 = 3N/2 Total money of A = 2 x 3N/2 = 3N Average money of three persons = 12000 ∴ Total money of three = 12000 x 3 ∴ 3N + (3N/2) + N = 12000 x 3 ⇒ (6N + 3N + 2N)/2 = 36000 ∴ N = (36000 x 2)/11 = 72000/11 Now, money of A = 3N = (3 x 72000)/11 = ₹ 216000/11

Question 15

A cricketer played 80 innings and scored an average of 99 runs. His score in the last inning is zero run. To have an average of 100 at the end, his score in the last innings should have been.

Accepted Answer

Let the required runs be N. According to the question, (80 x 99 + N) / 80 = 100 ⇒ 7920 + N = 8000 ⇒ N = 80

Question 16

The arithmetical mean of the scores of a group of students in a test was 52. The brightest 20% of them secured a mean score of 80 and the dullest 25%, a mean score of 31. The mean score of remaining 55% is

Accepted Answer

Let the total number of students be 100. Also, let the mean score = N According to the question, 20 x 80 + 25 x 31 + (100 - 20 - 25) x N = 100 x 52 ⇒ 1600 + 775 + (55) x N = 5200 ⇒ 55N = 5200 - 1600 - 775 ∴ N = 2825/55 = 51.36 = 51.4% So, mean score of remaining 55% is 51.4%

Question 17

The average weight of 4 men, A, B, C and D is 67 kg. The 5th man E is included and the average weight decreses by 2 kg. A is replaced by F. The weight of F is 4 kg more then E . Average weight decreases because of the replacement of A and now the av…

Accepted Answer

Sum of the weight of A, B, C and D = 67 x 4 = 268 kg and average weight of A, B , C , D and E = 67 - 2 = 65kg ∴ Sum of the weight of A, B, C, D and E = 65 x 5 = 325 kg ∴ Weight of E = 325 - 268 = 57 kg ∴ Weight of F = 57 + 4 = 61 kg Now, average weight of F, B, C, D and E = 64 kg ∴ Sum of the weight of F, B, C, D and E = 64 x 5 = 320 Kg ∴ Sum of the weights of B, C and D = 320 - 57 - 61 = 202 kg ∴ Weight of A = 268 - 202 = 66 kg

Question 18

5 yr ago, the average age of Rashmi and Surbhi was 20 yr, Now, the average age of Rashmi, Surbhi and Meeta is 30 yr. What will be the age of Meeta after 15 yr.?

Accepted Answer

Total age of Rashmi and Surbhi (5 yr ago) 20 x 2 = 40 yr. Total age of Rashmi and Surbhi (at present) = 40 + 5 + 5 = 50 yr. Total age of Rashmi, Surbhi and Meeta (at present) = 30 x 3 = 90 yr. ∴ Present age of Meeta = 90 - 50 = 40 yr. ∴ Age of meeta after 15yr. = 40 + 15 = 55 yr.

Question 19

The average weight of the students in four sections A, B, C and D is 60 kg. The average weight of the students of A, B, C and D individually are 45 kg, 50 kg, 72 kg and 80 kg, respectively. If the average weight of the students of section A and B to…

Accepted Answer

Let number of students in the sections A, B, C and D be a, b, c and d respectively. Then, total weight of students of sections A = 45a Total weight of students of section B = 50b Total weight of students of section C = 72c Total weight of students of section D = 80d According to the question, Average weight of students of sections of A and B = 48 kg ⇒ (45a + 50b)/(a + b) = 48 ⇒ 45a +50b = 48a + 48b ⇒ 3a = 2b ⇒ 15a = 10b And average weight of students of sections B and C = 60 kg ⇒ 50b + 72c = 60(b + c) ⇒ 10b = 12c Now, average weight of students of A, B, C and D = 60 kg ∴ 45a + 50b + 72c + 80d = 60(a + b + c + d) ⇒ 15a + 10b - 12c - 20d = 0 ⇒ 15a = 20d ⇒ a : d = 4 : 3

Question 20

A batsman makes a score of 87 runs in the 17th inning and thus increases his average by 3. Find his average after 17th inning?

Accepted Answer

Let the average after 17th innings = x Then average after 16th innings = (x - 3) Therefore 16 × (x - 3) + 87 = 17x Therefore x = 39

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