# RS Aggarwal Solutions Class 9 Ex 14D

## RS Aggarwal Class 9 Ex 14D Chapter 14

Q.1: The mean mark obtained by 7 students in a group is 226. If the marks obtained by six of them are 340, 180, 260, 56 275 and 307 respectively.  find the marks obtained by the seventh student.

Sol:

Mean marks of 7 students = 226

Sum of marks of seven students = (226×7$226\times 7$) = 1582

Marks obtained by 6 students = (340 + 180 + 260 + 56 + 275 + 307) = 1418

Therefore, marks obtained by seventh student

= [(Sum of marks of 7 students) – (marks obtained by 6 students)]

= (1582 – 1418) =164

Therefore, the Marks obtained by seventh student = 164

Q.2: The mean weight of a class of 34 students is 46.5 kg. If the weight of the teacher is included, the mean rises by 500 g. find the weight of the teacher.

Sol:

Mean weight of 34 students = 46.5 kg

Total weight of 34 students = 34×46.5$34\times 46.5$ kg =1581 kg

Mean weight of 34 students and the teacher = (46.5+0.5) kg =47 kg

Therefore, Total weight of 34 students and the teacher = 47×35$47\times 35$ kg =1645 kg

Therefore, the weight of the teacher = (1645- 1581) kg = 64 kg

Q.3: The mean weight of a class of 36 students is 41 kg. If one of the students leaves the class then the mean is decreased by 200 g. Find the weight of the student who left.

Sol:

Mean weight of 36 students = 41 kg

Total weight of 36 students = 41×36kg$41 \times 36\; kg$= 1476 kg

One student leaves the class mean is decreased by 200 g.

Therefore, the New mean = (41- 0.2) kg = 40.8 kg

Total weight of 35 students = 40.8 ×$\times$ 35 kg = 1428 kg

Therefore, the weight of the student who left = (1476 – 1428) kg = 48 kg

Q.4: The average weight of a class of 39 students is 40 kg. When a new student is admitted to the class, the average decreases by 200 g. Find the weight of the new student.

Sol:

Mean weight of 39 students = 40 kg

Total weight of 39 students = 40 ×$\times$ 39 = 1560 kg

One student joins the class mean is decreased by 200 g.

Therefore, New mean = (40- 0.2) kg = 39.8 kg

Total weight of 40 students = 39.8×40$39.8\times 40$ kg =1592 kg

Therefore, the weight of new student = Total weight of 40 students – Total weight of 39 students

= 1592-1560 = 32 kg

Q.5: The average monthly salary of 20 workers in an office is र7650. If the manager’s salary is added, the average salary becomes र8200 per month. What is the manager’s salary per month?

Sol:

Mean salary of 20 workers = Rs. 7650

Total salary of 20 workers = Rs (7650 ×$\times$ 20) = Rs. 153000

Now, if manager salary is added mean become Rs. 8200

Therefore, Total salary of 20 workers + manager’s salary = Rs. (8200 ×$\times$ 21)

= Rs. 172200

Therefore, Manager’s salary per month = Total salary of 20 workers and the manager – Total salary of 20 workers

= Rs(172200 – 153000) = Rs. 19200

Q.6: The average monthly wage of a group of 10 persons is  र9000. One member of the group, whose monthly wage is र8100,  leaves the group and is replaced by a new member whose monthly wage is र7200. Find the new monthly average wage.

Sol:

Mean monthly wage of 10 persons = Rs. 9000

Total monthly wage of 10 persons = Rs. (9000 ×$\times$ 10) = Rs. 90000

New mean monthly wage:

= (totalmonthlywageof10persons)(wagesofworkerwholeft)+(wagesofworkerwhojoined)10$\frac{(total\; monthly\; wage \;of \;10 \;persons)- (wages\; of \; worker \; who \; left )+ (wages\; of \; worker \; who\; joined)}{10}$

= Rs.90000Rs.8100+Rs.720010$\frac{Rs. 90000 – Rs. 8100 + Rs. 7200}{10}$

= Rs.8910010=8910$\frac{Rs.89100}{10}= 8910$

Therefore, the new monthly average wage = Rs 8910

Q.7: The average monthly consumption of petrol for a car for the first 7 months of a year is 330 liters, and for the next 5 months is 270 liters. What is the average consumption per month during the whole year?

Sol:

Mean consumption of petrol for the first 7 months = 330 liters

Total consumption of petrol for the first 7 months = 330×7$330\times 7$ liters = 2310 liters

Mean consumption of petrol for the next 5 months = 270 litres

Total consumption of petrol for the next 5 months = 270×5$270 \times 5$ = 1350 liters

Total consumption of petrol in one year = 2310 + 1350 = 3660 liters

Therefore, the Mean consumption of petrol= 366012=305$\frac{3660}{12}=305$ liters per month.

Q.8: Find the mean of 25 numbers if the mean of 15 of them is 18 and the mean of the remaining numbers is 13.

Sol:

Mean of 15 numbers = 18

Total sum of 15 numbers = 18×15$18 \times 15$= 270

Remaining number = 25 – 15 = 10

Mean of 10 numbers = 13

Total sum of 10 numbers = 13×10$13\times 10$ =130

Therefore, the Total sum of 25 numbers = (270+130 ) = 400

Therefore, the Mean of 25 numbers = 40025=16$\frac{400}{25}= 16$

Q.9: The mean weight of 60 students of a class is 52.75 kg. If the mean weight of 25 of them is 51 kg, find the mean weight of the remaining students.

Sol:

Mean weight of 60 students = 52.75 kg

Sum of weight of 60 students = 60×52.75$60 \times 52.75$ kg = 3165 kg

Mean weight of 25 students = 51 kg

Sum of weight of 25 students = 25×51$25 \times 51$ kg = 1275 kg

Remaining students = 60 – 25 = 35

Total weight of the sum of the remaining 35 students = Sum of weight of 60 students – Sum of weight of 25 students = (3165 – 1275) = 1890kg

Therefore, the Mean weight of 60 students = 189035=54$\frac{1890}{35}= 54$kg

Q.10: The average weight of 10 oarsmen in a boat is increased by 1.5 kg  when one of the crew  who weighs 58 kg is replaced by a new man. Find the weight of the new man.

Sol:

The increase in the average of 10 oarsmen = 1.5 kg

Total weight increased = 1.5×10$1.5 \times 10$ kg = 15 kg

Since, the man weighing 58 kg has been replaced,

Thus the weight of new man = (58 + 15) kg = 73 kg

Q.11: The mean of 8 numbers is 35. If a number is excluded then the mean is reduced by 3. Find the excluded number.

Sol:

Mean of 8 numbers = 35

Therefore, Total sum of 8 numbers = 35×8$35 \times 8$ = 280

Since One number is excluded, New mean = 35 – 3 = 32

Therefore, Total sum of 7 numbers = 32×7$32 \times 7$ = 224

The excluded number = Sum of 8 numbers – Sum of 7 numbers = 280 – 224 = 56

Q.12: The mean of 150 items was found to be 60. Later on, it was discovered that the values of two items were misread as 52 and 8 instead of 152 and 88 respectively. Find the correct mean.

Sol:

Mean of 150 items = 60

Total Sum of 150 items = 150×60$150 \times 60$ = 9000

Therefore, Correct sum of items = [(Sum of 150 items) – (Sum of wrong items) + ( Sum of right items)] = [9000 – (52 + 8) + (152 + 88)] = 9180

Therefore, the Correct mean = 9180150=61.2$\frac{9180}{150}= 61.2$

Q.13: The mean of 31 results is 60. If the mean of the first 16 results is 58 and that of the last 16 results is 62, find the 16th result.

Sol:

Mean of 31 results = 60

Total sum of 31 results = 31×60$31 \times 60$ = 1860

Mean of the first 16 results = 16×58$16 \times 58$ = 928

Total sum of the first 16 results = 16×58$16 \times 58$ = 928

Mean of the last 16 results = 62

Total sum of the last 16 results = 16×62$16 \times 62$ = 992

Therefore, The 16th result = 928 + 992 – 1860 = 1920-1860 = 60

Therefore, The 16th result = 60

Q.14: The mean of 11 numbers is 42. If the mean of the first 6 numbers is 37 and that of the last 6 numbers is 46, find the 6th number.

Sol:

Mean of 11 numbers = 42

Total sum of 11 numbers = 42×11$42 \times 11$ = 462

Mean of the first 6 numbers = 37

Total sum of first 6 numbers = 37×6$37 \times 6$ = 222

Mean of the last 6 numbers = 46

Total sum of last 6 numbers = 6×46$6\times 46$ = 276

Therefore, the 6th number = 276 + 222 – 462 = 498 – 462 = 36

Therefore, the 6th number = 36

Q.15: The mean weight of 25 students of a class is 52 kg. If the mean weight of the first 13 students of the class is 48 kg and that of the last 13 students is 55 kg, find the weight of the 13th student.

Sol:

Mean weight of 25 students = 25 kg

Total weight of 25 students = 52×25$52\times 25$ kg = 1300 kg

Mean of the first 13 students = 48 kg

Total weight of the first 13 students = 48×13$48\times 13$ kg = 624 kg

Mean of the last 13 students = 55 kg

Total weight of the last 13 students = 55×13$55\times 13$ kg =715 kg

Therefore, The weight of 13th student = Total weight of the first 13 students + Total weight of the last 13 students – Total weight of 25 students = 624 + 715 – 1300 kg

Q.16: The mean score of 25 observations is 80 and the mean score of another 55 observations is 60. Determine the mean score of the whole set of observations.

Sol:

Mean score of 25 observations = 80

Total score of 25 observation = 80×25=2000$80\times 25=2000$

Mean score of 55 observation = 60

Total score of 55 observations = 60×55=3300$60\times 55=3300$

Total no. of observations = 25+55 =80 observations

Therefore, the Total score = 2000+3200 = 5300

Therefore, the Mean score = 530080=66.25$\frac{5300}{80}= 66.25$

Q.17: Arun scored 36 marks in English, 44 marks in Hindi, 75 marks in mathematics and x marks in science. If he has secured an average of 50 marks, find the value of x.

Sol:

Average marks of 4 subjects = 50

Total marks of 4 subjects = 50×4=200$50\times 4 =200$

Therefore, 36+44+75+x = 200

155+x=200$\Rightarrow 155+x =200$
x=200155=45$\Rightarrow x =200 – 155 = 45$

Therefore, the value of x = 45

Q.18: The mean monthly salary paid to 75 workers in a factory is र5680. The mean salary of 25 of them is र5400 and that of 30 others is र5700. Find the mean salary of the remaining workers.

Sol:

Mean monthly salary of 75 workers = Rs. 5680

So, Total monthly salary of 75 workers = Rs. 5680×75=Rs.426000$5680 \times 75 = Rs.\; 426000$

Mean monthly salary of 25 workers = Rs. 5400

So, total monthly salary of 25 workers = Rs. 5400×25=Rs.135000$5400 \times 25 = Rs.\; 135000$

Mean monthly salary of 30 workers = Rs.5700

So, total monthly salary of 30 workers 5700×30=Rs.171000$5700 \times 30 = Rs.\; 171000$

Remaining workers = 75 – 55 = 20 workers

Total salary of remaining 20 workers = Rs. [426000-(135000+171000)]

= Rs. [426000 – 306000]

= Rs. [426000 – 306000]

= Rs. 120000

Therefore, the mean salary of the remaining 20 workers = Rs. 12000020$\frac{120000}{20}$ = Rs. 6000

Q.19: A ship sails out to an island at the rate of 15 km/h and sails back to the starting point at 10 km/h.  Find the average sailing speed for the whole journey.

Sol:

Let, the distance of marks from the starting point be x km.

Then, time taken by the ship to reach the mark = (x15)$\left ( \frac{x}{15} \right )$hours,   ( since time = distancespeed$\frac{distance }{speed}$ )

Time taken by the ship reaching the starting point from the marks = (x10)$\left ( \frac{x}{10} \right )$hours

Total time taken = x15+x15=x6$\frac{x}{15}+\frac{x}{15}=\frac{x}{6}$ hours

Total distance covered = x + x = 2x km.

Therefore, the average speed of the whole journey = 2x÷x6=2x×6x=12$2x \div \frac{x}{6}= \frac{2x\times 6}{x}=12$km/ hour.

Q.20: There are 50 students in a class, of which 40 are boys. The average weight of the class is 44kg and that of the girls is 40kg.Find the average weight of the boys.

Sol:

Total number of students = 50

Total number of girls = 50 – 40 = 10

Average weight of the class = 44 kg

Total weight of 50 students = 44×50$44\times 50$ kg = 2200 kg

Average weight of 10 girls = 40 kg

Total weight of 10 girls = 40 40×10$40\times 10$ kg = 400 kg

Therefore, total weight of 40 boys = 2200- 400 kg = 1800 kg

Therefore, the average weight of the boys = 180040=45kg$\frac{1800 }{40}= 45 \; kg$

#### Practise This Question

In which direction does the buoyant force acts?