ANSWERS AND EXPLANATIONS

Transcription

1 11 ANSWERS AND EXPLANATIONS EXERCISE 1 1. (a) Required average (c) Average score 1 [ ] 1 [44] 407. (a) Required average ( ) (e) Required average (d) Required average (e) Required average (a) Required average (e) Average 9. (e) a + b 11 11? a? b?? Average of a and b? 10. (b) 1a + 1b 114 or, 1 (a + b) 114 a + b Required average 11. (b) 1a + 1b 7 or, 1 (a + b) 7 a + b a b 4 7 a b Required average (c) 7a + 7b 1 or, 7(a + b) 1 or, a + b a b 1 Average (a) Assume the third number According to question or, or, or, (b) Age of the fourth friend years

2 1 AVERAGE 1. (e) Let the first number be Second number and the third number According to the question, or, Required difference (d) Let the five consecutive even numbers be, +, + 4, + and + respectively. According to the question, or or 0 0 or 40 4 B and E B E (e) Let the A According to the question E A E (a) Let the third number be First number and second number Third number Required difference (c) Let the consecutive odd numbers be, +, + 4, + and + According to the question or, or, A 7 and E Required product (d) Let A, According to the question, A + B + C + D + E + ( + ) + ( + 4) + ( + ) + ( ) B D (c) Let the other number is N. Then, X N XY N XY X. (a) Let the four consecutive odd nos. be, 1, + 1 and +. Now, 1 or, Lowest odd no. 9. (c) E? H? G? M E? S? M? H? 1 4 4? G S 0 4. (c) Age of the CT yrs.. (c) Here one boy is ecluded and final average of the group decreases. change in average is ( )ve 0.1 kg. Using the formula Sum of the quantities ecluded According to the question. or, or, 11 or, Change in no. of quantities Change in average Original average Final no. of quantities weight of the boy who left (1 4) ( ) 49.9 kg weight of the boy who left the class is 49.9 kg.. (c) Age of the teacher (7 1 14) years

3 1 years. 7. (b) total weight increases 1. kg 1 kg so the weight of new person (e) Required persons (e) Set the first, second and third no be F, S and T Respectively F + S S + T +1. Solving, we get F T (c) E + H ( ) 110; E + S ( ) 10 Reqd difference (b) Let the four consecutive even nos. be, +, + 4 and + respectively. Reqd difference +. (d) Let the three consecutive odd numbers be,, + respectively. According to question,??? 14? ? 0 an even number, which goes against our supposition.. (d) Correct average 7 ( ) (a) Suppose the consecutive odd numbers are :, +, + 4, + and + Therefore, the required difference + Note that answering the above question does not require the average of the five consecutive odd numbers.. (b) Let the average ependiture of all the nine be `. Then, 1 + ( + ) 9 or 104 or 1. Total money spent 9 ` (9 1) ` (c) Since average price of a pencil ` Price of 7 pencils ` 10 Price of 0 pens ` (10 10) ` 0 Average price of a pen 7. (b) Required average marks 0??` (d) Average of 0 numbers 0. Sum of 0 numbers (0 0) 0. It is quite possible that 19 of these numbers may be positive and if their sum is a, then 0th number is ( a). 9. (b) Sum of the remaining two numbers (.9 ) [(.4 ) + (. )].70 ( ) Required average (c) Total marks EXERCISE 1. (a) Given A + B 40 (i) C + B (ii) A + C 4 (iii) (i) + (ii) + (iii) A + B + C 0 (iv) from (i) and (iv), we get C 0 years B 1 years and A years. (b) Sum of present ages of the si members (17 ) years 10 years. Sum of present ages of the members (ecluding baby) (17 + ) years 100 years. Age of the baby years. (a) Let the number of persons, initially going for Picnic Sum of their ages 1 Also, years 4. (b) Let be the average score after 1 th innings 1 11 ( ) + 4. (c) Let there be pupils in the class. Total increase in marks 1. ( ) (a) Total runs in the first 10 overs 10. Run rate required in the remaining 40 overs 0. runs per over 40 40

4 14 AVERAGE 7. (b) Remaining pages Let average printing error in remaining pages Then, (b) Attendance on the fifth day (d) Income of months ` ( ) debt ` 10 debt Income of the man for net 4 months ` debt + ` 0 ` 70 + debt Income of 10 months ` 70 Average monthly income ` ` (c) Total runs Let his highest score be Then his lowest score (b) Now 000 ( 17) ( ) 0 10 ( ) ( ) ( ) (b) Required average marks (c) Let the average ependiture of all the nine be ` Then, Total money spent 9 9. ` (d) Let the number of wickets taken till the last match be. Then, (c) Let the number of innings played in England be. Total runs scored in England 4 Total runs scored for innings played in India 4 0 ( the number of innings played in India 4) Also, (b) Let the original average ependiture be `.. Then, 4( 1) Original ependiture `. ( 1) ` (b) Required average years. 7 1 (d) Let Arun s weight be X kg. According to Arun, < X < 7. According to Arun s brother, 0 < X < 70. According to Arun s mother, X <. The values satisfying all the above conditions are and 7. Required average 7 1. kg. 19. (a) Let the new man was younger than the director years and years ago, the sum of ages of board of directors S S 4 Then, years ago, average age of board of directors S 4 Now, S 4 S 4 years Shortcut Method : If the new young director would have been not substituted, then total age would have increased at present by 4 years. Therefore, the new man is 4 years younger keeping the average at present same as years ago. 0. (a) Sum of 10 numbers 40 Corrected sum of 10 numbers Hence, new average (a) Let be the increase in the average

5 (7? 4) 9? n? n? For to be a whole number 7 ( 9 ) should be divisible by (n + ) From the choices it can be said that and 7 are two such factors. But 7 does not lie within the range. number of students in class are.. (c) If you assume the first number to be a, naturally net number would be more than a and so on The numbers are : a, a +, a + 4, a +, a +. Hence, their average Sum/ (a + 4). (d) Sprouts 00 gms + Potatoes 1000 gms + Cauliflower 00 gms + Meat 700 gms 00 gms. Hence total cooked food 100 gms and is served among 4 people. Thus average weight is gms. 4. (c) Total mx, Sum of n numbers ny Average of remaining no. (mx ny) / (m n). (a) Let the ages of P, Q and R are a, b, c a + b 40, b + c and c + a 4 a + b + c 40?? 4? 0 a, b 1 and c 0. (d) The total age of the family at the birth of first child 1 While the total age of the couple at marriage 0. The years from marriage till the first child s birth 4 0 years. The total age of family at the birth of the second child years. 0 4 Second child was born years after the first. (1? )? 0 Similarly the twins were born 4 years. After the second child and today the twins are 4 years old. ( average age of the family became 1 years from 1 years) Age of eldest son years. Let p, m, r, s and b be the weights of boers P, M, R, S and B respectively. 1 From data : p r (1) b s () 4m p + b + r + s... () p + b m + s... (4) p + b + m + r + s 0... () From () and (); m 0 m 104 lb From (1) and (); p + b r + s () From (4) and () : r + s + 4 m + s r m lb From (1) : p r lb From () : b s b + s 0... (7) From (7) and (); b s 10 lb and b + s 0 b 19 Average of the weights of M and R lb Average of the weights of P, S and B lb. 7. (b) Let the lowest marks be Then, ( 9) lb and s (b) Let no. of students in classes X, Y and Z, y and z respectively. 7y 79 y 7y z and 1 y z We have to find From (i) 4 y or From (ii) y 4z or...(i)...(ii) 7y z y z z 4 y ; y Required average 4

6 1 AVERAGE (d) Average weight of the students in group D cannot be determined since we do not know the average weight of each student. The given data is insufficient to compare its average with other groups. 0. (c) If one student from group A is shifted to group B, still there is no effect on the whole class. In any case, the no. of students inside the class is same. Hence the average weight of the class remains same. 1. (c) Since all the students of the class have the same weight, then the average of weight of any group of any no. of students will be the same as that of each students weight. Hence, the average weight of D cannot be greater than average weight of A. EXERCISE 1. (c) Let the number of Mechanical engineering graduates be M and Electronic engineering graduates be E. Then.4M?.E?.1 M? E.4M +.E.1M +.1E 0.44E 0.7M E M M For E to be an integer, the least value will be 7.. (d) Total attendance for Day 1,, and 114 Total for Day, and 4 1 Total for Day and Hence attendance on Day (a) Given, 4. (b) (7? )? (1? )?(9? )?(107? )?(1? )? Required mean is, 10?? 1???? 0?? 1?? 7? 7? 0 7???? 7??...? ? 4.? 1??...? 10? 4 and 11? 1?...? 40? 10 1??...? 40? 10 1??...? ?? (b) Sum of temp. for Monday, Tuesday and Wednesday 1 Sum of temp. for Tuesday, Wednesday, an Thursday 0 10 Since temp. on Thursday Temp for Tue and Wed Temp of Monday (d) (b) Let average of team years Then, + + S 9 11 (i) where S 9 is the sum of ages of remaining players Also, Np + S 9 11( ), (ii) where Np is the sum of ages of new players (i) (ii) Np Np 1 Average age of new two players 1 1. years. (b) Let the average after 17th inning. Then, average after 1th inning ( ). 1( ) or (7 4) (a) Corrected mean (d) Let the number of papers be. Then, or or (c) Let average cost of petrol per litre be ` Then Rs 7.9 per litre (d) Let the average score of 19 innings be. Then,

7 The average score after 0th innings (b) Total weight of 4 students 4 40 kg Total weight of students who leave 4 40 kg Total weight of students who join 4 70 kg Therefore, new total weight of 4 students New average weight 70 kg (c) Here one boy is ecluded and final average of the group decreases. change in average is ( )ve 0.1 kg. Using the formula Sum of the quantities ecluded 17 Change in no. of quantities Change in average Original average Final no. of quantities weight of the boy who left (1 4) ( ) 49.9 kg weight of the boy who left the class is 49.9 kg. 1. (d) Total sales for the first eleven months 1, `.1000 Total sales for the whole year ` `.100 Value of sales during the last month ` (a),. (c)