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Examples(2)

June 3, 2020
Category: Examples(2)

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(8) If a, b, c, d are four consecutive odd integers, then give its average in the form of a.

Soln. A b c

Here, a, b, c, d are in A.P.

$\therefore$ Its average = $\frac{a+d}{2}$ = $\frac{b+c}{2}$ = b + 1 = c – 1 = d – 3 = a + 3

$\therefore$ Its required average ( in the form of a ) = a + 3V

(9) If p, q, r, s, t are five consecutive odd integers, then give its average in the form of t.

Soln. p q r s t

Here, p, q, r, s, t are in A.P.

Its average = r ( Middle term ) = p + 4 = q + 2 = s – 2 = t – 4

Its required average ( in the form of t ) = t – 4

(10) Find two numbers whose AM is 17 and GM is 8.

Soln. Let the two numbers be ‘a’ and ‘b’.

A.M. = 17

$\therefore \frac{a+b}{2}$ = 17 G.M. = 8

$\therefore\textup{ } \sqrt{ab}$ = 8

ab = 64 –––––– ( 1 )

a + b = 34 –––––– ( 2 )

i.e. the factors of 64 whose sum is 34 $\large \Leftrightarrow$ 32 & 2

(10) The average age of 20 children increased an orphanage by 9 months, when one orphan whose age is 5 years is replaced by a new boy. What is the age of the new boy?

Soln. Total increase in age = 20 $\times$ 9 months = 180 months = 15 years.

Age of the new boy = 5 + 15 = 20 years.

$\overline{x_{1}},\overline{x_{2}},\overline{x_{3}},...,\overline{x_{k}}$ are n arithmetic means of k groups with sizes respectively $n_{1},n_{2},n_{3},...,n_{k}$ then combined mean = $\overline{x}_{c}$ = $\frac{n_{1}\overline{x}_{1} \textup{ }+\textup{ }n_{2}\overline{x}_{2} \textup{ }+\textup{ }n_{3}\overline{x}_{3} \textup{ }+\textup{ } ...+n_{k}\overline{x}_{k} }{n_{1} +n_{2}+n_{3}+...+n_{k} }$

(11) There are two sections A and B of a class, consisting of 36 and 64 students respectively. If the average weight of section A is 50 kg and that of section B is 25 kg, find the average weight of the whole class.

Soln

Combined Mean = $\overline{x}_{c}$ = $\frac{n_{1}\overline{x}_{1}+n_{2}\overline{x}_{2}}{n_{1}+n_{2}}$

= $\frac{36(50)+64(25)}{36+64}$

= $\frac{1800+1600}{100}$

= $\frac{3400}{100}$ = 34 kg.

(12) There were 40 students in a hostel. Due to the admission of 10 new students, the expenses of the mess were increased by Rs. 200 per day while the average expenditure per head diminished by Rs. 2. What was the original average.

Soln. Let the average of 40 students = $\overline{x}_{1}$ = x, that of new 10 students = $\overline{x}_{2}$ = 20 and that of total 50 students = $\overline{x}_{c}$ = x – 2

Now, $\overline{x}_{c}$ = $\frac{n_{1}\overline{x}_{1}+n_{2}\overline{x}_{2}}{n_{1}+n_{2}}$ $\therefore$ x – 2 = $\frac{40(x)+10(20)}{40+10}$ = $\frac{40x+200}{50}$

$\therefore$ 50 x – 100 = 40 x + 200 $\therefore$ 10 x = 300 $\therefore$ x = 30 Rs.

(13) Ten persons went to a hotel for taking their meals. Nine of them spent Rs.15 each on their meals and the tenth spent Rs.9 more than the average expenditure of all the ten. What was the total money spent by them?

Soln. Let the average expenditure of all the ten be Rs. x.

Then, 15 $\times$ 9 + (x + 9) = 10x or 9x = 144 or x = 16.

Total money spent = 10x = Rs. (10 $\times$ 16) = Rs. 160.

(14) The average weight of A, B, C is 50 kg. If the average weight of A and B be 45 kg and that of B and C be 47 kg, find the weight of B.

Soln. Let A, B and C represent their individual weights. Then,

A + B + C = (50 $\times$ 3) kg = 150 kg.

A + B = (45 $\times$ 2) kg = 90 kg and B + C = (47 $\times$ 2) kg = 94 kg.

$\therefore$ B = (A + B) + (B + C) – (A + B + C) = (90 + 94 – 150) kg = 34 kg.

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