Total sale for 5 months = Rs. (6435 + 6927 + 6855 + 7230 + 6562) = Rs. 34009.
$\therefore$ Required sale = Rs. [$\left (6500 \times 6\right)$ - 34009 ]
= Rs. (39000 - 34009)
= Rs. 4991.
If a car covers a certain distance at $x$ kmph and an equal distance at $y$ kmph. Then,
average speed of the whole journey = $\dfrac{2xy}{x+y}$ kmph.
By using the same formula, we can find out the average speed quickly.
Average speed
$=\dfrac{2 \times 84 \times 56}{84 + 56}=\dfrac{2 \times 84 \times 56}{140}$
$=\dfrac{2 \times 21 \times 56}{35}=\dfrac{2 \times 3 \times 56}{5}$
$=\dfrac{336}{5}=67.2$
Let P, Q and R represent their respective monthly incomes. Then, we have:
P + Q = $\left(5050 \times 2\right)$ = 10100 .... (i)
Q + R = $\left(6250 \times 2\right)$ = 12500 .... (ii)
P + R = $\left(5200 \times 2\right)$ = 10400 .... (iii)
Adding (i), (ii) and (iii), we get: 2(P + Q + R) = 33000 or P + Q + R = 16500 .... (iv)
Subtracting (ii) from (iv), we get P = 4000.
$\therefore$ Ps monthly income = Rs. 4000.
Total weight of students in division A = $ 36 \times 40$
Total weight of students in division B = $ 44 \times 35$
Total students = 36 + 44 = 80
Average weight of the whole class
$= \dfrac{\left(36 \times 40\right)+\left(44 \times 35\right)}{80} $
$= \dfrac{\left(9 \times 40\right)+\left(11\times 35\right)}{20} $
$= \dfrac{\left(9 \times 8\right)+\left(11\times 7\right)}{4} $
$= \dfrac{72+77}{4}\\$
$= \dfrac{149}{4}\\$
=$37.25$
Let the average after 17 innings = $x$
Total runs scored in 17 innings = 17$x$
Average after 16 innings = $\left(x-3\right)$
Total runs scored in 16 innings = 16$\left(x-3\right)$
Total runs scored in 16 innings + 87 = Total runs scored in 17 innings
=> 16 $ \left(x-3 \right) + 87 = 17x$
=> 16x - 48 + 87 = 17$x$
=>$x$ = 39
Runs scored in the first 10 overs = $ 10 \times 3.2 = 32$
Total runs = 282
Remaining runs to be scored = 282 - 32 = 250
Remaining overs = 40
Run rate needed = $\dfrac{250}{40} = 6.25$
average age of 36 students in a group is 14
Sum of the ages of 36 students = $ 36 \times 14 $
When teachers age is included to it, the average increases by one
=> average = 15
Sum of the ages of 36 students and the teacher = $ 37 \times 15$
Hence teachers age
$= 37 \times 15 - 36 \times 14$
$= 37 \times 15 - 14(37 - 1)$
$= 37 \times 15 - 37 \times 14 + 14$
= 37(15 - 14) + 14
=37 + 14
= 51
Sum of the present ages of husband, wife and child = $\left(27 \times 3 + 3 \times 3\right)$ years = 90 years.
Sum of the present ages of wife and child = $\left(20 \times 2 + 5 \times 2\right)$ years = 50 years.
$\therefore$ Husbands present age = (90 - 50) years = 40 years.
| Total quantity of petrol consumed in 3 years | =$ \left(\dfrac{4000}{7.50} +\dfrac{4000}{8} +\dfrac{4000}{8.50} \right) $ litres |
| = 4000$ \left(\dfrac{2}{15} +\dfrac{1}{8} +\dfrac{2}{17} \right) $ litres | |
| =$ \left(\dfrac{76700}{51} \right) $ litres |
Total amount spent = Rs. $\left(3 \times 4000\right)$ = Rs. 12000.
| $\therefore$ Average cost = Rs.$ \left(\dfrac{12000 \times 51}{76700} \right) $= Rs.$ \dfrac{6120}{767} $ = Rs. 7.98 |
Let A, B, C represent their respective weights. Then, we have:
A + B + C = $\left(45 \times 3\right)$ = 135 .... (i)
A + B = $\left(40 \times 2\right)$ = 80 .... (ii)
B + C = $\left(43 \times 2\right)$ = 86 ....(iii)
Adding (ii) and (iii), we get: A + 2B + C = 166 .... (iv)
Subtracting (i) from (iv), we get : B = 31.
$\therefore$ Bs weight = 31 kg.