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37. The average temperature of the first three days is $27^\circ $C and that of the next three days is $29^\circ $C. If the average of the whole week is $28.5^\circ $C, the temperature of the last day of the week is :
a. $10.5^\circ C$
b. $21^\circ C$
c. $31.5^\circ C$
d. $42^\circ C$

Answer: C

Explanation:
Let the seventh day temperature be x. Then total temperature for the whole week is
${3 \times 27 + 3 \times 29 + 1 \times x = 7 \times 28.5}$
x = 31.5

38. The average age of 30 students in a class is 12 years. The average age of a group of 5 of the students is 10 years and that of another group of 5 of them is 14 years. The average of the remaining students is :
a. 8 years
b. 10 years
c. 12 years
d. 14 years

Answer: C

Explanation:
Let it be x . Then :
${5 \times 10 + 5 \times 14 + 20 \times x = 30 \times 12}$
20x = 360 - 120 or 20x = 240 or x = 12

39. The average of 8 numbers is 21. If each of the number is multiplied by 8, the average of the new set of numbers is :
a. 8
b. 21
c. 29
d. 168

Answer: D

Explanation:
Average of new numbers = $(21 \times 8) = 168$

40. The average of 50 numbers id 38. If two numbers, namely 45 and 55 are discarded, the average of the remaining numbers is :
a. 36.5
b. 37.0
c. 37.5
d. 37.52

Answer: C

Explanation:
Total of 50 numbers = $(50 \times 38) = 1900$
Total of 48 numbers =(1900 - (45 + 55)] = 1800
Required average = $\displaystyle\frac{{1800}}{{48}} = \displaystyle\frac{{225}}{6} = 37.5$

41. The average height of 30 girls out of a class of 40 is 160 cm. and that of the remaining girls is 156 cm. The average height of the whole class is :
a. 158 cms
b. 158.5 cms
c. 159 cms
d. 159.5 cms

Answer: C

Explanation:
Average height of the whole class =$\left( {\displaystyle\frac{{36 \times 160 + 10 \times 156}}{{40}}} \right) = 159$ cms

42. A batsman had a certain average of runs for 16 innings. In the 17th innings, he made a score of 87 runs thereby increasing his average by 3. What is his average after 17 innings?
a. 39
b. 38
c. 37
d. 36

Answer: A

Explanation:
Assume his initial average = $x$
His total runs after 16 innings = 16$x$
After scoring 87 runs his average got increased by 3 to $x$ + 3
So his total runs after 17 innings = 17 × ($x$+3)
But it was given that the difference in the total scores after 16 innings and 17 innings = 87
Therefore ${\rm{17}} \times {\rm{(x + 3) - 16x = 87 }} \Rightarrow {\rm{x = 36}}$
His new average = 36 + 3 = 39

Alternate method:
His 87 runs in the 17th innings contributed to all the seventeen innings to increase the average by 3. So he must have scored 17 × 3 = 51 runs extra to maintain the average. So his previous average will be 87 - 51= 36. Present average = 39

43. Some consecutive natural numbers, starting with 1, are written on the board. Now, one of the numbers was erased and the average of the remaining numbers is 800/39. Find the number which was erased.
a. 24
b. 20
c. 18
d. 16

Answer: B

Explanation:
We know that average of n consecutive numbes average = $\displaystyle\frac{{\displaystyle\frac{{n \times (n + 1)}}{2}}}{n} = \frac{{(n + 1)}}{2}$
If the given n is sufficiently large, the average does not change much even though we exclude one or two numbers from it. So the approximate number of observations is almost double to the average (Remember: the average of consecutive numbers almost lies in the middle)
The approximate average is 800/39 = Approx 20. So the initial numbers may be nearer to 40.
In this question it is actually 40 as from the denominator of the new average 800/39. The initial numbers are 40.
Sum of 40 consecutive numbers = $\displaystyle\frac{{40 \times (40 + 1)}}{2} = 820$
Sum of 39 numbers = average x number of observations = $\displaystyle\frac{{800}}{{39}} \times 39$ = 800
So the number excluded = 820 - 800 = 20

44. Two vessels contain mixture of milk and water in the ratio of 5 : 2 and 3 : 1 respectively. Find the ratio of milk and water in the new solution, if two mixtures are mixed in equal amount.
a. 41 : 15
b. 39 : 17
c. 39 : 14
d. 15 : 41

Answer: A

Explanation:
Concentration of milk in the first vessel = $\dfrac{5}{{2 + 5}} = \dfrac{5}{7}$
Concentration of milk in the second vessel = $\dfrac{3}{{3 + 1}} = \dfrac{3}{4}$
Let us use weighted average formula by taking one unit of both these solutions.
Final concentration of milk = $\dfrac{{mx + ny}}{{m + n}} = \dfrac{{1\left( {\dfrac{5}{7}} \right) + 1\left( {\dfrac{3}{4}} \right)}}{{1 + 1}}$ = $\left( \vcenter{\dfrac{{\dfrac{{20 + 21}}{{28}}}}{2}} \right) = \dfrac{{41}}{{56}}$
So ratio of milk and water = 41:(56-41) = 41:15

Alternative Method:
Sum of the ratios are 5 + 2, 3 + 1 or 7 and 4.
LCM of sum of the ratios i.e. LCM of 7 and 4 is 28.
Therefore, We assume that 28 litres of mixture is taken from each vessel.
Now, milk in first vessel = $\displaystyle\frac{{\rm{5}}}{{\rm{7}}}$ x 28 = 20 litres
Therefore, Water in the first vessel = 28 - 20 = 8 litres
And milk in second vessel = $\displaystyle\frac{{\rm{3}}}{{\rm{4}}}$ x 28 = 21 litres
Therefore, Water in second vessel = 28 - 21 = 7 litres
Total quantity of milk in the resultant mixture = 20 + 21 = 41.
Total quantity of water in the resultant mixture = 8 + 7 = 15.
Therefore, Ratio of milk and water in the new solution = 41 : 15.

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