Average - Study Mode

[#181] There are five boxes in cargo hold. The weight of the first box is 200 kg and the weight of the second box is 20% higher than the weight of the third box, whose weight is 25% higher than the first box's weight. The fourth box at 350 kg is 30% lighter than the fifth box. Find the difference in the average weight of the four heaviest boxes and the four lightest boxes.
Correct Answer

(B) 75 kg

Explanation

Solution: Step 1: Find the weight of the third box. The third box is 25% heavier than the first box (200 kg). 25% of 200 kg = (25/100) * 200 kg = 50 kg Weight of the third box = 200 kg + 50 kg = 250 kg Step 2: Find the weight of the second box. The second box is 20% heavier than the third box (250 kg). 20% of 250 kg = (20/100) * 250 kg = 50 kg Weight of the second box = 250 kg + 50 kg = 300 kg Step 3: Find the weight of the fifth box. The fourth box (350 kg) is 30% lighter than the fifth box. This means the fourth box is 70% of the fifth box's weight. Let the weight of the fifth box be x. 0.7x = 350 kg x = 350 kg / 0.7 = 500 kg Step 4: Arrange the weights in ascending order. The weights are: 200 kg, 250 kg, 300 kg, 350 kg, 500 kg Step 5: Calculate the average weight of the four heaviest boxes. Four heaviest boxes: 250 kg, 300 kg, 350 kg, 500 kg Total weight = 250 + 300 + 350 + 500 = 1400 kg Average weight = 1400 kg / 4 = 350 kg Step 6: Calculate the average weight of the four lightest boxes. Four lightest boxes: 200 kg, 250 kg, 300 kg, 350 kg Total weight = 200 + 250 + 300 + 350 = 1100 kg Average weight = 1100 kg / 4 = 275 kg Step 7: Find the difference in the averages. Difference = 350 kg - 275 kg = 75 kg Therefore, the correct option is B: 75 kg

[#182] The average of 7 consecutive odd number is A. If next 4 and previous 3 odd numbers to these 7 odd numbers are also included, then what is the new average of these 14 consecutive odd numbers?
Correct Answer

(D) A + 1

Explanation

Solution: Average of 7 consecutive odd numbers = A xa0 (Given) Then 7 number consecutive odd numbers = - - - A - - - According to the question, If next 4 and previous 3 odd numbers are also included the numbers of terms = 7 + 3 + 4 = 14, and effected sum of numbers = 2 × 7 = 14 ∴ Increment of average $$ = frac{{14}}{{14}} = 1$$ ∴ Required average = A + 1

[#183] The average of sixteen numbers is 48. The average of the first six of these numbers is 45 and that of the last seven numbers is 53. The seventh and the eighth numbers are, respectively, 3 and 7 greater than the ninth number. What is the average of the ninth and seventh numbers?
Correct Answer

(B) 40.5

Explanation

Solution: Sum of 16 numbers = 16 × 48 = 768 Sum of first 6 numbers = 6 × 45 = 270 Sum of last 7 numbers = 7 × 53 = 371 [x08egin{array}{*{20}{c}}
{{7^{{ ext{th}}}}}&{{8^{{ ext{th}}}}}&{,,,{9^{{ ext{th}}}}} \
{left( {x + 3}
ight)}&{left( {x + 7}
ight)}&{left( x
ight)}
end{array}] $$eqalign{
& Rightarrow 3x = 127 - 10 cr
& Rightarrow x = 39 cr
& herefore { ext{Average of}},{{ ext{9}}^{{ ext{th}}}},{ ext{and}},{{ ext{7}}^{{ ext{th}}}},{ ext{number}} cr
& = frac{{x + left( {x + 3}
ight)}}{2} cr
& = frac{{2x + 3}}{2} cr
& = x + 10.5 cr
& = 39 + 1.5 cr
& = 40.5 cr} $$

[#184] What is the average of all numbers between 8 and 74 which are divisible by 7?
Correct Answer

(C) 42

Explanation

Solution: $$eqalign{
& 14,,21,......,70 cr
& mathop {8 - - - - - - - 74}limits_{{ ext{Divisible by 7}}} cr
& { ext{Number of items}} = frac{{{ ext{Lat term}} - { ext{First term}}}}{{{ ext{Common Difference}}}} + 1 cr
& = frac{{70 - 14}}{7} + 1 cr
& = 9 cr
& { ext{Average of all number divisible by 7}} cr
& Rightarrow frac{n}{{2 imes n}}left[ {a + l}
ight] cr
& Rightarrow frac{{84}}{2} cr
& Rightarrow 2 cr} $$

[#185] Three numbers are such that if the average of any two of them is added to the third number, the sums obtained are 164, 158 and 132 respectively. What is the average of the original three numbers?
Correct Answer

(A) $$75frac{2}{3}$$

Explanation

Solution: $$eqalign{
& frac{{a + b}}{2} + c = 164 cr
& frac{{b + c}}{2} + a = 158 cr
& frac{{c + a}}{2} + b = 132 cr
& overline {frac{{2left( {a + b + c}
ight)}}{2} + a + b + c = 454} cr
& a + b + c = 227 cr
& { ext{Average}} = frac{{a + b + c}}{3} = frac{{227}}{3} = 75frac{2}{3} cr} $$