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Quantitative Aptitude Test - 15

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Quantitative Aptitude Test - 15
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Weekly Quiz Competition
  • Question 1
    4 / -1

    If in a race of 500 meters, A beats by B by 5 seconds or 50 meters, find the speed of A.

    Solution

    A beats B by 5 seconds or 50 meters which means B was 50 meters behind when A finished the race.

    And B reaches after 5 seconds which means B travels 50 meters in 5 seconds so he has a speed of 50/5 = 10 m/s

    So B takes 500/10 = 50 seconds to complete the race and we can say A takes 50 - 5 = 45 seconds.

    So speed of A = 500/45 m/s = 100/9 m/s = 11.11 m/s

  • Question 2
    4 / -1

    Directions For Questions

    Direction: Study the bar graph and answer the following question.

    The Bar graph shows the quantity of different types of mangoes sold (in kg) by a shop in May 2017 and May 2018.

    ...view full instructions

    The quantity of Langra mangoes sold in May 2017 is what percent of the quantity of Alphonso and Neelam mangoes sold in May 2018 together?

    Solution

    Quantity of Langra mangoes sold in \(2017=180 {~kg}\)

    Quantity of Alphonso mangoes sold in \(2018=160 {~kg}\)

    Quantity of Neelam mangoes sold in \(2018=80 {~kg}\)

    \(\therefore\) Required percentage \(=\frac{180}{(160+80)} \times 100= 75\%\)

  • Question 3
    4 / -1

    Directions For Questions

    Direction: Study the bar graph and answer the following question.

    The Bar graph shows the quantity of different types of mangoes sold (in kg) by a shop in May 2017 and May 2018.

    ...view full instructions

    What is the difference between the quantity of all mangoes sold in May 2017 and 2018?

    Solution

    Total quantity of mangoes sold in May \(2017=(120+180+75+\) \(250+60)=685 {~kg}\)

    Total quantity of mangoes sold in May \(2018=(160+200+60+\) \(225+80)=725 {~kg}\)

    \(\therefore\) Required difference \(=(725-685)= 40 kg\)

  • Question 4
    4 / -1

    Directions For Questions

    Direction: Study the bar graph and answer the following question.

    The Bar graph shows the quantity of different types of mangoes sold (in kg) by a shop in May 2017 and May 2018.

    ...view full instructions

    What is the percentage decrease in sales of Himsagar mango in May 2018 with respect to May 2017?

    Solution

    Quantity of Himsagar sold in May \(2017=250 {~kg}\)

    Quantity of Himsagar sold in May \(2018=225 {~kg}\)

    \(\therefore\) Percentage decrease in sales \(=\frac{(250-225)}{250} \times 100\) \(=10 \%\)

  • Question 5
    4 / -1

    Directions For Questions

    Direction: Study the bar graph and answer the following question.

    The Bar graph shows the quantity of different types of mangoes sold (in kg) by a shop in May 2017 and May 2018.

    ...view full instructions

    In the case of Dasheri mango, the quantity sold in May 2018 was 80% of the total sales that year. What ratio between the total quantity of Dasheri sold in 2018 and Himsagar sold in May 2018?

    Solution

    Quantity of Dasheri mango sold in May \(2018=60 {~kg}\)

    Let the total sale in the 2018 year be \(x\).

    According to the question,

    \(60=80 \%\) of \(x\)

    \(\Rightarrow 60=\frac{80}{100} \times x\)

    \(\Rightarrow 60 \times \frac{100}{80}=x\)

    \(\therefore x=75 {~kg}\)

    Total quantity of Dasheri mango sold in \(2018=75 {~kg}\)

    Quantity of Himsagar sold in May \(2018=225 {~kg}\)

    \(\therefore\) Required Ratio \(=75: 225=1: 3\)

  • Question 6
    4 / -1

    Directions For Questions

    Direction: Study the bar graph and answer the following question.

    The Bar graph shows the quantity of different types of mangoes sold (in kg) by a shop in May 2017 and May 2018.

    ...view full instructions

    What is the average quantity of Mango sold by the shop in June 2018, if average sales is increased by 20% than May 2018?

    Solution

    Total quantity of mangoes sold in May \(2018=(160+200+60+225+80)=725 {~kg}\)

    \(\therefore\) Average quantity of mango sold in May \(2018=\frac{725}{5}=145 {~kg}\)

    \(\therefore\) Average quantity of mango sold in June 2018,

    \(=145+145 \times \frac{20}{100}\)

    \(=(145+29)\)

    \(=174 {~kg}\)

  • Question 7
    4 / -1

    A boat is 154km away from the shore. It has a leak which admits 9 tonnes of water in 22 minutes. When 184 tonnes of water enters the boat, the boat would sink. But the pump can throw out 1 tonne in every 5 minutes. The minimum average speed at which the boat may just reach the shore before it begins to sink.

    Solution

    In 1 minute the amount of water (in tonnes) which enters into the boat = 9/22

    In 1 minute the amount of the water which the pump can throw out = 1/5 tonnes

    The net inflow of water per minute = 9/22 - 1/5 = 23/110 tonnes

    Time taken to accumulate 184 tonnes of water = 184/(23/110) = 880 minutes

    So the boat must sail at = 154/(880/60) = 10.5kmph

  • Question 8
    4 / -1

    A hawker purchased oranges at the rate of 4 oranges in a rupee, but he sells at the rate of 5 oranges in a rupee. His loss is:

    Solution

    Assume he took 20 oranges (LCM of 4,5)

    CP= Rs.5 SP=Rs.4

    Loss = 1/5 X 100 = 20%

  • Question 9
    4 / -1

    The following pie chart gives the percentage distribution for the favorite color of the people of a colony. Total number of people living in the colony is less than 500 but more than 300.

    Find the degree for the people for whom the favorite color is red.

    Solution

    Red is the favorite color of 10%

    10% is 1/10th so the angle is 1/10th of 360 degree which is 36 degree.

  • Question 10
    4 / -1

    A vessel contains 50 litres of paint. 20% of this paint is taken out and the same is repeated two more times. An oil of amount equal to the paint taken out is mixed with the remaining paint in the vessel. What is the ratio of the paint and the oil in the vessel?

    Solution

    Let 50L = 100%.

    Given that the process is taken three times

    So paint left = 100*80*80*80/1000000 = 51.2%

    And paint removed = oil added = 100 - 51.2 = 48.8%

    Required ratio = 51.2:48.8 = 64:61

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