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Quantitative Aptitude Test - 11

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Quantitative Aptitude Test - 11
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Weekly Quiz Competition
  • Question 1
    1 / -0

    A right circular cone has a height \(8\) cm. If the radius of its base is \(6\) cm, then what is its total surface area?

    Solution
    Height \(=8\) m and radius \(=6\) cm
    Slant height \(l=\sqrt{8^{2}+6^{2}}=10 \) cm
    Total surface area of cone \(=π r^{2}+π r l\)
    \(= π(36+60)\)
    \(= 96 π\) cm\(^{2}\)
  • Question 2
    1 / -0

    Direction: The question below is followed by two statements I and II. You have to determine whether the data given in the statement is sufficient for answering the question. You should use the data and your knowledge of Mathematics to choose the best possible answer.

    Ishita bought 2000 mobiles for 50000 find his profit %.

    I. Ishita gives away 25% of mobiles free.

    II. 10% mobiles sold in the market for 25% discount and rest at 33.33% discount.

    Solution

    Statement I:

    Number of mobiles sold by Ishita in the market = Bought mobiles × (1−25% of the free giveaway)

    ⇒ 2000 × (1−25%) = 1500

    Statement II:

    From this statement, we cannot say at what price the product was sold at the market so we cannot find the profit %.

    ∴ Both the statements together are not sufficient to answer the question.

  • Question 3
    1 / -0

    A sold goods to B at a loss of \(5 \%\) but B sells it for Rs. 570 and he gains \(20 \%\). What is the cost price of the goods for A?

    Solution

    Given:

    Selling price of B\(=\) Rs. 570

    Gain \(\%\) for \(B=20\)

    Loss for A \(=5 \%\)

    Let the cost price for B be Rs. \({x}\).

    Profit = S.P. - C.P.

    \(\Rightarrow\left(\frac{20 x}{100}\right)=570-x\)

    \(\Rightarrow \frac{120 x}{100}=570\)

    \(\Rightarrow x=\) Rs. 475

    Rs. \(475\) is the selling price for A.

    Loss = C.P. - S.P.

    \(\Rightarrow 5 \%\) of \(x=x-475\)

    \(\Rightarrow \frac{95 x}{100}=475\)

    \(\Rightarrow x=\) Rs. \(500\)

    ∴ Cost price of the good for A isRs. \(500\).

  • Question 4
    1 / -0

    Directions For Questions

    Direction: The following pie-chart shows the sources of funds to be collected by the National Highways Authority of India (NHAI) for its Phase II projects. Study the pie chart and answers the question that follows.

    Sources of funds to be arranged by NHAI for Phase II projects (in crores Rs.)

    ...view full instructions

    Near about 20% of the funds are to be arranged through:

    Solution

    Given:

    Total funds = 11486 + 5252 + 4910 + 6000 + 29952

    = Rs. 57600crores

    20% of the total funds to be arranged

    = Rs. (20% of 57600) crores

    =57600×20100

    =Rs. 11520 croresRs. 11486 crores

    Near about 20% of the funds are to be arranged through external assistance.

  • Question 5
    1 / -0

    Directions For Questions

    Direction: The following pie-chart shows the sources of funds to be collected by the National Highways Authority of India (NHAI) for its Phase II projects. Study the pie chart and answers the question that follows.

    Sources of funds to be arranged by NHAI for Phase II projects (in crores Rs.)

    ...view full instructions

    If NHAI could receive a total of Rs. 9695 crores as External Assistance, by what percent (approximately) should it increase the Market Borrowing to arrange for the shortage of funds?

    Solution

    Given:

    Funds required through External Assistance = Rs. 11486 crores

    Funds receivethrough External Assistance =Rs. 9695 crores

    Shortage of funds arranged through External Assistance =Rs. (11486 - 9695) crores

    = Rs. 1791 crores

    Increase required in Market Borrowing = Rs. 1791 crores

    Funds arranged through Market Borrowing = Rs. 29952 crores

    Percentage increase required =179129952×100

    = 5.98%6%

  • Question 6
    1 / -0

    Directions For Questions

    Direction: The following pie-chart shows the sources of funds to be collected by the National Highways Authority of India (NHAI) for its Phase II projects. Study the pie chart and answers the question that follows.

    Sources of funds to be arranged by NHAI for Phase II projects (in crores Rs.)

    ...view full instructions

    If the toll is to be collected through an outsourced agency by allowing a maximum 10% commission, how much amount should be permitted to be collected by the outsourced agency, so that the project is supported with Rs. 4910 crores?

    Solution

    Given:

    Funds required from Toll for projects = Rs. 4910crores

    Amount permitted = (Funds required from Toll for projects of Phase II) +(10% of these funds)

    = 4910 + (10% of 4910)

    = 4910 +4910×10100

    = 4910 + 491

    = Rs. 5401 crores

  • Question 7
    1 / -0

    Directions For Questions

    Direction: The following pie-chart shows the sources of funds to be collected by the National Highways Authority of India (NHAI) for its Phase II projects. Study the pie chart and answers the question that follows.

    Sources of funds to be arranged by NHAI for Phase II projects (in crores Rs.)

    ...view full instructions

    The central angle corresponding to Market Borrowing is:

    Solution

    Given:

    Total funds = 11486 + 5252 + 4910 + 6000 + 29952

    = Rs. 57600 crores

    Investment required byMarket Borrowing =Rs. 29952crores

    Central angle for pie chart

    =2995257600×360°

    =1078272057600

    = 187.2°

  • Question 8
    1 / -0

    Directions For Questions

    Direction: The following pie-chart shows the sources of funds to be collected by the National Highways Authority of India (NHAI) for its Phase II projects. Study the pie chart and answers the question that follows.

    Sources of funds to be arranged by NHAI for Phase II projects (in crores Rs.)

    ...view full instructions

    The approximate ratio of the funds to be arranged through Toll and that through Market Borrowing is:

    Solution

    Given:

    Investment required by Market Borrowing = Rs. 29952 crores

    Investment required by Toll = Rs. 4910 crores

    Required ratio =491029952

    =16.116

    =1 : 6

  • Question 9
    1 / -0

    Find the sum of ages of A, B, C, D, and E

    I. Average of A, C, and D is 34.

    II. Average A, B, and E is 45.

    III. The sum of ages of B and E is 90. 

    Solution

    Given:

    The average of A, C, and D is 34.

    The average A, B, and E is 45.

    The sum of ages of B and E is 90. 

    Average \(=\frac{\text { Total sum }}{\text { Total numbers }}\)

    Statement (I):

    The average of A, C, and D = 34

    The total sum of A, C, and D = 34 × 3 = 102

    Statement (II):

    The average A, B, and E = 45

    The total sum of A, B, and E = 45 × 3 = 135 

    Statement (III):

    The sum of ages of B and E = 90 

    From statement (I) and (III):

    The total sum of A, B, C, D, and E = 102 + 90 = 192

    ∴ The total sum of A, B, C, D, and E is 192.

    Thus, The only statement I and III together are sufficient to answer the question.

  • Question 10
    1 / -0

    In certain years a sum of money is doubled itself at \(6.25 \%\) simple interest p.a then the required time in years will be?

    Solution

    Given:

    In certain years a sum of money is doubled itself at \(6.25 \%\) simple interest per annum.

    Formula:

    Amount (A) = Principal (P) + Simple Interest (SI)

    Simple Interest \((\mathrm{SI})=\frac{P \times N \times R}{100}\)

    Where,

    P = Principal

    N = Time Duration

    R = Rate of Interest

    SI = Simple Interest

    Given, In certain years a sum of money is doubled itself

    Let the time required and principal be \(t\) years and Rs. \(p\) respectively.

    Amount \(=2 p\)

    As, Amount \(=S . I+p\)

    \(\Rightarrow S . I=p\)

    We know that \(\mathrm{SI}=\frac{\mathrm{p} \times \mathrm{r} \times \mathrm{t}}{100}\)

    Given, \(r=6.25 \%\)

    \(\Rightarrow p=\frac{p \times r \times t}{100}\)

    \(\Rightarrow t=\frac{100}{r}\)

    \(\Rightarrow t=\frac{100}{6.25}\)

    \(=16\) years

    \(\therefore\) The required time in years will be \(16\) years.

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