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General Test - 20

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General Test - 20
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  • Question 1
    4 / -1

    P and Q invested in a business. The profit earned was divided in the ratio 2 : 3. If P invested Rs 40000, the amount invested by Q is:

    Solution

    P and \(Q\) invested in a business. The profit earned was divided in the ratio \(2: 3\). If P invested Rs. 40000.

    Concept used:

    Amount invested \(P\) to \(Q=\) ratio of their profit

    Calculation:

    The amount of \(P\) invested Rs. 40000 and amount invested by \(Q\) is \(y\).

    Amount invested \(P\) to \(Q=\) ratio of their profit

    \(\Rightarrow \frac{40000}{y}=\frac{2}{3}\)

    \(\Rightarrow y=40000 \times \frac{3}{2}\)

    \(\Rightarrow y=R s .60000\)

    \(\therefore\) The amount invested by \(Q\) is Rs. \(60000 .\)

  • Question 2
    4 / -1

    At what rate percent per annum will Rs. 8000 amount to Rs. 8820 in 2 years at compound interest?

    Solution

    Given:

    Principal = Rs. 8000

    Amount after 2 years = Rs. 8820

    As interest is compounded annually for 2 years, we have to square root both principal and amount and compare them to find the interest.

    Compound interest\(=\)Amount \(=\) Principal\(\left(1 + \frac{\text{Rate}}{100}\right)^t\)

    Let, the rate of interest be r %.

    Accordingly,

    \(8000 \times\left(1+\frac{r}{ 100}\right)^{2}=8820\)

    \(\Rightarrow \frac{8820} { 8000}=\left(1+\frac{r}{ 100}\right)^{2}\)

    \(\Rightarrow \frac{441}{400}=\left(1+\frac{r}{ 100}\right)^2\)

    \(\Rightarrow \frac{21}{20}=1+\frac{r}{100}\)

    \(\Rightarrow \frac{r}{100}=\frac{1}{20}\)

    \(\Rightarrow \mathrm{r}=5\)

    \(\therefore\) Rate of Interest is \(5 \%\) per annum.

  • Question 3
    4 / -1

    A certain sum of money fetched an interest of Rs.1800 at 15% per annum simple interest for 2 years. Find the sum.

    Solution

    Let the sum of money be Rs. 100x.
    Simple Interest at15% per annum for2 years=2×(15% of100x)= Rs.30x
    So, according to the question,

    30x=1800
    x=60
    Therefore, Sum= Rs.100x= Rs.(100×60)= Rs.6000

  • Question 4
    4 / -1

    A trader buys an article at 80% of its marked price and sells it at a 10% discount on its marked price. His percentage profit is:

    Solution

    Let M.P. of the article be Rs. 100

    C.P. of the article for trader \(=100 \times \frac{80}{100}=80\)

    S.P. of the article \(=100 \times \frac{90}{100}=90\)

    Required percentage \(=\frac{(90-80)}{80} \times 100=12 \frac{1}{2} \%\)

  • Question 5
    4 / -1

    1 bar of chocolate costs Rs. 80 but containing 6 bars of the same chocolate costs Rs. 400. What is the effective discount (in %) on the box?

    Solution

    Cost of 6 bars of chocolate at the rate of Rs. 80 per bar \(=6 \times 80=\) Rs. 480

    Selling price of 6 bars \(=\) Rs. 400

    Discount \(=\frac{(480-400)}{480} \times 100\)\(=16.67 \%\)

  • Question 6
    4 / -1

    An insurance company pays Rs. 3,81,800 after a new car worth Rs. 4,60,000 gets completely damaged in an accident. Whatpercent of the car was insured?

    Solution

  • Question 7
    4 / -1

    Solution

  • Question 8
    4 / -1

    Pipe P and Q can fill a tank in 5 hours and 12 hours respectively. If pipe P is opened at 8:00 a.m. and pipe Q is opened at 10:10 a.m. then, at what time will the tank be full?

    Solution

    Given:

    Pipe P fills a tank = 5 hours

    Pipe Q fills a tank = 12 hours

    Calculation:

    Let the total capacity of a tank = 60 units

    Tank filled by pipe P in 1 hour = 605

    = 12 units

    Tank filled by pipe Q in 1 hour =6012

    = 5 units

    Tank filled by pipe P from 8:00 a.m. to 10:00 a.m. = 24 units

    Tank filled by pipe P from 10:00 a.m. to 10:10 a.m. = 2 units

    Remaining tank to be filled = 60 – 26

    = 34 units

    Tank filled in 1 hour, when both the pipes are open (from 10: 10 a.m. to 11:10 a.m.) =17 units

    Tank filled in next 1 hour, when both the pipes are opened from 11:10 a.m. to 12:10 p.m. = 17 units.

    ∴ The tank will be filled at 12:10 P.m.

  • Question 9
    4 / -1

    Solution

  • Question 10
    4 / -1

    The value of \(\frac{18.43 \times 18.43-6.57 \times 6.57}{11.86}\) is:

    Solution

    \(\frac{18.43 \times 18.43-6.57 \times 6.57}{11.86}\)

    \(= \frac{(18.43)^2-(6.57)^2}{18.43-6.57}\)

    \(= \frac{(18.43-6.57)(18.43+6.57)}{18.43-6.57}\)\(= 25\)

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