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Comparing Quantities Test - 1

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Comparing Quantities Test - 1
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  • Question 1
    1 / -0

    The price of a scooter was Rs 34,000 last year. It has increased by 20% this year. What is the price now?

    Solution

     

    Explanation:

    Purchasing price of the scooter is = ₹ 34,000

    Now, increasing value=

    34,000 × 20/100

    = 340 × 20

    = ₹6800

    So, total cost = ₹(34,000 + 6,800)

     = ₹ 40,800

     

  • Question 2
    1 / -0

    A picnic is being planned in a school for Class VII. Girls are 60% of the total number of students and are 18 in number. Find the ratio of the number of girls to the number of boys in the class.

    Solution

    Explanation:

    let the total number of students be = x

    60% of x=18 (Girls Ratio & Number given in the question)

    x = 18/60 ×100

    x=30

    Total no of students=30
    No. of Boys =30-18 =12

    No. of Girls : No. of Boys
    18 : 12 = 3 2

     

  • Question 3
    1 / -0

    A shopkeeper purchased 200 bulbs for Rs 10 each. However 5 bulbs were fused and had to be thrown away. The remaining were sold at Rs 12 each. Find the gain or loss %.

    Solution

     

    Explanation:

    Total Purchase Price of Bulbs = 200 × 10

    = Rs.2,000

    Number of fused Bulbs = 5

     Number of remaining Bulbs = Total - fused = 200-5 =195

    Selling Price of Remaining Bulbs = 195 × 12

    Total Selling Price                     = 2340

    Gain                                       =2,340-2,000 =Rs 340

    Gain (%)                                  =   Gain / PurchasePrice ×  100

                                                  =    340 / 2000 × 100

                                                  =17%

     

     

  • Question 4
    1 / -0

    During a sale, a shop offered a discount of 10 % on the Marked Prices of all the items. What would a customer have to pay for a pair of jeans marked at Rs 1450 and two shirts marked at Rs 850 each?

    Solution

    Explanation:

    Marked Price of Jeans = ₹1,450

    Marked Price of two Shirts =₹(850×2)

                                           = ₹1,700

    Total Marked Price=₹(1450+1700)

                                           = ₹3,150

    Let the Marked Price be     =₹100

    Discount (10%)                 = 10 /100 × 100

                                           =₹10

    Paid by customer =₹(100−10)                             

      =₹90

    If Marked Price is ₹100then amount paid by customer is =₹90

    If Marked Price is Rs. 1 then amount paid by customer is   = 90/100

    If Marked Price is Rs.3150 then paid by customer  = 90/100 × 3150

     =₹2,835

     

     

                                                                         

  • Question 5
    1 / -0

    Calculate the Amount on Rs 10,800 for 3 years at 12.5 % per annum Compounded annually.

    Solution

    Explanation:

    A( Amount) = ?

    P(Principal) = Rs.10,800

    n(Time) = 3Years

    r (Rate) = 12.5%

    A = P (1+r/100)n

    A = 10,800(1+12.5/100)3=10,800(45/40)3

    = Rs.15,377.34

     

     

  • Question 6
    1 / -0

    ________ means comparing two quantities.

    Solution

    Explanation:

    A ratio is a relationship between two numbers indicating how many times the first number contains the second.

  • Question 7
    1 / -0

    Rohan bought a second hand refrigerator for Rs 2,500, then spent Rs 500 on its repairs and sold it for Rs 3,300. Find his gain or loss per cent.

    Solution

    Explanation:

     Purchase amount =₹2500

    After that, spent ₹500 for repairing.

    Total amount= ₹3000

    Then, sold it at ₹3300

    Gain = ₹300

    Gain %= 300/3000×100%

                 = 10%

     

  • Question 8
    1 / -0

    Find the ratio of speed of a cycle 15 km per hour to the speed of scooter 30 km per hour.

    Solution

     

    Explanation:

    Speed of the cycle/hour :Speed of the scooter/hour

    =15:30

    = 1:2

     

  • Question 9
    1 / -0

    The population of a city was 20,000 in the year 1997. It increased at the rate of 5%% p.a. Find the population at the end of the year 2000.

    Solution

     

    Explanation:

    Population is calculated as compound Interest Formula

    A =P (1+r8/100)n

    Intial Population (P) =20,000

    Rate(r)                       =5%

    Time(n)                      = 3 Years

    A (Population at the end of the year 2000) = 20,000(1+5/100)3

                                                                  = 23,152.5 ( But persons cannot be taken in points)

                                                                  =23,153

     

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