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Comparing Quantities Test - 34

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Comparing Quantities Test - 34
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  • Question 1
    1 / -0
    A scientific calculator is available at Universal shop in Hazratganz at 20%20\% discount and the same is available at only 15%15\% discount at Bhootnath market. Pooja has just sufficient amount of Rs.800Rs.\,800 to purchase it at Universal shop, Hazratganz. What is the amount that Pooja has less than the required amount to purchase it at Bhootnath?
    Solution
    Let cost of calculator is Rs.100Rs. 100
     calculator is available at Universal shop in Hazratganj at 20%20\% discount then sell cost =10020=80 RS=100-20=80\ RS
    Pooja paid Rs.800Rs. 800 at Universal shop 
    So actual cost of calculator =80080×100=1000=\dfrac{800}{80}\times 100= 1000Rs
    same as 15% discount at Bhootnath market
    Then selling price of calculator at bhootnath market =85100×1000=850=\dfrac{85}{100}\times 1000= 850Rs
    amount that Pooja has less paid =850800=Rs.50=850-800=Rs.50


  • Question 2
    1 / -0
    The list price of a slipper is Rs. 160. The  discount offered is 10%.Find  the selling Price of a slipper.
    Solution
    List price of slipper =160=160
    discount =10%=10\%
    10%10\% of 160=10100×160=16160=\cfrac{10}{100}\times160=16
    Selling Price of slipper aftter  discount=16016=144160-16=144

  • Question 3
    1 / -0
    If the makeup percentage of an article is 50%  and discount is also 20%, then the profit percentage will be ..........
    Solution
    Let us assume cost price =Rs100Rs 100
    Makeup percentage of an article = 5050%
    Marked price =100+100×50100=150100+100 \times \dfrac{50}{100}=150
    Discount = 2020%

    Selling price\text{Selling price}=Marked priceMarked price×Discount\text{Marked price}-\text{Marked price} \times \text{Discount} %

    S=150150×20100S=150-150\times \dfrac{20}{100}

    S=15030=120S=150-30=120

    Profit Percentage\text{Profit Percentage}= (Selling Price - Cost Price)cost price\dfrac{(\text{Selling Price - Cost Price})}{\text{cost price}} ×100\times 100

    PP%=(120100)100×100\dfrac{(120-100)}{100} \times 100

    P %=2020 %

    Answer (D) 2020%

  • Question 4
    1 / -0
    By selling a book for Rs. 10, the publisher loses 111\displaystyle \frac{1}{11} of what it costs him. His cost price is ....
    Solution
    Let's assume cost price =x=x
    Selling price =Rs. 10=\text{Rs.}\ 10
    Loss=111.x=\dfrac{1}{11}.x
    Cost Price=Selling Price + Loss\text{Cost Price=Selling Price + Loss}
    x=10+x11x=10+\dfrac { x }{ 11 }

    10x11=10\dfrac { 10x }{ 11 } =10
    x=Rs. 11x=\text{Rs.}\ 11
    Answer (C) Rs. 11


  • Question 5
    1 / -0
    Samant brought microwave oven and paid 10% less than the original price. He sold it with 30% profit on the price he had paid. What percentage of profit did Samant earn on the original price?
    Solution
    Lets assume original price of the microwave oven = Rs. 100
    So, buying price for samant = Rs. 90
    Profit %=30%=p×100Costprice\% = 30\% = \dfrac{p\times100}{Cost price} 
    30%=p×1009030\% = \dfrac{p\times100}{90}
    30×90100=p{30\times90}{100}=p
    profit=27profit=27
    Selling price =90+27=117 Rs
    Profit =117-100=17 Rs
    profit %=17×100100=17\%=\dfrac{17\times100}{100} = 17%

  • Question 6
    1 / -0
    Latika bought a radio set and spent Rs. 110 on its repairs. She then sold it to Garima at 20% profit, Garima sold it to Megha at a loss of 10% and Megha sold it for Rs. 1188 at a profit of 10%. What is the amount for which Latika bought the radio set?
    Solution
    Let's assume the amount for which Latika bought the radio set=x
    So, As per question,
    Latika's selling price=(x+110)*1.20
    Megha's buying price=(x+110)*1.20*0.90
    Megha's selling price= (x+110)×1.20×0.90×1.10=1188(x+110)\times 1.20\times 0.90\times 1.10=1188
    1.19x+130.68=11881.19x+130.68=1188
    x=888.50x=888.50
    Answer (B) Rs. 890
  • Question 7
    1 / -0
    The taxi charges in a city comprise of a fixed charge, together with the charge of the distance covered. For a journey of 16 km, the charges paid are Rs. 156 and for a journey of 24 km, the charge paid are Rs, 204. What will a person have to pay for travelling a distance of 30 km?
    Solution
    Let's take fixed charge=F
    charge of the distance covered=x
    So, As per question, 
    F+16x=156F+16x=156.................eq1
    F+24x=204F+24x=204.......................eq2
    Eq2-eq1
    8x=488x=48
    x=6x=6
    from eq 1
    F=15696=60F=156-96=60
    Charge a distance of 30 km=
    F+30x=60+306=240RsF+30x=60+30*6=240Rs
    Answer (B) 
    Rs. 240
  • Question 8
    1 / -0
    A person has enough money to buy 2525 cycles worth Rs.500Rs.\,500 each. How many cycles will the person be able to buy, if each cycle now costs Rs.125Rs.\,125 more ?
    Solution
    Total money=25*500=12500
    New cost of cycle=500+125=625
    No of cycle person is able to buy
    N=Totalmoneycostof1cycleN=\frac { Total\quad money }{ cost\quad of\quad 1\quad cycle }
    N=12500625=20N=\frac { 12500 }{ 625 } =20
    Answer (C) 20
  • Question 9
    1 / -0
    A trader marks his goods at 20% above the cost price. He sold half the stock at the marked price one quarter at a discount of 20% on the marked price and the rest at a discount of 40% on the marked price. His total gain is
    Solution
    Let the CPCP be xx
    MP=120%×x=1.2xMP=120\%\times x=1.2x
    Gain while selling half stocks of goods=0.5(1.2xx)=0.1x=0.5(1.2x-x)=0.1x
    Gain while selling remaining one quarter stocks of goods=0.25[(80%×1.2x)x]=0.01x=0.25[(80\%\times 1.2x)-x]=-0.01x
    Gain while selling remaining stocks of goods=0.25[(60%×1.2x)x]=0.07x=0.25[(60\%\times 1.2x)-x]=-0.07x
    Net gain=(0.10.010.07)x=0.02x=(0.1-0.01-0.07)x=0.02x
    Gain%=gainCP×100\%=\cfrac{gain}{CP}\times 100
        0.02xx×100=2%\implies\cfrac{0.02x}{x}\times 100=2\%


  • Question 10
    1 / -0
    A tempo is insured to the extent of 45\displaystyle \frac{4}{5} of its original value. If the premium on it at the rate of 1.3 percent amounts to Rs 910 the original value of the tempo is
    Solution

    Let the original value of the tempo be Rs. x. Then,

    1.3 Percent  of 4/5 of x = 910
     1310×1100×45×x=910\Rightarrow  \frac{13}{10}\times\frac{1}{100}\times\frac{4}{5}\times x=910
    x=(910×10×100×513×4)=87500\Rightarrow x = \left(\frac{910\times10\times100\times5}{13\times4}\right)=87500


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