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Comparing Quantities Test - 60

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Comparing Quantities Test - 60
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  • Question 1
    1 / -0
    The market price of a mixer is $$2300$$ rupees. A customer purchased it for $$Rs.1955$$. Find percentage of discount offered to the customer.
    Solution
    Let the Discount amount be $$x$$,

    Market price $$=Rs. \ 2300$$

    Selling price $$=Rs.\ 1995$$

    Market price $$=$$ selling price$$(1-\dfrac{x}{100})$$

     $$\Rightarrow 2300-\dfrac{x}{100}\times2300=1995$$

    $$\Rightarrow 2300-1995=23x$$

    $$\Rightarrow x=13.26$$

    Hence, discount percentage  $$=13.26$$ %
  • Question 2
    1 / -0
    $$45 \% \text { of } 1500 + 35 \% \text { of } 1700 = ? \% \text { of } 3175$$
    Solution

    $$45\%$$ of $$1500$$ is given as 

    $$\dfrac{45}{100}\times 1500$$

    $$\implies 45\times 15=675$$

    $$35\%$$ of $$1700$$ is given as 

    $$\dfrac {35}{100}\times 1700$$

    $$\implies 35\times 15=595$$

    Total is $$675+595=1270$$

    Let the percent be $$x$$ 

    $$\implies \dfrac x{100}\times 3175=1270$$

    $$\implies x=\dfrac {1270}{3175}\times 100$$

    $$\implies x=40\%$$ 

  • Question 3
    1 / -0
    If a banana's cost is $$Rs. 1.25$$ and apple's cost is $$Rs. 1.75$$ what will be the cost of $$2$$ Dozen of Banana and $$3$$ Dozen of apple?
    Solution
    $$\because$$ Cost of one banana $$=Rs.1.25$$
    $$\therefore 2$$ dozens banana's cost $$=24\times 1.25=Rs.30$$

    $$\because$$ One apple's cost $$=Rs.1.75$$
    $$\therefore 3$$ dozens banana's cost $$=36\times 1.75=Rs.63$$

    Total cost $$=63+30=Rs.93$$


    Hence, option (A) is the correct option
  • Question 4
    1 / -0
    If market price of a article is $$Rs.1900$$, and selling price is $$Rs.1750$$, then find the discount percentage.
    Solution
    Given, 
    Marked price $$=Rs.1900$$
    Selling price $$=Rs.1750$$

    $$Discount\%=\dfrac{Marked\ Price -\ Selling\ Price}{Marked \ price}\times 100\%$$

    $$\Rightarrow Discount\%=\dfrac{1900 -1750}{1900}\times 100\%$$

    $$\Rightarrow Discount\%=\dfrac{150}{1900}\times 100\%$$

    $$\Rightarrow Discount\%=7.89 $$ %
  • Question 5
    1 / -0
    If $$A$$ is $$150$$% of $$B$$ then $$B$$ is what percent of $$A+B$$?
    Solution
    $$A=\dfrac{150}{100}B$$

    $$A+B=\dfrac{150}{100}B+B=\dfrac{250}{100}B=2.5B$$

    B is $$\dfrac{B}{A+B}\%  of A+B$$

    $$\dfrac{1}{2.5}\times100$$

    $$=40\%$$
    Option C is correct


  • Question 6
    1 / -0
    $$60 \% \text { of } 264$$ is the same as-
    Solution
    $$60\% \ of \ 264 =\dfrac{60}{100} \times 264 =158.4$$

    Also, $$15\% \ of \ 1056 =\dfrac{15}{100} \times 1056 = 158.4$$

    Hence, $$60\% \ of \ 264 = 15\% \ of \ 1056$$
  • Question 7
    1 / -0
    Ram purchases a chair at $$Rs.70$$ and spent $$Rs.17$$ on its repair and $$50$$ paise on cartage. If he sold the chair at $$Rs.100$$ then his approximate margin of profit will be ?
    Solution
    $$Profit ~\%=\dfrac{SP-CP}{CP}\times100$$

                     $$=\dfrac{100-87.5}{87.5}\times100$$

                     $$=14.3\%$$
  • Question 8
    1 / -0
    What would be the $$C.I$$ obtained on an amount of $$12500$$ at the rate of $$12$$ $$p.c.p.a.$$ after $$2$$ years? 
    Solution
    The  formula of for  compound interest is 
    $$C.I. = P\left( 1+\cfrac{r}{100} \right)^{n}-P$$
    $$=12,500 \left( 1+\cfrac{12}{100} \right)^{2}-12500$$
    $$= 12500\times \frac{112}{100}\times \frac{112}{100}-12500$$
    $$=3180$$
  • Question 9
    1 / -0
    If $$11$$ Mango are bought for $$Rs.10$$ and sold at $$10$$ for $$Rs.11$$. What was Gain or Loss $$\%$$?
    Solution

  • Question 10
    1 / -0
    The cost price of $$20$$ articles is the same as the selling price of X articles. If the profit is $$25$$%, then the value of X is :
    Solution
    Let the Cost price for one article be $$CP$$

    The Cost price of $$20$$ articles is $$20CP$$

    The Selling price be $$SP$$

    The Selling price is $$X\times SP$$

    According to the question 

    $$20CP=X\times SP\cdots(1)$$

    The profit is $$25\%$$ 

    $$\implies $$ Profit is $$\dfrac{25}{100}\times CP=\dfrac {CP}4$$

    $$\implies SP-CP=\dfrac{CP}4$$

    $$4SP-4CP=CP$$

    $$4SP=5CP\cdots(2)$$

    Dividing $$(1)$$ with $$(2)$$

    $$\dfrac{20CP}{5CP}=\dfrac{X\times SP}{4SP}

    $$\implies 4=\dfrac{X}4$$

    $$\implies X=16$$
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