Question : A merchant changed his trade discount from 25% to 15%. This would increase his selling price by:
Option 1: $3\frac{1}{3}$%
Option 2: $6\frac{1}{6}$%
Option 3: $13\frac{1}{3}$%
Option 4: $16\frac{1}{3}$%
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Correct Answer: $13\frac{1}{3}$%
Solution : Given: A merchant changed his trade discount from 25% to 15%. Let the marked price be Rs. 100. Now for a 25% discount the selling price is = (100 – 25) = Rs. 75 and for a 15% discount, the selling price is = (100 – 15) = Rs. 85 So, the increase in the selling price = ($\frac{85-75}{75}$×100) = $\frac{40}{3}$ = $13\frac{1}{3}$% Hence, the correct answer is $13\frac{1}{3}$%.
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Question : Let $x=\frac{5 \frac{3}{4}-\frac{3}{7} \times 15 \frac{3}{4}+2 \frac{2}{35} \div 1 \frac{11}{25}}{\frac{3}{4} \div 5 \frac{1}{4}+5 \frac{3}{5} \div 3 \frac{4}{15}}$. When $y$ is added to $x$, the result is $\frac{7}{13}$. What is the value of $y$?
Option 1: $\frac{1}{13}$
Option 2: $\frac{9}{13}$
Option 3: $\frac{2}{13}$
Option 4: $\frac{4}{13}$
Question : If the cost price of 28 oranges is equal to the selling price of 24 oranges, then the profit percentage is:
Option 1: $16 \frac{2}{3}$%
Option 2: $16 \frac{1}{3}$%
Option 3: $18 \frac{2}{3}$%
Option 4: $18 \frac{1}{3}$%
Question : If $2=x+\frac{1}{1+\frac{1}{5+\frac{1}{2}}}$, then the value of $x$ is equal to:
Option 1: $\frac{14}{13}$
Option 2: 1
Option 3: $\frac{13}{15}$
Option 4: $\frac{15}{13}$
Question : If $(x-\frac{1}{3})^2+(y-4)^2=0$, then what is the value of $\frac{y+x}{y-x}$?
Option 1: $\frac{11}{13}$
Option 2: $\frac{13}{11}$
Option 3: $\frac{16}{9}$
Option 4: $\frac{9}{16}$
Question : If $(4 y-\frac{4}{y})=13$, find the value of $(y^2+\frac{1}{y^2})$.
Option 1: $12 \frac{11}{16}$
Option 2: $10 \frac{9}{16}$
Option 3: $12 \frac{9}{16}$
Option 4: $8 \frac{9}{16}$
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