Tag: maths

Questions Related to maths

Find a single discount equivalent to the successive discounts of $10\%, 20\%$ and $20\%$ (in percent).

maths money math different discounts successive discounts problems on discounts

  1. $42.1\%$

  2. $42.4\%$

  3. $42.8\%$

  4. $45\%$

Show answer
Correct Option: B
Explanation:

$\Rightarrow$  Single equivalent discount for successive discount of $10\%$ and $20\%$.

$\Rightarrow$  $[10+20-\dfrac{20\times 10}{100}]\%=28\%$
$\Rightarrow$   Single equivalent discount for $28\%$  and $20\%$

$\Rightarrow$   $[28+20-\dfrac{28\times 20}{100}]\%=42.4\%$

After getting three equal successive discounts percentages over a marked price of Rs. $1000$ a customer has to pay $729$ for an article. What is the rate of each of the successive discounts ?

maths money math different discounts successive discounts problems on discounts

  1. $10\%$

  2. $20\%$

  3. $30\%$

  4. $40\%$

Show answer
Correct Option: A
Explanation:

Let $x$ is the factor by which successive discount was given.

Thus $1000\times x\times x\times x=729$
$\Rightarrow$ $x^3=\dfrac{729}{1000}$
$\Rightarrow$ $x=\dfrac{9}{10}$
$\Rightarrow$ $x=0.9\approx 10\%$

Pepsi and Coke, there are two companies , selling the packs of cold-drinks. For the same selling price Pepsi gives two successive discounts of $10$ % and $25$ %. While coke sells it by giving two successive discounts of $15$ % and $20$ %. what is the ratio of their marked price?

maths money math different discounts successive discounts problems on discounts

  1. $143 : 144$

  2. $43 : 44$

  3. $135 : 134$

  4. $136 : 135$

Show answer
Correct Option: D
Explanation:

$SP=MP-$discount$\times MP$

After first discount,
$(SP)'=(1-discount _1)\times MP$

After second discount,
$SP=(1-discount _1)(1-discount _2)\times MP$

For pepsi
$(SP) _1=(1-0.1)(1-0.25)\times (MP) _1$

For coke
$(SP) _2=(1-0.15)(1-0.2)\times (MP) _2$

We know that $(SP) _1=(SP) _2$
$\cfrac{(MP) _1}{(MP) _2}=\cfrac{0.85\times0.8}{0.9\times 0.75}=136/135=136:135$

$50$% discount + $20$% discount = ____%discount

maths money math different discounts successive discounts problems on discounts

  1. $60$

  2. $65$

  3. $40$

  4. $70$

Show answer
Correct Option: A
Explanation:

$\Rightarrow$  Let $x$ be the first discount $50\%$ and $y$ be the second discount $20\%$.


$\Rightarrow$   Total Discount = $(x+y-\dfrac{xy}{100})\%$

$\Rightarrow$   Total discount = $(50+20-\dfrac{50\times 20}{100})\%$

$\Rightarrow$   Total discount = $(70-\dfrac{1000}{100})\%=60\%$

$\therefore$    $50\%$ discount + $20\%$ discount = $60\%$ discount.

Find a single discount equivalent to following successive discounts of $20\%$, $10\%$ and $50\%$ in percent.

maths money math different discounts successive discounts problems on discounts

  1. $54\%$

  2. $64\%$

  3. $74\%$

  4. $84\%$

Show answer
Correct Option: B
Explanation:

$\Rightarrow$  Single discount equivalent for successive discounts of $20\%$ and $10\%$.

$\Rightarrow$   $(20+10-\dfrac{20\times 10}{100})\%=28\%$

$\Rightarrow$  Single discount equivalent for successive discounts of $28\%$ and $50\%$.
$\Rightarrow$   $(28+50-\dfrac{28\times 50}{100})\%=64\%$

$50$% discount + $50$% discount = ____%discount

maths money math different discounts successive discounts problems on discounts

  1. $75$

  2. $50$

  3. $100$

  4. $60$

Show answer
Correct Option: A
Explanation:

$\Rightarrow$  Let $x$ be the first discount $50\%$ and $y$ be the second discount $50\%$.


$\Rightarrow$   Total Discount = $(x+y-\dfrac{xy}{100})\%$

$\Rightarrow$   Total discount = $(50+50-\dfrac{50\times 50}{100})\%$

$\Rightarrow$   Total discount = $(100-\dfrac{2500}{100})\%=75\%$

$\therefore$    $50\%$ discount + $50\%$ discount = $75\%$ discount.

After getting three equal successive discounts percentages over a marked price of Rs. $1000$ a customer has to pay $512$ for an article. What is the rate of each of the successive discounts ?

maths money math different discounts successive discounts problems on discounts

  1. $5$%

  2. $10$%

  3. $15$%

  4. $20$%

Show answer
Correct Option: D
Explanation:

Let the discount price be$x$.

After the first discount, price$=(1-x)1000$
After the second discount, price$=(1-x)(1-x)1000=(1-x)^21000$
After the third discount, price$=(1-x)(1-x)(1-x)1000=(1-x)^3 1000$
According to the question 
$(1-x)^3 1000=512\(1-x)^3=512/1000\(1-x)=8/10\x=1/5$
Required percentage is $(1/5)\times 100=20\%$.

$50$% discount + $40$% discount = ____%discount

maths money math different discounts successive discounts problems on discounts

  1. $60$

  2. $70$

  3. $80$

  4. $90$

Show answer
Correct Option: B
Explanation:

$\Rightarrow$  Let $x$ be the first discount $50\%$ and $y$ be the second discount $40\%$.


$\Rightarrow$   Total Discount = $(x+y-\dfrac{xy}{100})\%$

$\Rightarrow$   Total discount = $(50+40-\dfrac{50\times 40}{100})\%$

$\Rightarrow$   Total discount = $(90-\dfrac{2000}{100})\%=70\%$

$\therefore$    $50\%$ discount + $40\%$ discount = $70\%$ discount.

Find a single discount equivalent to following successive discounts of $50\%$, $10\%$ and $20\%$ in percent.

maths money math different discounts successive discounts problems on discounts

  1. $54\%$

  2. $64\%$

  3. $74\%$

  4. $84\%$

Show answer
Correct Option: B
Explanation:

$\Rightarrow$  Single discount equivalent for successive discounts of $50\%$ and $10\%$.

$\Rightarrow$   $(50+10-\dfrac{50\times 10}{100})\%=55\%$

$\Rightarrow$  Single discount equivalent for successive discounts of $55\%$ and $20\%$.
$\Rightarrow$   $(55+20-\dfrac{55\times 20}{100})\%=64\%$

If the difference between a discount of 25% and two successive discounts of 15% and 10% is ' 63, then the marked price of the article is

maths money math different discounts successive discounts problems on discounts

  1. Rs. $4200$

  2. Rs. $6400$

  3. Rs. $2100$

  4. Rs. $3200$

Show answer
Correct Option: A
Explanation:

Let the marked price be $M$


After $25\%$  discount price will be $0.75M$

And the price after 2 successive discounts of $15\%$ and $10\%=0.85M\times 0.9=0.765M$.

Difference $\Rightarrow 0.765 M-0.75M=0.015M=63\implies M=4200$