A shopkeeper marks his goods at such a price that after allowing a discount of 12.5 % on the marked price, he still earns a profit of 10%. The marked price of an article which costs him Rs. 4900 is -

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A shopkeeper marks his goods at such a price that after allowing a discount of 12.5 % on the marked price, he still earns a profit of 10%. The marked price of an article which costs him Rs. 4900 is -

ARs. 5390

BRs. 5490

CRs. 6160

DRs. 6260

Nithin ? undefined

Correct answer is: Rs. 6160

Explanation:

\(\eqalign{ & {\text{C}}{\text{.P}}{\text{. = Rs}}{\text{. 4900}}{\text{.}} \cr & {\text{S}}{\text{.P}}{\text{. = 110}}\% {\text{ of Rs}}{\text{.4900}} \cr & {\text{ = Rs}}{\text{.}}\left( {\frac{{110}}{{100}} \times 4900} \right) \cr & = {\text{Rs}}{\text{. 5390}}{\text{.}} \cr & {\text{Let marked price be Rs}}{\text{.}}x \cr & {\text{Then,}} \cr & {\text{ = 87}}\frac{1}{2}\% {\text{ of }}x{\text{ = 5390}} \cr & \Rightarrow \left( {\frac{{175}}{2} \times \frac{1}{{100}} \times x} \right) = 5390 \cr & \Rightarrow x = \left( {\frac{{5390 \times 8}}{7}} \right) = 6160 \cr & \therefore {\text{Marked price}} = {\text{Rs}}{\text{.6160}} \cr}\)

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To m litres of a m% solution of acid. x litres of water is muted to yield (m - 10)% solution of acid. If m > 25. then x equals

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To m litres of a m% solution of acid. x litres of water is muted to yield (m - 10)% solution of acid. If m > 25. then x equals

A$\frac{10m}{m-10}$

B$\frac{5m}{m-10}$

C$\frac{2m}{m-10}$

D$\frac{m}{m-10}$

Nithin ? undefined

Correct answer is: \(\frac{10m}{m-10}\)

Explanation:

Tricky approach

\(\text{ Acid = }\frac{m}{100} \times m = \frac{m^{2}}{100} \)

\(\text{New percentage =} \frac{m^{2} \times 100}{ \left(m+x\right) \times 100 }\)

\(\implies \frac{m^{2}}{m+x}=m-10 \)

\(\implies m^{2} - 10m + mx - 10x = m^{2}\)
\(\implies x \left(m-10\right)=10m\)

\(\implies x = \frac{10m}{m-10}\)

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The cost of a diamond varies directly as the square of its weight. A diamond broke into four pieces with their weights in the ratio of 1:2:3:4. If the loss in total value of the diamond was the ₹ 70,000, what was the price of the original diamond?

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The cost of a diamond varies directly as the square of its weight. A diamond broke into four pieces with their weights in the ratio of 1:2:3:4. If the loss in total value of the diamond was the ₹ 70,000, what was the price of the original diamond?

A₹ 1,00,000

B₹ 1,40,000

C₹ 1,50,000

D₹ 1,75,000

Nithin ? undefined

Correct answer is: ₹ 1,00,000

Explanation:

\(Cost ∝ (weight)^2\)

Total weight of original diamond = 1k + 2k + 3k + 4k = 10k, where k is positive integer

Cost of original diamond = \((10k)^2\) = \(100k^2\)

Total cost of four broken pieces = \(k^2\)(1 + 4 + 9 + 16) = 30\(k^2\)

The value of broken pieces is 70,000 less than the original diamond.

Difference in cost = 70,000

\(100k^2 - 30k^2 = 70,000\)

\(70k^2 = 70,000\)

\(k^2 = 1,000\)

Cost of original diamond = \(100k^2 = 100 × 1,000\)

= 1,00,000

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A watch is sold at a profit of 20%. If both the cost price and the selling price of the watch are decreased by Rs. 100, the profit would be 5% more. Original cost price of the watch is -

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A watch is sold at a profit of 20%. If both the cost price and the selling price of the watch are decreased by Rs. 100, the profit would be 5% more. Original cost price of the watch is -

ARs. 450

BRs. 500

CRs. 550

DRs. 600

Nithin ? Aug 10 '2022 at 20:26

Correct answer is:  Rs. 500

Explanation:

\(\eqalign{ & {\text{Let C}}{\text{.P}}{\text{. be Rs}}{\text{. }}x \cr & {\text{Profit}} = 20\% \cr & {\text{S}}{\text{.P}}{\text{.}} = 120\% {\text{ of Rs}}{\text{. }}x \cr & \,\,\,\,\,\,\,\,\,\,\,\, = {\text{Rs}}{\text{.}}\frac{{6x}}{5} \cr & {\text{New C}}{\text{.P}}{\text{.}} = {\text{Rs}}.\left( {x - 100} \right) \cr & {\text{New S}}{\text{.P}}{\text{.}} = {\text{Rs}}.\left( {\frac{{6x}}{5} - 100} \right) \cr} \)

\(\eqalign{ & {\text{Profit = }} \cr & {\text{Rs}}.\left[ {\left( {\frac{{6x}}{5} - 100} \right) - \left( {x - 100} \right)} \right] \cr & = {\text{Rs}}.\frac{x}{5} \cr & \therefore \frac{x}{5} \times \frac{1}{{\left( {x - 100} \right)}} \times 100 = 25 \cr & \Rightarrow 20x = 25\left( {x - 100} \right) \cr & \Rightarrow 5x = 2500 \cr & \Rightarrow x = 500 \cr} \)

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The ages of Melwyn and Louis are in the ratio of 7 : 10 respectively. After 6 years the ratio of their ages will be 17 : 23. What is the difference in their ages ?

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The ages of Melwyn and Louis are in the ratio of 7 : 10 respectively. After 6 years the ratio of their ages will be 17 : 23. What is the difference in their ages ?

A8 years

B4 years

C12 years

D10 years

Akilesh Kharvi ? Jun 28 '2020 at 21:18

Correct answer is: 12 years

Explanation:

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