A and B are two alloys of gold and copper prepared by mixing metals in the proportions 7 : 2 and 7 : 11 respectively. If equal quantities of the alloys are melted to form a third alloy C, proportion of gold and copper in C will be

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A and B are two alloys of gold and copper prepared by mixing metals in the proportions 7 : 2 and 7 : 11 respectively. If equal quantities of the alloys are melted to form a third alloy C, proportion of gold and copper in C will be

A5 : 7

B5 : 9

C9 : 5

D7 : 5

Nithin ? undefined

Correct answer is: 7 : 5

Explanation:

Gold in C = \(\left( \frac{7}{9} + \frac{7}{18} \right)\)

\(= \frac{21}{18} = \frac{7}{6}\)

Copper in C =  \( \left( \frac{2}{9} + \frac{11}{18} \right)\)

\(= \frac{15}{18} = \frac{5}{6}\)

Gold : Copper = \(\frac{7}{6} : \frac{5}{6}\)

= 7 : 5

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To x litres of an x% solution of acid, y litres of water is added to get (x -10)% solution of acid. If x > 20, then value of y is

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To x litres of an x% solution of acid, y litres of water is added to get (x -10)% solution of acid. If x > 20, then value of y is

A$\frac{x^2}{100}$

B$\frac{10x}{x-10}$

C$\frac{10x}{x+10}$

D$\frac{10x^2}{x-10}$

Nithin ? undefined

Correct answer: \(\frac{10x}{x-10}\)

Explanation:

In litres of solution,

Acid = \(\frac{x^2}{100}\)

Water = \(x- \frac{x^2}{100}\)

\(\therefore \frac{\frac{x^2}{100}}{x+y} \times 100 = x-10\)

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

\(\implies x^2 =y^2-10x-y(x-10)\)

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

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The price of a diamond is proportional to the square of its mass, which is measured in carats. A 6 carat diamond was broken into two parts and total price of the two pieces is $\frac{5}{8}$ th of the price of original diamond. The masses (in carat) of two pieces are

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The price of a diamond is proportional to the square of its mass, which is measured in carats. A 6 carat diamond was broken into two parts and total price of the two pieces is $\frac{5}{8}$ th of the price of original diamond. The masses (in carat) of two pieces are

A3.5 and 2.5

B5 and 1

C4.5 and 1.5

D3.2 and 2.8

Nithin ? undefined

Correct answer is: 4.5 and 1.5

Explanation:

Actual price of diamond = \( k \times 6^{2}\)

= 36k where k is constant of proportionately.
New price of diamond

\(=kx^{2}_{1} + kx^{2}_{2}\)

\(= 36k \times \frac{5}{8} = k \left(x^{2}_{1} + x^{2}_{2}\right)\)

\(= \frac{45}{2} = x^{2}_{1}+x^{2}_{2}\)

\(= 22.5 = x^{2}_{1}+x^{2}_{2}\)

\(= \left(4.5\right)^{2}+ \left(1.5\right)^{2} = x^{2}_{1}+x^{2}_{2}\)

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The mean of n observations is x1. If the first observation is increased by 1, second by 2, and so on, then their mean is x2 . The value of x2 - x1 is

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The mean of n observations is x1. If the first observation is increased by 1, second by 2, and so on, then their mean is x2 . The value of x2 - x1 is

An

B$\frac{n}{2}+1$

C$\frac{n \times (n+1)}{2}$

D$\frac{(n+1)}{2}$

Akilesh Kharvi ? Jun 28 '2020 at 21:45

Correct answer is: \(\frac{(n+1)}{2}\)

Explanation:

 

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A person leaves a place A to place B at 6 a.m. and reaches place B at 10 a.m. Another person leaves B at 7.30 a.m. and reaches A at 11 a.m. They will meet each other at

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A person leaves a place A to place B at 6 a.m. and reaches place B at 10 a.m. Another person leaves B at 7.30 a.m. and reaches A at 11 a.m. They will meet each other at

A7.40 a.m.

B8.20 a.m.

C8.40 a.m.

D9.20 a.m.

akilesh kharvi ? Jun 20 '2020 at 12:7

Correct answer is: 8.40 a.m.

Explanation:

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