A shopkeper placed on display some shirts each with a marked price. He then posted a board $\frac{1}{4}$ off on shirts". If the cost of a shirt was $\frac{2}{3}$ of the price at which it was actually sold, the ratio of cost and marked price 01' shirt was

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INSURANCE EXAMS (UIIC) 2012 click here to view all questions

A shopkeper placed on display some shirts each with a marked price. He then posted a board $\frac{1}{4}$ off on shirts". If the cost of a shirt was $\frac{2}{3}$ of the price at which it was actually sold, the ratio of cost and marked price 01' shirt was

A1 : 2

B1 : 3

C2 : 3

D3 : 4

Nithin ? undefined

Correct answer is: 1 : 2

Explanation:

Let. S.P. of shirt = x

\(\text{C.P = }\frac{2x}{3}\)

Therefore, Marked price

\(= \frac{4}{3} \times x = \frac{4x}{3}\)

Therefore, Required ratio = \(\frac{2x}{3} : \frac{4x}{3} = 1 : 2\)

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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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INSURANCE EXAMS (OIC) 2012 click here to view all questions

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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Abhijit invested in three schemes A, B and C the amounts in the ratio of 2 : 3 : 4 respectively. If the schemes offered interest @ 20 p.c.p.a. 16 p.c.p.a. and 15 p. c.p. a. respectively, what will be the respective ratio of the amounts after one year ?

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INSURANCE EXAMS (LIC) 2008 click here to view all questions

Abhijit invested in three schemes A, B and C the amounts in the ratio of 2 : 3 : 4 respectively. If the schemes offered interest @ 20 p.c.p.a. 16 p.c.p.a. and 15 p. c.p. a. respectively, what will be the respective ratio of the amounts after one year ?

A10 : 8 : 5

B12 : 14 : 15

C12 : 15 : 22

DNone of these

Nithin ? undefined

Correct answer is: None of these

Explanation:

Let the amount invested in schemes A, B aand C in Rs. respectively be 2x, 3x and 4x

Amount after 1 year in scheme :

\(A = \frac{120}{100} \times 2x = 2.4 x\)

\(B = \frac{116}{100} \times 3x = 3.48x\)

\(C = \frac{115}{100}\times 4x =4.6x\)

\(\implies \text{Required ratio =} 2.4: 3.48:4.6\)

\(= 60:87:115\)

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The angles of a quadrilateral are in ratio of 3 : 5 : 9 : 7. The second largest angle of the quadrilateral is equal to the largest angle of a triangle. One of the angles of the triangle is $25^∘$. What is the value-of second largest angle of the triangle?

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INSURANCE EXAMS (UIIC) 2011 click here to view all questions

The angles of a quadrilateral are in ratio of 3 : 5 : 9 : 7. The second largest angle of the quadrilateral is equal to the largest angle of a triangle. One of the angles of the triangle is $25^∘$. What is the value-of second largest angle of the triangle?

A$60^∘$

B$50^∘$

C$40^∘$

D$20^∘$

Nithin ? undefined

Correct answer is: \(50^o\)

Explanation:

Sum of ratios = 3 + 5 + 9 + 7 = 24

Second largest angle of quadrilateral = \(\frac{7}{24} \times 360 = 105 ^{\circ}\)

Largest angle of triangle
Therefore, Third angle of triangle = \(\frac{7}{24} \times 360 = 105 ^{\circ}\)

\(=180 ^{\circ} - 105 ^{\circ} - 25 ^{\circ}\)

\(= 50^∘ = \text{second largest angle of triangle.}\)

 

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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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INSURANCE EXAMS (New India Assurance) 2009 click here to view all questions

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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The sides of a triangle are in the ratio $\frac{1}{2}: \frac{1}{3}: \frac{1}{4} and its perimeter is 104 cm. The length of the longest side (in cm) is -

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INSURANCE EXAMS (LIC) 2009 click here to view all questions

The sides of a triangle are in the ratio $\frac{1}{2}: \frac{1}{3}: \frac{1}{4}$ and its perimeter is 104 cm. The length of the longest side (in cm) is -

A26

B32

C48

D52

Nithin ? undefined

Correct answer is: 48

Explanation:

\(\eqalign{ & {\text{Ratio of sides}} \cr & = \frac{{\text{1}}}{{\text{2}}}:\frac{{\text{1}}}{{\text{3}}}:\frac{{\text{1}}}{{\text{4}}} \cr & = 6:4:3 \cr} \)

Let the sides be 6x, 4x and 3x
Then,
⇒ 6x + 4x + 3x = 104
or 13x = 104
or x = 8
∴ Longest side = 6x = (6 × 8)m = 48m

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The sum of three numbers is 136. If the ratio between first and second be 2 : 3 and that between second and third be 5 : 3, then the second number is:

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INSURANCE EXAMS (LIC) 2009 click here to view all questions

The sum of three numbers is 136. If the ratio between first and second be 2 : 3 and that between second and third be 5 : 3, then the second number is:

A40

B48

C52

D60

Nithin ? undefined

Correct answer is: 60
Explanation:

Let the three numbers be a, b and c respectively.
Now.
a : b = 2 : 3
b : c = 5 : 3

Therefore, a : b : c =  \(\Large 2 \times 5:3 \times 5:3 \times 3\)

= 10 : 15 : 9
According to the question,
a + b + c = 136
=> 10x + 15x + 9x = 136

\(\implies 34x = 136 \implies x = \frac{136}{34}=4\)
\(\therefore b = 15x = 15×4 = 60\)

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