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## Arithmetic: Ratio and Proportion Test-2

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Question 1 |

The sum of the squares of three numbers is 532 and the ratio of the first to the second and also ratio of the second to the third number is 3 : 2. What is the second number?

15 | |

14 | |

12 | |

17 |

Question 1 Explanation:

From the given conditions we have

a:b=3:2

b:c=3:2

so therefore a:b:c = 9:6:4

let the numbers are

a= 9p

b=6p

c=4p

Therefore = (9P)

81P

133p

p = 2

Therefore b = 6 x 2 = 12

a:b=3:2

b:c=3:2

so therefore a:b:c = 9:6:4

let the numbers are

a= 9p

b=6p

c=4p

Therefore = (9P)

^{2 }+ (6P)^{2}+ (4P)^{2}= 53281P

^{2 }+ 36P^{2}+ 16P^{2}= 532133p

^{2}= 532p = 2

Therefore b = 6 x 2 = 12

Question 2 |

The prime cost of an article is three times the value of the raw material used. The cost of raw materials increases in the ratio of 5 : 12 and manufacturing expenses in the ratio 4 : 5. The article, which originally cost Rs. 6, will new cost?

Rs. 10 | |

Rs. 17 | |

Rs. 20.50 | |

None of these |

Question 2 Explanation:

We are given by the statement that the prime cost of an article is three times the value of the raw material used.

Also we are given by the statement that the original cost of the article = Rs. 6

Therefore the cost of the original raw material = Rs. 2

Now cost is increased in the ratio 5 :12

So the cost will be = 2 x 12/5 = Rs. 4.80

Original manufacturing expenses = 6 – 2 = 4

Also given that the original manufacturing expense increases in the ratio 4:5 = 4 x 5/4 = Rs. 5

Therefore finally the new cost of the article is = 4.80 + 5 = Rs. 9.80

Also we are given by the statement that the original cost of the article = Rs. 6

Therefore the cost of the original raw material = Rs. 2

Now cost is increased in the ratio 5 :12

So the cost will be = 2 x 12/5 = Rs. 4.80

Original manufacturing expenses = 6 – 2 = 4

Also given that the original manufacturing expense increases in the ratio 4:5 = 4 x 5/4 = Rs. 5

Therefore finally the new cost of the article is = 4.80 + 5 = Rs. 9.80

Question 3 |

The sum of the reciprocal of the ages of two brothers is five times the difference of the reciprocal of their ages. If the ratio of the product of their ages to the sum of their ages is 14.4 : 1, find their ages.

36 and 24 yr | |

24 and 20 yr | |

18 and 15 yr | |

12 and 9 yr |

Question 3 Explanation:

Solution : Let the ages of two brothers a,b

From the condition given we can write it as

1/a + 1/b = 5 (1/a – 1/b)

b + a = 5 (b – a)

6a = 4b

a/b = 2/3 ……………………………………..1

also

ab/(a+b) = 14.4/1

ab = 14.4 (a+b)……………………............2

from both the equations

a = 24

b = 36

From the condition given we can write it as

1/a + 1/b = 5 (1/a – 1/b)

b + a = 5 (b – a)

6a = 4b

a/b = 2/3 ……………………………………..1

also

ab/(a+b) = 14.4/1

ab = 14.4 (a+b)……………………............2

from both the equations

a = 24

b = 36

Question 4 |

If 1 is added to the age of the elder sister, then the ratio of the ages of two sisters becomes 0.5 : 1, but if 2 is subtracted from the age of the younger one, the ratio becomes 1 : 3. Find the age of the two sisters.

8 and 5 yr | |

11 and 6 yr | |

9 and 5 yr | |

8 and 6 yr |

Question 4 Explanation:

Let the ages of two sisters a, b

a/(b+1) = 0.5/1

2a = b +1……………………………………1

(a – 2)/b = 1/3

3a – 6 = b …………………………………….2

From both the equations

a = 5

and b = 9

so the ages of both the sisters

a = 5 and b = 9 yrs

a/(b+1) = 0.5/1

2a = b +1……………………………………1

(a – 2)/b = 1/3

3a – 6 = b …………………………………….2

From both the equations

a = 5

and b = 9

so the ages of both the sisters

a = 5 and b = 9 yrs

Question 5 |

A vessel contains two liquids P and G in the ratio 5:3. If 16L of the mixture is removed and the same quantity of liquid G is added , the a ratio becomes 3:5 . What quantity does the vessel hold?

35 | |

45 | |

40 | |

50 |

Question 5 Explanation:

Let the quantity of two vessels is 8a

So the according to the ratio the quantity becomes = 5a and 3a L

Now initially the quantity P removed = 5/(5+3) x 16 = 10 L

And the quantity of G which is removed = 3/(5+3) x 16 = 6 L Now (5a – 10)/3a – 6 + 16 = 3/5

25a - 50 = 9a + 30

a = 5

Therefore the quantity that the vessel hold = 8 x 5 = 40 L

So the according to the ratio the quantity becomes = 5a and 3a L

Now initially the quantity P removed = 5/(5+3) x 16 = 10 L

And the quantity of G which is removed = 3/(5+3) x 16 = 6 L Now (5a – 10)/3a – 6 + 16 = 3/5

25a - 50 = 9a + 30

a = 5

Therefore the quantity that the vessel hold = 8 x 5 = 40 L

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