Robert Bosch Ratio & Proportion Quiz 2

Answer: d.
Explanation:
Total money = 20 + 40 = 60
In this question the total money is given
so we need to multiply the value of the coins in the ratio.
10 × 0.10 x + 8 × 0.25x + 9 × 0.50x = 60
x + 2x + 4.5x = 60
7.5 x = 60
x = 8
Number of 50 paise coins = 9 × 8 = 72

Answer: a.
Explanation:
Let the number of seats for Mathematics, Physics and Biology be 5x, 7x and 8x respectively.
Number of increased seats are (140% of 5x), (150% of 7x) and (175% of 8x).
(\frac{140}{100}\times5x),(\frac{150}{100}\times7x)and(\frac{175}{100}\times8x)
7x,(\frac{21}{2}x),14x
14x : 21x : 28x
2 : 3 : 4.

Answer:
Explanation:
Let the number of boys and girls in the college be 7x and 8x respectively.
Their increased number is (120% of 7x) and (110% of 8x).
(\frac{120}{100}\times7x) and (\frac{110}{100}\times8x)
(\frac{42}{5}x) and (\frac{44}{5}x)= 21:22
Required Ratio = 21 : 22

Answer: d.
Explanation:
Let Aurther = 2k, Billy = 3k and Chris= 5k.
Aurther's new salary = 115 / 100 of 2k,
(\frac{115}{100}\times 2k)=\frac{23k}{10}
Billy's new salary = 110 / 100 of 3k,
(\frac{110}{100}\times 3k)=\frac{33k}{10}
Chris's new salary = 120 / 100 of 5k,
(\frac{120}{100}\times 5k)=6k
NewRatio=(\frac{23k}{10}:\frac{33k}{10}:6k) = 23:33:60

Answer: a.
Explanation:
Let the number of seats forEngineering, Arts and Architecture be 5x, 7x and 8x respectively.
Number of increased seats are (140% of 5x), (150% of 7x) and (175% of 8x).
(\frac{140}{100}\times5x),(\frac{150}{100}\times7x)and(\frac{175}{100}\times8x)
7x,(\frac{21}{2}x),14x
= 14x : 21x : 28x
= 2 : 3 : 4.

Answer:
Explanation:
Let the first number be x and the second number be y.
According to the question,
60% of x = 40% of y,
=(\frac{x}{y}) = \frac{40}{60}
Ratio = x : y = 2 : 3.

Answer: c.
Explanation:
In 400 gm alloy,
Titanium = (⅝) x 400 = 250 gm, Aluminium = 400 – Titanium = 150 gm
If x gm of Aluminium be mixed, then
=(\frac{250}{150+x}) = \frac{5}{4}
750 + 5x = 1000
⇒ 5x = 1000 – 750 = 250
⇒ x = 50 gm.

Answer: b.
Explanation:
Let,
Marks obtained by Galahad theory and total marks of theory be ‘7M’ and ‘10M’ respectively

So, total marks (theory + practical) = 10M + 2M = 12M
Marks obtained by Galahad (theory + practical) = 7M + 2M = 9M
Required Ratio = \frac{9M}{12M}=\frac{3}{4}

Answer:
Explanation:
Let the First-year income of Sansa = 2x
And, Second-year income of Sansa = 3x
Now, Acc. To Question:
the 2nd year income = 45000
So, x = 45000 / 3 = 15000
Therefore, 1st year income = 2 x 15000 = 30000
Similarly,
Let the 1st Year expenditure = 5y
And Acc. to Question :
1st Year expenditure =25000
So, y = 25000/5 = 5000
Second-year expenditure = 9y= 9* 5000 = 45000
Apply formula:
Income = expenditure + savings, savings = income - expenditure
Therefore, 1st year savings = 30000- 25000 = 5000
Similarly,
2nd year savings = 45000- 45000 = 0
Now, total saving in two years = first-year savings + second-year savings,
Total saving = 5000 + 0 = 5000.

Answer: b.
Explanation:
Let the total number of kids be 21x and 26x
=\frac{(21x+8)}{26x+8}=\frac{5}{6}
6(21x + 8) = 5(26x + 8)
126x + 48 = 130x + 40
x = 2
So, kids will be 42 and 52.
If 6 is subtracted, then kids will be 36 and 46.
Required ratio = 36 : 46. i.e. 18 : 23

So, numbers will be 42 and 52.

If 6 is subtracted, then numbers will be 36 and 46.

Required ratio = 36 : 46. i.e. 18 : 23