In a mixture \(60 litres,\) the ratio of milk and water\( 2 : 1.\) If this ratio is to be \(1 : 2,\) then the quanity of water to be further added is:
Answer: A
Quantity of milk \(= (60 \times \frac{2}{3}) littres = 40 littres.\)
Quantity of water in it \(= (60 - 40) littres = 20 littres.\)
New ratio \(= 1:2\)
Let quantity of water to be added further be \(x \) \(littres.\)
Then, milk : water \(= (\frac{40}{20 + x})\)
Now, \((\frac{40}{20 + x}) = \frac{1}{2}\)
\(\Rightarrow 20 + x = 80\)
\(\Rightarrow x = 60.\)
\(\therefore\) Quantity of water to be added \(= 60\) \(littres.\)
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In a bag, there are coins of \(25 p, 10 p\) and \(5 p\) in the ratio of \(1 : 2 : 3.\) If there is \(Rs. 30\) in all, how many \(5 p\) coins are there?
Answer: C
Let the number of \(25 p, 10 p\) and \(5 p\) coins be \(x, 2x, 3x \) respectively.
Then, sum of there value \(= Rs. (\frac{25x}{100} + \frac{10 \times 2x}{100} + \frac{5 \times 3x}{100}) = Rs. \frac{60x}{100}\)
\(\therefore \frac{60x}{100} = 30 \Leftrightarrow x = \frac{300 \times 100}{60} = 50.\)
Heance, the number of \(5 p\) coins \(= (3 \times 50) = 150.\)
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If \(40\)% of a number is equal to two-third of another number, what is the ratio of first number to the second number?
Answer: C
Let \(40\)% of \(A = \frac{2}{3} B\)
Then, \(\frac{40A}{100} = \frac{2B}{3}\)
\(\Rightarrow \frac{2A}{5} = \frac{2B}{3}\)
\(\Rightarrow \frac{A}{B} = (\frac{2}{3} \times \frac{5}{2}) = \frac{5}{3}\)
\(\therefore A : B = 5 : 3.\)
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If \(0.75 : x :: 5 : 8,\) then \(x\) is equal to:
Answer: C
\((x \times 5) = (0.75 \times 8 )\)
\(\Rightarrow x = (\frac{6}{5}) = 1.20\)
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\(A\) and \(B\) together have \(RS. 1210.\) If \(\frac{4}{15}\) of \(A,s\) amount is equal to \(\frac{2}{5}\) of \(B's\) amount, how much amount does \(B\) have?
Answer: C
\(\frac{4}{12}A = \frac{2}{5}B\)
\(\Rightarrow A = (\frac{2}{5} \times \frac{15}{4})B\)
\(\Rightarrow A = \frac{3}{2} B\)
\(\Rightarrow \frac{A}{B} = \frac{3}{2}\)
\(\Rightarrow A : B = 3 : 2.\)
\(\therefore B's\) share \(= RS. (1210 \times \frac{2}{5}) = Rs. 484\)
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Two number are in the ratio \(3 : 5.\) If \(9\) is subtracted from each, the new numbers are in the ratio \(12 : 23.\) The smaller number is:
Answer: B
Let the numbers be \(3x\) and \(5x.\)
Then, \(\frac{3x - 9}{5x - 9} = \frac{12}{23}\)
\(\Rightarrow 23(3x - 9) = 12(5x - 9)\)
\(\Rightarrow 9x = 99\)
\(\Rightarrow x = 11.\)
\(\therefore\) The smaller number \(= (3 \times 11) = 33.\)
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The fourth proportional to \(5, 8, 15\) is:
Answer: D
Let the fourth proportional to \(5, 8 , 15\) be \(x.\)
Then, \(5 : 8 : 15 : x\)
\(\Rightarrow 5x = (8 \times 15)\)
\(x = \frac{(8 \times 15)}{5} = 24.\)
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Two numbers are respectively \(20\)% and \(50\)% more than a \(3rd\) number. The ratio of the two numbers is:
Answer: B
Let the \(3rd\) number be \(x.\)
Then, first number \(= 120\)% of \(x = \frac{120x}{100} = \frac{6x}{5}\)
Second number \(= 150 \)% of \(x = \frac{150x}{100} = \frac{3x}{2}\)
\(\therefore\) Ratio of first two numbers \(= (\frac{6x}{5} : \frac{3x}{2}) = 12x : 15x = 4:5\)
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The ratio of the number of boys and girls in a college is \(7 : 8.\) If the percentage increase in the number of boys and girls be \(20\)% and \(10\)% respectively, what will be the new ratio?
Answer: A
Let the number of boyes and girls in the collage be \(7x\) and \(8x\) respectively. Their increased number is \((120\)% of \(7x)\) and \((110\)% of \(8x)\).
\(\Rightarrow (\frac{120}{100} \times 7x)\) and \((\frac{110}{100} \times 8x)\)
\(\Rightarrow \frac{42x}{5} \) and \(\frac{44x}{5}\)
\(\therefore\) The required ratio \(= (\frac{42x}{5} : \frac{44x}{5}) = 21 :22.\)
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Seats for Mathematics, Physics and Biology in a school are in the ratio \( 5 : 7 : 8.\) There is a proposal to increase these seats by \(40\)%, \(50\)% and \(75\)% respectively. What will be the ratio of increased seats?
Answer: C
Let the number of seats for Mathematics, Physics and Biology be \(5x, 7x\) and \(8x\)
Number of increased seats are \((140\)% of \(5x),\) \((150\)% of \(7x)\) and \((175\)% of \(8x).\)
\(\Rightarrow (\frac{140}{100} \times 5x), (\frac{150}{100} \times 7x)\) and \((\frac{175}{100} \times 8x)\)
\(\Rightarrow 7x, \frac{21x}{2}\) and \(14x.\)
\(\therefore\) The required ratio \(= 7x : \frac{21x}{2} : 14x\)
\(\Rightarrow 14x : 21x : 28x\)
\(\Rightarrow 2 : 3 : 4.\)
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