There are two pipes which are functioning simultaneouly to fill a tank in 12 hours, if one pipe fills the tank 10 hours faster than other then how many hours second pipe will take to fill the tank ?
Answer: A
Lets suppose tank got filled by first pipe in X hours,
So, second pipe will fill the tank in X + 10 hours.
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If one is added to the numerator of the fraction it becomes one. If one is added to the denominator of the fraction it becomes 1/2. The fraction is?
Answer: C
Let the fraction be x/y.
=> (x+1)/y = 1
=> x+1 = y-(1) x/(y-1) = 1/2
=> 2x = (y+1)-(2)
=>Equating (1) and (2)
=> x+1 = 2x-1
=>x=2
Substituting the value of x in (1)
=>y=3
=>2/3
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If x + 5y = 16 and x= -3y, then y =
Answer: C
Substitute \( x = -3y \) into the first equation.
\[ -3y + 5y = 16 \quad\Longrightarrow\quad 2y = 16 \]
\[ y = 8 \]
Checking: \( x = -24 \), and \( -24 + 40 = 16 \). The \( -24 \) on offer is the value of \( x \), not \( y \).
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Which of the following fraction is greater than 2/3 and less than 4/5 ?
Answer: C
Convert everything to decimals and compare.
\[ \frac{2}{3} = 0.667, \qquad \frac{4}{5} = 0.800 \]
So \( \tfrac{3}{4} \) is the only one strictly between the two bounds.
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Which of the following fraction is the smallest?
Answer: B
Convert each to a decimal.
\[ \frac{12}{14} = 0.857, \qquad \frac{13}{19} = 0.684 \]
\[ \frac{17}{21} = 0.810, \qquad \frac{7}{8} = 0.875 \]
The smallest is \( \tfrac{13}{19} \). It is also the only one whose numerator is well below three quarters of its denominator, which is a quick way to spot it.
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The product of two fractions is 3/4 and their quotient is 6/18. If the denominator of one is 1/3 of the other's numerator, then find the pair of fractions?
Answer: B
Let A = x/y, and B be two fractions
It is given that A×B = 3/4
Therefore x/y × B = 3/4
B = 3y/4x
Therefore, A = x/y and B = 3y/4x
It is given that A/B = (x/y) / (3y/4x) = 6/18
Therefore x/y × 4x/3y = 6 /18
Therefore 4x2/3y2 = 6/18
x2/y2 = 18/(4×18)
Therefore, x2/y2 = 1/4
x/y = 1/2
B = 3y/4x = 3/4 (1/(x/y)) = 3/4 [1/(1/2)] = 6/4
Therefore A = 1/2 and B = 3/2
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4/7 of 2/3 of 5/6 of 5/8 of 1008 is
Answer: A
Work from the right, cancelling as you go.
\[ 1008 \times \frac{5}{8} = 630 \]
\[ 630 \times \frac{5}{6} = 525 \]
\[ 525 \times \frac{2}{3} = 350 \]
\[ 350 \times \frac{4}{7} = 200 \]
Multiplying all four fractions first gives \( \tfrac{200}{1008} \), and \( 1008 \times \tfrac{200}{1008} = 200 \) - the same answer.
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A class eats 2/5 of chocolates on 1st day. On the 2nd day they eat 3/4 of the remainder. How many chocolates were there initially if still 75 chocolates are left?
Answer: A
Let there be 100 chocolates
215th of 100 40 chocolates eaten on first day
314th of the remaining 314th of (100-40) 45 chocolates
Chocolates left 100-40-45 15 chocolates
15 chocolates are left when 100 are there
75 chocolates are left when ? chocolates are there? = 500 Chocolates
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If a fraction’s denominator is decreased by 80% and numerator is increased by 300%, the fraction becomes 2/9. What is the fraction?
Answer: C
Let the fraction be \( \tfrac{n}{d} \). Decreasing the denominator by 80% leaves \( 0.2d \); increasing the numerator by 300% makes it \( 4n \).
\[ \frac{4n}{0.2d} = \frac{20n}{d} = \frac{2}{9} \]
\[ \frac{n}{d} = \frac{2}{9 \times 20} = \frac{2}{180} = \frac{1}{90} \]
An increase of 300% makes a quantity four times as large, not three - the step that decides the answer.
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Three pipes A, B and C can fill a tank from empty to full in 30 minutes, 20 minutes, and 10 minutes respectively. When the tank is empty, all the three pipes are opened. A, B and C discharge chemical solutions P,Q and R respectively. What is the proportion of the solution R in the liquid in the tank after 3 minutes?
Answer: B
Work out how much each pipe contributes in 3 minutes.
\[ P = \frac{3}{30} = \frac{1}{10}, \qquad Q = \frac{3}{20}, \qquad R = \frac{3}{10} \]
Over the common denominator 20:
\[ \frac{2}{20} + \frac{3}{20} + \frac{6}{20} = \frac{11}{20} \text{ total} \]
\[ \text{proportion of R} = \frac{6/20}{11/20} = \frac{6}{11} \]
The tank is only \( \tfrac{11}{20} \) full at that point, so the proportions are taken against that total rather than against a full tank.
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