Q1
Find the sum of \( 132 \) and \( 121 \pmod{23} \).
Answer: 0
To find the sum modulo 23, we add the numbers and then find the remainder when divided by 23.
\[ (132 + 121) \pmod{23} = 253 \pmod{23} \]Now, we divide 253 by 23:
\[ 253 = 23 \times 11 + 0 \]Since the remainder is 0, we have:
\[ 253 \equiv 0 \pmod{23} \]Q2
Find \( 7^6 \pmod{3} \).
Answer: 1
First, we find the remainder of the base 7 when divided by 3:
\[ 7 = 2 \times 3 + 1 \Rightarrow 7 \equiv 1 \pmod{3} \]Now raise both sides to the power of 6:
\[ 7^6 \equiv 1^6 \pmod{3} \] \[ 7^6 \equiv 1 \pmod{3} \]Q3
Find the value of \( x \), given that \( x \equiv 23 \pmod{7} \); if \( 21 \leq x < 31 \).
Answer: 23 and 30
First, simplify \( 23 \pmod{7} \):
\[ 23 = 3 \times 7 + 2 \Rightarrow 23 \equiv 2 \pmod{7} \]So, the given condition becomes:
\[ x \equiv 2 \pmod{7} \]This means \( x \) leaves a remainder of 2 when divided by 7. We look for integers satisfying this in the range \( 21 \leq x < 31 \).
- \( 21 \equiv 0 \pmod{7} \)
- \( 22 \equiv 1 \pmod{7} \)
- \( \mathbf{23 \equiv 2 \pmod{7}} \) (Matches)
- ...
- \( \mathbf{30 \equiv 2 \pmod{7}} \) (Matches, since \( 30 = 4 \times 7 + 2 \))
Therefore, the possible values for \( x \) are 23 and 30.
Q4
Find the remainder when \( 226 \times 369 \times 122 \times 461 \times 1025 \) is divided by 8.
Answer: 4
We find the remainder of each factor modulo 8 individually:
- \( 226 = 8 \times 28 + 2 \Rightarrow 226 \equiv 2 \pmod{8} \)
- \( 369 = 8 \times 46 + 1 \Rightarrow 369 \equiv 1 \pmod{8} \)
- \( 122 = 8 \times 15 + 2 \Rightarrow 122 \equiv 2 \pmod{8} \)
- \( 461 = 8 \times 57 + 5 \Rightarrow 461 \equiv 5 \pmod{8} \)
- \( 1025 = 8 \times 128 + 1 \Rightarrow 1025 \equiv 1 \pmod{8} \)
Now, multiply the remainders:
\[ \begin{aligned} \text{Product} &\equiv 2 \times 1 \times 2 \times 5 \times 1 \pmod{8} \\ &\equiv 20 \pmod{8} \end{aligned} \]Finally, find \( 20 \pmod{8} \):
\[ 20 = 8 \times 2 + 4 \Rightarrow 20 \equiv 4 \pmod{8} \]Q5
Evaluate \( (16 \times 29) \pmod{7} \).
Answer: 2
First, reduce each number modulo 7:
\[ \begin{aligned} 16 &= 7 \times 2 + 2 \Rightarrow 16 \equiv 2 \pmod{7} \\ 29 &= 7 \times 4 + 1 \Rightarrow 29 \equiv 1 \pmod{7} \end{aligned} \]Now multiply the remainders:
\[ 16 \times 29 \equiv 2 \times 1 \equiv 2 \pmod{7} \]Q6
If \( x \pmod{9} = 2 \), find all the possible values of \( x \); where \( 0 < x < 47 \).
Answer: 2, 11, 20, 29, 38
The condition \( x \pmod{9} = 2 \) means \( x \) can be written in the form \( 9k + 2 \) for some integer \( k \).
We substitute values for \( k \) starting from 0 to find \( x \) within the range \( 0 < x < 47 \):
- \( k = 0 \Rightarrow x = 9(0) + 2 = 2 \)
- \( k = 1 \Rightarrow x = 9(1) + 2 = 11 \)
- \( k = 2 \Rightarrow x = 9(2) + 2 = 20 \)
- \( k = 3 \Rightarrow x = 9(3) + 2 = 29 \)
- \( k = 4 \Rightarrow x = 9(4) + 2 = 38 \)
- \( k = 5 \Rightarrow x = 9(5) + 2 = 47 \) (Not strictly less than 47, so we stop here)
The possible values are 2, 11, 20, 29, and 38.
Q7
Find the positive integers less than 50 forming the equivalence class 4 for modulo 6.
Answer: 4, 10, 16, 22, 28, 34, 40, 46
The equivalence class 4 modulo 6 contains all integers \( x \) such that:
\[ x \equiv 4 \pmod{6} \]This can be written as \( x = 6k + 4 \), where \( k \) is an integer. We find positive integers less than 50:
- \( k = 0 \Rightarrow x = 4 \)
- \( k = 1 \Rightarrow x = 10 \)
- \( k = 2 \Rightarrow x = 16 \)
- \( k = 3 \Rightarrow x = 22 \)
- \( k = 4 \Rightarrow x = 28 \)
- \( k = 5 \Rightarrow x = 34 \)
- \( k = 6 \Rightarrow x = 40 \)
- \( k = 7 \Rightarrow x = 46 \)
- \( k = 8 \Rightarrow x = 52 \) (Greater than 50)
Q8
What time will it be after 200 hours, if the present time is 5:00 am?
Answer: 1:00 pm
Time cycles every 24 hours. We need to find the remainder of 200 when divided by 24.
\[ 200 = 24 \times 8 + 8 \]The remainder is 8 hours. So, the time will be 8 hours after 5:00 am.
\[ 5:00 \text{ am} + 8 \text{ hours} = 13:00 \text{ hours} \]13:00 is 1:00 pm in the 12-hour format.
Q9
There are 81 boxes with 21 articles in each. When we rearrange all of the articles so that each box has 5 articles, how many articles will be left out without a box?
Answer: 1 article
The total number of articles is \( 81 \times 21 \). We need to find the remainder when this product is divided by 5.
\[ \text{Total Articles} = 81 \times 21 \]Using modular arithmetic properties:
\[ \begin{aligned} 81 &\equiv 1 \pmod{5} \\ 21 &\equiv 1 \pmod{5} \end{aligned} \]Therefore:
\[ 81 \times 21 \equiv 1 \times 1 \equiv 1 \pmod{5} \]So, 1 article will be left out.
Q10
Find \( 3^{128} \pmod{7} \).
Answer: 2
We look for a power of 3 that is close to a multiple of 7 (ideally 1 or -1).
- \( 3^1 = 3 \)
- \( 3^2 = 9 \equiv 2 \pmod{7} \)
- \( 3^3 = 27 \equiv 6 \equiv -1 \pmod{7} \)
Since \( 3^3 \equiv -1 \pmod{7} \), squaring both sides gives \( 3^6 \equiv 1 \pmod{7} \).
Now we express the exponent 128 in terms of 6:
\[ 128 = 6 \times 21 + 2 \]Substitute this into the expression:
\[ \begin{aligned} 3^{128} &= 3^{6 \times 21 + 2} \\ &= (3^6)^{21} \times 3^2 \\ &\equiv (1)^{21} \times 3^2 \pmod{7} \\ &\equiv 1 \times 9 \pmod{7} \\ &\equiv 9 \pmod{7} \\ &\equiv 2 \pmod{7} \end{aligned} \]EXERCISE-2 (Deleted from Syllabus)
EXERCISE-3
Q1
In what ratio must rice at ₹ 69 per kg be mixed with rice ₹ 100 per kg so that the mixture be worth ₹ 80 per kg?
Answer: 20 : 11
We use the rule of alligation to solve this problem.
Cost of cheaper rice (\(c\)) = ₹ 69 per kg
Cost of dearer rice (\(d\)) = ₹ 100 per kg
Mean price (\(m\)) = ₹ 80 per kg
According to the rule of alligation:
\[ \frac{\text{Quantity of cheaper}}{\text{Quantity of dearer}} = \frac{d - m}{m - c} \] \[ \text{Ratio} = (100 - 80) : (80 - 69) \] \[ \text{Ratio} = 20 : 11 \]Q2
The average salary per head of the entire staff of a small factory including the supervisor and labours is ₹ 5750. The average salary per head of the supervisor is ₹ 20,000 and that of the labours is ₹ 5000. Find the number of labours in the factory if there are 4 supervisors.
Answer: 76
Let the number of labours be \( x \).
Given:
- Number of supervisors = 4
- Average salary of supervisors = ₹ 20,000
- Average salary of labours = ₹ 5,000
- Average salary of entire staff = ₹ 5,750
- Total number of staff = \( x + 4 \)
We can form the equation based on the total salary:
\[ \text{Total Salary} = (\text{Avg of entire staff}) \times (\text{Total staff}) \] \[ 5750(x + 4) = (20000 \times 4) + (5000 \times x) \]Solving for \( x \):
\[ 5750x + 23000 = 80000 + 5000x \] \[ 5750x - 5000x = 80000 - 23000 \] \[ 750x = 57000 \] \[ x = \frac{57000}{750} = 76 \]Therefore, there are 76 labours in the factory.
Q3
A container contains 70 \( l \) of orange squash. The squash being too concentrated 7 \( l \) of squash was taken out from this container and replaced by water. This process was repeated thrice to reduce the concentration of squash. How much quantity of orange squash is left in the container?
Answer: 45.927 \( l \)
We use the formula for repeated replacement:
\[ \text{Final Quantity} = \text{Initial Quantity} \left(1 - \frac{y}{x}\right)^n \]Where:
- Initial Quantity (\( x \)) = 70 \( l \)
- Quantity taken out (\( y \)) = 7 \( l \)
- Number of operations (\( n \)): The process is performed once, and then repeated "thrice" (3 times). Total operations \( n = 1 + 3 = 4 \).
Substituting the values:
\[ \text{Final Quantity} = 70 \left(1 - \frac{7}{70}\right)^4 \] \[ = 70 \left(1 - 0.1\right)^4 \] \[ = 70 \times (0.9)^4 \] \[ = 70 \times 0.6561 \] \[ = 45.927 \text{ liters} \]Q4
Cost of two types of pulses is ₹ 55 per kg and ₹ 90 per kg. If both the pulses are mixed together in the ratio 2:3, what should be the price of the mixed variety of pulses per kg?
Answer: ₹ 76 per kg
Let the price of the mixture be \( P \).
We can calculate the weighted average price:
\[ P = \frac{(2 \times 55) + (3 \times 90)}{2 + 3} \] \[ P = \frac{110 + 270}{5} \] \[ P = \frac{380}{5} \] \[ P = 76 \]The price of the mixed variety is ₹ 76 per kg.
Q5
A shopkeeper has 1 quintal of wheat, part of which she sells at 18% gain and the rest at 28% gain. In total she gains 24 %. Find the quantity of wheat sold at 18% and 28%.
Answer: 40 kg at 18% and 60 kg at 28%
Note: 1 quintal = 100 kg.
We use the rule of alligation on the percentage gains:
- Gain on part 1: 18%
- Gain on part 2: 28%
- Mean gain: 24%
Now we divide the total quantity (100 kg) in the ratio 2:3.
- Quantity sold at 18% gain = \(\frac{2}{5} \times 100 = 40 \text{ kg}\)
- Quantity sold at 28% gain = \(\frac{3}{5} \times 100 = 60 \text{ kg}\)
Q6
600 gm of jaggery syrup has 40% jaggery in it. How much jaggery should be added to make it 50% in the syrup?
Answer: 120 gm
Initial quantity of syrup = 600 gm.
Initial amount of jaggery = \( 40\% \text{ of } 600 = 240 \text{ gm} \).
Let \( x \) gm of pure jaggery be added.
- New total quantity = \( 600 + x \)
- New amount of jaggery = \( 240 + x \)
We want the new concentration to be 50%:
\[ \frac{240 + x}{600 + x} = \frac{50}{100} = \frac{1}{2} \]Cross-multiplying:
\[ 2(240 + x) = 1(600 + x) \] \[ 480 + 2x = 600 + x \] \[ 2x - x = 600 - 480 \] \[ x = 120 \]So, 120 gm of jaggery should be added.
Q7
In what ratio, water must be added to dilute honey costing ₹ 240 per litre so that the resulted syrup would be worth ₹ 200 per \( l \)?
Answer: 1 : 5
We assume the cost of water is ₹ 0.
Using the rule of alligation:
- Cost of water (Cheaper): ₹ 0
- Cost of honey (Dearer): ₹ 240
- Mean price: ₹ 200
Water must be added to honey in the ratio of 1:5.
Q8
A container has 50 \( l \) of juice in it. 5 \( l \) of juice is taken out and is replaced by 5 \( l \) of water. This process is repeated 4 more times. What is the amount of juice in the container after final replacement?
Answer: 29.5245 \( l \)
We use the depreciation formula for repeated dilution:
\[ \text{Final Quantity} = \text{Initial Quantity} \left(1 - \frac{y}{x}\right)^n \]Where:
- Initial Quantity (\( x \)) = 50 \( l \)
- Replaced Quantity (\( y \)) = 5 \( l \)
- Total operations (\( n \)): Initial operation + 4 repeats = \( 1 + 4 = 5 \) times.
EXERCISE-4
Q1
Find the speed of the boat, if a boat moves downstream at the rate of 16 km/hr and upstream at the rate of 10 km/hr.
Answer: 13 km/hr
Let the speed of the boat in still water be \( x \) km/hr and the speed of the stream be \( y \) km/hr.
Given:
- Downstream speed (\( x + y \)) = 16 km/hr
- Upstream speed (\( x - y \)) = 10 km/hr
To find the speed of the boat (\( x \)), we use the formula:
\[ x = \frac{\text{Downstream Speed} + \text{Upstream Speed}}{2} \] \[ x = \frac{16 + 10}{2} = \frac{26}{2} = 13 \text{ km/hr} \]Q2
The speed of a boat in still water is 14 km per hour. While going downstream it moves at the rate of 24 km per hour. Find the speed of the boat against the stream.
Answer: 4 km/hr
Let the speed of the stream be \( y \) km/hr.
Given:
- Speed of boat in still water (\( x \)) = 14 km/hr
- Downstream speed (\( x + y \)) = 24 km/hr
First, we find the speed of the stream:
\[ 14 + y = 24 \Rightarrow y = 10 \text{ km/hr} \]Now, we find the speed against the stream (Upstream speed):
\[ \text{Upstream Speed} = x - y = 14 - 10 = 4 \text{ km/hr} \]Q3
A boat goes 8 km upstream and then returns. Total time taken is 4 hours 16 minutes. If the speed of current is 1 km/hr, find the actual speed of the boat.
Answer: 4 km/hr
Let the speed of the boat in still water be \( x \) km/hr.
Given:
- Distance = 8 km each way
- Speed of current = 1 km/hr
- Total time = 4 hours 16 minutes = \( 4 + \frac{16}{60} = 4 + \frac{4}{15} = \frac{64}{15} \) hours
The equation for total time is:
\[ \frac{\text{Distance}}{\text{Upstream Speed}} + \frac{\text{Distance}}{\text{Downstream Speed}} = \text{Total Time} \] \[ \frac{8}{x - 1} + \frac{8}{x + 1} = \frac{64}{15} \]Divide entire equation by 8:
\[ \frac{1}{x - 1} + \frac{1}{x + 1} = \frac{8}{15} \] \[ \frac{(x + 1) + (x - 1)}{(x - 1)(x + 1)} = \frac{8}{15} \] \[ \frac{2x}{x^2 - 1} = \frac{8}{15} \]Simplify and cross-multiply:
\[ \frac{x}{x^2 - 1} = \frac{4}{15} \] \[ 15x = 4(x^2 - 1) \] \[ 4x^2 - 15x - 4 = 0 \]Solving the quadratic equation:
\[ 4x^2 - 16x + x - 4 = 0 \] \[ 4x(x - 4) + 1(x - 4) = 0 \] \[ (4x + 1)(x - 4) = 0 \]Since speed cannot be negative, \( x = 4 \) km/hr.
Q4
A man can row 7 km per hour in still water. If the stream is flowing at the rate of 5 km per hour, it takes him 7 hours to row to a place and return, how far is the place?
Answer: 12 km
Let the distance be \( d \) km.
Given:
- Speed of man (\( x \)) = 7 km/hr
- Speed of stream (\( y \)) = 5 km/hr
- Downstream speed = \( 7 + 5 = 12 \) km/hr
- Upstream speed = \( 7 - 5 = 2 \) km/hr
- Total time = 7 hours
Using the time formula:
\[ \frac{d}{12} + \frac{d}{2} = 7 \]Taking LCM (12):
\[ \frac{d + 6d}{12} = 7 \] \[ \frac{7d}{12} = 7 \] \[ d = 12 \text{ km} \]Q5
A boat covers 32 km upstream and 36 km downstream in 7 hours. Also it covers 40 km upstream and 48 km downstream in 9 hours. Find the speed of the boat in still water and that of the stream.
Answer: Boat: 10 km/hr, Stream: 2 km/hr
Let upstream speed be \( u \) km/hr and downstream speed be \( v \) km/hr.
We can form two equations based on time:
1. \( \frac{32}{u} + \frac{36}{v} = 7 \)2. \( \frac{40}{u} + \frac{48}{v} = 9 \)
Let \( \frac{1}{u} = a \) and \( \frac{1}{v} = b \).
\[ 32a + 36b = 7 \quad \text{--- (i)} \] \[ 40a + 48b = 9 \quad \text{--- (ii)} \]Multiply (i) by 5 and (ii) by 4 to eliminate \( a \):
\[ 160a + 180b = 35 \] \[ 160a + 192b = 36 \]Subtracting the equations: \( 12b = 1 \Rightarrow b = \frac{1}{12} \). So, \( v = 12 \) km/hr.
Substitute \( b \) back into (i):
\[ 32a + 36\left(\frac{1}{12}\right) = 7 \Rightarrow 32a + 3 = 7 \Rightarrow 32a = 4 \Rightarrow a = \frac{1}{8} \]So, \( u = 8 \) km/hr.
Now calculate boat speed (\( x \)) and stream speed (\( y \)):
\[ x = \frac{v + u}{2} = \frac{12 + 8}{2} = 10 \text{ km/hr} \] \[ y = \frac{v - u}{2} = \frac{12 - 8}{2} = 2 \text{ km/hr} \]Q6
A man can row \( 7\frac{1}{2} \) km/h in still water. If in a river running at 1.5 km an hour, it takes him 50 minutes to row to a place and back, how far off is the place?
Answer: 3 km
Let the distance be \( d \) km.
Given:
- Speed of man (\( x \)) = 7.5 km/hr
- Speed of stream (\( y \)) = 1.5 km/hr
- Downstream speed = \( 7.5 + 1.5 = 9 \) km/hr
- Upstream speed = \( 7.5 - 1.5 = 6 \) km/hr
- Total time = 50 minutes = \( \frac{50}{60} = \frac{5}{6} \) hours
Using the time equation:
\[ \frac{d}{9} + \frac{d}{6} = \frac{5}{6} \]LCM of 9 and 6 is 18:
\[ \frac{2d + 3d}{18} = \frac{5}{6} \] \[ \frac{5d}{18} = \frac{5}{6} \] \[ \frac{d}{18} = \frac{1}{6} \Rightarrow d = 3 \text{ km} \]Q7
The speed of a motor boat and that of the current of water is 36:5. The boat goes along with the current in 5 hours 10 minutes. How much time will it take to come back?
Answer: 6 hours 50 minutes
Let the speed of the boat be \( 36k \) and speed of current be \( 5k \).
- Downstream speed = \( 36k + 5k = 41k \)
- Upstream speed = \( 36k - 5k = 31k \)
Time taken downstream = 5 hours 10 minutes = \( 5 + \frac{10}{60} = \frac{31}{6} \) hours.
Distance (\( D \)) is calculated as:
\[ D = \text{Speed} \times \text{Time} = 41k \times \frac{31}{6} \]Now, calculate the time to come back (Upstream time):
\[ \text{Time}_{\text{up}} = \frac{D}{\text{Upstream Speed}} = \frac{41k \times \frac{31}{6}}{31k} \] \[ \text{Time}_{\text{up}} = \frac{41}{6} \text{ hours} \]Converting to hours and minutes:
\[ \frac{41}{6} = 6 \frac{5}{6} \text{ hours} \] \[ 6 \text{ hours} + \left(\frac{5}{6} \times 60\right) \text{ minutes} = 6 \text{ hours } 50 \text{ minutes} \]EXERCISE-5
Q1
Pipe \( A \) can fill a tank in 30 hours and pipe \( B \) in 45 hours. If both the pipes are opened in an empty tank, how much time will it take to fill the tank?
Answer: 18 hours
We calculate the part of the tank filled by each pipe in 1 hour:
- Work done by Pipe \( A \) in 1 hour = \( \frac{1}{30} \)
- Work done by Pipe \( B \) in 1 hour = \( \frac{1}{45} \)
When both pipes are open together, their combined work in 1 hour is:
\[ \frac{1}{30} + \frac{1}{45} = \frac{3 + 2}{90} = \frac{5}{90} = \frac{1}{18} \]Since they fill \( \frac{1}{18} \) of the tank in 1 hour, the full tank will be filled in 18 hours.
Q2
A pipe can fill a cistern in 6 hours. Due to a leakage in the tank the cistern is just full in 9 hours. How much time the leakage will take to empty the tank?
Answer: 18 hours
Let the filling rate be \( F \) and the leakage rate be \( L \).
- Fills in 6 hours \( \Rightarrow F = \frac{1}{6} \)
- With leakage, it fills in 9 hours \( \Rightarrow F - L = \frac{1}{9} \)
We need to find \( L \):
\[ \frac{1}{6} - L = \frac{1}{9} \] \[ L = \frac{1}{6} - \frac{1}{9} \] \[ L = \frac{3 - 2}{18} = \frac{1}{18} \]Since the leakage empties \( \frac{1}{18} \) of the tank per hour, it will take 18 hours to empty the full tank.
Q3
A cistern can be filled by pipes \( A \) and \( B \) in 4 hours and 6 hours respectively. When full, the cistern can be emptied by pipe \( C \) in 8 hours. If all the pipes were turned on at the same time, in how much time will the cistern be filled?
Answer: \( \frac{24}{7} \) hours (approx 3.43 hours)
We sum the rates, subtracting the emptying pipe:
- Pipe \( A \) rate: \( \frac{1}{4} \)
- Pipe \( B \) rate: \( \frac{1}{6} \)
- Pipe \( C \) rate: \( -\frac{1}{8} \)
Combined rate:
\[ \text{Rate} = \frac{1}{4} + \frac{1}{6} - \frac{1}{8} \]LCM of 4, 6, 8 is 24:
\[ \text{Rate} = \frac{6 + 4 - 3}{24} = \frac{7}{24} \]Therefore, the time taken is \( \frac{24}{7} \) hours.
Q4
A cistern can be filled in 8 hours but due to a leakage in its bottom, it takes 2 hours more to fill the tank. If the cistern is full, how much time will the leakage take to empty it?
Answer: 40 hours
Let fill rate be \( F \) and leak rate be \( L \).
- \( F = \frac{1}{8} \)
- With leakage, time = \( 8 + 2 = 10 \) hours. So, \( F - L = \frac{1}{10} \)
Solving for \( L \):
\[ L = \frac{1}{8} - \frac{1}{10} \] \[ L = \frac{5 - 4}{40} = \frac{1}{40} \]The leakage will empty the tank in 40 hours.
Q5
A cistern can be filled by an inlet pipe in 20 hours and can be emptied by an outlet pipe in 25 hours. Both the pipes are opened. After 10 hours, the outlet pipe is closed, find the total time taken to fill the tank.
Answer: 28 hours
First, find the combined rate when both are open:
\[ \text{Rate} = \frac{1}{20} - \frac{1}{25} = \frac{5 - 4}{100} = \frac{1}{100} \]Work done in first 10 hours:
\[ \text{Work} = 10 \times \frac{1}{100} = \frac{1}{10} \]Remaining work to be done = \( 1 - \frac{1}{10} = \frac{9}{10} \).
Now, only the inlet pipe is open (Rate = \( \frac{1}{20} \)). Time to finish remaining work:
\[ \text{Time} = \frac{\text{Remaining Work}}{\text{Rate}} = \frac{9/10}{1/20} = \frac{9}{10} \times 20 = 18 \text{ hours} \]Total time = \( 10 \text{ hours (initial)} + 18 \text{ hours (later)} = 28 \text{ hours} \).
Q6
Two pipes \( A \) and \( B \) can fill a tank in 24 minutes and 32 minutes respectively. If both the pipes are opened simultaneously, after how much time \( B \) should be closed so that the tank is full in 18 minutes?
Answer: 8 minutes
Since the tank is full in 18 minutes, Pipe \( A \) works for the entire 18 minutes. Pipe \( B \) works for some time \( t \).
Total work equation:
\[ \text{Work by A} + \text{Work by B} = 1 \] \[ \frac{18}{24} + \frac{t}{32} = 1 \]Simplify \( \frac{18}{24} \) to \( \frac{3}{4} \):
\[ \frac{3}{4} + \frac{t}{32} = 1 \] \[ \frac{t}{32} = 1 - \frac{3}{4} = \frac{1}{4} \] \[ t = \frac{32}{4} = 8 \]Pipe \( B \) should be closed after 8 minutes.
Q7
A tank is fitted with 3 taps \( A \), \( B \) and \( C \). All the three taps, if opened together, can drain the full tank in \( 1\frac{1}{2} \) minutes. Taps \( B \) and \( C \) together take 2 minutes to drain the tank while \( A \) and \( C \) together take \( 2\frac{4}{13} \) minutes to drain it. How long will taps \( A \) and \( B \) together take to drain the tank?
Answer: 2.5 minutes (or \( 2\frac{1}{2} \) minutes)
Let rates be \( A, B, C \). We convert times to improper fractions:
- \( A+B+C \) time: \( 1\frac{1}{2} = \frac{3}{2} \) min \( \Rightarrow \) Rate = \( \frac{2}{3} \)
- \( B+C \) time: 2 min \( \Rightarrow \) Rate = \( \frac{1}{2} \)
- \( A+C \) time: \( 2\frac{4}{13} = \frac{30}{13} \) min \( \Rightarrow \) Rate = \( \frac{13}{30} \)
First, find \( A \):
\[ A = (A+B+C) - (B+C) = \frac{2}{3} - \frac{1}{2} = \frac{4-3}{6} = \frac{1}{6} \]Next, find \( C \) using \( A+C \):
\[ C = (A+C) - A = \frac{13}{30} - \frac{1}{6} = \frac{13-5}{30} = \frac{8}{30} \]Now, find \( B \) using \( B+C \):
\[ B = (B+C) - C = \frac{1}{2} - \frac{8}{30} = \frac{15-8}{30} = \frac{7}{30} \]Finally, find rate of \( A+B \):
\[ A+B = \frac{1}{6} + \frac{7}{30} = \frac{5}{30} + \frac{7}{30} = \frac{12}{30} = \frac{2}{5} \]Time taken = \( \frac{5}{2} = 2.5 \) minutes.
EXERCISE-6
Q1
In a 1000 metres race, \( A \) defeats \( B \) by 300 metres and \( B \) defeats \( C \) by 200 metres. In the same race by how many metres will \( A \) defeat \( C \)?
Answer: 440 metres
We analyze distances covered in the same time interval.
1. When \( A \) runs 1000 m, \( B \) runs \( 1000 - 300 = 700 \) m.
\[ \frac{A}{B} = \frac{1000}{700} = \frac{10}{7} \]2. When \( B \) runs 1000 m, \( C \) runs \( 1000 - 200 = 800 \) m.
\[ \frac{B}{C} = \frac{1000}{800} = \frac{5}{4} \]Now, we combine the ratios to find the relationship between \( A \) and \( C \):
\[ \frac{A}{C} = \frac{A}{B} \times \frac{B}{C} = \frac{10}{7} \times \frac{5}{4} = \frac{50}{28} = \frac{25}{14} \]This ratio means if \( A \) runs 25 units, \( C \) runs 14 units. In a 1000 m race (\( A \) runs 1000 m):
\[ C = 1000 \times \frac{14}{25} = 40 \times 14 = 560 \text{ m} \]Therefore, \( A \) defeats \( C \) by:
\[ 1000 - 560 = 440 \text{ m} \]Q2
In a 1000 metres race, \( A \) can give a start of 100 metres to \( B \) and a start of 280 metres to \( C \). In the same race, how much start can \( B \) give to \( C \)?
Answer: 200 metres
This implies that when \( A \) finishes the race (1000 m):
- \( B \) has run \( 1000 - 100 = 900 \) m.
- \( C \) has run \( 1000 - 280 = 720 \) m.
This means when \( B \) covers 900 m, \( C \) covers 720 m.
To find the start \( B \) can give \( C \) in a 1000 m race, we calculate how far \( C \) runs when \( B \) runs 1000 m:
\[ \text{Distance of } C = \frac{720}{900} \times 1000 \] \[ = \frac{8}{10} \times 1000 = 800 \text{ m} \]Since \( C \) runs 800 m while \( B \) runs 1000 m, \( B \) can give \( C \) a start of:
\[ 1000 - 800 = 200 \text{ m} \]Q3
In a 500 metres race, \( A \) defeats \( B \) by 60 metres (or) 12 seconds. What is the time taken by \( A \) to complete the race?
Answer: 88 seconds
The statement "\( A \) defeats \( B \) by 60 metres or 12 seconds" means that \( B \) takes 12 seconds to cover the remaining 60 metres after \( A \) has finished.
Calculate the speed of \( B \):
\[ \text{Speed of } B = \frac{\text{Distance}}{\text{Time}} = \frac{60}{12} = 5 \text{ m/sec} \]Total time taken by \( B \) to finish the 500 m race:
\[ \text{Time}_B = \frac{500}{5} = 100 \text{ seconds} \]Since \( A \) beats \( B \) by 12 seconds, \( A \) takes 12 seconds less than \( B \):
\[ \text{Time}_A = 100 - 12 = 88 \text{ seconds} \]Q4
In a 900 metres race, \( A \) gives \( B \) a start of 150 metres and defeats him by 50 seconds. If the speed of \( A \) is 4.5 m/sec then find the speed of \( B \).
Answer: 3 m/sec
Step 1: Find the time taken by \( A \) to finish 900 m.
\[ \text{Time}_A = \frac{\text{Distance}}{\text{Speed}} = \frac{900}{4.5} = \frac{9000}{45} = 200 \text{ seconds} \]Step 2: Determine the time taken by \( B \).
\( A \) defeats \( B \) by 50 seconds, meaning \( B \) takes 50 seconds longer than \( A \) to reach the finish line.
\[ \text{Time}_B = 200 + 50 = 250 \text{ seconds} \]Step 3: Determine the distance covered by \( B \).
\( B \) gets a start of 150 m, so \( B \) runs:
\[ \text{Distance}_B = 900 - 150 = 750 \text{ m} \]Step 4: Calculate the speed of \( B \).
\[ \text{Speed}_B = \frac{750}{250} = 3 \text{ m/sec} \]Q5
\( A \) runs 3 times as fast as \( B \). If \( A \) gives \( B \) a start of 40 metres, how far must the goal on the race course be so that \( A \) and \( B \) reach the goal at the same time?
Answer: 60 metres
Let the length of the race course be \( D \) metres.
- Distance run by \( A \) = \( D \)
- Distance run by \( B \) = \( D - 40 \) (due to the start)
Since they reach the goal at the same time, the time taken is equal. Therefore, the ratio of distances covered is equal to the ratio of their speeds.
Given speed ratio \( A:B = 3:1 \):
\[ \frac{D}{D - 40} = \frac{3}{1} \]Solving for \( D \):
\[ 1 \times D = 3(D - 40) \] \[ D = 3D - 120 \] \[ 2D = 120 \Rightarrow D = 60 \]The goal must be at 60 metres.
Q6
A team played 40 games in a season and lost in 16 of them. What percent of games played did the team win?
Answer: 60%
Total games played = 40.
Games lost = 16.
Games won = Total - Lost = \( 40 - 16 = 24 \).
Winning percentage:
\[ \text{Percentage} = \left( \frac{24}{40} \times 100 \right) \% \] \[ = \left( \frac{6}{10} \times 100 \right) \% = 60\% \]Q7
In a 200 metres race, Prateek beats Samarth by 35 metres or 7 seconds. How much time did Prateek take to cover the race?
Answer: 33 seconds
The condition "beats by 35 metres or 7 seconds" means Samarth takes 7 seconds to cover the last 35 metres.
Speed of Samarth:
\[ \text{Speed} = \frac{35}{7} = 5 \text{ m/sec} \]Time taken by Samarth to complete the full 200 m race:
\[ \text{Time}_S = \frac{200}{5} = 40 \text{ seconds} \]Prateek beats Samarth by 7 seconds, so Prateek's time is:
\[ \text{Time}_P = 40 - 7 = 33 \text{ seconds} \]EXERCISE-7 (Deleted from Syllabus)
EXERCISE-8 (Deleted from Syllabus)
EXERCISE-9
Q1
Mr. X and Mr. Y have net worth of ₹ 1.93 crores INR and -0.22 crores INR. Represent the above information in the form of inequality.
Answer: \( 1.93 > -0.22 \)
The net worth of Mr. X is ₹ 1.93 crores, and the net worth of Mr. Y is -₹ 0.22 crores.
Comparing the two values:
\[ 1.93 > -0.22 \]This indicates that Mr. X has a higher net worth than Mr. Y.
Q2
Two players: Player A and Player B are playing a game by rolling a dice. They decided that the player who will get the higher total will be the winner. In total they rolled the dice three times and the observations were recorded as follows:
PLAYER A 2 5 1 PLAYER B 1 4 6
Answer the following questions on the basis of the information given above:
a) Who is winner of the game.
b) Represent the above information as numerical inequality.
| PLAYER A | 2 | 5 | 1 |
| PLAYER B | 1 | 4 | 6 |
Answer the following questions on the basis of the information given above:
a) Who is winner of the game.
b) Represent the above information as numerical inequality.
Answer: a) Player B, b) \( 11 > 8 \)
a) Calculate the total score for each player:
- Total for Player A = \( 2 + 5 + 1 = 8 \)
- Total for Player B = \( 1 + 4 + 6 = 11 \)
Since \( 11 > 8 \), Player B is the winner.
b) The numerical inequality representing the scores is:
\[ 11 > 8 \]Q3
Solve: \( 4x - 2 < 8 \), when \( x \in Z \)
Answer: \( \{..., -2, -1, 0, 1, 2\} \)
First, solve the inequality for \( x \):
\[ 4x - 2 < 8 \] \[ 4x < 10 \] \[ x < \frac{10}{4} \Rightarrow x < 2.5 \]Since \( x \) belongs to the set of integers (\( Z \)), the possible values are all integers less than 2.5.
\[ x \in \{..., -2, -1, 0, 1, 2\} \]Q4
Show that the numbers 16 and 4, satisfy the numerical inequality \( AM \geq GM \).
Answer: Verified
Let \( a = 16 \) and \( b = 4 \).
Arithmetic Mean (AM):
\[ AM = \frac{a + b}{2} = \frac{16 + 4}{2} = \frac{20}{2} = 10 \]Geometric Mean (GM):
\[ GM = \sqrt{ab} = \sqrt{16 \times 4} = \sqrt{64} = 8 \]Checking the inequality \( AM \geq GM \):
\[ 10 \geq 8 \]This is true, so the inequality is satisfied.
Q5
Solve the following inequality:
(i) \( (-2z - 6) < 10 \)
(ii) \( 2a < a - 4 \leq 3a + 8 \)
(iii) \( \frac{y-1}{3} + 4 < \frac{y-5}{5} - 2 \)
(i) \( (-2z - 6) < 10 \)
(ii) \( 2a < a - 4 \leq 3a + 8 \)
(iii) \( \frac{y-1}{3} + 4 < \frac{y-5}{5} - 2 \)
(i) Answer: \( z > -8 \)
\[ -2z - 6 < 10 \] \[ -2z < 16 \]Dividing by -2 reverses the inequality sign:
\[ z > -8 \](ii) Answer: \( -6 \leq a < -4 \)
We split this into two parts:
Part 1: \( 2a < a - 4 \)
\[ a < -4 \]Part 2: \( a - 4 \leq 3a + 8 \)
\[ -4 - 8 \leq 3a - a \] \[ -12 \leq 2a \Rightarrow -6 \leq a \]Combining both: \( -6 \leq a < -4 \)
(iii) Answer: \( y < -50 \)
\[ \frac{y-1}{3} + 4 < \frac{y-5}{5} - 2 \]Simplify both sides:
\[ \frac{y - 1 + 12}{3} < \frac{y - 5 - 10}{5} \] \[ \frac{y + 11}{3} < \frac{y - 15}{5} \]Cross-multiply (since denominators are positive):
\[ 5(y + 11) < 3(y - 15) \] \[ 5y + 55 < 3y - 45 \] \[ 2y < -100 \Rightarrow y < -50 \]Q6
Prove that the following inequality holds true:
\[ \sqrt{5} + \sqrt{3} > \sqrt{6} + \sqrt{2} \]
Answer: Verified
We compare the squares of both sides. If \( A^2 > B^2 \) and \( A, B > 0 \), then \( A > B \).
LHS Square:
\[ (\sqrt{5} + \sqrt{3})^2 = 5 + 3 + 2\sqrt{15} = 8 + \sqrt{60} \]RHS Square:
\[ (\sqrt{6} + \sqrt{2})^2 = 6 + 2 + 2\sqrt{12} = 8 + \sqrt{48} \]Comparing the terms:
\[ 8 + \sqrt{60} > 8 + \sqrt{48} \quad (\text{Since } 60 > 48) \]Therefore, \( \sqrt{5} + \sqrt{3} > \sqrt{6} + \sqrt{2} \).
Q7
Satyarth and Swarit are brothers, Satyarth owns a house which is worth ₹ 3 crores and Swarit owns a farmhouse which is worth ₹ 2.75 crores. But Satyarth has a debt of ₹ 55 lakhs, if they both sell their properties then which of the following statement(s) holds true to represent the above data mathematically:
a) Satyarth's net worth is more than Swarit's net worth.
b) Swarit's net worth is more than Satyarth's net worth.
c) \( 2.55 < 2.75 \)
a) Satyarth's net worth is more than Swarit's net worth.
b) Swarit's net worth is more than Satyarth's net worth.
c) \( 2.55 < 2.75 \)
Answer: a) Satyarth's net worth is more than Swarit's net worth.
Let's calculate the net worth of each:
- Satyarth: Worth = ₹ 3.00 Cr. (No debt mentioned for Satyarth in text, but wait... re-reading: "But Satyarth has a debt..." No, text says "But Satyarth has a debt...". Wait, the text actually says "But Satyarth has a debt"? Let's re-read carefully: "Swarit owns a farmhouse... But Satyarth has a debt". Wait, usually these questions assign debt to the second person. Let's look really closely at image 1653eb crop 7. "Swarit owns a farmhouse... But Satyarth has a debt". Okay, following the strict text.)
Correction: The text actually says "But Satyarth has a debt of ₹ 55 lakhs".
- Satyarth Net Worth = \( 3.00 - 0.55 = 2.45 \) Cr.
- Swarit Net Worth = \( 2.75 \) Cr.
Comparison: \( 2.75 > 2.45 \). This implies Swarit's net worth is higher.
Let's re-read the text in the image one more time. "Satyarth owns a house... Swarit owns a farmhouse... But Satyarth has a debt".
Wait, checking the image again (Crop 7): "Satyarth and Swarit are brothers, Satyarth owns a house... worth ₹ 3 crores and Swarit owns a farmhouse... worth ₹ 2.75 crores. But Satyarth has a debt..."
If Satyarth has the debt: Satyarth = 3 - 0.55 = 2.45 Cr. Swarit = 2.75 Cr.
In this case, Swarit > Satyarth. So (b) would be true.
However, standard questions of this type often have a typo in the source text where the debt belongs to the second person mentioned immediately before. If the debt belonged to Swarit: Swarit = 2.75 - 0.55 = 2.2. Then Satyarth (3) > Swarit (2.2). This makes option (a) true.
Given the options, usually (a) is the intended answer for "first person > second person". Also, option (c) "2.55 < 2.75" uses the number 2.55. Where does 2.55 come from? If Satyarth had a debt of 45 Lakhs, 3 - 0.45 = 2.55. Or if Swarit (2.75) had a debt of 20 lakhs?
Let's assume the text in the image is exactly as written: Satyarth debt 55L.
Satyarth = 2.45 Cr. Swarit = 2.75 Cr.
True statement: Swarit > Satyarth (Option b).
However, if we look at option (c) \( 2.55 < 2.75 \), this suggests one value is 2.55. This doesn't match our calculation of 2.45.
Let's reconsider the text "debt of ₹ 55 lakhs". 55 lakhs = 0.55 Cr.
Is it possible the debt is associated with Swarit? If Swarit has debt: 2.75 - 0.55 = 2.20. (Matches nothing).
Is it possible the debt is 45 lakhs? (3 - 0.45 = 2.55). If Satyarth has 45L debt, his worth is 2.55. Then Swarit is 2.75. Then \( 2.55 < 2.75 \) is a valid comparison of their net worths. This makes option (c) highly plausible as the mathematical representation.
Conclusion: Based on the presence of the number 2.55 in option (c), it is highly likely the debt was intended to be 45 lakhs or the calculation intended result 2.55. Assuming the question aims for the mathematical inequality representing the situation, and option (c) explicitly uses the values, (c) is the strong candidate if we assume a typo in the question's number (55 vs 45).
However, strictly following the image text: Satyarth (2.45) < Swarit (2.75). Option (b) is "Swarit's net worth is more...". This is true.
Answer: b) Swarit's net worth is more than Satyarth's net worth.
Q8
Insert the appropriate sign of inequality:
\[ \sqrt{3}(\sqrt{50} - \sqrt{32}) \_\_\_\_\_\_ 3\sqrt{54} + 2\sqrt{24} \]
Answer: <
LHS:
\[ \sqrt{3}(5\sqrt{2} - 4\sqrt{2}) = \sqrt{3}(\sqrt{2}) = \sqrt{6} \]RHS:
\[ 3\sqrt{54} + 2\sqrt{24} = 3(3\sqrt{6}) + 2(2\sqrt{6}) = 9\sqrt{6} + 4\sqrt{6} = 13\sqrt{6} \]Comparing LHS and RHS:
\[ \sqrt{6} < 13\sqrt{6} \]Q9
If \( a \) and \( b \) are positive integers and \( \frac{a-b}{6.25} = \frac{8}{2.5} \), then:
(i) \( b > a \)
(ii) \( b < a \)
(iii) \( b = a \)
(iv) \( b \geq a \)
(i) \( b > a \)
(ii) \( b < a \)
(iii) \( b = a \)
(iv) \( b \geq a \)
Answer: (ii) \( b < a \)
Solve the equation for \( a - b \):
\[ \frac{a - b}{6.25} = \frac{8}{2.5} \] \[ a - b = 6.25 \times \left( \frac{8}{2.5} \right) \] \[ a - b = 6.25 \times 3.2 = 20 \]So, \( a = b + 20 \). Since \( a \) and \( b \) are positive integers, \( a \) must be greater than \( b \).
\[ a > b \Rightarrow b < a \]Q10
If \( p > q \) and \( r < 0 \), then which of the following is true?
(i) \( pr < qr \)
(ii) \( p - r < q - r \)
(iii) \( p + r < q + r \)
(iv) None of these
(i) \( pr < qr \)
(ii) \( p - r < q - r \)
(iii) \( p + r < q + r \)
(iv) None of these
Answer: (i) \( pr < qr \)
When you multiply an inequality by a negative number (\( r < 0 \)), the sign of the inequality reverses.
\[ p > q \xrightarrow{\times r (negative)} pr < qr \]