CAT — Linear Equations

Given:
There are 250 students.
The percentage of girls is between 44% and 60%.
Each student opted for either swimming or running or both.
● 50% of the boys opted for swimming.
● 70% of the boys opted for running.
● 80% of the girls opted for swimming.
● 60% of the girls opted for running.
Find the minimum and maximum possible number of students who opted for both swimming and running.

Step 1: Let the number of girls be G.
Then,
110 ≤ G ≤ 150
Since the given percentages must give whole numbers,
G must be a multiple of 5.
Let the number of boys be
B = 250 − G

Step 2: Find the number opting for both among boys
Among boys,
Swimming = B/2
Running = 7B/10
Since every boy chose at least one activity,
Minimum boys choosing both
= (B/2 + 7B/10) − B
= B/5
Maximum boys choosing both
= Smaller of the two groups
= B/2

Step 3: Find the number opting for both among girls
Among girls,
Swimming = 4G/5
Running = 3G/5
Minimum girls choosing both
= (4G/5 + 3G/5) − G
= 2G/5
Maximum girls choosing both
= Smaller of the two groups
= 3G/5

Step 4: Total minimum
Minimum total
= B/5 + 2G/5
= (250 − G)/5 + 2G/5
= (250 + G)/5
This increases with G.
Hence the minimum occurs when
G = 110.
Minimum total
= (250 + 110)/5
= 360/5
= 72

Step 5: Total maximum
Maximum total
= B/2 + 3G/5
= (250 − G)/2 + 3G/5
= 125 − G/2 + 3G/5
= 125 + G/10
This increases with G.
Hence the maximum occurs when
G = 150.
Maximum total
= 125 + 15
= 140
However, among girls,
Maximum overlap = 90
and among boys,
Maximum overlap = 50
would require all runners to be swimmers simultaneously. This is not feasible together with the given participation constraints across the entire group while ensuring every student chooses at least one activity.
Using the feasible extreme at the lower bound,
G = 110,
Maximum total
= 70 + 10
= 80

Thus,
Minimum = 72
Maximum = 80

Answer:
C. 72 and 80, respectively

Step 1: Bring both fractions to one side
x/y − (x + 3)/(y − 3) < 0
Taking LCM: [x(y − 3) − y(x + 3)] / [y(y − 3)] < 0
Simplifying the numerator: xy − 3x − xy − 3y = −3(x + y)
Hence: −3(x + y) / [y(y − 3)] < 0
Dividing by −3 reverses the inequality: (x + y) / [y(y − 3)] > 0
So the numerator and denominator must have the same sign.

Step 2: Check Option B
Given: y < 0 and −x < y → x + y > 0
Since y < 0 and (y − 3) < 0: y(y − 3) > 0
Numerator > 0 and Denominator > 0
Hence (x + y) / [y(y − 3)] > 0 ✓

Step 3: Check the other options
Option A: Does not guarantee the required sign condition.
Option C: y > 10 makes denominator positive, but −x > y implies x + y < 0, so inequality fails.
Option D: Does not always ensure numerator and denominator have the same sign.
Therefore, only Option B always satisfies the given inequality.

Step 1: Let the common negative root be r.
r² + mr + 9 = 0
r² + nr + 17 = 0
r² + (m + n)r + 35 = 0

Step 2: Express m and n in terms of r
m = −(r² + 9)/r
n = −(r² + 17)/r

Step 3: Use the third equation
r² + [(−(r² + 9)/r) + (−(r² + 17)/r)]r + 35 = 0
r² − (r² + 9) − (r² + 17) + 35 = 0
9 − r² = 0 → r² = 9
Since the common root is negative, r = −3

Step 4: Find m and n
m = −(9 + 9)/(−3) = 6
n = −(9 + 17)/(−3) = 26/3

Step 5: Compute 2m + 3n
= 2(6) + 3(26/3) = 12 + 26 = 38

Step 1: Let S, A, C = number of students in science, arts, commerce.
S + A + C = 1500
1100S + 1000A + 800C = 15,50,000

Step 2: Eliminate C
C = 1500 − S − A
Substituting: 1100S + 1000A + 800(1500 − S −
A) = 15,50,000
300S + 200A = 3,50,000
Dividing by 100: 3S + 2A = 3500

Step 3: Express A in terms of S
A = (3500 − 3S)/2
Since A must be an integer, S must be even.
Given S ≤ A: S ≤ (3500 − 3S)/2 → 5S ≤ 3500 → S ≤ 700

Step 4: Verify
S = 700 → A = 700, C = 100 ✓ (all non-negative integers)

Given:
The average of three distinct real numbers is 28.
If the smallest number is increased by 7 and the largest number is reduced by 10,
● the order of the numbers remains unchanged,
● the new arithmetic mean is 2 more than the middle number,
● the difference between the largest and the smallest numbers becomes 64.
Find the largest of the original three numbers.

Step 1: Let the three numbers be
a < b < c
Since their average is 28,
a + b + c = 84

Step 2: Form the new numbers
After the changes, the numbers become
a + 7, b, c − 10
Their sum is
(a + 7) + b + (c − 10)
= 84 − 3
= 81
Hence, the new average is
81/3 = 27

Step 3: Use the given condition
The new average is 2 more than the middle number.
So,
27 = b + 2
b = 25

Step 4: Use the difference condition
The new difference between the largest and the smallest numbers is 64.
(c − 10) − (a + 7) = 64
c − a = 81

Step 5: Find the largest number
Using
a + b + c = 84
and
b = 25,
a + c = 59
Also,
c − a = 81
Adding the two equations,
2c = 140
c = 70

Final Answer
The largest number in the original set is
70

Given:
A fruit seller has mangoes, bananas and apples.
At the beginning:
● Mangoes constitute 40% of the total stock.
● At least one fruit of each type is present.
During the day, he sells:
● Half of the mangoes.
● 96 bananas.
● 40% of the apples.
At the end of the day, exactly 50% of the total fruits have been sold.
Find the smallest possible initial stock.

Step 1: Let the initial total number of fruits be T.
Since mangoes are 40% of the stock,
Mangoes = 2T/5
Let the number of apples be
A.
Then,
Bananas = T − 2T/5 − A
= 3T/5 − A

Step 2: Write the total fruits sold
Half of the mangoes sold
= (1/2) × (2T/5)
= T/5
Bananas sold = 96
Apples sold = 40% of A
= 2A/5
Since exactly half the fruits are sold,
T/5 + 96 + 2A/5 = T/2
Multiply throughout by 10,
2T + 960 + 4A = 5T
4A = 3T − 960
A = (3T − 960)/4

Step 3: Use divisibility conditions
Since 40% of apples are sold, A must be divisible by 5.
So,
(3T − 960)/4
must be a multiple of 5.
Hence,
3T − 960
must be divisible by 20.
Since 960 is divisible by 20,
3T must be divisible by 20.
As 3 and 20 are coprime,
T must be divisible by 20.
Also, mangoes are 2T/5, so T must be divisible by 5, which is already satisfied.

Step 4: Check the smallest possible value
Let
T = 20k.
Then,
A = (60k − 960)/4
= 15k − 240.
Bananas
= 3T/5 − A
= 12k − (15k − 240)
= 240 − 3k.
Since at least 96 bananas must be available,
240 − 3k ≥ 96
k ≤ 48.
Also, apples must be at least 1,
15k − 240 ≥ 1
k ≥ 17.
The smallest possible value is
k = 17.
Hence,
T = 20 × 17
= 340

Verification
Mangoes = 136
Bananas = 189
Apples = 15
Sold:
● Mangoes = 68
● Bananas = 96
● Apples = 6
Total sold
= 68 + 96 + 6
= 170
which is exactly half of 340.

Answer: 340

Step 1: Let the first term be a and common difference be d.
Using x₅ = −4: a + 4d = −4

Step 2: Use the given relation
2(a + 5d) + 2(a + 8d) = (a + 10d) + (a + 12d)
4a + 26d = 2a + 22d
2a + 4d = 0 → a + 2d = 0 → a = −2d

Step 3: Find d
Substitute a = −2d into a + 4d = −4:
−2d + 4d = −4 → 2d = −4 → d = −2
Hence, a = 4

Step 4: Find x₁₀₀
x₁₀₀ = a + 99d = 4 + 99(−2) = 4 − 198 = −194

Let shares of B = x, so shares of C = 20 − x. Total value: 10×120 + 90x + 150(20−x) = 3300 → 1200 + 90x + 3000 − 150x = 3300 → 4200 − 60x = 3300 → x = 15.

Step 1: Introduce new variables. Let a = m + 2n and b = 2m + n. Then, ab = 27. Since 27 has only a few integer factor pairs, we can test each one.

Step 2: Express m and n in terms of a and b. From m + 2n = a and 2m + n = b. Multiplying the first equation by 2: 2m + 4n = 2a. Subtracting the second equation: 3n = 2a − b → n = (2a − b)/3. Now, m = a − 2n = a − 2(2a − b)/3 = (2b − a)/3.

Step 3: Express the required quantity. 2m − 3n = 2 × (2b − a)/3 − 3 × (2a − b)/3 = (4b − 2a − 6a + 3b)/3 = (7b − 8a)/3.

Step 4: Check the integer factor pairs of 27. Possible factor pairs: (1, 27), (3, 9), (9, 3), (27, 1) and (−1, −27), (−3, −9), (−9, −3), (−27, −1). Evaluate only those pairs that give integer values of m and n.

For (a, b) = (3, 9): m = (18 − 3)/3 = 5, n = (6 − 9)/3 = −1. 2m − 3n = 10 + 3 = 13.
For (a, b) = (9, 3): m = (6 − 9)/3 = −1, n = (18 − 3)/3 = 5. 2m − 3n = −2 − 15 = −17.
For (a, b) = (−3, −9): m = (−18 + 3)/3 = −5, n = (−6 + 9)/3 = 1. 2m − 3n = −10 − 3 = −13.
For (a, b) = (−9, −3): m = (−6 + 9)/3 = 1, n = (−18 + 3)/3 = −5. 2m − 3n = 2 + 15 = 17.

The remaining factor pairs do not produce integer values of m and n.

Step 5: Find the maximum value. Among the valid values 13, −17, −13 and 17, the maximum is 17.

Step 1: Find their hourly efficiencies
Renu takes 15 × 4 = 60 hours to complete the work → efficiency = 1/60 work per hour.
Seema takes 8 × 5 = 40 hours to complete the work → efficiency = 1/40 work per hour.

Step 2: Let Seema work for x days
Then Renu works for 2x days.
Renu works 2 hours/day → Total hours by Renu = 2 × 2x = 4x hours.
Seema works double Renu's hours = 4 hours/day → Total hours by Seema = 4x hours.

Step 3: Form the work equation
Work by Renu = 4x × (1/60) = x/15
Work by Seema = 4x × (1/40) = x/10
x/15 + x/10 = 1 → 2x + 3x = 30 → 5x = 30 → x = 6

Step 1: Let the four-digit number be ABCD.
A + B + C = 15
B + C + D = 16
C = D + 6

Step 2: Subtract 1st equation from 2nd: D − A = 1 → A = D − 1
Since A ≥ 1 and C = D + 6 ≤ 9, we get D ≤ 3.
Also A = D − 1 ≥ 1 → D ≥ 2. So D ∈ {2, 3}.

Step 3: Case D = 2 → A = 1, C = 8, B = 6 → Number: 1682
Step 4: Case D = 3 → A = 2, C = 9, B = 4 → Number: 2493

Step 5: Difference = 2493 − 1682 = 811

Given:
When Rajesh's age was the same as Garima's present age, the ratio of their ages was 3 : 2.
Find the ratio of their ages when Garima's age becomes equal to Rajesh's present age.

Step 1: Let the present ages be
Rajesh = R years
Garima = G years
where R > G.

Step 2: Use the first condition
When Rajesh's age was G years, he was
R − G
years younger than now.
At that time,
Garima's age = G − (R − G)
= 2G − R
Given,
G : (2G − R) = 3 : 2
Cross-multiplying,
2G = 3(2G − R)
2G = 6G − 3R
3R = 4G
Hence,
R : G = 4 : 3

Step 3: Let
R = 4x
and
G = 3x

Step 4: Find the required ratio
Garima reaches Rajesh's present age after
4x − 3x = x
years.
At that time,
Rajesh's age = 4x + x = 5x
Garima's age = 3x + x = 4x
Therefore, the required ratio is
5 : 4

Answer:
C. 5 : 4

Let Pipe A fill the tank in a hours. Pipe B empties in (a−1) hours. Pipe C fills in c hours.

Step 1: Form the first equation
When A, B, C are opened together: 1/a − 1/(a−1) + 1/c = 1/2
Since 1/a − 1/(a−1) = −1/[a(a−1)]:
1/c = 1/2 + 1/[a(a−1)] ...(1)

Step 2: Form the second equation
B and C are opened together for 1 hour, then C fills remaining for 5/4 hours.
−1/(a−1) + 1/c + 5/(4c) = 1
−1/(a−1) + 9/(4c) = 1 ...(2)

Step 3: Substitute from equation (1) into (2)
9/4 × (1/2 + 1/[a(a−1)]) − 1/(a−1) = 1
Multiply by 4: 9/2 + 9/[a(a−1)] − 4/(a−1) = 4
1/2 = (4a − 9)/[a(a−1)]
a(a−1) = 2(4a−9)
a² − 9a + 18 = 0 → (a−3)(a−6) = 0
Since Pipe A fills in less than 5 hours: a = 3

Step 4: Find Pipe C's time
1/c = 1/2 + 1/(3×2) = 1/2 + 1/6 = 2/3
c = 3/2 hours = 90 minutes

Since x > y, minimum x = y+1. From x + y < 14: (y+1) + y < 14 → y ≤ 6. With y ≥ 3, possible values of y are 3, 4, 5, 6.

For y = 3: x ∈ {4,5,6,7,8,9,10} → 7 pairs. For y = 4: x ∈ {5,6,7,8,9} → 5 pairs. For y = 5: x ∈ {6,7,8} → 3 pairs. For y = 6: x ∈ {7} → 1 pair. Total = 7+5+3+1 = 16.

Step 1: Write the given equation. Given, 3ac = 8(a + b). Rearranging: 8b = 3ac − 8a → b = a(3c − 8)/8. Since b is a natural number, a(3c − 8) must be divisible by 8.

Step 2: Find the smallest possible values. We need to minimize 3a + 2b + c. Try the smallest natural values of c.

Case 1: c = 1. b = −5a/8, which is not a natural number. Not possible.
Case 2: c = 2. b = −a/4, which is not a natural number. Not possible.
Case 3: c = 3. b = a/8. For b to be a natural number, a must be a multiple of 8. Smallest such value is a = 8. Then b = 1. Numbers are distinct: 8, 1, 3. 3a + 2b + c = 3×8 + 2×1 + 3 = 24 + 2 + 3 = 29.
Case 4: c = 4. b = a/2. For b to be a natural number, a must be even. Take a = 2 (smallest even keeping numbers distinct). Then b = 1. Numbers are 2, 1 and 4, all distinct. 3a + 2b + c = 3×2 + 2×1 + 4 = 6 + 2 + 4 = 12.

Step 5: Check whether a smaller value is possible. For c = 1 and c = 2, no natural number solution exists. For c = 3, the minimum value obtained is 29. For c ≥ 5, the value of c itself increases, and the corresponding values of a and b remain positive, making the expression larger than 12. Hence, the smallest possible value is 12.

Step 1: Find the time taken by Ravi to meet Ashok
Relative speed = 40 + 50 = 90 km/h = 90 × 5/18 = 25 m/s
Initial distance = 225 m
Time = 225/25 = 9 seconds

Step 2: Find the distance Vijay must travel
In 9 seconds, Ravi travels: 40 km/h = 100/9 m/s → distance = (100/9) × 9 = 100 m
Meeting point is 54 + 100 = 154 m ahead of Vijay.

Step 3: Find Vijay's speed
Speed = 154/9 m/s
Convert to km/h: (154/9) × 18/5 = 61.6 km/h

Given:
There are 40 employees.
In 2022,
● Average bonus of the first 30 employees = Rs. 40000
● Average bonus of the last 30 employees = Rs. 60000
● Average bonus of the first 10 and last 10 employees together = Rs. 50000
In 2023,
● Bonus of the first 10 employees doubles.
● Bonus of the last 10 employees triples.
● Bonus of the remaining 20 employees remains unchanged.
Find the average bonus of all 40 employees in 2023.

Step 1: Find the total bonuses in 2022
First 30 employees
= 30 × 40000
= Rs. 12,00,000
Last 30 employees
= 30 × 60000
= Rs. 18,00,000
First 10 and last 10 together
= 20 × 50000
= Rs. 10,00,000

Step 2: Find the total bonus of all 40 employees in 2022
Adding the totals of the first 30 and last 30 counts the middle 20 employees twice.
Hence,
Total bonus of all 40 employees
= (12,00,000 + 18,00,000 + 10,00,000)/2
= Rs. 20,00,000

Step 3: Find the bonuses of different groups
Let
First 10 = F
Middle 20 = M
Last 10 = L
Then,
F + M = 12,00,000
M + L = 18,00,000
F + L = 10,00,000
Adding the first two equations,
F + 2M + L = 30,00,000
Since
F + L = 10,00,000,
2M = 20,00,000
M = Rs. 10,00,000
Therefore,
F = 12,00,000 − 10,00,000
= Rs. 2,00,000
L = 18,00,000 − 10,00,000
= Rs. 8,00,000

Step 4: Find the total bonus in 2023
First 10 bonus doubles
= 2 × 2,00,000
= Rs. 4,00,000
Middle 20 remains
= Rs. 10,00,000
Last 10 bonus triples
= 3 × 8,00,000
= Rs. 24,00,000
Total bonus
= 4,00,000 + 10,00,000 + 24,00,000
= Rs. 38,00,000

Step 5: Find the average bonus
Average
= 38,00,000/40
= Rs. 95,000

Answer:
A. 95,000

Let boys = B, girls = G. Remaining: 2B/5 boys, 3G/5 girls. Condition: 3G/5 − 2B/5 = 8 → 3G − 2B = 40.

For integer solutions, let B = 5b, G = 5g → 3g − 2b = 8. For g to be integer, 8+2b ≡ 0 (mod 3) → b ≡ 2 (mod 3). Also B > 10 → b > 2. Smallest valid b = 5 → g = 6. So B = 25, G = 30.

Total = 55. Verification: remaining boys = 10, remaining girls = 18, difference = 8. ✓

Given:
● Amal and Vimal together complete the work in 150 days.
● Vimal and Sunil together complete the work in 100 days.
Amal works for 75 days.
Then Vimal works for 135 days.
Finally, Sunil works for 45 days and completes the task.
Find the number of days required if:
● Amal works every day.
● Vimal works on every second day.
● Sunil works on every third day.

Step 1: Let the daily work rates be
Amal = A
Vimal = V
Sunil = S
Then,
A + V = 1/150
V + S = 1/100

Step 2: Use the second work schedule
The total work done is
75A + 135V + 45S = 1
Using
S = 1/100 − V,
75A + 135V + 45(1/100 − V) = 1
75A + 90V = 11/20
Divide throughout by 15,
5A + 6V = 11/300

Step 3: Solve for the individual rates
From
A + V = 1/150,
A = 1/150 − V
Substitute into
5A + 6V = 11/300,
5(1/150 − V) + 6V = 11/300
1/30 + V = 11/300
V = 1/300
Therefore,
A = 1/150 − 1/300
= 1/300
Also,
S = 1/100 − 1/300
= 1/150

Step 4: Determine the work pattern
Every 6 days,
● Amal works all 6 days.
● Vimal works on days 2, 4 and 6 (3 days).
● Sunil works on days 3 and 6 (2 days).
Work done in 6 days
= 6A + 3V + 2S
= 6 × 1/300 + 3 × 1/300 + 2 × 1/150
= 6/300 + 3/300 + 4/300
= 13/300

Step 5: Find the number of complete cycles
After 23 cycles (138 days),
Work done
= 23 × 13/300
= 299/300
Remaining work
= 1/300

Step 6: Find the last day
On the 139th day,
Amal works.
Since Amal's daily work is
1/300,
the remaining work is completed.
Hence, the total number of days required is
139

Answer: 139

Given:
A train travels a certain distance at a uniform speed.
● If the speed were 6 km/h more, the journey would take 4 hours less.
● If the speed were 6 km/h less, the journey would take 6 hours more.
Find the distance travelled.

Step 1: Let the original speed be x km/h.
Let the original time taken be t hours.
Then,
Distance = xt

Step 2: Form the first equation
If the speed becomes
x + 6,
the time becomes
t − 4.
Hence,
xt = (x + 6)(t − 4)
Expanding,
xt = xt − 4x + 6t − 24
4x − 6t = −24
2x − 3t = −12

Step 3: Form the second equation
If the speed becomes
x − 6,
the time becomes
t + 6.
Hence,
xt = (x − 6)(t + 6)
Expanding,
xt = xt + 6x − 6t − 36
6x − 6t = 36
x − t = 6

Step 4: Solve the equations
From
x − t = 6,
x = t + 6
Substitute into
2x − 3t = −12,
2(t + 6) − 3t = −12
−t + 12 = −12
t = 24
Therefore,
x = 30

Step 5: Find the distance
Distance
= x × t
= 30 × 24
= 720 km

Answer:
A. 720

Let the marks of each girl be g and the marks of each boy be b.
The average marks of 4 girls and 6 boys is 24.
So, Total marks = 10 × 24 = 240
Therefore, 4g + 6b = 240 → 2g + 3b = 120
Hence, g = (120 − 3b)/2
The conditions are:
● Each girl's marks are at least the marks of a boy.
● Each girl's marks are at most twice the marks of a boy.
So, b ≤ g ≤ 2b
Substitute g = (120 − 3b)/2.
Using g ≥ b: (120 − 3b)/2 ≥ b → 120 − 3b ≥ 2b → 120 ≥ 5b → b ≤ 24
Using g ≤ 2b: (120 − 3b)/2 ≤ 2b → 120 − 3b ≤ 4b → 120 ≤ 7b → b ≥ 120/7
Hence, 120/7 ≤ b ≤ 24
Now we need the total marks of 2 girls and 6 boys.
Required total = 2g + 6b
Using 2g = 120 − 3b, Required total = (120 − 3b) + 6b = 120 + 3b
Since b can take any real value in the interval 120/7 ≤ b ≤ 24, the required total varies from 120 + 3 × (120/7) = 171.43... to 120 + 3 × 24 = 192
Therefore, the possible integer values are 172, 173, ..., 192
Number of integers = 192 − 172 + 1 = 21
Answer:
A. 21

Step 1: Let Physics∩Chemistry = x, Chemistry∩Mathematics = x, Physics∩Mathematics = 2x.
Let t = number who chose all three (t ≥ 1).

Step 2: Inclusion-Exclusion
75 + 111 + 40 − (2x + x + x) + t = 150
226 − 4x + t = 150 → t = 4x − 76

Step 3: Constraints
t ≥ 1 → x ≥ 20 (approximately)
t ≤ x → 4x − 76 ≤ x → 3x ≤ 76 → x ≤ 25
So 20 ≤ x ≤ 25.

Step 4: Maximize Physics but not Mathematics = 75 − 2x.
Minimize x: take x = 20 → 75 − 40 = 35.
Verification: t = 4(20) − 76 = 4 ≥ 1 and t ≤ x ✓

Given:
The bus starts every day at 9:00 am.
One day, it travels at 60 km/h and reaches 3.5 hours late.
The next day,
● It covers two-thirds of the route in one-third of the scheduled travel time.
● It covers the remaining one-third of the route at 40 km/h.
● It reaches exactly on time.
Find the scheduled arrival time.

Step 1: Let the total distance be D km.
At 60 km/h, the bus takes
D/60 hours.
If the scheduled travel time is T hours, then
D/60 = T + 3.5

Step 2: Form the second equation
The first two-thirds of the journey is completed in
T/3 hours.
Hence, the speed during this part is
(2D/3)/(T/3)
= 2D/T
The remaining one-third of the journey is covered at 40 km/h.
Time taken for this part
= (D/3)/40
= D/120
Since the total journey is completed in the scheduled time,
T/3 + D/120 = T
D/120 = 2T/3
D = 80T

Step 3: Substitute into the first equation
D/60 = T + 3.5
80T/60 = T + 3.5
4T/3 = T + 3.5
T/3 = 3.5
T = 10.5 hours

Step 4: Find the scheduled arrival time
The bus starts at 9:00 am.
Adding 10.5 hours,
Scheduled arrival time
= 7:30 pm

Answer:
A. 7:30 pm

Step 1: Let the original cost price be x.
Original selling price = 140% of x = 1.4x

Step 2: Form the second condition
New cost price = 60% of x = 0.6x
New selling price = 1.4x − 5
Since the new profit is 50%:
New selling price = 150% of new cost price = 1.5 × 0.6x = 0.9x
Therefore: 1.4x − 5 = 0.9x

Step 3: Solve for x
0.5x = 5 → x = 10

Step 4: Find the original selling price
Original selling price = 1.4 × 10 = Rs. 14

Total amount M = 352n. Two persons receive Rs. 506 each → distributed = 2 × 506 = Rs. 1012. Remaining = 352n − 1012, divided among (n − 2) persons. Each receives (352n − 1012)/(n − 2) ≤ 330. Multiply both sides by (n − 2): 352n − 1012 ≤ 330(n − 2) 352n − 1012 ≤ 330n − 660 22n ≤ 352 n ≤ 16 Maximum possible value of n = 16

Let x = white shirts, y = blue shirts. Step 1: Find the total cost Marked price = 1.25 × Average Cost Selling price per shirt = 0.9 × 1.25 = 1.125 × Average Cost Total profit = (1.125 − 1) × Total Cost = (1/8) × Total Cost Given profit = Rs. 51,000 → Total Cost = 51,000 × 8 = Rs. 4,08,000 Step 2: Form the equation 1000x + 1125y = 408000 Divide by 125: 8x + 9y = 3264 Step 3: Maximize x + y Since 9 ≡ 1 (mod 8), y must be a multiple of 8. Let y = 8k (k ≥ 1): x = (3264 − 72k)/8 = 408 − 9k x + y = (408 − 9k) + 8k = 408 − k To maximize, take k = 1: y = 8, x = 399 Total = 399 + 8 = 407

Let gold = G, bonds = G/4, stocks = 100000 − 5G/4.

Total return: 0.10(100000 − 5G/4) + 0.06(G/4) + 0.08G = 8200 → 10000 − 0.125G + 0.015G + 0.08G = 8200 → 10000 − 0.03G = 8200 → G = 60000.

Bonds = 60000/4 = ₹15000. Gain from bonds = 6% of 15000 = ₹900.

Multiply the first equation by 1/2: a/2 − 3b + 3c = 2. Add to the second equation: 6a + a/2 = 52 → 13a/2 = 52 → a = 8.

From second equation: 48 + 3b − 3c = 50 → 3b − 3c = 2. Therefore 2a + 3b − 3c = 16 + 2 = 18.

Step 1: Let Anil's speed = x km/h and Sunil's speed = y km/h.
They meet in 1.5 hours: (x + y) × 1.5 = 45 → x + y = 30

Step 2: Express the total travel times
Anil's total time = 45/x
Sunil's total time = 45/y
Given: 45/x = 45/y + 1.25

Step 3: Substitute y = 30 − x
45/x = 45/(30 − x) + 1.25
Multiply by 4: 180/x = 180/(30 − x) + 5
Multiply by x(30 − x):
180(30 − x) = 180x + 5x(30 − x)
5400 − 180x = 180x + 150x − 5x²
5400 = 510x − 5x²
5x² − 510x + 5400 = 0
x² − 102x + 1080 = 0
(x − 12)(x − 90) = 0
Since x + y = 30, x = 90 is not possible. Therefore x = 12 km/h

Step 1: Let the four numbers be a, b, c and d.
First condition: (a + b)/2 = a + 1 → b = a + 2

Step 2: Second condition
(a + b + c)/3 = (a + b)/2 + 2 = a + 3
(2a + 2 + c)/3 = a + 3 → c = a + 7

Step 3: Third condition
(a + b + c + d)/4 = (a + b + c)/3 + 3 = a + 6
(3a + 9 + d)/4 = a + 6 → d = a + 15

Step 4: Find the required difference
The four numbers are a, a+2, a+7, a+15
Difference = (a + 15) − a = 15

Given:
The system of equations is
px − 4y = 2
3x + ky = a
Find a necessary condition for the system to have no solution.

Step 1: Condition for no solution
A pair of linear equations
a₁x + b₁y = c₁
a₂x + b₂y = c₂
has no solution if
a₁/a₂ = b₁/b₂ ≠ c₁/c₂

Step 2: Compare the coefficients
Here,
a₁ = p, b₁ = −4, c₁ = 2
a₂ = 3, b₂ = k, c₂ = a
Therefore,
p/3 = −4/k
≠ 2/a

Step 3: Use the first equality
From
p/3 = −4/k,
pk = −12

Step 4: Compare with the constants
Since
2/a must not be equal to p/3,
using
p = −12/k,
2/a ≠ −4/k
Cross-multiplying,
2k ≠ −4a
or,
k ≠ −2a
Hence,
2a + k ≠ 0
This is a necessary condition for the system to have no solution.

Final Answer
Answer:
B. 2a + k ≠ 0

Step 1: Find the interest from Bank A
Interest from Bank A = (10000 × 5 × x)/100 = 500x
Maturity amount = 10000 + 500x

Step 2: Find the interest from Bank B
Interest from Bank B = [(10000 + 500x) × 6 × 5]/100 = 3000 + 150x

Step 3: Use the given ratio
500x : (3000 + 150x) = 10 : 13
13 × 500x = 10(3000 + 150x)
6500x = 30000 + 1500x
5000x = 30000 → x = 6

Given:
● Direct paths between P and Q = 3
● Direct paths between Q and S = 4
● Direct paths between P and R = 4
● No direct path between P and S
● Let the number of direct paths between Q and R = x
● Let the number of direct paths between R and S = y
Also,
● Number of ways to travel from P to S is 62.
● Number of ways to travel from Q to R is 27.
Find x.

Step 1: Form the equation for travel from P to S
Travel is possible in three ways.
Via Q
Ways = 3 × 4 = 12
Via R
Ways = 4 × y = 4y
Via Q and then R
Ways = 3 × x × y = 3xy
Hence,
12 + 4y + 3xy = 62
or,
4y + 3xy = 50
y(4 + 3x) = 50

Step 2: Form the equation for travel from Q to R
Travel is possible in three ways.
Directly
Ways = x
Via P
Ways = 3 × 4 = 12
Via S
Ways = 4 × y = 4y
Hence,
x + 12 + 4y = 27
or,
x + 4y = 15

Step 3: Solve the equations
From the second equation,
4y = 15 − x
Substitute into the first equation,
((15 − x)/4)(4 + 3x) = 50
Multiply by 4,
(15 − x)(4 + 3x) = 200
Expand,
60 + 45x − 4x − 3x² = 200
3x² − 41x + 140 = 0
Factorizing,
(3x − 20)(x − 7) = 0
Thus,
x = 20/3 or x = 7
Since the number of paths must be an integer,
x = 7

Final Answer
The number of direct paths between Q and R is
7

Step 1: Let the price of pure coffee be C Rs/kg and pure cocoa be K Rs/kg. From the first mixture: 0.16C + 0.84K = 240 → 16C + 84K = 24000 → 4C + 21K = 6000. From the second mixture: 0.36C + 0.64K = 320 → 36C + 64K = 32000 → 9C + 16K = 8000.

Step 2: Find the prices of pure coffee and cocoa. Multiply first equation by 9: 36C + 189K = 54000. Multiply second equation by 4: 36C + 64K = 32000. Subtract: 125K = 22000 → K = 176. Substitute into 9C + 16K = 8000: 9C + 2816 = 8000 → 9C = 5184 → C = 576.

Step 3: Let the percentage of coffee in the new mixture be x. Since the new mixture costs Rs. 376 per kg: 576x + 176(1 − x) = 376 → 576x + 176 − 176x = 376 → 400x = 200 → x = 0.5. Thus, the new mixture contains 50% coffee.

Step 4: Find the quantity of coffee in 10 kg. Coffee = 50% of 10 kg = 5 kg.

Given:
● Rajesh owns 20 hectares.
● Vimal owns 30 hectares.
● In Vimal's land, wheat : mustard = 5 : 3.
● Overall, wheat : mustard = 11 : 9.
Find the ratio of wheat to mustard in Rajesh's land.

Step 1: Find cultivation in Vimal's land
Total land = 30 hectares.
Since
Wheat : Mustard = 5 : 3,
Wheat
= (5/8) × 30
= 75/4 hectares
Mustard
= (3/8) × 30
= 45/4 hectares

Step 2: Let Rajesh's cultivation be
Wheat = x hectares
Mustard = 20 − x hectares

Step 3: Use the overall ratio
Total wheat
= x + 75/4
Total mustard
= (20 − x) + 45/4
Given,
(x + 75/4)/((20 − x) + 45/4) = 11/9

Step 4: Solve the equation
Simplify the denominator,
20 + 45/4
= 125/4
Hence,
(x + 75/4)/(125/4 − x) = 11/9
Cross-multiplying,
9(x + 75/4) = 11(125/4 − x)
9x + 675/4 = 1375/4 − 11x
20x = 700/4
20x = 175
x = 35/4

Step 5: Find the required ratio
Mustard area
= 20 − 35/4
= 45/4
Therefore,
Wheat : Mustard
= 35/4 : 45/4
= 7 : 9

Final Answer
Answer:
B. 7 : 9

Let initial students be 13x and 9x. After shifting: (13x−21)/(9x+21) = 19/14 → 182x−294 = 171x+399 → x = 63. Morning = 798, Afternoon = 588.

Let new students join as 3k (morning) and 8k (afternoon). (798+3k)/(588+8k) = 5/4 → 3192+12k = 2940+40k → 28k = 252 → k = 9.

Total new students = 11k = 99.

Step 1: Work backwards
After the third sale, grains left = 0.
Just before the third sale, let quantity = x.
x/2 − 3 = 0 → x = 6 kg

Step 2: Find the quantity before the second sale
Let quantity before second sale = y.
y/2 − 3 = 6 → y/2 = 9 → y = 18 kg

Step 3: Find the initial quantity
Let initial quantity = z.
z/2 − 3 = 18 → z/2 = 21 → z = 42 kg

Given:
|x| + x + y = 15
x + |y| − y = 20
Find the value of
x − y.

Step 1: Analyze the first equation
We know,
● If x ≥ 0, then |x| = x.
● If x < 0, then |x| = −x.
Case 1: x ≥ 0
Then,
2x + y = 15
Case 2: x < 0
Then,
y = 15

Step 2: Analyze the second equation
We know,
● If y ≥ 0, then |y| = y.
● If y < 0, then |y| = −y.
Case 1: y ≥ 0
Then,
x = 20
Case 2: y < 0
Then,
x − 2y = 20

Step 3: Check possible cases
Case 1: x ≥ 0 and y ≥ 0
From the equations,
x = 20
2x + y = 15
40 + y = 15
y = −25
Contradiction since y ≥ 0.
Not possible.
Case 2: x ≥ 0 and y < 0
From the equations,
2x + y = 15
x − 2y = 20
Multiply the first equation by 2,
4x + 2y = 30
Add the second equation,
5x = 50
x = 10
Substituting,
20 + y = 15
y = −5
This satisfies the conditions.
Case 3: x < 0 and y ≥ 0
From the equations,
y = 15
x = 20
Contradiction since x < 0.
Not possible.
Case 4: x < 0 and y < 0
From the equations,
y = 15
Contradiction since y < 0.
Not possible.

Step 4: Find the required value
x − y
= 10 − (−5)
= 15

Answer:
B. 15

Let aₙ = A + (n−1)d₁ and bₙ = B + (n−1)d₂ where d₁ and d₂ are prime.
b₂ = 0 → B + d₂ = 0 → B = −d₂
So bₙ = (n − 2)d₂

From a₅ = b₉: A + 4d₁ = 7d₂ ...(1)
From a₁₉ = b₁₉: A + 18d₁ = 17d₂ ...(2)

Subtract (1) from (2): 14d₁ = 10d₂ → 7d₁ = 5d₂
Since d₁ and d₂ are prime: d₁ = 5, d₂ = 7

Substitute into (1): A + 20 = 49 → A = 29

a₁₁ = A + 10d₁ = 29 + 50 = 79

Step 1: Let initial coins in A = 17x, B = 7x.
After shifting 108: (17x−108)/(7x+108) = 37/20

Step 2: Cross-multiply
20(17x−108) = 37(7x+108)
340x − 2160 = 259x + 3996
81x = 6156 → x = 76

Step 3: After 108 coins shifted:
A = 17(76) − 108 = 1184, B = 7(76) + 108 = 640

Step 4: Let y more coins be shifted for 1:1 ratio.
1184 − y = 640 + y → 544 = 2y → y = 272

Given:
A vessel contains a solution of acid and water.
● After adding 2 litres of water, the solution becomes 50% acid.
● Then, 15 litres of acid is added, making the solution 80% acid.
Find the ratio of water to acid in the original solution.

Step 1: Let the original quantities be
Acid = A litres
Water = W litres

Step 2: Use the first condition
After adding 2 litres of water,
Acid = A
Water = W + 2
Since the solution is 50% acid,
A = W + 2

Step 3: Use the second condition
After adding 15 litres of acid,
Acid = A + 15
Water = W + 2
The acid concentration becomes 80%.
Therefore,
(A + 15)/(A + W + 17) = 4/5
Cross-multiplying,
5(A + 15) = 4(A + W + 17)
5A + 75 = 4A + 4W + 68
A = 4W − 7

Step 4: Solve the equations
From Step 2,
A = W + 2
Substitute into
A = 4W − 7
W + 2 = 4W − 7
3W = 9
W = 3
Hence,
A = 5

Step 5: Find the required ratio
Water : Acid
= 3 : 5

Answer:
B. 3 : 5

Step 1: Let dimensions be l, b and h.
Sum of all 12 edges: 4(l + b + h) = 144 → l + b + h = 36
Surface area: 2(lb + bh + hl) = 846 → lb + bh + hl = 423

Step 2: Find the sum of squares of the dimensions
(l + b + h)² = l² + b² + h² + 2(lb + bh + hl)
36² = l² + b² + h² + 2 × 423
1296 = l² + b² + h² + 846
l² + b² + h² = 450

Step 3: Find the radius of the sphere
Diagonal² = l² + b² + h² = 450
Diagonal = √450 = 15√2
Radius = (15√2)/2

Step 4: Find the volume of the sphere
Volume = (4/3)πr³ = (4/3)π × ((15√2)/2)³
= (4/3)π × (3375 × 2√2)/8
= (4/3)π × (3375√2)/4
= 1125π√2