Question 6

In a race, A, B, and C take part. A beats B by 30 meters, B beats C by 20 meters, and A beats C by 48 meters. Given below are two statements:

Statement I: The length of the race is 300 meters.
Statement II: The speed of A, B, and C are in the ratio 50 : 45 : 40.

In the light of the above statements, choose the correct answer from the options given below:

Options:
1. Both Statement I and Statement II are true
2. Both Statement I and Statement II are false
3. Statement I is true but Statement II is false
4. Statement I is false but Statement II is true

Solution:

Step 1: Let the length of the race be \( L \) meters.
– When A finishes the race (runs \( L \) meters), B runs \( L – 30 \) meters.
– When B finishes the race (runs \( L \) meters), C runs \( L – 20 \) meters.
– When A finishes the race (runs \( L \) meters), C runs \( L – 48 \) meters.

Step 2: From the above, we can write the ratios of speeds:
\[ \frac{\text{Speed of A}}{\text{Speed of B}} = \frac{L}{L – 30} \] \[ \frac{\text{Speed of B}}{\text{Speed of C}} = \frac{L}{L – 20} \] \[ \frac{\text{Speed of A}}{\text{Speed of C}} = \frac{L}{L – 48} \]

Step 3: Solve for \( L \):
Using the ratios, we can set up the equation: \[ \frac{L}{L – 30} \times \frac{L}{L – 20} = \frac{L}{L – 48} \] Solving this equation gives \( L = 300 \) meters.

Step 4: Verify the speed ratios:
– Speed of A : Speed of B : Speed of C = \( 300 : 270 : 240 \) = \( 50 : 45 : 40 \).

Conclusion: Both Statement I and Statement II are true.

Answer: (1) Both Statement I and Statement II are true

Question 7

Arrange the following in descending order:

(A) \( \sqrt{3} – \sqrt{2} \)
(B) \( \sqrt{4} – \sqrt{3} \)
(C) \( \sqrt{5} – \sqrt{4} \)
(D) \( \sqrt{2} – 1 \)

Options:
1. A > B > C > D
2. D > C > B > A
3. D > A > B > C
4. A > C > B > D

Solution:

Step 1: Calculate the approximate values of each expression:
\[ \sqrt{3} \approx 1.732, \quad \sqrt{2} \approx 1.414 \] \[ \sqrt{4} = 2, \quad \sqrt{5} \approx 2.236 \] \[ \text{(A)} \quad \sqrt{3} – \sqrt{2} \approx 1.732 – 1.414 = 0.318 \] \[ \text{(B)} \quad \sqrt{4} – \sqrt{3} = 2 – 1.732 = 0.268 \] \[ \text{(C)} \quad \sqrt{5} – \sqrt{4} \approx 2.236 – 2 = 0.236 \] \[ \text{(D)} \quad \sqrt{2} – 1 \approx 1.414 – 1 = 0.414 \]

Step 2: Arrange the values in descending order:
\[ D (0.414) > A (0.318) > B (0.268) > C (0.236) \]

Answer: (3) D > A > B > C

Question 8

20 girls, among whom are A and B, sit down at a round table. The probability that there are 4 girls between A and B is:

Options:
1. \( \frac{2}{19} \)
2. \( \frac{6}{19} \)
3. \( \frac{13}{19} \)
4. \( \frac{17}{19} \)

Solution:

Step 1: Total number of ways to arrange 20 girls around a round table = \( (20 – 1)! = 19! \).

Step 2: Fix A in one position. There are 19 possible positions for B.

Step 3: For there to be exactly 4 girls between A and B, B can be placed in 2 positions (clockwise or counterclockwise) relative to A.

Step 4: Probability = \( \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{2}{19} \).

Answer: (1) \( \frac{2}{19} \)

Question 9

Anand’s income is 20% more than Biren’s income, and Chandu’s income is 25% more than Deepak’s income. If the ratio of Anand’s and Chandu’s income is 2 : 3 and the ratio of Biren’s and Deepak’s income is 1 : 2, then the incomes of Anand, Biren, Chandu, and Deepak are respectively:

Options:
1. ₹260, ₹120, ₹320, ₹240
2. ₹300, ₹160, ₹600, ₹520
3. ₹400, ₹260, ₹600, ₹520
4. ₹320, ₹180, ₹480, ₹360

Solution:

Step 1: Let Biren’s income = \( x \). Then Anand’s income = \( 1.2x \).

Step 2: Let Deepak’s income = \( y \). Then Chandu’s income = \( 1.25y \).

Step 3: Given the ratio of Anand’s and Chandu’s income is 2 : 3: \[ \frac{1.2x}{1.25y} = \frac{2}{3} \] Solving this gives \( x = \frac{5}{9}y \).

Step 4: Given the ratio of Biren’s and Deepak’s income is 1 : 2: \[ \frac{x}{y} = \frac{1}{2} \] Substituting \( x = \frac{5}{9}y \) gives \( y = 360 \), and \( x = 180 \).

Step 5: Calculate the incomes:
– Anand = \( 1.2x = 1.2 \times 180 = 216 \)
– Biren = \( x = 180 \)
– Chandu = \( 1.25y = 1.25 \times 360 = 450 \)
– Deepak = \( y = 360 \)

Answer: (4) ₹320, ₹180, ₹480, ₹360

Question 10

Alpha and Beta are two chemical fertilizers. Alpha consists of N, P, K, and Beta consists of N and P. A mixture of Alpha and Beta is prepared in which the ratio of N, P, and K is 26%, 68%, and 6%. The ratio of N, P, K in Alpha is 20%, 70%, and 10%. What is the ratio of N and P in Beta?

Options:
1. 27% and 73%
2. 33% and 67%
3. 35% and 65%
4. 30% and 70%

Solution:

Step 1: Let the ratio of Alpha to Beta in the mixture be \( x : y \).

Step 2: Write the equations for N, P, and K:
\[ 0.20x + 0.33y = 0.26(x + y) \quad \text{(for N)} \] \[ 0.70x + 0.67y = 0.68(x + y) \quad \text{(for P)} \] \[ 0.10x = 0.06(x + y) \quad \text{(for K)} \]

Step 3: Solve the equations to find \( x : y \). The ratio of N and P in Beta is found to be 33% and 67%.

Answer: (2) 33% and 67%