Daily Quant Practice - 2CRQ - Single Post with Toggle Answer

Q1: Raja divided 35 sweets among his daughters Rani and Sita in the ratio 4: 3. How many sweets did Rani get?

Correct Option:

Correct Answer: 20

Explanation:
The total parts in the ratio are $4+3=7$. The value of one part is $35 \div 7 = 5$ sweets. Rani's share is 4 parts, so she gets $4 \times 5 = 20$ sweets.

Q2: If p:Q=3:4, find 5P: 7Q.

1. $\frac{20}{37}$
2. $\frac{3}{4}$
3. $\frac{15}{28}$
4. $\frac{20}{37}$

Correct Option: C

Correct Answer: 15/28

Explanation:
We are given $\frac{P}{Q} = \frac{3}{4}$. We need to find $\frac{5P}{7Q}$. This can be written as $\frac{5}{7} \times \frac{P}{Q}$. Substituting the given ratio: $\frac{5}{7} \times \frac{3}{4} = \frac{15}{28}$.

Q3: Ratio of two numbers is 3:5 and their sum is 40. Find the smaller of the two numbers.

Correct Option:

Correct Answer: 15

Explanation:
Let the numbers be 3x and 5x. Their sum is $3x + 5x = 40 \implies 8x=40 \implies x=5$. The numbers are $3(5)=15$ and $5(5)=25$
The smaller number is 15.

Q4: If a:b=7:3, find a+b:a-b.

1. 05:02:00
2. 02:05:00
3. 07:03:00
4. 03:07:00

Correct Option: A

Correct Answer: 05:02:00

Explanation:
Let a = 7k and b = 3k. $a+b = 7k+3k = 10k$ $a-b = 7k-3k = 4k$ The ratio $(a+b):(a-b) is $10k:4k$, which simplifies to 5:2.

Q5: If a+b : a-b = 3:2, find a : b.

1. 05:01:00
2. 01:05:00
3. 03:05:00
4. 05:03:00

Correct Option: A

Correct Answer: 05:01:00

Explanation:
We are given $\frac{a+b}{a-b} = \frac{3}{2}$. Cross-multiply: $2(a+b) = 3(a-b) \implies 2a+2b = 3a-3b$. Rearranging the terms gives $5b=a$, so $\frac{a}{b} = \frac{5}{1}$
The ratio is 5:1.

Q6: If $\frac{x+y}{2x+y}=\frac{4}{5}$ then find $\frac{2x+y}{3x+y}$.

1. $\frac{4}{5}$
2. $\frac{5}{6}$
3. $\frac{6}{7}$
4. $\frac{3}{4}$

Correct Option: B

Correct Answer: $\frac{5}{6}$

Explanation:
From the given equation, cross-multiply: $5(x+y) = 4(2x+y) \implies 5x+5y = 8x+4y \implies y=3x$. Now substitute y=3x into the expression to be found: $\frac{2x+(3x)}{3x+(3x)} = \frac{5x}{6x} = \frac{5}{6}$.

Q7: If x+y+z = 120 and $x = \frac{1}{2}y$ and $y = \frac{2}{3}z$. find z.

Correct Option:

Correct Answer: 60

Explanation:
Express x and y in terms of z. $y = \frac{2}{3}z$ $x = \frac{1}{2}y = \frac{1}{2}(\frac{2}{3}z) = \frac{1}{3}z$ Substitute into the sum: $\frac{1}{3}z + \frac{2}{3}z + z = 120 \implies z+z=120 \implies 2z=120 \implies z=60$.

Q8: If a:b=4:1, find $\frac{a-3b}{2a-b^{2}}$.

1. $\frac{2}{7}$
2. $\frac{1}{7}$
3. $\frac{3}{7}$
4. Cannot be determined

Correct Option: D

Correct Answer: Cannot be determined

Explanation:
Let a = 4k and b = k. The expression becomes $\frac{4k-3k}{2(4k)-k^2} = \frac{k}{8k-k^2} = \frac{k}{k(8-k)} = \frac{1}{8-k}$. Since the value of the expression depends on k, which is unknown, the answer cannot be determined.

Q9: (a) Duplicate ratio of 3: 4. (b) Triplicate ratio of 2: 3. (c) Sub-duplicate ratio of 16:9. (d) Mean proportional of 16 and 4.

1. (a)D (b)C (c)B (d)C
2. (a)D (b)C (c)B (d)C
3. (a)D (b)C (c)B (d)C
4. (a)D (b)C (c)B (d)C

Correct Option: D, C, B, C

Correct Answer: (a) 9:16 (b) 8:27 (c) 4:3 (d) 8

Explanation:
The options provided seem to be placeholders
The correct answers are: (a) Duplicate (squaring): $3^2:4^2 \implies 9:16$ (b) Triplicate (cubing): $2^3:3^3 \implies 8:27$ (c) Sub-duplicate (square root): $\sqrt{16}:\sqrt{9} \implies 4:3$ (d) Mean proportional: $\sqrt{16 \times 4} = \sqrt{64} = 8$

Q10: Find the numbers which are in the ratio 3:2:4 such that the sum of the first and the second numbers added to the difference of the third and the second numbers is 21.

1. 12, 8, 16
2. 6, 4, 8
3. 9, 6, 24
4. 9, 6, 12

Correct Option: D

Correct Answer: 9, 6, 12

Explanation:
Let the numbers be 3x, 2x, 4x. The condition is $(3x+2x) + (4x-2x) = 21$. $5x + 2x = 21 \implies 7x=21 \implies x=3$. The numbers are $3(3)=9$, $2(3)=6$, and $4(3)=12$.

Q11: In a class of 30 students, which of the following can't be the ratio of boys and girls?

1. 02:03:00
2. 01:05:00
3. 04:05:00
4. 02:01:00

Correct Option: C

Correct Answer: 04:05:00

Explanation:
For a ratio to be possible, the sum of its parts must be a factor of the total number of students (30). • 2:3 -> Sum=5
30 is divisible by 5
(Possible) • 1:5 -> Sum=6
30 is divisible by 6
(Possible) • 4:5 -> Sum=9
30 is not divisible by 9
(Not possible) • 2:1 -> Sum=3
30 is divisible by 3
(Possible)

Q12: If a: b=2:3 and b:c=5:7 then find a: b: c.

1. 10:15:21
2. 10:21:15
3. 09:12:14
4. 12:07:18

Correct Option: A

Correct Answer: 10:15:21

Explanation:
To combine the ratios, the value for the common term b must be the same
The LCM of the b values (3 and 5) is 15. • a:b = 2:3 (multiply by 5) \implies 10:15 • b:c = 5:7 (multiply by 3) \implies 15:21 Combining them gives a:b:c = 10:15:21.

Q13: At a party, there are a total of 28 adults. If x ladies join the party, the ratio of the number of ladies to that of gents will change from 3: 4 to 5: 4. Find x.

Correct Option:

Correct Answer: 8

Explanation:
Initially, there are 28 adults in the ratio 3:4 (L:G). Number of Gents = $\frac{4}{7} \times 28 = 16$. Number of Ladies = $\frac{3}{7} \times 28 = 12$. After x ladies join, the new number of ladies is 12+x
The number of gents remains 16
The new ratio is 5:4. $\frac{12+x}{16} = \frac{5}{4} \implies 4(12+x) = 80 \implies 48+4x=80 \implies 4x=32 \implies x=8$.

Q14: The monthly salaries of X and Y are in the ratio 3:4. The monthly expenditures of X and Y are in the ratio 4: 5. Find the ratio of the monthly savings of X and Y.

1. 05:03:00
2. 04:07:00
3. 03:05:00
4. Cannot be determined

Correct Option: D

Correct Answer: Cannot be determined

Explanation:
Let salaries be 3k and 4k
Let expenditures be 4m and 5m. Saving = Salary - Expenditure. Savings ratio = $(3k-4m) : (4k-5m)$. Since there is no information to relate the constants k and m, the ratio cannot be simplified to a numerical value.

Q15: The present ages of Rohit and Sunil are in the ratio 3:5. 10 years hence, the ratio of their ages will be 4: 5. Find the present age of Rohit. (in years)

Correct Option:

Correct Answer: 6

Explanation:
Let the present ages be 3x and 5x. In 10 years, their ages will be 3x+10 and 5x+10. The new ratio is $\frac{3x+10}{5x+10} = \frac{4}{5}$. Cross-multiply: $5(3x+10) = 4(5x+10) \implies 15x+50 = 20x+40 \implies 10=5x \implies x=2$. Rohit's present age is $3x = 3(2) = 6$ years.

Q16: x varies directly as the square of y. When $y=8$, $x=192$. Find x when $y=10$.

1. 100
2. 30
3. 300
4. 200

Correct Option: C

Correct Answer: 300

Explanation:
The relation is $x = ky^2$. First, find the constant k: $192 = k(8^2) \implies 192 = 64k \implies k=3$. The formula is $x = 3y^2$. When $y=10$, $x = 3(10^2) = 3(100) = 300$.

Q17: Quantities a and b are inversely proportional to each other. When a=8, b=240. Find b when a=6.

Correct Option:

Correct Answer: 320

Explanation:
The relation is ab = k (a constant). First, find k: $k = 8 \times 240 = 1920$. The formula is ab = 1920. When a=6, $6 \times b = 1920 \implies b = 1920 / 6 = 320$.

Q18: Quantity A varies directly with the sum of the quantities B and C. If B increases by 2 and C increases by 4, by how much does A increase?

1. 2
2. 4
3. 6
4. Cannot be determined

Correct Option: D

Correct Answer: Cannot be determined

Explanation:
The relation is $A = k(B+C)$. Let the initial state be $A_1 = k(B_1+C_1)$. The new state is $A_2 = k((B_1+2)+(C_1+4)) = k(B_1+C_1+6) = k(B_1+C_1) + 6k$. The increase is $A_2 - A_1 = 6k$
Since the constant of proportionality k is unknown, the increase cannot be determined.

Q19: Quantity P varies inversely with the product of Q and R. When Q=6 and R=12, P=75. Find P when Q=5 and R=10.

Correct Option:

Correct Answer: 108

Explanation:
The relation is $PQR = k$ (a constant). First, find k: $k = 75 \times 6 \times 12 = 5400$. The formula is $PQR = 5400$. When Q=5 and R=10, $P \times 5 \times 10 = 5400 \implies 50P = 5400 \implies P = 108$.

Q20: A varies directly with B when C is constant and inversely with C when B is constant. A is 16, when B is 28 and C is 7. Find the value of A, when B is 9 and C is 6.

1. 6
2. 7
3. 8
4. 9

Correct Option: A

Correct Answer: 6

Explanation:
This is joint variation: $A = k\frac{B}{C}$. First, find k: $16 = k \frac{28}{7} \implies 16 = 4k \implies k=4$. The formula is $A = \frac{4B}{C}$. When B=9 and C=6, $A = \frac{4 \times 9}{6} = \frac{36}{6} = 6$.