Daily Quant Practice - 1CRQ - Set 1

Q1: Solve for x. $3(x+4)+8=5x$

Correct Option:

Correct Answer: 10

Explanation: Step 1: Distribute the 3 on the left side: $3x + 12 + 8 = 5x$. Step 2: Combine the constant terms: $3x + 20 = 5x$. Step 3: Subtract $3x$ from both sides: $20 = 2x$. Step 4: Divide by 2: $x = 10$.

Q2: Solve the following pair of equations for x and y respectively. $12x-10y-2=0$ and $10x-10y+20=0$

1. 13, 11
2. 11, 13
3. 12, 10
4. 10, 12

Correct Option: B

Correct Answer: (11, 13)

Explanation: Step 1: Arrange the equations: (i) $12x - 10y = 2$ (ii) $10x - 10y = -20$ Step 2: Subtract equation (ii) from (i): $(12x - 10y) - (10x - 10y) = 2 - (-20)$, which simplifies to $2x = 22$, so $x = 11$. Step 3: Substitute $x=11$ into equation (i): $12(11) - 10y = 2$, which gives $132 - 10y = 2$. Step 4: Solve for y: $130 = 10y$, so $y = 13$.

Q3: Thrice a number exceeds three-fourth of it by 36. Find the number.

Correct Option:

Correct Answer: 16

Explanation: Step 1: Let the number be x. The equation is $3x - \frac{3}{4}x = 36$. Step 2: Find a common denominator: $\frac{12x - 3x}{4} = 36$. Step 3: Simplify the numerator: $\frac{9x}{4} = 36$. Step 4: Solve for x: $9x = 144$, so $x = 16$.

Q4: Ashok's age, 30 years hence, will be twice his age five years ago. Find his present age. (in years)

Correct Option:

Correct Answer: 40

Explanation: Step 1: Let Ashok's present age be x. Step 2: His age in 30 years will be $x + 30$. His age five years ago was $x - 5$. Step 3: Set up the equation: $x + 30 = 2(x - 5)$. Step 4: Solve for x: $x + 30 = 2x - 10$, which simplifies to $40 = x$.

Q5: Three pens and four erasers cost ₹18. Four pens and three erasers cost ₹17. Find the cost of 14 pens and 14 erasers.

1. 70
2. 60
3. 50
4. 40

Correct Option: A

Correct Answer: 70

Explanation: Step 1: Let the cost of a pen be p and an eraser be e. (i) $3p + 4e = 18$ (ii) $4p + 3e = 17$ Step 2: Add both equations: $7p + 7e = 35$. Step 3: Divide by 7: $p + e = 5$. Step 4: We need the cost of 14 pens and 14 erasers, which is $14p + 14e = 14(p + e)$. Step 5: Substitute $p+e=5$: $14(5) = 70$.

Q6: The cost of two dosas and three idlis is ₹46. The cost of a dosa and two idlis is ₹26. Find the cost of four dosas and four idlis (in ₹).

Correct Option:

Correct Answer: 80

Explanation: Step 1: Let the cost of a dosa be d and an idli be i. (i) $2d + 3i = 46$ (ii) $d + 2i = 26$ Step 2: From (ii), $d = 26 - 2i$. Substitute this into (i): $2(26 - 2i) + 3i = 46$. Step 3: Solve for i: $52 - 4i + 3i = 46$, which gives $i = 6$. Step 4: Substitute $i=6$ into (ii) to find d: $d + 2(6) = 26$, so $d = 14$. Step 5: Cost of four dosas and four idlis: $4d + 4i = 4(14) + 4(6) = 56 + 24 = 80$.

Q7: The digits of a two-digit number differ by six. Find the difference of the number and the number formed by reversing its digits.

Correct Option:

Correct Answer: 54

Explanation: Step 1: Let the number be $10x + y$. The reversed number is $10y + x$. We are given `$

Q8: The difference between a three-digit number and the number formed by reversing its digits is divisible by

1. 9
2. 11
3. Both 9 and 11
4. Neither 9 nor 11

Correct Option: C

Correct Answer: Both 9 and 11

Explanation: Step 1: Let the number be $100x + 10y + z$. The reversed number is $100z + 10y + x$. Step 2: The difference is $(100x + 10y + z) - (100z + 10y + x) = 99x - 99z = 99(x - z)$. Step 3: The result, $99(x-z)$, is always a multiple of 99. Since $99 = 9 \times 11$, the difference is divisible by both 9 and 11.

Q9: The first and the last digits of a three-digit number differ by 4. Find the difference of the number and the number formed by reversing its digits.

Correct Option:

Correct Answer: 396

Explanation: Step 1: Let the number be $100x + 10y + z$. We are given `$

Q10: Five sharpeners and six erasers cost ₹28. Six sharpeners and five erasers cost ₹27. Find the cost (in ₹) of each sharpener and each eraser respectively.

1. 3,2
2. 2,3
3. 1,4
4. 4, 1

Correct Option: B

Correct Answer: (2, 3)

Explanation: Step 1: Let sharpener cost be s and eraser cost be e. (i) $5s + 6e = 28$ (ii) $6s + 5e = 27$ Step 2: Add (i) and (ii): $11s + 11e = 55 \implies s + e = 5$. Step 3: Subtract (ii) from (i): $-s + e = 1$. Step 4: Add the two new equations: $2e = 6 \implies e = 3$. Step 5: Substitute $e=3$ into $s+e=5$ to get $s=2$. Thus, the cost is ₹2 for a sharpener and ₹3 for an eraser.

Q11: How many pairs of x and y satisfy $3x+6y=18$ and $9x+18y=57$?

1. 2
2. 1
3. 0
4. None of these

Correct Option: C

Correct Answer: 0

Explanation: Step 1: For a system $a_1x + b_1y = c_1$ and $a_2x + b_2y = c_2$, we check the ratio of coefficients. Step 2: $a_1/a_2 = 3/9 = 1/3$. Step 3: $b_1/b_2 = 6/18 = 1/3$. Step 4: $c_1/c_2 = 18/57 = 6/19$. Step 5: Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the lines are parallel and have no solution.

Q12: How many pairs of x and y satisfy the equations $4x+6y=16$ and $6x+9y=24$?

1. 0
2. 1
3. ∞
4. None of these

Correct Option: C

Correct Answer: ∞

Explanation: Step 1: Check the ratio of coefficients. Step 2: $a_1/a_2 = 4/6 = 2/3$. Step 3: $b_1/b_2 = 6/9 = 2/3$. Step 4: $c_1/c_2 = 16/24 = 2/3$. Step 5: Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, the lines are coincident (the same line) and have infinitely many solutions.

Q13: How many pairs of x and y satisfy the equations $6x+5y=16$ and $8x+7y=22$?

1. 0
2. 1
3. ∞
4. None of these

Correct Option: B

Correct Answer: 1

Explanation: Step 1: Check the ratio of coefficients. Step 2: $a_1/a_2 = 6/8 = 3/4$. Step 3: $b_1/b_2 = 5/7$. Step 4: Since $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, the lines intersect at a single point, meaning there is exactly one unique solution.

Q14: Three chocolates, four biscuits and five cakes cost ₹34. Six chocolates and eight biscuits cost ₹38. Find the cost of each cake (in ₹).

Correct Option:

Correct Answer: 3

Explanation: Step 1: Let the costs be c, b, and k. (i) $3c + 4b + 5k = 34$ (ii) $6c + 8b = 38$ Step 2: Divide equation (ii) by 2: $3c + 4b = 19$. Step 3: Substitute this result into equation (i): $19 + 5k = 34$. Step 4: Solve for k: $5k = 15 \implies k = 3$.

Q15: The sum of a two-digit number and its reverse is k times the sum of its digits. Find the value of k.

1. 9
2. 10
3. 11
4. Cannot be determined

Correct Option: C

Correct Answer: 11

Explanation: Step 1: Let the number be $10x + y$. The sum of its digits is $x + y$. Step 2: The reversed number is $10y + x$. Step 3: The sum of the number and its reverse is $(10x + y) + (10y + x) = 11x + 11y = 11(x + y)$. Step 4: We are given $11(x + y) = k(x + y)$. Step 5: Dividing both sides by $(x+y) gives $k = 11$.