Daily Quant Practice - 1CRQ - Single Post with Pagination
Q1: Solve for x.
$3(x+4)+8=5x$
Correct Option:
Correct Answer: 10
Explanation: 1: Distribute the 3 on the left side: $3x + 12 + 8 = 5x$. 2: Combine the constant terms: $3x + 20 = 5x$. 3: Subtract $3x$ from both sides: $20 = 2x$. 4: Divide by 2: $x = 10$.
Q2: Solve the following pair of equations for x and y respectively.
$12x-10y-2=0$ and $10x-10y+20=0$
13, 11
11, 13
12, 10
10, 12
Correct Option: B
Correct Answer: (11, 13)
Explanation: 1: Arrange the equations:
(i) $12x - 10y = 2$
(ii) $10x - 10y = -20$ 2: Subtract equation (ii) from (i): $(12x - 10y) - (10x - 10y) = 2 - (-20)$, which simplifies to $2x = 22$, so $x = 11$. 3: Substitute $x=11$ into equation (i): $12(11) - 10y = 2$, which gives $132 - 10y = 2$. 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: 1: Let the number be x The equation is $3x - \frac{3}{4}x = 36$. 2: Find a common denominator: $\frac{12x - 3x}{4} = 36$. 3: Simplify the numerator: $\frac{9x}{4} = 36$. 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: 1: Let Ashok's present age be x. 2: His age in 30 years will be $x + 30$ His age five years ago was $x - 5$. 3: Set up the equation: $x + 30 = 2(x - 5)$. 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.
70
60
50
40
Correct Option: A
Correct Answer: 70
Explanation: 1: Let the cost of a pen be p and an eraser be e (i) $3p + 4e = 18$
(ii) $4p + 3e = 17$ 2: Add both equations: $7p + 7e = 35$. 3: Divide by 7: $p + e = 5$. 4: We need the cost of 14 pens and 14 erasers, which is $14p + 14e = 14(p + e)$. 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: 1: Let the cost of a dosa be d and an idli be i (i) $2d + 3i = 46$
(ii) $d + 2i = 26$ 2: From (ii), $d = 26 - 2i$ Substitute this into (i): $2(26 - 2i) + 3i = 46$. 3: Solve for i: $52 - 4i + 3i = 46$, which gives $i = 6$. 4: Substitute $i=6$ into (ii) to find d: $d + 2(6) = 26$, so $d = 14$. 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: 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
9
11
Both 9 and 11
Neither 9 nor 11
Correct Option: C
Correct Answer: Both 9 and 11
Explanation: 1: Let the number be $100x + 10y + z$ The reversed number is $100z + 10y + x$. 2: The difference is $(100x + 10y + z) - (100z + 10y + x) = 99x - 99z = 99(x - z)$. 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: 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.
3,2
2,3
1,4
4, 1
Correct Option: B
Correct Answer: (2, 3)
Explanation: 1: Let sharpener cost be s and eraser cost be e (i) $5s + 6e = 28$
(ii) $6s + 5e = 27$ 2: Add (i) and (ii): $11s + 11e = 55 \implies s + e = 5$. 3: Subtract (ii) from (i): $-s + e = 1$. 4: Add the two new equations: $2e = 6 \implies e = 3$. 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 $9x+18y=57$?
2
1
0
None of these
Correct Option: C
Correct Answer: 0
Explanation: 1: For a system $a_1x + b_1y = c_1$ and $a_2x + b_2y = c_2$, we check the ratio of coefficients. 2: $a_1/a_2 = 3/9 = 1/3$. 3: $b_1/b_2 = 6/18 = 1/3$. 4: $c_1/c_2 = 18/57 = 6/19$. 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 $6x+9y=24$?
0
1
∞
None of these
Correct Option: C
Correct Answer: ∞
Explanation: 1: Check the ratio of coefficients. 2: $a_1/a_2 = 4/6 = 2/3$. 3: $b_1/b_2 = 6/9 = 2/3$. 4: $c_1/c_2 = 16/24 = 2/3$. 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 $8x+7y=22$?
0
1
∞
None of these
Correct Option: B
Correct Answer: 1
Explanation: 1: Check the ratio of coefficients. 2: $a_1/a_2 = 6/8 = 3/4$. 3: $b_1/b_2 = 5/7$. 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: 1: Let the costs be c, b, and k (i) $3c + 4b + 5k = 34$
(ii) $6c + 8b = 38$ 2: Divide equation (ii) by 2: $3c + 4b = 19$. 3: Substitute this result into equation (i): $19 + 5k = 34$. 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.
9
10
11
Cannot be determined
Correct Option: C
Correct Answer: 11
Explanation: 1: Let the number be $10x + y$ The sum of its digits is $x + y$. 2: The reversed number is $10y + x$. 3: The sum of the number and its reverse is $(10x + y) + (10y + x) = 11x + 11y = 11(x + y)$. 4: We are given $11(x + y) = k(x + y)$. 5: Dividing both sides by $(x+y) gives $k = 11$.
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