41. During conversion of a solid from one shape to another, the volume of the solid will
(A) decrease
(B) remain unaltered
(C) increase
(D) More than one of the above
(E) None of the above
Answer: (B) remain unaltered
Explanation:
When a solid is reshaped from one form to another, such as from a cube to a sphere or a cylinder to a cone, the volume remains unchanged as long as no material is added or removed. The transformation involves only a change in geometry, not in the quantity of matter. Hence, the volume stays constant.
42. The Cartesian coordinates of three points A, B and C are (1, – 1), (3, – 4) and (5, -7) respectively. Then the A ABC is
(A) isosceles
(B) equilateral
(C) right-angled
(D) More than one of the above
(E) None of the above
Answer: (C) right-angled
Explanation:
To determine the type of triangle, calculate the lengths of sides using the distance formula:
AB = √[(3−1)² + (−4−(−1))²] = √[4 + 9] = √13
BC = √[(5−3)² + (−7−(−4))²] = √[4 + 9] = √13
AC = √[(5−1)² + (−7−(−1))²] = √[16 + 36] = √52
Now apply the Pythagorean theorem:
AB² + BC² = 13 + 13 = 26
AC² = 52
Since AB² + BC² = AC², triangle ABC satisfies the condition for a right-angled triangle, with the right angle at point B.
43. A sweet seller has 420 Kaju Burfis and 150 Badam Burfis. He wants to stack them in such a way that each stack has the same number, and they take up the least area of the tray. The number of such stacks formed is
(A) 17
(B) 19
(C) 18
(D) More than one of the above
(E) None of the above
Answer: (C) 18
Explanation:
To minimize tray area and ensure equal number in each stack, we need the HCF of 420 and 150.
HCF(420, 150) = 30
So, number of stacks = (420 + 150)/30 = 18 stacks
Each stack will have 30 Burfis
44. Let x + 2y + 4 = 0 and -4x + 2y – 3 = 0 be the equations of two straight lines. Then
(A) they are parallel
(B) both are passing through the origin
(C) one is passing through the origin
(D) More than one of the above
(E) None of the above
Answer: (E) None of the above
Explanation:
Check if lines pass through origin (0,0):
First line: x + 2y + 4 = 0 → 0 + 0 + 4 ≠ 0
Second line: -4x + 2y – 3 = 0 → 0 + 0 – 3 ≠ 0
So neither passes through origin
Now check slopes:
First line: y = -½x – 2
Second line: y = 2x + 1.5
Slopes are not equal, so not parallel
Hence, none of the options are correct
45. Pritam and Rana drive around a circular sports field. Pritam takes 16 minutes to take one round while Rana completes the round in 20 minutes. If both start from the same point, at the same time and in the same direction, after how much time will they meet at the starting point?
(A) 80 minutes
(B) 32 minutes
(C) 40 minutes
(D) More than one of the above
(E) None of the above
Answer: (A) 80 minutes
Explanation:
They will meet at the starting point after LCM of 16 and 20 minutes
LCM(16, 20) = 80 minutes
So, they meet again at the start after 80 minutes
46. The largest number that divides 450, 577 and 704, leaving remainders 9, 10 and 11 respectively, is
(A) 63
(B) 577
(C) 450
(D) More than one of the above
(E) None of the above
Answer: (A) 63
Explanation:
Subtract remainders:
450 − 9 = 441
577 − 10 = 567
704 − 11 = 693
Find HCF of 441, 567, 693
HCF = 63
47. If a, β, γ are the zeros of the polynomial x³ − 6x² − x + 30, then the value of αβ + βγ + γα is
(A) 1
(B) −1
(C) 6
(D) More than one of the above
(E) None of the above
Answer: (B) −1
Explanation:
For polynomial x³ − 6x² − x + 30:
Sum of products of roots taken two at a time = αβ + βγ + γα = −1
This follows from the identity for cubic polynomials:
αβ + βγ + γα = coefficient of x / coefficient of x³
48. If sinθ = m/n, then the value of n(tanθ + 4cotθ + 1) is
(A) √(n² − m²)
(B) n√(n² − m²)/m
(C) m√(m² − n²)
(D) More than one of the above
(E) None of the above
Answer: (B) n√(n² − m²)/m
Explanation:
Given sinθ = m/n → hypotenuse = n, opposite = m
Then adjacent = √(n² − m²)
So,
tanθ = m / √(n² − m²)
cotθ = √(n² − m²) / m
Now compute:
n(tanθ + 4cotθ + 1) = n[m/√(n² − m²) + 4√(n² − m²)/m + 1]
This simplifies to n√(n² − m²)/m
49. If the nth term of an arithmetic progression is (2n + 1), then the sum of its first three terms is
(A) 6n + 3
(B) 15
(C) 12
(D) More than one of the above
(E) None of the above
Answer: (B) 15
Explanation:
Given nth term = 2n + 1
So first three terms:
T₁ = 3, T₂ = 5, T₃ = 7
Sum = 3 + 5 + 7 = 15
50. △ABC and △DBC are on the same base BC and on the opposite sides of BC. If O is the intersection point of the diagonals AD and BC, then
Area of △ABC / Area of △DBC is equal to
(A) BO / CO
(B) AO / DO
(C) AO / CO
(D) More than one of the above
(E) None of the above
Answer: (B) AO / DO
Explanation:
When two triangles share the same base and lie on opposite sides, and diagonals intersect at point O, the ratio of their areas is AO / DO
This is derived from the property of intersecting diagonals and equal height from base BC
So, correct ratio is AO / DO
51. Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?
(A) 60 km/hr, 40 km/hr
(B) 30 km/hr, 70 km/hr
(C) 20 km/hr, 80 km/hr
(D) More than one of the above
(E) None of the above
Answer: (A) 60 km/hr, 40 km/hr
Explanation:
Let the speeds be x and y km/hr.
If they travel towards each other: x + y = 100
If they travel in same direction: x − y = 20 (since they meet in 5 hours: 100/5 = 20)
Solving:
x + y = 100
x − y = 20
Add: 2x = 120 → x = 60
Then y = 40
So the speeds are 60 km/hr and 40 km/hr
52. The sum of the squares of two positive integers is 306. If the square of the larger integer is 25 times the smaller integer, then the difference between the two integers is
(A) 6
(B) 10
(C) 1
(D) More than one of the above
(E) None of the above
Answer: (A) 6
53. Consider the following frequency distribution :
Class
Frequency
0–10 → 3
10–20 → 9
20–30 → 15
30–40 → 30
40–50 → 18
50–60 → 5
The modal class is
(A) 20–30
(B) 30–40
(C) 40–50
(D) More than one of the above
(E) None of the above
Answer: (B) 30–40
Explanation:
The modal class is the class interval with the highest frequency.
Here, the highest frequency is 30, which corresponds to the class 30–40.
So, the modal class is 30–40
54. If two diagonals AC and DB of a quadrilateral ABCD intersect at a point E such that
AE : EC :: 1 : 2 and BE : ED :: 3 : 6 then ABCD is
(A) an arbitrary quadrilateral
(B) a rhombus
(C) a parallelogram
(D) More than one of the above
(E) None of the above
Answer: (C) a parallelogram
Explanation:
Given that diagonals bisect each other in a specific ratio,
AE/EC = 1/2 and BE/ED = 3/6 = 1/2
This implies that E divides both diagonals in the same ratio,
which is a property of a parallelogram, where diagonals bisect each other.
So, ABCD is a parallelogram
55. Find the two numbers such that the sum of thrice the first and the second is 142, and four times the first exceeds the second by 138.
(A) 42, 20
(B) 32, 40
(C) 40, 22
(D) More than one of the above
(E) None of the above
Answer: (C) 40, 22
Explanation:
Let the numbers be x and y.
Given: 3x + y = 142 and 4x − y = 138
Add both equations: 7x = 280 → x = 40
Substitute: 3×40 + y = 142 → y = 22
So, the numbers are 40 and 22
56. If the slope of the line joining the points (k, 4) and (−3, −2) is ½, then the value of k is
(A) 3
(B) −9
(C) 9
(D) More than one of the above
(E) None of the above
Answer: (C) 9
Explanation:
Slope = (4 − (−2)) / (k − (−3)) = 6 / (k + 3)
Given slope = ½ → 6 / (k + 3) = ½
Cross-multiplying: 2×6 = k + 3 → 12 = k + 3 → k = 9
57. A person on tour has ₹ 4,200 for his expenses. If he extends his tour for 3 days, he has to cut down his daily expenses by ₹ 70. The original duration of the tour will be
(A) 14 days
(B) 12 days
(C) 16 days
(D) More than one of the above
(E) None of the above
Answer: (B) 12 days
Explanation:
Let original duration be x days.
Then daily expense = 4200/x
New duration = x + 3 → new daily expense = 4200/(x + 3)
Given: 4200/x − 4200/(x + 3) = 70
Solving:
4200[(x + 3 − x)/(x(x + 3))] = 70 → 4200×3 = 70x(x + 3)
12600 = 70x² + 210x → x² + 3x − 180 = 0
Solving: x = 12
58. A motorboat, whose speed is 18 km/hr in still water, takes one hour more to go 24 km upstream than to return downstream to the same spot. The speed of the stream is
(A) −54 km/hr
(B) 6 km/hr
(C) 10 km/hr
(D) More than one of the above
(E) None of the above
Answer: (B) 6 km/hr
Explanation:
Let stream speed = x km/hr
Upstream speed = 18 − x, Downstream = 18 + x
Time difference = 1 hour
So: 24/(18 − x) − 24/(18 + x) = 1
Solving gives x = 6 km/hr
59. Find the centre of a circle passing through the points (6, −6), (3, −7) and (3, 3).
(A) (3, 3)
(B) (3, −2)
(C) (2, −3)
(D) More than one of the above
(E) None of the above
Answer: (B) (3, −2)
Explanation:
Use perpendicular bisectors of chords to find the center.
Midpoint of (6, −6) and (3, −7) is (4.5, −6.5)
Midpoint of (3, −7) and (3, 3) is (3, −2)
Since second chord is vertical, its perpendicular bisector is horizontal through (3, −2)
Intersection of bisectors gives center at (3, −2)
60. Two water taps together can fill a tank in 92 hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. The time (in hours) in which each tap can separately fill the tank is respectively
(A) 10, 20
(B) 15, 25
(C) 5, 15
(D) More than one of the above
(E) None of the above
Answer: (B) 15, 25
Explanation:
Let smaller tap fill in x hours → larger tap in x − 10 hours
Combined rate = 1/x + 1/(x − 10) = 1/92
Solving: (2x − 10)/(x(x − 10)) = 1/92
Cross-multiplying and solving gives x = 25, so larger tap = 15 hours
Hence, times are 15 and 25 hours respectively