SSC CGL 24 September 2025 · Shift 3 · Quantitative Aptitude — questions with answers · ShikshaSphere
Chapter 155 of 192
24 September 2025 · Shift 3
Quantitative Aptitude
25 questions ~25 min readFree
25 questions
A
51
What will take the place of "?" in the given equation:
=
A.7
B.8
C.9
D.4
Solution
Proportion with surds
Simplify the right side: = 11 and 10 + 1 = 11.
So the right side = = 1.
Hence = 1, i.e. ? = .
? = 8 — option (b).
52
What will be the ratio of the simple interest earned on a given amount at the same rate over 6 years compared to that earned over 10 years?
A.1 : 3
B.3 : 5
C.2 : 3
D.None of these
Solution
SI ratio
SI = ; with P and R fixed, SI is proportional to T.
So the ratio is simply 6 : 10.
Divide both by 2.
= 3 : 5 — option (b).
53
A train covers 360 km in 4 hours. What is its average speed in metres per second?
A.29 m/s
B.30 m/s
C.20 m/s
D.25 m/s
Solution
Speed conversion
Speed = = 90 km/h.
To convert km/h into m/s, multiply by .
90 × .
= 25 m/s — option (d).
54
Two ships positioned on opposite sides of a lighthouse observe the angles of elevation to its top as 30° and 45° respectively. Given that the height of the lighthouse is 200 metres, determine the distance between the two ships.
A.200( − 1) m
B. m
C.200( + 1) m
D.200( + 1) m
Solution
Two angles of elevation
For the 30° ship: tan30° = ⇒ BD = 200.
For the 45° ship: tan45° = ⇒ CD = 200.
The ships are on opposite sides, so the distance = BD + CD.
= 200 + 200 = 200( + 1) m — option (d).
55
A bathroom floor measures 4.5 m in length and 3 m in width. One of its walls, which is to be tiled fully, measures 2 m in length and 2.5 m in height. If square tiles of side 25 cm are used, and an extra 12% of tiles are purchased to cover cuts and wastage, how many tiles are needed in total?
A.90
B.70
C.95
D.80
Solution
Tiling a wall
Only the wall is to be tiled: area = 2 × 2.5 = 5 m².
Each tile is 0.25 m × 0.25 m = 0.0625 m², so plain requirement = = 80 tiles.
Extra 12% = 80 × = 9.6, i.e. 10 more tiles.
Total = 80 + 10 = 90 — option (a).
56
If two partners invest for the same period but share profits in the ratio 3 : 5, and the total profit is ₹24,000, what is the capital of the second partner if the first invested ₹12,000?
A.₹15,000
B.₹20,000
C.₹25,000
D.₹18,000
Solution
Capital from profit ratio
When the time is the same, profits are shared in the ratio of the capitals.
So capital ratio = 3 : 5, where 3 parts = ₹12,000.
1 part = ₹4,000.
Second partner = 5 × 4,000 = ₹20,000 — option (b).
57
If half the interior angle of a regular polygon is k times its exterior angle, express the number of sides (n) of the polygon in terms of k.
A.2k + 4
B.2k − 4
C.4k + 2
D.4k − 3
Solution
Polygon angles in terms of k
Let the exterior angle be x, so the interior angle is 180° − x.
Given (180 − x) = kx ⇒ 180 − x = 2kx ⇒ 180 = x(2k + 1), so x = .
n = = 360 × .
= 2(2k + 1) = 4k + 2 — option (c).
58
A man spends 80% of his income. Due to inflation his expenses increase by 10% and his income increases by 5%. What is the percentage change in his savings?
A.30% decrease
B.15% decrease
C.10% decrease
D.20% decrease
Solution
Change in savings
Take income = ₹100, so expenses = ₹80 and savings = ₹20.
New income = 100 × 1.05 = ₹105; new expenses = 80 × 1.10 = ₹88.
New savings = 105 − 88 = ₹17.
Change = × 100 = 15% decrease — option (b).
59
A square tile of area 289 cm² is cut into two equal rectangular tiles. What will be the perimeter of each new tile?
A.48 cm
B.52 cm
C.51 cm
D.50 cm
Solution
Square cut in two
Side of the square = = 17 cm.
Cutting it in half gives rectangles of 17 cm × 8.5 cm.
Perimeter = 2(17 + 8.5).
= 2 × 25.5 = 51 cm — option (c).
60
A box has 120 pencils in red, blue and black in the ratio 3 : 2 : 5. If 10 red and 20 black pencils are removed, what is the new ratio of red to black pencils?
A.1 : 2
B.3 : 5
C.13 : 15
D.13 : 20
Solution
Ratio after removal
Total 10 parts = 120, so 1 part = 12: red 36, blue 24, black 60.
After removal: red = 36 − 10 = 26 and black = 60 − 20 = 40.
Ratio = 26 : 40.
= 13 : 20 — option (d).
61
Two similar hexagons have their perimeters in the ratio 3 : 4. If the smaller hexagon has an area of 90 cm², what is the area of the larger hexagon?
A.140 cm²
B.155 cm²
C.160 cm²
D.175 cm²
Solution
Similar figures
For similar figures the areas are in the ratio of the SQUARES of corresponding lengths.
So area ratio = 3² : 4² = 9 : 16.
9 parts = 90 cm², so 1 part = 10 cm².
Larger area = 16 × 10 = 160 cm² — option (c).
62
A shopkeeper bought 200 toys at ₹50 each. He sold 80% of the toys at a 10% profit and the rest at a 10% loss. What is the overall profit or loss percentage?
A.8% profit
B.6% profit
C.5% loss
D.6% loss
Solution
Mixed profit and loss
Total cost = 200 × 50 = ₹10,000.
160 toys at +10%: 160 × 55 = ₹8,800; 40 toys at −10%: 40 × 45 = ₹1,800.
Total revenue = ₹10,600, so profit = ₹600.
× 100 = 6% profit — option (b).
63
The average of 15 results is 80. If the top two results are excluded, the average becomes 78. What is the average of the top two results?
A.90
B.100
C.93
D.83
Solution
Average of the top two
Total of 15 results = 15 × 80 = 1200.
Total of the remaining 13 = 13 × 78 = 1014.
Top two together = 1200 − 1014 = 186.
Average = = 93 — option (c).
64
A seller gives a 10% discount and charges 10% GST on the discounted price. On a ₹5,000 item, how much is payable?
A.₹4,900
B.₹4,950
C.₹4,980
D.₹5,000
Solution
Discount then GST
After a 10% discount: 5,000 × = ₹4,500.
GST of 10% is charged on this discounted price.
4,500 × .
= ₹4,950 — option (b).
65
What is the length of a diagonal of a rectangle with length 30 cm and width 16 cm?
A.32 cm
B.38 cm
C.36 cm
D.34 cm
Solution
Diagonal of a rectangle
Diagonal = .
= = .
= .
= 34 cm — option (d).
66
A shopkeeper offers a festival deal: "Flat ₹175 off on every purchase above ₹1,100." What is the effective discount percentage if a buyer purchases an item marked ₹1,400?
A.12.5%
B.15.5%
C.10.25%
D.11.25%
Solution
Flat discount as a percentage
Discount% = × 100.
= × 100.
= × 100.
= 12.5% — option (a).
67
Ankit can copy 50 pages in 10 hours; Ankit and Pankaj together can copy 300 pages in 40 hours. In how much time can Pankaj copy 25 pages?
A.10 hours
B.12 hours
C.15 hours
D.8 hours
Solution
Combined work rate
Ankit’s rate = = 5 pages per hour.
Together = = 7.5 pages per hour.
So Pankaj alone = 7.5 − 5 = 2.5 pages per hour.
Time for 25 pages = = 10 hours — option (a).
68
The average of three numbers is 44. The first number is 25% more than the second, and the third is 25% less than the second. What is the second number?
A.44
B.45
C.48
D.50
Solution
Three numbers from an average
Take the second number as 100 units; then the first is 125 and the third is 75.
So the three are in the ratio 125 : 100 : 75 = 5 : 4 : 3, a total of 12 units.
Sum of the numbers = 3 × 44 = 132, so 12 units = 132 and 1 unit = 11.
Second number = 4 × 11 = 44 — option (a).
69
If secA − tanA = , then what is the value of secA + tanA?
A.8
B.6
C.4
D.2
Solution
sec and tan identity
The identity sec²A − tan²A = 1 gives (secA − tanA)(secA + tanA) = 1.
So secA + tanA is the reciprocal of secA − tanA.
= 1 ÷ .
= 4 — option (c).
70
A 40-litre mixture of milk and water contains them in the ratio 3 : 1. Some quantity of water is added, changing the ratio to 3 : 5. What is the quantity of water added?
A.20 L
B.40 L
C.35 L
D.25 L
Solution
Adding water to a mixture
Milk = × 40 = 30 L and water = 10 L.
The milk stays 30 L, and in the new ratio 3 parts = 30 L, so 1 part = 10 L.
Water must now be 5 parts = 50 L.
Water added = 50 − 10 = 40 L — option (b).
If sin(2x) = cos(3x − 60°), what is the value of x?
A.30°
B.25°
C.18°
D.20°
Solution
sin = cos
sinA = cos(90° − A), so the two angles must be complementary.
2x + (3x − 60°) = 90°.
5x = 150°.
x = 30° — option (a).
73
If a triangle has a right angle, where is its orthocentre located?
A.Outside the triangle
B.On the midpoint of the hypotenuse
C.At the vertex of the right angle
D.Inside the triangle
Solution
Orthocentre
In a right triangle the two legs are themselves altitudes, being perpendicular to each other.
They meet at the right-angled vertex, and the third altitude passes through it as well.
So the orthocentre sits exactly at that vertex.
Hence option (c).
74
In a circle with centre O, chords AB and CD are equal in length. If the distance from O to AB is 16 cm, what is the distance from O to CD?
A.4 cm
B.8 cm
C.16 cm
D.12 cm
Solution
Equal chords
Equal chords of a circle are equidistant from the centre.
Since AB = CD, their distances from O must be the same.
The distance to AB is 16 cm.
So the distance to CD is also 16 cm — option (c).
75
A circle has a diameter of 18 centimetres. A line is tangent to the circle. What is the angle between the tangent and the diameter at the point of contact?
A.30°
B.45°
C.90°
D.180°
Solution
Tangent and radius
The radius drawn to the point of contact is always perpendicular to the tangent.
The diameter passes through the centre and the point of contact, so it lies along that radius.
The size of the diameter makes no difference to the angle.
The angle is 90° — option (c).
Finished reading? Now test yourself.
The same 25 questions, but with a timer and the answers hidden — 15 minutes.
Simplify the right side: = 11 and 10 + 1 = 11.
So the right side = = 1.
Hence = 1, i.e. ? = .
? = 8 — option (b).
SI = ; with P and R fixed, SI is proportional to T.
So the ratio is simply 6 : 10.
Divide both by 2.
= 3 : 5 — option (b).
Speed = = 90 km/h.
To convert km/h into m/s, multiply by .
90 × .
= 25 m/s — option (d).
For the 30° ship: tan30° = ⇒ BD = 200.
For the 45° ship: tan45° = ⇒ CD = 200.
The ships are on opposite sides, so the distance = BD + CD.
= 200 + 200 = 200( + 1) m — option (d).
Only the wall is to be tiled: area = 2 × 2.5 = 5 m².
Each tile is 0.25 m × 0.25 m = 0.0625 m², so plain requirement = = 80 tiles.
Extra 12% = 80 × = 9.6, i.e. 10 more tiles.
Total = 80 + 10 = 90 — option (a).
When the time is the same, profits are shared in the ratio of the capitals.
So capital ratio = 3 : 5, where 3 parts = ₹12,000.
1 part = ₹4,000.
Second partner = 5 × 4,000 = ₹20,000 — option (b).
Let the exterior angle be x, so the interior angle is 180° − x.
Given (180 − x) = kx ⇒ 180 − x = 2kx ⇒ 180 = x(2k + 1), so x = .
n = = 360 × .
= 2(2k + 1) = 4k + 2 — option (c).
Take income = ₹100, so expenses = ₹80 and savings = ₹20.
New income = 100 × 1.05 = ₹105; new expenses = 80 × 1.10 = ₹88.
New savings = 105 − 88 = ₹17.
Change = × 100 = 15% decrease — option (b).
Side of the square = = 17 cm.
Cutting it in half gives rectangles of 17 cm × 8.5 cm.
Perimeter = 2(17 + 8.5).
= 2 × 25.5 = 51 cm — option (c).
Total 10 parts = 120, so 1 part = 12: red 36, blue 24, black 60.
After removal: red = 36 − 10 = 26 and black = 60 − 20 = 40.
Ratio = 26 : 40.
= 13 : 20 — option (d).
For similar figures the areas are in the ratio of the SQUARES of corresponding lengths.
So area ratio = 3² : 4² = 9 : 16.
9 parts = 90 cm², so 1 part = 10 cm².
Larger area = 16 × 10 = 160 cm² — option (c).
Total cost = 200 × 50 = ₹10,000.
160 toys at +10%: 160 × 55 = ₹8,800; 40 toys at −10%: 40 × 45 = ₹1,800.
Total revenue = ₹10,600, so profit = ₹600.
× 100 = 6% profit — option (b).
Total of 15 results = 15 × 80 = 1200.
Total of the remaining 13 = 13 × 78 = 1014.
Top two together = 1200 − 1014 = 186.
Average = = 93 — option (c).
After a 10% discount: 5,000 × = ₹4,500.
GST of 10% is charged on this discounted price.
4,500 × .
= ₹4,950 — option (b).
Diagonal = .
= = .
= .
= 34 cm — option (d).
Discount% = × 100.
= × 100.
= × 100.
= 12.5% — option (a).
Ankit’s rate = = 5 pages per hour.
Together = = 7.5 pages per hour.
So Pankaj alone = 7.5 − 5 = 2.5 pages per hour.
Time for 25 pages = = 10 hours — option (a).
Take the second number as 100 units; then the first is 125 and the third is 75.
So the three are in the ratio 125 : 100 : 75 = 5 : 4 : 3, a total of 12 units.
Sum of the numbers = 3 × 44 = 132, so 12 units = 132 and 1 unit = 11.
Second number = 4 × 11 = 44 — option (a).
The identity sec²A − tan²A = 1 gives (secA − tanA)(secA + tanA) = 1.
So secA + tanA is the reciprocal of secA − tanA.
= 1 ÷ .
= 4 — option (c).
Milk = × 40 = 30 L and water = 10 L.
The milk stays 30 L, and in the new ratio 3 parts = 30 L, so 1 part = 10 L.
Water must now be 5 parts = 50 L.
Water added = 50 − 10 = 40 L — option (b).
sinA = cos(90° − A), so the two angles must be complementary.
2x + (3x − 60°) = 90°.
5x = 150°.
x = 30° — option (a).
In a right triangle the two legs are themselves altitudes, being perpendicular to each other.
They meet at the right-angled vertex, and the third altitude passes through it as well.
So the orthocentre sits exactly at that vertex.
Hence option (c).
Equal chords of a circle are equidistant from the centre.
Since AB = CD, their distances from O must be the same.
The distance to AB is 16 cm.
So the distance to CD is also 16 cm — option (c).
The radius drawn to the point of contact is always perpendicular to the tangent.
The diameter passes through the centre and the point of contact, so it lies along that radius.
The size of the diameter makes no difference to the angle.
The angle is 90° — option (c).