SSC CGL 19 September 2025 · Shift 2 · Quantitative Aptitude — questions with answers · ShikshaSphere
Chapter 91 of 192
19 September 2025 · Shift 2
Quantitative Aptitude
25 questions ~25 min readFree
25 questions
A
51
If + = 7 and − = 1, what is the value of a and b?
A.a = 4, b = 3
B.a = 9, b = 16
C.a = 16, b = 9
D.a = 3, b = 4
Solution
Simultaneous surds
Add the two equations: 2 = 8 ⇒ = 4 ⇒ a = 16.
Subtract them: 2 = 6 ⇒ = 3 ⇒ b = 9.
Check: 4 + 3 = 7 ✓ and 4 − 3 = 1 ✓.
a = 16, b = 9 — option (c).
52
Which of the following is the least 6-digit number which is NOT a perfect square?
A.100489
B.100000
C.101124
D.100788
Solution
Not a perfect square
The least 6-digit number is 100000 itself, so check that one first.
316² = 99856 and 317² = 100489, so no whole number squares to 100000.
Hence 100000 is the least 6-digit number that is not a perfect square.
= 100000 — option (b).
53
If 37.5% of p is equal to 25% of q, what is the ratio p : q?
A.1 : 4
B.4 : 1
C.3 : 2
D.2 : 3
Solution
Percentage equality
37.5% = and 25% = .
So = ⇒ = .
= .
p : q = 2 : 3 — option (d).
54
A grocer mixes two varieties of rice — one costing ₹40 per kg and the other ₹50 per kg — in the ratio 3 : 5. If he sells the mixed variety at ₹55.50 per kg, find his gain or loss percent.
A book originally priced at ₹800 is subject to two successive discounts: an initial markdown of 30%, followed by an additional 20% discount. Determine the final selling price of the book.
A.₹432
B.₹396
C.₹448
D.₹350
Solution
Two successive discounts
After the first discount: 800 × = ₹560.
After the second: 560 × .
= ₹448.
Hence option (c).
56
A retailer marked a commodity at 25% above its cost price and then gave a discount. If the sale finally yielded a profit of 20% on the cost price, what was the discount percentage?
If a² + b² = 152 and ab = 15, find the value of a + b.
A.9
B.6
C.10
D.8
Solution
a + b from a² + b²
ab = 15 gives the pairs (5, 3) or (15, 1).
For (5, 3): a³ + b³ = 125 + 27 = 152, which matches the given value.
So a = 5 and b = 3.
a + b = 8 — option (d).
58
At what rate of simple interest will ₹3,000 become ₹4,020 in 5 years?
A.8.5%
B.5.5%
C.8.8%
D.6.8%
Solution
SI — find rate
Interest in 5 years = 4,020 − 3,000 = ₹1,020.
Interest for one year = = ₹204.
Rate = × 100.
= 6.8% — option (d).
59
A person invested ₹8,000 in one scheme at 6% simple interest and ₹5,000 in another at 8% simple interest for the same duration. If he received a total interest of ₹2,640, what was the duration of the investments?
A.3.5 years
B.6 years
C.5 years
D.3 years
Solution
SI — find time
Interest per year: 8,000 × = ₹480 and 5,000 × = ₹400.
Together ₹880 per year.
880t = 2,640 ⇒ t = .
= 3 years — option (d).
60
A wheelchair ramp needs to reach a platform that is 2.5 metres high. If the ramp makes an angle of 30° with the level ground, what is the length of the ramp?
A.1.5 m
B.2 m
C.5 m
D.3 m
Solution
Ramp at 30°
sin30° = .
= .
Ramp = 2.5 × 2.
= 5 m — option (c).
61
From a point A on the ground, the angle of elevation to the top of a building is 30°. From a point B, which is directly between A and the foot of the building, the angle of elevation is 45°. What is the ratio of the distance of A from the foot to the distance of B from the foot?
A. : 3
B. : 1
C. :
D.2 : 3
Solution
Two elevation angles
Let the height be h. From A: tan30° = ⇒ DA = h.
From B: tan45° = ⇒ DB = h.
DA : DB = h : h.
= : 1 — option (b).
62
If the curved surface area of a sphere is 154 cm², find its diameter.
A.9 cm
B.8 cm
C.7 cm
D.5 cm
Solution
Sphere — diameter from CSA
Surface area of a sphere = 4πr² = 154.
4 × × r² = 154 ⇒ r² = = 12.25.
r = 3.5 cm.
Diameter = 2r = 7 cm — option (c).
63
A cone has height h and radius r. It is melted and recast into a smaller cone whose height is and radius is . What fraction of the original volume is unused?
A.
B.
C.
D.
Solution
Cone recast — unused part
Original volume V = πr²h.
Smaller cone = π()²() = × πr²h = .
Unused = V − .
= V — option (b).
64
If a regular hexagon and a square have the same side length, what will be the ratio of their areas?
A.3 : 2
B.2 : 3
C. : 1
D.1 : 6
Solution
Hexagon : square area
Area of a regular hexagon = a².
Area of a square = a².
Ratio = a² : a².
= 3 : 2 — option (a).
65
A prism has an equilateral triangular base with a side length of 6 cm. Its height starts at 1 cm and increases by 1 cm for each subsequent layer, forming 5 layers in total. What is the total volume of the prism? (Use = 1.732)
A.207.84 cm³
B.233.82 cm³
C.259.80 cm³
D.303.10 cm³
Solution
Prism with AP heights
Base area = a² = × 36 = 9 cm².
The five layers give a total height of 1 + 2 + 3 + 4 + 5 = 15 cm.
Volume = base area × height = 9 × 15 = 135.
= 135 × 1.732 = 233.82 cm³ — option (b).
66
What is the equation of the line through (3, −5) with slope −2?
A.y = −2x − 5
B.y = −2x + 1
C.y = −2x − 2
D.y = −2x + 3
Solution
Equation of a line
Point-slope form: y − y₁ = m(x − x₁).
y + 5 = −2(x − 3) = −2x + 6.
y = −2x + 6 − 5.
y = −2x + 1 — option (b).
67
If every exterior angle of a regular polygon is 72°, how many sides does the polygon have?
A.4
B.5
C.6
D.8
Solution
Exterior angle
The exterior angles of any polygon add up to 360°.
Number of sides = .
= .
= 5 — option (b).
68
In a triangle PQR, the centroid is at G(3, 1). If the vertices P and Q are at (5, −2) and (−1, 3) respectively, what are the coordinates of vertex R?
A.(5, 2)
B.(2, 4)
C.(1, 3)
D.(2, 3)
Solution
Third vertex from centroid
Centroid = (, ).
x: 3 = ⇒ 9 = 4 + x ⇒ x = 5.
y: 1 = ⇒ 3 = 1 + y ⇒ y = 2.
R = (5, 2) — option (a).
69
Evaluate: ( + )² − ( − )²
A.4
B.2
C.6
D.8
Solution
Difference of squares
The identity (a + b)² − (a − b)² = 4ab.
Here a = and b = .
4 × × = 4.
Hence option (a).
70
Two circles with radii r₁ and r₂ have their centres separated by a distance d. If the length of a transverse common tangent equals the distance between the centres, which of the following is true?
A.r₁ × r₂ = 0
B.(r₁ − r₂) = 0
C.(r₁ + r₂) = 0
D.r₁ = 2r₂
Solution
Transverse tangent = d
Transverse common tangent: L = .
Given L = d, so d² = d² − (r₁ + r₂)².
That leaves (r₁ + r₂)² = 0.
So (r₁ + r₂) = 0 — option (c).
71
A chord of a circle has a length of 2 cm. The distance of the chord from the centre is 8 cm. What is the radius of the circle?
A.6.89 cm
B.6 cm
C.9.52 cm
D.9 cm
Solution
Chord — radius
Half the chord = cm.
r² = ()² + 8² = 17 + 64.
r² = 81.
r = 9 cm — option (d).
72
If sinA + cosA = Y, then find the value of sin²A + cos²A + 2 sinA cosA.
A.Y
B.Y²
C.1
D.Y + 1
Solution
Expand (sinA + cosA)²
Note that sin²A + cos²A + 2 sinA cosA is exactly (sinA + cosA)².
And sinA + cosA is given as Y.
So the expression equals Y².
Hence option (b).
If x + y + z = 0 and x = 0.5, y = 0.7, z = −1.2, then what is (x³ + y³ + z³) ÷ (3xyz)?
A.−3
B.1
C.0
D.−2
Solution
a³+b³+c³ when sum = 0
When x + y + z = 0, the identity gives x³ + y³ + z³ = 3xyz.
So the expression becomes .
The actual values are not even needed.
= 1 — option (b).
75
If p = 0.7 and q = 0.05, find the value of .
A.64
B.27
C.59
D.65
Solution
Common factor 3³
(3p)³ + (3q)³ = 27p³ + 27q³ = 27(p³ + q³).
So the ratio = .
The bracket cancels.
= 27 — option (b).
Finished reading? Now test yourself.
The same 25 questions, but with a timer and the answers hidden — 15 minutes.
Add the two equations: 2 = 8 ⇒ = 4 ⇒ a = 16.
Subtract them: 2 = 6 ⇒ = 3 ⇒ b = 9.
Check: 4 + 3 = 7 ✓ and 4 − 3 = 1 ✓.
a = 16, b = 9 — option (c).
Cost of 8 kg = 40×3 + 50×5 = 120 + 250 = ₹370.
Selling price of 8 kg = 8 × 55.50 = ₹444.
Profit = 444 − 370 = ₹74.
Gain% = × 100 = 20% — option (b).
After the first discount: 800 × = ₹560.
After the second: 560 × .
= ₹448.
Hence option (c).
Take CP = 100 ⇒ MP = 125 and SP = 120.
Discount = MP − SP = 125 − 120 = 5.
Discount% = × 100.
= 4% — option (b).
Interest per year: 8,000 × = ₹480 and 5,000 × = ₹400.
Together ₹880 per year.
880t = 2,640 ⇒ t = .
= 3 years — option (d).
sin30° = .
= .
Ramp = 2.5 × 2.
= 5 m — option (c).
Let the height be h. From A: tan30° = ⇒ DA = h.
From B: tan45° = ⇒ DB = h.
DA : DB = h : h.
= : 1 — option (b).
Surface area of a sphere = 4πr² = 154.
4 × × r² = 154 ⇒ r² = = 12.25.
r = 3.5 cm.
Diameter = 2r = 7 cm — option (c).
Original volume V = πr²h.
Smaller cone = π()²() = × πr²h = .
Unused = V − .
= V — option (b).
Area of a regular hexagon = a².
Area of a square = a².
Ratio = a² : a².
= 3 : 2 — option (a).
Base area = a² = × 36 = 9 cm².
The five layers give a total height of 1 + 2 + 3 + 4 + 5 = 15 cm.
Volume = base area × height = 9 × 15 = 135.
= 135 × 1.732 = 233.82 cm³ — option (b).
Point-slope form: y − y₁ = m(x − x₁).
y + 5 = −2(x − 3) = −2x + 6.
y = −2x + 6 − 5.
y = −2x + 1 — option (b).
The exterior angles of any polygon add up to 360°.
Number of sides = .
= .
= 5 — option (b).
Centroid = (, ).
x: 3 = ⇒ 9 = 4 + x ⇒ x = 5.
y: 1 = ⇒ 3 = 1 + y ⇒ y = 2.
R = (5, 2) — option (a).
Transverse common tangent: L = .
Given L = d, so d² = d² − (r₁ + r₂)².
That leaves (r₁ + r₂)² = 0.
So (r₁ + r₂) = 0 — option (c).
Half the chord = cm.
r² = ()² + 8² = 17 + 64.
r² = 81.
r = 9 cm — option (d).
sinx = cosx gives tanx = , so x = 60°.
sin60° = and cos60° = .
sin⁴x + cos⁴x = + = .
= — option (b).
(3p)³ + (3q)³ = 27p³ + 27q³ = 27(p³ + q³).
So the ratio = .
The bracket cancels.
= 27 — option (b).