SSC CGL Practice Test 2026

About the SSC CGL exam

The Staff Selection Commission - Combined Graduate Level (SSC CGL) is a government examination administered in India. It is designed to assess candidates' knowledge and proficiency across 5 key subject areas including Quantitative Aptitude, English Language, General Intelligence & Reasoning, General Awareness. This exam is widely recognized as an essential benchmark for students and professionals.

Use short SSC CGL practice sets to identify topics to revisit. Choose a subject and difficulty, answer questions, then review the explanations. New accounts get 30 question credits; further new questions use paid credits. Retaking previously seen questions is free.

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8 free SSC CGL practice questions (2026)

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1. If the 6-digit number 5x423y is completely divisible by 88, then find the value of (5x - 2y).

   • 36
   • 38
   • 40
   • 32
   Show answer & explanation

   Answer: A. 36

   Since the number is divisible by 88, it must be divisible by both 8 and 11. For divisibility by 8, the last three digits 23y must be divisible by 8, which gives y = 2 (as 232 is divisible by 8). For divisibility by 11, the difference between the sum of digits at odd places and even places must be 0 or a multiple of 11. Sum of odd places = 5 + 4 + 3 = 12. Sum of even places = x + 2 + y = x + 2 + 2 = x + 4. Therefore, 12 - (x + 4) = 8 - x. For this to be 0, x must be 8. Thus, 5x - 2y = 5(8) - 2(2) = 40 - 4 = 36.

   Topic: Quantitative Aptitude · medium

2. If x^4 + 1/x^4 = 194, then find the value of x^3 + 1/x^3 (where x > 0).

   • 76
   • 52
   • 64
   • 36
   Show answer & explanation

   Answer: B. 52

   Add 2 on both sides: x^4 + 1/x^4 + 2 = 196 => (x^2 + 1/x^2)^2 = 14^2 => x^2 + 1/x^2 = 14. Again, add 2 on both sides: x^2 + 1/x^2 + 2 = 16 => (x + 1/x)^2 = 4^2 => x + 1/x = 4. Now, x^3 + 1/x^3 = (x + 1/x)^3 - 3(x + 1/x) = 4^3 - 3(4) = 64 - 12 = 52.

   Topic: Quantitative Aptitude · easy

3. In a right-angled triangle, the sides containing the right angle are 15 cm and 20 cm. Find the ratio of the inradius to the circumradius of this triangle.

   • 2 : 5
   • 3 : 5
   • 1 : 2
   • 3 : 7
   Show answer & explanation

   Answer: A. 2 : 5

   The hypotenuse of the triangle is sqrt(15^2 + 20^2) = 25 cm. For a right-angled triangle, Inradius (r) = (a + b - c)/2 = (15 + 20 - 25)/2 = 5 cm. Circumradius (R) = hypotenuse/2 = 25/2 = 12.5 cm. The ratio r : R = 5 : 12.5 = 2 : 5.

   Topic: Quantitative Aptitude · medium

4. From the top of a 75 m high tower, the angles of depression of two points on the ground, on opposite sides of the tower and in a straight line with its base, are 30° and 45°. Find the distance between the two points.

   • 75(sqrt(3) + 1) m
   • 75(sqrt(3) - 1) m
   • 50(sqrt(3) + 1) m
   • 150(sqrt(3) - 1) m
   Show answer & explanation

   Answer: A. 75(sqrt(3) + 1) m

   Let the height of the tower be h = 75 m. Let the distances of the two points from the base of the tower be d1 and d2. From the right triangles, tan 30° = h / d1 => d1 = 75 * sqrt(3) m. Also, tan 45° = h / d2 => d2 = 75 * 1 = 75 m. Total distance between the two points on opposite sides is d1 + d2 = 75*sqrt(3) + 75 = 75(sqrt(3) + 1) m.

   Topic: Quantitative Aptitude · medium

5. A shopkeeper marks his goods 40% above the cost price and allows a discount of 25% on the marked price. Find his net gain or loss percent.

   • 5% loss
   • 5% gain
   • 10% gain
   • 10% loss
   Show answer & explanation

   Answer: B. 5% gain

   Let CP = 100. Marked Price (MP) = 140. Selling Price (SP) after 25% discount = 140 * 0.75 = 105. Since SP > CP, there is a gain of 105 - 100 = 5%. Net gain percentage is 5%.

   Topic: Quantitative Aptitude · easy

6. A thief is spotted by a policeman from a distance of 250 m. When the policeman starts the chase, the thief also starts running. Assuming the speed of the thief to be 10 km/h and that of the policeman to be 12 km/h, how far will the thief have run before he is overtaken?

   • 1000 m
   • 1250 m
   • 1500 m
   • 800 m
   Show answer & explanation

   Answer: B. 1250 m

   Relative speed = 12 - 10 = 2 km/h = 2 * (5/18) = 5/9 m/s. Distance to be covered = 250 m. Time taken to overtake = 250 / (5/9) = 450 seconds. Distance covered by the thief in 450 seconds = Speed * Time = (10 * 5/18) * 450 = 1250 m.

   Topic: Quantitative Aptitude · medium

7. The difference between compound interest (compounded annually) and simple interest on a certain sum of money for 3 years at 10% per annum is ₹155. Find the principal sum.

   • ₹4,000
   • ₹5,000
   • ₹6,000
   • ₹4,500
   Show answer & explanation

   Answer: B. ₹5,000

   The formula for the difference between CI and SI for 3 years is D = P * (r/100)^2 * ((300+r)/100). Here, D = 155 and r = 10. So, 155 = P * (10/100)^2 * (310/100) => 155 = P * (1/100) * (31/10) => 155 = P * (31/1000) => P = (155 * 1000) / 31 = 5 * 1000 = ₹5,000.

   Topic: Quantitative Aptitude · medium

8. A and B can complete a work in 15 days and 10 days respectively. They started doing the work together, but after 2 days B had to leave and A alone completed the remaining work. In how many total days was the whole work completed?

   • 10 days
   • 8 days
   • 12 days
   • 15 days
   Show answer & explanation

   Answer: C. 12 days

   Let total work be LCM(15, 10) = 30 units. Efficiency of A = 2 units/day; Efficiency of B = 3 units/day. Combined efficiency = 5 units/day. Work done together in 2 days = 5 * 2 = 10 units. Remaining work = 30 - 10 = 20 units. Time taken by A to complete remaining work = 20 / 2 = 10 days. Total days = 2 + 10 = 12 days.

   Topic: Quantitative Aptitude · easy

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SSC CGL Exam Subjects

Available practice subjects: Quantitative Aptitude, English Language, General Intelligence & Reasoning, General Awareness, Statistics. Select a subject for focused preparation. Coverage is not a guarantee of the complete official syllabus.

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