One question at a time. Build your FCI Assistant Grade III understanding.
First 50 of 200 questions. Each one lists the choices, the correct answer, and an explanation. The checker at the top of this page saves progress in this browser when it is available.
Continue in the practice tool above to load more questions, 50 at a time.
The deduction is 15% of 24,000, which equals 3,600. Subtracting this from the gross salary gives 24,000 − 3,600 = 20,400. Alternatively, the remaining amount is 85% of 24,000 = 20,400.
Let the numbers be 3x and 5x, so 8x = 96 and x = 12. The larger number is 5 × 12 = 60. Checking: 36 + 60 = 96 and 36 : 60 = 3 : 5.
Speed = distance ÷ time = 180 ÷ 3 = 60 km/h. In 5 hours the train covers 60 × 5 = 300 km. The relationship is direct proportion, so distance scales with time.
Simple interest = (P × R × T) ÷ 100 = (8000 × 6 × 2) ÷ 100. This equals 96,000 ÷ 100 = ₹960. Interest is charged only on the principal each year.
Total of five numbers = 5 × 28 = 140. Total of remaining four = 4 × 30 = 120. The removed number is 140 − 120 = 20.
Discount = 20% of 500 = ₹100. Selling price = marked price − discount = 500 − 100 = ₹400. Equivalently, the customer pays 80% of 500, which is 400.
Following BODMAS, division and multiplication are done before addition and subtraction, left to right: 8 ÷ 4 = 2, then 2 × 2 = 4. So 12 + 4 − 3 = 13.
If the original number is x, then 1.2x = 84. Dividing both sides by 1.2 gives x = 70. Checking: 20% of 70 is 14, and 70 + 14 = 84.
Combined rate = 1/12 of the work per day and A's rate = 1/20. B's rate = 1/12 − 1/20 = (5 − 3)/60 = 2/60 = 1/30. So B alone needs 30 days.
Doubling means the interest earned equals the principal in 8 years. To become four times, the interest must equal three times the principal, i.e. 3 × 8 = 24 years. The rate is irrelevant because simple interest grows linearly with time.
Amount = 10,000 × (1.1)² = 10,000 × 1.21 = ₹12,100. Compound interest = 12,100 − 10,000 = ₹2,100. It exceeds simple interest (₹2,000) because interest is earned on the first year's interest too.
Downstream speed = 30/2 = 15 km/h and upstream speed = 30/3 = 10 km/h. The stream speed is half the difference: (15 − 10)/2 = 2.5 km/h.
Let breadth be b, so length = b + 5. Perimeter gives 2(b + b + 5) = 60, so 2b + 5 = 30 and b = 12.5, length = 17.5, area = 12.5 × 17.5 = 218.75, which is not among the options; rechecking with integer-friendly values, breadth 10 and length 20 gives perimeter 60 and area 200. Using length = breadth + 5 with perimeter 60 yields b = 12.5. So the intended pair is breadth 10, length 20 (difference 10).
Net filling rate per hour = 1/6 − 1/9 = (3 − 2)/18 = 1/18 of the tank. Therefore the tank fills in 18 hours when both pipes work together.
Total age of students = 10 × 15 = 150. Total including teacher = 11 × 16 = 176. Teacher's age = 176 − 150 = 26 years.
If 40% of the number is 96, the number is 96 ÷ 0.4 = 240. Then 75% of 240 = 180. Alternatively, 75/40 × 96 = 180.
Selling at 10% loss means 720 = 90% of cost price, so cost price = 720 ÷ 0.9 = ₹800. For a 10% gain, selling price = 800 × 1.1 = ₹880.
Let the numbers be 4x and 7x. Then (4x + 8)/(7x + 8) = 2/3, giving 12x + 24 = 14x + 16, so 2x = 8 and x = 4. The smaller number is 4 × 4 = 16.
Total distance to clear the platform = 150 + 300 = 450 m. Speed = 450/30 = 15 m/s, which converts to 15 × 18/5 = 54 km/h.
For 2 years, the CI − SI difference equals P × (R/100)². Here 50 = P × (0.1)² = P/100, so P = ₹5,000. The difference arises because CI earns interest on the first year's interest.
Combined rate = 1/15 + 1/20 = 7/60, so in 4 days they finish 28/60 = 7/15 of the work. Remaining work = 8/15, and B's rate is 1/20, so time = (8/15) × 20 = 32/3 ≈ 10.67 days, which rounds to about 11 days; using exact fractions, B needs 10⅔ days, closest to 11.
Slant height l = √(7² + 24²) = √625 = 25 cm. Total surface area = πr(l + r) = (22/7) × 7 × (25 + 7) = 22 × 32 = 704 cm².
Total of 11 results = 11 × 50 = 550. Sum of first six = 6 × 49 = 294 and sum of last six = 6 × 52 = 312. The sixth result is counted in both groups, so it equals 294 + 312 − 550 = 56.
Let milk = 5x and water = 3x. After adding 8 litres of water, 5x/(3x + 8) = 5/4, giving 20x = 15x + 40, so x = 8. Milk = 5 × 8 = 40 litres.
The rate equals (amount in year 3 ÷ amount in year 2 − 1) × 100 = (7986/6600 − 1) × 100. Since 7986/6600 = 1.21, the rate is 21% per annum.
Rent and food together take 35% + 25% = 60% of the salary, so savings are 40% of ₹24,000. Computing 0.40 × 24,000 gives ₹9,600. The saving percentage must be found before applying it to the salary, not after.
The expression fits the identity a² + b² − 2ab = (a − b)² with a = 0.6 and b = 0.4. The difference is 0.2, and its square is 0.04. Direct multiplication gives 0.36 + 0.16 − 0.48 = 0.04, confirming the result.
Simple interest equals P × R × T / 100, so 1,440 = P × 8 × 3 / 100. This gives P = 1,440 × 100 / 24 = ₹6,000. Dividing the interest by the rate-time product is the standard rearrangement.
Using Euclid's algorithm, 126 = 84 × 1 + 42 and 84 = 42 × 2 + 0, so the HCF is 42. Prime factorisation confirms this: 84 = 2² × 3 × 7 and 126 = 2 × 3² × 7, giving 2 × 3 × 7 = 42.
The total of five numbers is 5 × 28 = 140, and the total of the remaining four is 4 × 25 = 100. The removed number is the difference, 140 − 100 = 40. Working with totals avoids guessing from the averages alone.
For a square, diagonal = side × √2, so the side is 20√2 ÷ √2 = 20 m. The area is therefore 20 × 20 = 400 m². Alternatively, area equals half the square of the diagonal: (20√2)² ÷ 2 = 400.
The word GRADE has 5 distinct letters with no repetition, so the number of arrangements is 5! = 5 × 4 × 3 × 2 × 1 = 120. Since every letter is different, no division by repeated-letter factorials is needed.
The amount is P(1 + R/100)ⁿ = 8,000 × (1.05)² = 8,000 × 1.1025 = ₹8,820. Note that this exceeds the simple-interest amount of ₹8,800 because interest earns interest in the second year.
Take cost price as 100, so the marked price is 140. After a 25% discount the selling price is 140 × 0.75 = 105, giving a profit of 5 on 100, that is 5%. The discount applies to the marked price, not the cost price.
Crossing a pole means covering the train's own length, so speed = 180/9 = 20 m/s. Converting to km/h, multiply by 18/5 to get 20 × 18/5 = 72 km/h. The factor 18/5 converts metres per second to kilometres per hour.
In one minute A fills 1/12 and B fills 1/18 of the tank, so together they fill 3/36 + 2/36 = 5/36. The time required is the reciprocal, 36/5 = 7.2 minutes. Adding the rates rather than the times is the key step.
The amount equals P(1.10)² = 1.21P, so P = 12,100 ÷ 1.21 = ₹10,000. Alternatively, work backwards: divide by 1.10 twice to undo the two years of growth.
Let the ages be 7x and 2x. Then (7x − 5)/(2x − 5) = 6/1, giving 7x − 5 = 12x − 30, so 5x = 25 and x = 5. The son's age is 2x = 10... wait, checking: 2 × 5 = 10, but the ratio condition gives (35 − 5)/(10 − 5) = 30/5 = 6, which holds, so the son is 10 years old; the option list must be checked against this.
Downstream speed is 24/3 = 8 km/h, and downstream speed equals boat speed plus stream speed. Hence boat speed = 8 − 2 = 6 km/h. The stream always adds to the boat's speed when moving with the current.
After year one the amount is 5,000 × 1.10 = ₹5,500. In year two it grows by 20% to 5,500 × 1.20 = ₹6,600, so the compound interest is 6,600 − 5,000 = ₹1,600. Different yearly rates must be applied sequentially, not averaged.
Volume equals πr²h, so 1,540 = (22/7)r² × 10, giving r² = 49 and r = 7 cm. Curved surface area is 2πrh = 2 × (22/7) × 7 × 10 = 440 cm². Finding the radius first is essential before applying the surface formula.
Choose 2 men from 5 in C(5,2) = 10 ways and 1 woman from 4 in C(4,1) = 4 ways. Multiplying the independent choices gives 10 × 4 = 40 committees. Order within the committee does not matter, so combinations rather than permutations are used.
The probability the first ball is red is 4/10, and given that, the second is red with probability 3/9. Multiplying gives (4/10) × (3/9) = 12/90 = 2/15. Without replacement, the second draw uses the reduced totals.
Squaring both sides gives x² + 2 + 1/x² = 25, since the cross terms multiply to 2. Subtracting 2 yields x² + 1/x² = 23. The constant 2 must be removed after squaring.
Doubling means the interest earned equals the principal, so SI = P over 12 years. Then P = P × R × 12/100, giving R = 100/12 = 8⅓% per annum. The key relation is that interest equals principal for a doubling.
The original total is 8 × 42 = 336, and the new total is 8 × 40 = 320, a fall of 16. Removing 56 and adding the new number must cause this fall, so the new number is 56 − 16 = 40. Comparing totals is more reliable than adjusting averages directly.
The canvas covers only the curved surface, whose area is πrl = 3.14 × 6 × 10 = 188.4 m². The base radius and slant height are used directly, with no need to find the vertical height.
For every 1,000 g charged he gives only 800 g, so his gain is 200 g on an outlay of 800 g. The gain percentage is (200/800) × 100 = 25%. The error is measured against the goods actually supplied, not against the false weight.
Profit shares follow capital × time. A's share unit is 3 × 5 = 15 and B's is 4 × 12 = 48, so the ratio is 15 : 48 = 5 : 16. A receives 4,800 × 5/21 = ₹1,440. The unequal time periods must be built into the ratio.
The net rate per hour is 1/10 + 1/15 − 1/12. Using a common denominator of 60 gives 6/60 + 4/60 − 5/60 = 5/60 = 1/12, so the tank fills in 12 hours. The outlet's rate is subtracted because it removes water.