RRB NTPC Algebra and Geometry 1 — Questions and Answers

Question 1: If 2x + 3 = 11, what is the value of x?

3
4 (Correct answer)
5
6

Correct answer: 4

2x + 3 = 11 → 2x = 11 - 3 = 8 → x = 4.

Solving the linear equation: 2x + 3 = 11. Step 1: Subtract 3 from both sides: 2x = 11 - 3 = 8. Step 2: Divide both sides by 2: x = 8/2 = 4. Verification: 2(4) + 3 = 8 + 3 = 11 ✓. Linear equations are fundamental to RRB NTPC algebra. The process is: (1) move constants to one side, (2) move variable terms to the other side, (3) divide by the coefficient of the variable.

Question 2: The angles of a triangle are in the ratio 2:3:4. What is the measure of the largest angle?

60°
70°
80° (Correct answer)
90°

Correct answer: 80°

Let angles be 2x, 3x, 4x. Sum = 180°. So 9x = 180° → x = 20°. Largest angle = 4x = 80°.

Let the three angles be 2x°, 3x°, and 4x°. Property: Sum of angles of a triangle = 180°. 2x + 3x + 4x = 180 → 9x = 180 → x = 20. Three angles: 2×20 = 40°, 3×20 = 60°, 4×20 = 80°. Largest angle = 80°. Verification: 40° + 60° + 80° = 180° ✓. The largest angle is opposite the longest side. Since 80° < 90°, this is an acute-angled triangle. This type of problem tests both algebra and basic geometry.

Question 3: What is the area of a circle with radius 7 cm? (Use π = 22/7)

44 cm²
154 cm² (Correct answer)
176 cm²
88 cm²

Correct answer: 154 cm²

Area = πr² = (22/7) × 7² = (22/7) × 49 = 22 × 7 = 154 cm².

Area of circle = πr², where r = 7 cm and π = 22/7. Area = (22/7) × 7² = (22/7) × 49 = 22 × 7 = 154 cm². The cancellation 49/7 = 7 makes calculation easy. Circumference = 2πr = 2 × (22/7) × 7 = 44 cm. Key formulas for circles: Area = πr²; Circumference = 2πr = πd; Area = (πd²)/4 where d is diameter. Using π = 22/7 is standard for RRB NTPC unless otherwise specified.

Question 4: If x + y = 10 and x - y = 4, what is the value of xy?

21 (Correct answer)
24
20
16

Correct answer: 21

Adding equations: 2x = 14 → x = 7. Subtracting: 2y = 6 → y = 3. So xy = 7×3 = 21.

System of equations: x + y = 10 ... (i); x - y = 4 ... (ii). Adding (i) and (ii): 2x = 14 → x = 7. Substituting in (i): 7 + y = 10 → y = 3. Therefore xy = 7 × 3 = 21. Alternative using identity: (x+y)² = x² + 2xy + y² → 100 = x² + 2xy + y²; (x-y)² = x² - 2xy + y² → 16 = x² - 2xy + y². Subtracting: 100 - 16 = 4xy → 84 = 4xy → xy = 21. Verification: x + y = 7+3 = 10 ✓; x - y = 7-3 = 4 ✓.

Question 5: The perimeter of a square is 48 cm. What is its area?

100 cm²
112 cm²
120 cm²
144 cm² (Correct answer)

Correct answer: 144 cm²

Perimeter = 4 × side → 48 = 4s → s = 12 cm. Area = s² = 12² = 144 cm².

Perimeter of square = 4 × side. Given: Perimeter = 48 cm → 4 × side = 48 → side = 12 cm. Area of square = side² = 12² = 144 cm². Verification: Perimeter = 4 × 12 = 48 ✓. Key formulas for square: Perimeter = 4a; Area = a²; Diagonal = a√2. For this square, Diagonal = 12√2 ≈ 16.97 cm. Quick relationship: If perimeter = P, then side = P/4 and Area = (P/4)² = P²/16 = 48²/16 = 2304/16 = 144 cm².

Question 6: In a right-angled triangle, the hypotenuse is 13 cm and one leg is 5 cm. What is the length of the other leg?

8 cm
10 cm
11 cm
12 cm (Correct answer)

Correct answer: 12 cm

By Pythagoras theorem: a² + b² = c². 5² + b² = 13² → b² = 169 - 25 = 144 → b = 12 cm.

Pythagoras Theorem: In a right triangle, hypotenuse² = leg₁² + leg₂². Given: Hypotenuse (c) = 13 cm, one leg (a) = 5 cm. 5² + b² = 13² → 25 + b² = 169 → b² = 144 → b = 12 cm. This is the famous 5-12-13 Pythagorean triple. Common Pythagorean triples to memorize for RRB NTPC: (3,4,5), (5,12,13), (8,15,17), (7,24,25), (6,8,10), (9,12,15) — note multiples of basic triples also work.

RRB NTPC Algebra and Geometry 1

If 2x + 3 = 11, what is the value of x?