Railways Exams RRB ALP PDF Sarkari Exam Sarkari Exam RRB ALP PDF Questions Available Here at: https://www.easy-quizzz.com/in/sarkari-exam/railways-exams/rrb-alp-test/quiz.html Enrolling now you will get access to 150 questions in a unique set ofRRB ALP Question 1 In the sequence 5, 9, 18, 34, 59, __, each term after the first is formed by adding a value to the preceding term. Determine the next term. Options: A. 92 B. 94 C. 95 D. 96 Answer: C Explanation: Compare consecutive terms: the differences are 4, 9, 16, and 25, which are consecutive squares 2², 3², 4², and 5². The next difference is therefore 6² = 36. Adding it to the last term gives 59 + 36 = 95. The nearby options can result from extending the differences incorrectly, such as adding 33 or 35 instead of the next square. Question 2 Evaluate the expression, applying the standard order of operations: 2.4 ÷ 0.6 × (1.5 0.5) + (3/4 ÷ 1/2). Options: A. 5.5 B. 6.5 C. 4.5 D. 7.5 Sarkari Exam Sarkari Exam RRB ALP PDF https://www.easy-quizzz.com/in/ Answer: A Explanation: First evaluate the brackets: 1.5 0.5 = 1. Then perform division and multiplication from left to right: 2.4 ÷ 0.6 × 1 = 4. Also, 3/4 ÷ 1/2 means 3/4 × 2 = 1.5. Finally, add the two results: 4 + 1.5 = 5.5. Multiplying before dividing, rather than following left-to-right order, would produce an incorrect result. Question 3 In a circuit, a 2 resistor is connected in series with a parallel combination of 3 and 6 resistors across an ideal 6 V battery. What current flows through the 6 resistor? Options: A. 0.5 A B. 1.0 A C. 1.5 A D. 3.0 A Answer: A Explanation: The parallel pair has equivalent resistance (3 × 6)/(3 + 6) = 2 . In series with the 2 resistor, the total resistance is 4 , so the battery current is 6/4 = 1.5 A. The series resistor drops 1.5 × 2 = 3 V, leaving 3 V across each parallel branch. Thus the current through the 6 resistor is 3/6 = 0.5 A. Question 4 An object is represented in first-angle orthographic projection. Relative to the front view, where should the top view and the right-side view be placed, respectively? Options: A. Below the front view; to its left B. Above the front view; to its right C. Below the front view; to its right D. Above the front view; to its left Answer: A Explanation: In first-angle projection, the object is considered between the observer and the projection plane. Consequently, each view is placed on the side opposite the direction from which it is viewed: the top view goes below the front view, and the right-side view goes to the left of it. Placing both on the corresponding viewing sides is characteristic of third-angle projection, not first-angle projection. Sarkari Exam Sarkari Exam RRB ALP PDF https://www.easy-quizzz.com/in/ Question 5 A machinist measures a shaft with a vernier caliper. The main-scale reading is 12.40 mm, the sixth vernier division coincides, the least count is 0.02 mm, and the instrument has a positive zero error of 0.04 mm. What is the corrected diameter? Options: A. 12.48 mm B. 12.56 mm C. 12.44 mm D. 12.52 mm Answer: A Explanation: The observed reading is the main-scale reading plus the vernier contribution: 12.40 + (6 × 0.02) = 12.52 mm. A positive zero error means the instrument reads too high, so it must be subtracted from the observed reading. Thus the corrected diameter is 12.52 0.04 = 12.48 mm. Adding the zero error instead would increase, rather than correct, the measurement bias. Question 6 A right-triangular plate has perpendicular sides in the ratio 3:4 and a perimeter of 60 cm. Its area is: Options: A. 150 cm² B. 120 cm² C. 180 cm² D. 300 cm² Answer: A Explanation: A right triangle with perpendicular sides in the ratio 3:4 has side lengths 3k and 4k, with hypotenuse 5k by the 3-4-5 relationship. Its perimeter is therefore 12k, so 12k = 60 and k = 5. The perpendicular sides are 15 cm and 20 cm; the area is (1/2) × 15 × 20 = 150 cm². Using the perimeter as a side or omitting the one-half factor gives an incorrect result. Question 7 A polished iron strip is immersed in blue copper(II) sulfate solution. After some time, a reddish deposit forms and the solution becomes pale green. Which equation best accounts for both observations? Sarkari Exam Sarkari Exam RRB ALP PDF https://www.easy-quizzz.com/in/ Options: A. Fe + CuSO FeSO + Cu B. Cu + FeSO CuSO + Fe C. 2Fe + 3CuSO Fe(SO) + 3Cu D. Fe + CuSO FeCu + SO Answer: A Explanation: Iron is more reactive than copper, so it displaces copper from copper(II) sulfate. The deposited reddish solid is copper, while the pale green solution contains iron(II) sulfate. The reaction is a single-displacement redox reaction: iron atoms lose electrons to form Fe² ions, and Cu² ions gain those electrons to form copper metal. The equation is balanced with one iron atom and one copper ion. Question 8 An acute angle has tan = 3/4. What is the exact value of sin 2? Options: A. 24/25 B. 7/25 C. 3/5 D. 12/25 Answer: A Explanation: Since is acute and tan = 3/4, a right-triangle ratio gives sin = 3/5 and cos = 4/5. Then sin 2 = 2 sin cos = 2(3/5)(4/5) = 24/25. The value 7/25 is associated with cos 2, not sin 2; using tan directly in the double- angle formula also gives 2 tan /(1 + tan² ) = 24/25. Question 9 After a potted plant has been kept in darkness long enough to remove stored starch, part of one leaf is covered with opaque paper and the plant is placed in sunlight. After the leaf is tested with iodine, which result best supports the role of light in photosynthesis? Options: A. Both covered and uncovered regions turn blue-black because the plant was previously exposed to light. B. The uncovered region turns blue-black, while the covered region remains yellow-brown. Sarkari Exam Sarkari Exam RRB ALP PDF https://www.easy-quizzz.com/in/ C. The covered region turns blue-black, while the uncovered region remains yellow-brown. D. Neither region turns blue-black because starch is produced only at night. Answer: B Explanation: Destarching removes starch made before the experiment, so any starch detected afterward was produced during the test. The uncovered region receives light and can photosynthesise, producing sugars that are stored as starch; iodine then turns blue-black. The covered region lacks light and does not produce starch under these conditions. The plant’s earlier exposure to light does not invalidate the test because stored starch was first depleted. Question 10 An insulated container holds 0.20 kg of water at 80°C. It is mixed with 0.30 kg of water at 20°C. Neglecting the container’s heat capacity, the final temperature is Options: A. 44°C B. 40°C C. 50°C D. 60°C Answer: A Explanation: Because both samples are water and the container is insulated, heat lost by the warmer sample equals heat gained by the cooler one. Their specific heats therefore cancel: the final temperature is the mass- weighted average, [(0.20 × 80) + (0.30 × 20)] / 0.50 = 44°C. A simple average of 50°C would be valid only for equal masses; the larger mass here starts at the lower temperature. Would you like to see more? Don't miss ourRRB ALP PDF file at: https://www.easy-quizzz.com/in/sarkari-exam/railways-exams/rrb-alp-test-pdf/ Sarkari Exam Sarkari Exam RRB ALP PDF https://www.easy-quizzz.com/in/