Download PDF – RRB PH UT31 – General Arithmetic Practice Test : Mathematics & Quantitative Aptitude

RRB PH UT31 – General Arithmetic Practice Test : Mathematics & Quantitative Aptitude

Q1. At simple interest, 3 years at a rate 2% higher earns ₹360 extra. Find the principal.

A) Rs. 5000

B) Rs. 4500

C) Rs. 6000

D) Rs. 7200

Q1. 3 वर्ष के साधारण ब्याज में दर 2% बढ़ने पर ₹360 अतिरिक्त ब्याज मिलता है। मूलधन ज्ञात करें।

A) 5000 रुपये

B) 4500 रुपये

C) 6000 रुपये

D) 7200 रुपये

Answer: (C)

Let the principal be P. The extra interest fetched is entirely due to the 2% higher rate over 3 years: Extra Interest = (P × Extra Rate × Time) / 100 => 360 = (P × 2 × 3) / 100 => 360 = 6P / 100. Solving for P: P = (360 × 100) / 6 = 60 × 100 = Rs. 6000.

माना मूलधन P है। प्राप्त होने वाला अतिरिक्त ब्याज 3 वर्षों में 2% अधिक दर के कारण है: अतिरिक्त ब्याज = (P × अतिरिक्त दर × समय) / 100 => 360 = (P × 2 × 3) / 100 => 360 = 6P / 100। हल करने पर: P = (360 × 100) / 6 = 6000 रुपये।

Q2. Downstream and upstream rowing speeds are 14 and 8 km/h. Find the current speed.

A) 3.0 km/h

B) 2.5 km/h

C) 5.5 km/h

D) 6.0 km/h

Q2. धारा के अनुकूल और प्रतिकूल नाव की गति 14 और 8 km/h है। जलधारा की गति ज्ञात करें।

A) 3.0 किमी/घंटा

B) 2.5 किमी/घंटा

C) 5.5 किमी/घंटा

D) 6.0 किमी/घंटा

Answer: (A)

Let the downstream speed be 'u' and upstream speed be 'v'. The speed of the stream/current is given by: Speed of stream = (Downstream speed - Upstream speed) / 2 = (14 - 8) / 2 = 6 / 2 = 3.0 km/h. (The speed of the boat in still water is (14 + 8) / 2 = 11 km/h).

माना धारा के अनुकूल चाल 'u' और प्रतिकूल चाल 'v' है। धारा की चाल = (अनुकूल चाल - प्रतिकूल चाल) / 2 = (14 - 8) / 2 = 6 / 2 = 3.0 किमी/घंटा होगी। (शांत जल में नाव की चाल (14 + 8) / 2 = 11 किमी/घंटा होगी)।

Q3. Father and son’s present ages are 5:2. Four years ago, father was three times the son’s age. Find father’s present age.

A) 35 years

B) 45 years

C) 50 years

D) 40 years

Q3. पिता-पुत्र की वर्तमान आयु 5:2 है। 4 वर्ष पहले पिता पुत्र की आयु का 3 गुना था। पिता की वर्तमान आयु ज्ञात करें।

A) 35 वर्ष

B) 45 वर्ष

C) 50 वर्ष

D) 40 वर्ष

Answer: (D)

Let the present ages of father and son be 5x and 2x years respectively. Four years ago, their ages were (5x - 4) and (2x - 4). According to the condition: 5x - 4 = 3 × (2x - 4) => 5x - 4 = 6x - 12 => 6x - 5x = 12 - 4 => x = 8. Therefore, the present age of the father = 5 × 8 = 40 years (and the son's age = 2 × 8 = 16 years).

माना पिता और पुत्र की वर्तमान आयु क्रमशः 5x और 2x वर्ष है। 4 वर्ष पूर्व उनकी आयु (5x - 4) और (2x - 4) वर्ष थी। प्रश्नानुसार: 5x - 4 = 3 × (2x - 4) => 5x - 4 = 6x - 12 => x = 8। अतः पिता की वर्तमान आयु = 5 × 8 = 40 वर्ष होगी (पुत्र की आयु 16 वर्ष होगी)।

Q4. A cylinder has radius:height = 2:3 and volume 12,936 cm³. Find its total surface area (π = 22/7).

A) 1540 cm²

B) 2156 cm²

C) 2772 cm²

D) 3080 cm²

Q4. बेलन की त्रिज्या:ऊँचाई = 2:3 और आयतन 12,936 cm³ है। कुल पृष्ठीय क्षेत्रफल ज्ञात करें (π = 22/7)।

A) 1540 सेमी²

B) 2156 सेमी²

C) 2772 सेमी²

D) 3080 सेमी²

Answer: (D)

Let radius r = 2x and height h = 3x. Volume = pi × r^2 × h = (22/7) × (2x)^2 × (3x) = (22/7) × 12x^3 = 12936. This gives x^3 = (12936 × 7) / (22 × 12) = 343, so x = 7. Thus, r = 2 × 7 = 14 cm and h = 3 × 7 = 21 cm. Total Surface Area (TSA) = 2 × pi × r × (r + h) = 2 × (22/7) × 14 × (14 + 21) = 2 × 22 × 2 × 35 = 3080 cm².

माना त्रिज्या r = 2x और ऊँचाई h = 3x है। बेलन का आयतन = pi × r^2 × h = (22/7) × (2x)^2 × (3x) = (22/7) × 12x^3 = 12936। हल करने पर x^3 = 343, जिससे x = 7 प्राप्त होता है। अतः r = 14 सेमी और h = 21 सेमी। कुल पृष्ठीय क्षेत्रफल (TSA) = 2 × pi × r × (r + h) = 2 × (22/7) × 14 × (14 + 21) = 88 × 35 = 3080 सेमी²।

Q5. A sum amounts to ₹756 in 2 years and ₹873 in 3.5 years at simple interest. Find the annual rate.

A) 13%

B) 11%

C) 12.5%

D) 14%

Q5. एक राशि साधारण ब्याज पर 2 वर्ष में ₹756 और 3.5 वर्ष में ₹873 हो जाती है। वार्षिक ब्याज दर ज्ञात करें।

A) 13%

B) 11%

C) 12.5%

D) 14%

Answer: (A)

Simple interest for 1.5 years = Rs. 873 - Rs. 756 = Rs. 117. Simple interest per year = 117 / 1.5 = Rs. 78. Simple interest for 2 years = 78 × 2 = Rs. 156. Principal sum = Rs. 756 - Rs. 156 = Rs. 600. Annual rate of interest R = (Simple Interest × 100) / (Principal × Time) = (78 × 100) / (600 × 1) = 13% per annum.

1.5 वर्ष का साधारण ब्याज = 873 - 756 = 117 रुपये। 1 वर्ष का साधारण ब्याज = 117 / 1.5 = 78 रुपये। 2 वर्ष का साधारण ब्याज = 78 × 2 = 156 रुपये। मूलधन (P) = 756 - 156 = 600 रुपये। वार्षिक ब्याज दर (R) = (ब्याज × 100) / (मूलधन × समय) = (78 × 100) / (600 × 1) = 13% वार्षिक।

Q6. Find CI on ₹16,000 for 1.5 years at 10% p.a., compounded half-yearly.

A) Rs. 2400

B) Rs. 2500

C) Rs. 2640

D) Rs. 2522

Q6. ₹16,000 पर 10% वार्षिक दर से 1.5 वर्ष का चक्रवृद्धि ब्याज ज्ञात करें (अर्धवार्षिक संयोजन)।

A) 2400 रुपये

B) 2500 रुपये

C) 2640 रुपये

D) 2522 रुपये

Answer: (D)

When compounded semi-annually: Rate r = 10% / 2 = 5% per half-year; Time n = 1.5 × 2 = 3 half-years. Amount A = P × (1 + r/100)^n = 16000 × (1 + 5/100)^3 = 16000 × (21/20)^3 = 16000 × (9261 / 8000) = 2 × 9261 = Rs. 18,522. Compound Interest = Amount - Principal = 18,522 - 16,000 = Rs. 2,522.

ब्याज अर्धवार्षिक संयोजित होने पर: दर r = 10% / 2 = 5% प्रति छमाही; समय n = 1.5 × 2 = 3 छमाही। मिश्रधन A = P × (1 + r/100)^n = 16000 × (1 + 5/100)^3 = 16000 × (21/20)^3 = 16000 × (9261 / 8000) = 18522 रुपये। चक्रवृद्धि ब्याज = मिश्रधन - मूलधन = 18522 - 16000 = 2522 रुपये।

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