RRB NTPC Mathematics Questions: 31st July_00.1
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RRB NTPC Mathematics Questions: 31st July

Dear aspirants,

As you all know, RRB NTPC Exam will be held in the month of August-September 2019. So we are here to help you with the Mathematics subject. We are providing daily quantitative aptitude quizzes, which will help you to score good marks in this section. We aim to provide the best study material to our readers with exam level questions to help them get used to the recent pattern. Attempt this quiz and check your preparation.

Q1. From a solid cylinder of height 10 cm and radius of the base 6 cm, a cone of same height and same base is removed. The volume of the remaining solid is:
एक ठोस सिलिंडर से जिसकी ऊंचाई 10से.मी और आधार त्रिज्या 6 से.मी है उसमे से समान ऊंचाई और समान आधार त्रिज्या वाला एक शंकु निकाला जाता है. शेष भाग का आयतन कितना है?
(a) 240π cu.cm
(b) 5280 cu.cm
(c) 620π cu.cm
(d) 360π cu.cm

S1. Ans.(a)
Sol. Volume of cylinder = πr²h
= (6)² × 10 × π
= 360 π cm³
Volume of cone = 1/3πr²h
=1/3×π×36×10
= 120 π cm³
Volume of Remaining Solid
= 360 π – 120 π
= 240 π cm³

Q2. What part of a ditch, 48 metres long 16.5 metres broad and 4 metres deep can be filled by the sand got by digging a cylindrical tunnel of diameter 4 metres and length 56 metres? (use π = 22/7)
एक 48 मीटर लंबे, 16.5 मीटर चौड़े और 4 मीटर गहरे गढ्ढे का कितना हिस्सा 4 मीटर व्यास और 56 मीटर लंबाई की बेलनाकार सुरंग की खुदाई से प्राप्त मिट्टी से भरा जा सकता है? (π = 22/7 का उपयोग करें)
(a) 1/9
(b) 2/9
(c) 7/9
(d) 8/9

S2. Ans.(b)
Sol. Volume of ditch = lbh
= 48 × 16.5 × 4
= 48 × 66
Volume of sand = πr²h
= 22/7 × 4 × 56
= 22 × 4 × 8
Part of Ditch filled
=(22×4×8)/(48×66)
=2/9

Q3. The size of a rectangular piece of paper is 100 cm × 44 cm. A cylinder is formed by rolling the paper along its length. The volume of the cylinder is [use π = 22/7]
कागज के आयताकार टुकड़े का आकार 100 सेमी × 44 सेमी है. कागज को इसकी लंबाई के साथ रोल करके एक सिलेंडर बनाया जाता है. सिलेंडर का आयतन क्या है? (π = 22/7 का उपयोग करें)
(a) 4400 cm³
(b) 15400 cm³
(c) 35000 cm³
(d) 144 cm³

S3. Ans.(b)
Sol. 2πr = 44
2×22/7×r=44
r = 7 cm
h = 100 cm
Volume of cylinder = πr²h
=22/7×49×100
= 15400 cm³

Q4. A cylindrical rod of iron whose height is eight times its radius is melted and cast into spherical balls each of half the radius of the cylinder. The number of such spherical balls is
लोहे की एक बेलनाकार छड़ी जिसका ऊंचाई इसकी त्रिज्या से आठ गुना है और सिलेंडर के प्रत्येक आधी त्रिज्या में गोलाकार गेंदों को डाला जाता है. ऐसी गोलाकार गेंदों की संख्या है
(a) 12
(b) 16
(c) 24
(d) 48

S4. Ans.(d)
Sol. Let radius of cylindrical rod = r
Height of cylindrical rod = 8r
Radius of spherical ball = r/2
Number of spherical balls

RRB NTPC Mathematics Questions: 31st July_50.1

Q5. A circus tent is cylindrical up to a height of 3 m and conical above it. If its diameter is 105 m and the slant height of the conical part is 63 m, then the total area of the canvas required to make the tent is [take π = 22/7]
एक सर्कस तम्बू 3 मीटर की ऊंचाई तक बेलनाकार है और इसके ऊपर शंकुधारी है. यदि इसका व्यास 105 मीटर है और शंकु भाग की तिरछी ऊंचाई 63 मीटर है, तो तम्बू बनाने के लिए आवश्यक कैनवास के कुल क्षेत्रफल है (π = 22/7 का उपयोग करें)
(a) 11385 m²
(b) 10395 m²
(c) 9900 m²
(d) 990 m²

S5. Ans.(a)
Sol. Total area of canvas required
= Surface area of cylindrical part + Surface area of conical part
= 2πr₁h₁ + πr₂l₂
= π × 105 × 3 + π × 105/2 × 63
= 11385 m²

Q6. Water flows at the rate of 10 meters per minute from a cylindrical pipe 5 mm in diameter. How long it take to fill up a conical vessel whose diameter at the base is 30 cm and depth 24 cm?
एक बेलनाकार पाइप से 5 मिमी व्यास से 10 मीटर प्रति मिनट की दर से पानी बहता है. एक शंक्वाकार पात्र को भरने में कितना समय लगता है जिसके आधार का व्यास 30 सेमी और गहराई 24 सेमी है?
(a) 28 minutes 48 seconds/ 28 मिनट 48 सेकंड
(b) 51 minutes 24 seconds/51 मिनट 24 सेकंड
(c) 51 minutes 39 seconds/51 मिनट 39 सेकंड
(d) 28 minutes 36 seconds/28 मिनट 36 सेकंड

S6. Ans.(a)
Sol.

RRB NTPC Mathematics Questions: 31st July_60.1

Q7. A spherical copper ball, whose diameter is 9 cm. is melted and converted into a wire having diameter equal to 2 mm. Find the length of the wire.
एक व्यास 9 सेमी वाली गोलाकार तांबे की गेंद को पिघलाया जाता है और एक तार में परिवर्तित किया जाता है जिसका व्यास 2 मिमी के बराबर है. तार की लंबाई कितनी होगी?
(a) 1512 m
(b) 1152 m
(c) 2512 m
(d) 121.5 m

S7. Ans.(d)
Sol.

RRB NTPC Mathematics Questions: 31st July_70.1

Q8. Some portion of a cylindrical pot is filled with water. Radius of base of the pot is 6 cm. A sphere of radius 3 cm. is dropped. If the sphere completely submerges in the water, how much water level is increased?
बेलनाकार पॉट का कुछ हिस्सा पानी से भरा हुआ है. बर्तन के आधार का त्रिज्या 6 सेमी है. त्रिज्या 3 सेमी का एक गोला उसमें गिराया जाता है. यदि गोला पूरी तरह से पानी में डूब जाता है, पानी का स्तर कितना बढ़ गया है?
(a) 5 cm
(b) 8 cm
(c) 1 cm
(d) 12 cm

S8. Ans.(c)
Sol.

RRB NTPC Mathematics Questions: 31st July_80.1

Q9. Two rectangular sheets of paper each 30 cm × 18 cm are made into two right circular cylinders, one by rolling the paper along its length and the other along the breadth. The ratio of the volumes of the two cylinders thus formed, is:
30 सेमी × 18 सेमी की कागज की दो आयताकार शीट, से वृत्तीय बेलन बनाये जाते है, एक पेपर को इसकी लंबाई के साथ रोल किया जाता है और दूसरी को चौड़ाई के साथ रोल किया जाता है. इस प्रकार बनाए गए दो बेलनों के आयतन का अनुपात कितना है:
(a) 2 : 1
(b) 3 : 2
(c) 4 : 3
(d) 5 : 3

S9. Ans.(d)
Sol.

RRB NTPC Mathematics Questions: 31st July_90.1

Q10. If a cubic cm of cast iron weights 21 gms, then the weight of a cast iron pipe of length 1 m with a bore of 3 cm and in which the thickness of the metal is 1 cm, is:
यदि एक से.मी के लोहे के घन का भार 21 ग्राम है, तो 3से.मी की व्यास और 1 से.मी चौड़ाई की धांतु वाले 1 मी लंबे लोहे के पाइप का भार ज्ञात कीजिये?
(a) 46.2 kg
(b) 24.2 kg
(c) 26.4 kg
(d) 18.6 kg

S10. Ans.(a)
Sol.

RRB NTPC Mathematics Questions: 31st July_100.1

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