# RRB NTPC Mathematics Quiz : 1st October

Dear aspirants,

As you all know, RRB NTPC Exam will be held in the month of September-October 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. The perimeters of a square and a rectangle are equal. If their area be ‘A’ m² and ‘B’ m² then correct statement is
एक वर्ग और एक आयात का परिमाप बराबर है. यदि उनका क्षेत्रफल ‘A’ m² और ‘B’ m² है तो सही कथन है:
(a) A < B
(b) A ≤ B
(c) A > B
(d) A ≥ B

S1. Ans.(c)
Sol. 4a = 2(l + b)
a=1/2 (l+b)
Area of square, A = a²
= ¼ (l² + b² + 2lb)
Area of Rectangle, B = lb
A > B

Q2. The diagonal of a cuboid of length 5 cm, width 4 cm and height 3 cm is
एक घनाब जिसकी लंबाई 5से.मी चौड़ाई 4से.मी और लंबाई 3से.मी है, उसका विकर्ण है:
(a) 5√2 cm
(b) 2√5 cm
(c) 12 cm
(d) 10 cm

S2. Ans.(a)
Sol. Diagonal = √(l²+b²+h² )
=√(25+16+9)
=√50
=5√2 cm

Q3. A rectangle with one side 4 cm is inscribed in a circle of radius 2.5 cm. The area of the rectangle is:
एक आयात जिसकी एक भुजा 4से.मी है और वह एक 2.5से.मी कि त्रिज्या वाले वृत के अंदर निर्मित है. आयात का क्षेत्रफल है:
(a) 8 cm²
(b) 12 cm²
(c) 16 cm²
(d) 20 cm²

S3. Ans.(b)
Sol.
Diagonal of Rectangle = Diameter of circle
= 5 cm
25 = b² + 16
b = 3 cm
area of rectangle = lb
= 12 cm²

Q4. The area of the largest sphere (in cm²) that can be drawn inside a square of side 18 cm is
एक 18से.मी वाली भुजा वाले वर्ग के अंदर अधिकतम कितने क्षेत्रफल वाला गोला बनाया जा सकता है?
(a) 972 π
(b) 11664 π
(c) 36 π
(d) 288 π

S4. Ans.(a)
Sol. Radius of the largest sphere
=18/2=9 cm
Volume of sphere = 4/3 πr³
=4/3×81×9×π
= 972 π

Q5. The area of the circle with radius Y is W. The difference between the areas of the bigger circle (radius with radius Y) and that of the smaller circle (with radius X) is W’. So X/Y is equal to
Y त्रिज्या वाले एक वृत का क्षेत्रफल W है. बड़े वृत (त्रिज्या Y के साथ) और छोटे वृत (त्रिज्या X के साथ) के क्षेत्रफल के मध्य का अंतर W’ है. तो X/Y बराबर है:
(a) √(1-w’/w)
(b) √(1+w’/w)
(c) √(1+w/w’ )
(d) √(1-w/w’)

S5. Ans.(a)
Sol. w = π y²
Bigger Area – Smaller Area = w’
πy²-πx²=w’
1-(πx²)/(πy² )=w’/(πy² )
1-x²/y² =w’/w
x/y=√(1-w’/w)

Q6. An elephant of length 4 m is at one corner of a rectangular cage of size (16 m × 30 m) and faces towards the diagonally opposite corner. If the elephant starts moving towards the diagonally opposite corner it takes 15 seconds to reach this corner. Find the speed of the elephant
एक हाथी जिसकी लंबाई 4मी है वह एक (16 मी × 30 मी) वाले आयातकार वाले केज के कोने पर सामने वाले कोने कि ओर मुख करके खड़ा है. यदि हाथी विकर्णत: विपरीत कोने कि ओर चलना शुरू करता है तो उसे कोने तक पहुचने में 15 सेकंड लगते हैं. हाथी कि गति ज्ञात कीजिये:
(a) 1 m/sec
(b) 2 m/sec
(c) 1.87 m/sec
(d) 1.5 m/sec

S6. Ans.(b)
Sol. Diagonal = √(16)²+(30)²
=√(256+900)
=√1156
= 34 m
Distance that elephant has to cover = 34 – 4= 30 m
Speed of elephant = 30/15 m/sec= 2 m/sec

Q7. If the area of three adjacent faces of a rectangular box which meet in the corner are 12 cm², 15 cm² and 20 cm² respectively. Then the volume of the box is
यदि आयातकार बॉक्स के तीन आसन्न मुख जो कि कोनो पर मिलते हैं उनका क्षेत्रफल क्रमश: 12 से.मी ², 15 से.मी ² और 20 से.मी² है. तो बॉक्स का आयतन है:
(a) 3600 cm³
(b) 300 cm³
(c) 60 cm³
(d) 180 cm³

S7. Ans.(c)
Sol. Volume of box
=√(12×15×20)
=√3600  cm³
= 60 cm³

Q8. The ratio between the length and the breadth of a rectangular park is 3 : 2. If a man cycling along the boundary of the park at the speed of 12 km/hour completes one round in 8 minutes, then the area of the park is
आयताकार पार्क की लंबाई और चौड़ाई के बीच का अनुपात 3: 2 है. यदि कोई व्यक्ति 12 किमी/घंटा की गति से पार्क की बाउंड्री के साथ साइकिल चलाता और 8 मिनट में एक चक्कर पूरा करता है, तो पार्क का क्षेत्रफल कितना है?
(a) 153650 m²
(b) 135600 m²
(c) 153600 m²
(d) 156300 m²

S8. Ans.(c)
Sol. L : B = 3 : 2
Let length ⇒ 3x
Distance covered by cyclist = 2(3x + 2x) = 10x
ATQ,
10x = 12 × 8/60
x = 8/50 km
=8/50×1000 m
= 160 m
Length = 480 m
Breadth = 320 m
Area = 480 × 320
= 153600 m²

Q9. If the radius of a right circular cylinder open at both the ends, is decreased by 25% and the height of the cylinder is increased by 25%. Then the curved surface area of the cylinder thus formed is:
यदि एक दोनों ओर से खुले हुए सम वृताकार कि त्रिज्या को 25% से घटा दिया जाता है और इसकी ऊंचाई को 25% से बढ़ाया जाता है. तो निर्मित सिलिंडर का प्रष्ठ सतह क्षेत्रफल कितना है:
(a) remains unaltered
(b) is increased by 25%
(c) is increased by 6.25%
(d) is decreased by 6.25%

S9. Ans.(d)
Sol. Curved Surface are = 2πrh
New curved surface are
=2π×75/100×r×125/100×h
=2π×3/4×r×5/4×h
=15/8 πrh
Decrease om area
=2πrh-15/8 πrh
=1/8 πrh
% Decrease = (πrh/8)/2πrh×100
=100/16
= 6.25%

Q10. A cylindrical pencil of diameter 1.2 cm has one of its end sharpened into a conical shape of height 1.4 cm. The volume of the material removed is
एक बेलनाकार पेंसिल जिसकी व्यास 1.2 से.मी है उसके एक कोने को 1.4से.मी कि ऊंचाई वाले शंक्वाकार आकर में चील दिया जाता है. उसमें से हटाए गई सामग्री का आयतन कितना है?
(a) 1.056 cm³
(b) 4.224 cm³
(c) 1.089 cm³
(d) 42.24 cm³

S10. Ans.(b)
Sol. Volume of are removed
= πr²h – 1/3 πr²h
=  2/3 πr² h
=2/3×22/7×1.44 ×1.4
=88.704/21
= 4.224 cm³

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