RRB NTPC Mathematics Questions: 30th 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.The height of a right prism with a square base is 15 cm. If the area of the total surfaces of the prism is 608 sq. cm, its volume is
वर्गाकार आधार के साथ एक लम्ब प्रिज्म की ऊंचाई 15 से.मी. है. यदि प्रिज्म का कुल सतह क्षेत्रफल 608 वर्ग से.मी है, तो इसका आयतन कितना है?
(a) 910 cm³
(b) 920 cm³
(c) 960 cm³
(d) 980 cm³

S1.Ans(c)
Sol. Total surface area of Prism = Curved surface area + 2 × area of base
608 = perimeter of base × height + 2 × area of base
608=4x×15+2x²
x²+30x-304=0
x=8
Volume of prism = Area of base × height
= 8 × 8 ×15 = 960 cm³

Q2.If the diagonals of a rhombus are 8 and 6, then the square of its side is
एक समचतुर्भुज के विकर्ण 8 और 6 हैं. इसकी भुजाओं का वर्ग क्या होगा?
(a) 25
(b) 55
(c) 64
(d) 36

S2.Ans(a)
Sol.
D1² + D2² = 4a²
64+36 = 4 a²
a2 = 25

Q3.The volume of a solid hemisphere is 19404 cm³. Its total surface area is
एक ठोस गोलार्द्ध का आयतन 19404 सेमी³ है। इसका कुल पृष्ठ का क्षेत्रफल क्या होगा?
(a) 4158 cm²
(b) 2858 cm²
(c) 1738 cm²
(d) 2038 cm²

S3.Ans(a)
Sol.
2/3 πr³=19404
r=21 cm
Total surface area =3πr²=4158 cm³

Q4. ABCD is a rhombus whose side AB = 4 cm and ∠ABC = 120°, then the length of diagonal BD is equal to:
ABCD एक विषमकोण है जिसकी भुजा AB= 4 सेमी और ∠ABC = 120 डिग्री है, तो विकर्ण BD की लंबाई किसके बराबर है:
(a) 1 cm/सेमी
(b) 2 cm/सेमी
(c) 3 cm/सेमी
(d) 4 cm/सेमी

S4. Ans.(d)
Sol. According to the question.

Given: ∠B = 120°
In a rhombus diagonal are angle bisector and diagonal cut at right triangle.
∴sin⁡ 30°=P/H=BO/AB=1/2=BO/4
BO = 2 cm
∴ BD = 2 × BO
= 2 × 2 = 4 cm

Q5.If (x+1/x)²=3.Then the value of (x^72+x^66+x^54+x^36+x^24+x^6+1) is
यदि (x+1/x)²=3 है. तो (x^72+x^66+x^54+x^36+x^24+x^6+1) का मान ज्ञात करें.
(a) 1
(b) 2
(c) 3
(d) 4

S5.Ans(a)
Sol.
(x+1/x)²=3
x+1/x=√3 then, Power of difference 6 is cancel out.
x^6=-1
So,
x^72+x^66+x^54+x^36+x^24+x^6+1
=(x^6 )^12+(x^6 )^11+(x^6 )^9+(x^6 )^6+(x^6 )^4+x^6+1
=1

Q6.If a+b+c=0, then the value of (a²+b²+c²)/(a²-bc ) is
यदि a+b+c=0 है, तो (a²+b²+c²)/(a²-bc ) का मान ज्ञात करें.
(a) 0
(b) 1
(c) 2
(d) 3

S6.Ans(c)
Sol.
a+b+c=0
b+c=-a
Squaring on both sides.
(b+c)²=a²
a²+b²+c²+2bc=2a²
a²+b²+c²=2a²-2bc=2(a²-bc)
⇒ ((a²+b²+c² ))/((a²-bc) )=2

Q7.If n=7+4√3. then the value of (√n+1/√n) is
यदि n=7+4√3 है. तो (√n+1/√n) का मान ज्ञात करें.
(a) 2√3
(b) 4
(c) –4
(d) -2√3

S7.Ans(b)
Sol.
n=7+4√3=4+3+2×2×√3=(2+√3)²
√n=2+√3
1/√n=2-√3
(√n+1/√n)=4

Q8.If a+b+c=6, a²+b²+c²=14 and a³+b³+c³=36, then the value of abc is
यदि a+b+c=6, a²+b²+c²=14 और a³+b³+c³=36, तो abc का मान ज्ञात करें.
(a) 3
(b) 6
(c) 9
(d) 12

S8.Ans(b)
Sol.
(a+b+c)²=a²+b²+c²+2(ab+bc+ca)
36=14+2(ab+bc+ca)
ab+bc+ca=11 …………..(i)
a³+b³+c³-3abc=(a+b+c)(a²+b²+c²-ab-bc-ca)
=36-3abc=6(14-11)
36-3abc=84-66=18
3abc=36-18=18
abc=6

Q9.If a,b are rational numbers and (a-1) √2+3=b√2+a, the value of (a+b) is
यदि a, b तर्कसंगत संख्याएँ है और (a-1) √2+3=b√2+a, तो (a+b) का मान ज्ञात करें.
(a) –5
(b) 3
(c) –3
(d) 5

S9.Ans(d)
Sol.
(a-1) √2+3=b√2+a
a=3, a-1=b
3-1=b⇒b=2
a+b=5

Q10.The graph of the linear equation 3x+4y=24 is a straight line intersecting x-axis and y-axis at the points A and B respectively. P(2, 0) and Q (0, 3/2) are points on the sides OA and OB respectively of ∆OAB, where O is the origin of the co-ordinate system. Given that AB = 10 cm, then PQ =
रैखिक समीकरण का ग्राफ A और B बिंदुओं पर क्रमशः x-अक्ष और y-अक्ष को प्रतिच्छेद करता है. P(2,0) और Q (0, 3/2) त्रिभुज OAB की क्रमश: भुजा OA और OB पर स्थित बिंदु हैं, जहाँ O निर्देशांक प्रणाली का केंद्र है. यह दिया गया है कि AB= 10 सेमी है, तो PQ=,
(a) 20 cm
(b) 2.5 cm
(c) 40 cm
(d) 5 cm

S10.Ans(b)
Sol.
OP = √((2-0)²+ (0-0)² ) = 2
OQ = √((0-0)²+ (3/2-0)² ) = 3/2

PQ =√(OP ²+OQ ² )
=√(4+(3/2)² ) = √(25/4)=2.5 cm

Important Links for RRB NTPC Recruitment 2019

June Current Affairs PDF

×

Download success!

Thanks for downloading the guide. For similar guides, free study material, quizzes, videos and job alerts you can download the Adda247 app from play store.

Thank You, Your details have been submitted we will get back to you.

Leave a comment

Your email address will not be published. Required fields are marked *

×
Login
OR

Forgot Password?

×
Sign Up
OR
Forgot Password
Enter the email address associated with your account, and we'll email you an OTP to verify it's you.


Reset Password
Please enter the OTP sent to
/6


×
CHANGE PASSWORD