# RRB NTPC Mathematics Quiz : 20th August

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. Factorise x² + 3x ­-18
x² + 3x ­ 18 गुणनखंड कीजिये
(a) (x+18)(x­1)
(b) (x­1)(x+18)
(c) (x+6)(x­3)
(d) (x­6)(x+3)
S1. Ans.(c)
Sol. x² + 3x – 18
⇒ x² + 6x – 3x – 18
⇒ x (x + 6) – 3(x + 6)
⇒ (x – 3) (x + 6)

Q2. Rohit walks at 17 km/hr and Ruchira cycles at 22 km/hr towards each other. What was the distance between them when they started if they meet after 44 minutes?
रोहित 17कि.मी/घंटा पर चलता है और रुचिरा 22कि.मी/घंटा पर साइकिल चलाती है और वे एकदूसरे कि ओर बढ़ रहे हैं. यदि वे 44 मिनट बाद मिलते हैं तो आरम्भ में उनके मध्य कि दूरी कितनी थी?
(a) 42.9 km
(b) 35.8 km
(c) 21.5 km
(d) 28.6 km

S2. Ans.(d)
Sol. Time = Distance/Relative speed
44/60=x/39
(44×39)/60=x
28.6 km = x

Q3. If sum of the roots of a quadratic equation is –7 and product of the roots is 12. Find the quadratic equation.
यदि एक द्विघात समीकरण का योग -7 है और गुणनखंडो का मूल 12 है. तो द्विघात समीकरण ज्ञात कीजिये.
(a) x² + 7x + 12 = 0
(b) x² – 7x + 12 = 0
(c) x² – 7x – 12 = 0
(d) x² + 7x – 12 = 0

S3. Ans.(a)
Sol. x² – (sum of roots) x + product of the roots = 0
x² – (–7)x + 12 = 0
x² + 7x + 12 = 0

Q4. 20% discount is offered on an item. By applying a promo code the customer wins 10% cash back. What is the effective discount?
एक वास्तु पर 20% कि छूट दी जाती है. एक प्रोमो कोड लगाने पर कस्टमर 10% कैश बेक जीतता है. प्रभावी छूट क्या है?
(a) 30.8 percent
(b) 30 percent
(c) 12 percent
(d) 28 percent

S4. Ans.(d)
Sol. Effective discount
= –20 – 10 + 2
= 28%

Q5. Reduce 2530/1430 to lowest terms.
2530/1430 को न्यूनतम पदों तक कम कीजिये.
(a) 47/17
(b) 23/13
(c) 47/19
(d) 29/17

S5. Ans.(b)
Sol. 2530/1430
= 23/13

Q6. If 5x – 3 ≥ 3 + x/2 and 4x – 2 ≤ 6 + x; then x can take which of the following values?
यदि 5x – 3 ≥ 3 + x/2 और 4x – 2 ≤ 6 + x; तो x को निम्लिखित में कौन सा मान प्राप्त होगा?
(a) 1
(b) 2
(c) –1
(d) –2

S6. Ans.(b)
Sol. 5x – 3 ≥ 3 + x/2
10x – 6 ≥ 6 + x
9x ≥ 12
x ≥ 4/3
x ≥ 1.33 …(i)
4x – 2 ≤ 6 + x
3x ≤ 8
x ≤ 8/3
x ≤ 2.66 …(ii)
from (i) & (ii)
1.33 ≤ x ≤ 2.66
x = 2

Q7. The first and last terms of an arithmetic progression are 32 and –43. If the sum of the series is –88, then it has how many terms?
एक अंकगणितीय प्रगति के पहले और अंतिम पद 32 और -43 हैं. यदि श्रंखला का योग -88 है, तो इसमें कितने पद है?
(a) 16
(b) 15
(c) 17
(d) 14

S7. Ans.(a)
Sol. Ist term = 32
Last term = –43
Sum=n/2 [1st term+last term]
-88=n/2 [32-43]
-176=n[-11]
n = 16

Q8. The difference between simple and compound interests compounded annually on a certain sum of money for 2 years at 18% per annum is Rs 81. The sum is ____.
2 वर्षों के लिए 18% प्रति वार्षिक की दर से निश्चित राशि पर, संयोजित होने वाले साधारण और चक्रवृद्धि ब्याज के बीच का अंतर 81 रु है. योग है____
(a) Rs 2500
(b) Rs 5000
(c) Rs 10000
(d) Rs 7500

S8. Ans.(a)
Sol. If difference of S.I & C.I for 2 years is given than we can use the following formula
C.I-S.I=P(R/100)²
81=(P×18×18)/10000
P = Rs 2500

Q9. In what ratio is the segment joining (12, –1) and (–3, 4) divided by the Y­axis?
(12, –1) और (–3, 4) को जोड़ने वाले रेखाखंड को  Y अक्ष किस अनुपात में विभाजित करता है?
(a) 4:1
(b) 1:4
(c) 4:3
(d) 3:4

S9. Ans.(a)
Sol. Points (12, –1) & (–3, 4)
if the line segment joining above points is divided by y axis then x = 0
Let it divides it is the ratio m : n
x=(mx₂+nx₁)/(m+n)
0=(m(-3)+12n)/(m+n)
3m = 12n
m : n = 4 : 1

Q10. The line passing through (4,3) and (y,0) is parallel to the line passing through (–1,–2) and (3,0). Find y?
(4,3) और (y,0) से गुजरने वाली रेखा (–1,–2) और (3,0) से गुजरने वाली रेखा के समान है.  Y ज्ञात कीजिए?
(a) –1
(b) –2
(c) 2
(d) –5

S10. Ans.(b)
Sol. Slope of line passing through (4, 3) (y, 0) is
m1=(y2-y1)/(x2-x1)
=(0-3)/(y-4)
=(-3)/(y-4)
Slope of line passing through (–1, –2) (3, 0) is
m2=(y2-y1)/(x2-x1 )
=(0-(-2))/(3-(-1) )
=2/4
If two lines are parallel then, there slopes we equal
m₁ = m₂
(-3)/(y-4)=1/2
–6 = y – 4
y = –2

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