Important AP & GP Questions For RRB ALP CBT 2 : 21st November 2018

 


Dear students, you know that QUANT is a part of getting points and every chapter is important. Therefore, we are providing 10 questions of mathematics. Solve all these quizzes every day so that you can improve your accuracy and speed. We also provide lots of quant questions. So you can practice that chapter which takes more time to solve the questions.

प्रिय पाठकों, आप सभी जानते हैं कि संख्याताम्क अभियोग्यता का भाग बहुत ही महत्वपूर्ण है. इसलिए हम आपको संख्यात्मक अभियोग्यता कि 10 प्रश्नों कि प्रश्नोत्तरी प्रदान कर रहे हैं. इन सभी प्रश्नोत्तरी को दैनिक रूप से हल कीजिये ताकि आप अपनी गति और सटीकता में वृद्धि कर सकें. हम आपको अन्य कई संख्यात्मक अभियोग्यता के प्रश्न प्रदान करेंगे. ताकि आप पाठ्यक्रम अनुसार उन्हें हल कर पायें.

Q1. 9, 99, 999, 9999 . . . . . . . t₁₀₀ , Find the Sum ? 
9, 99, 999, 9999 . . . . . . . t₁₀₀ , योग ज्ञात कीजिए?
(10¹⁰¹-910)/9
(10¹⁰²-910)/9
(10¹⁰³-910)/9
(10¹⁰⁴-910)/9
Solution:
S1. Ans.(a) Sol. 9 + 99 + 999 + 9999 + . . . . . . . . . . . . . upto 100 terms
= 10 – 1 + 10² – 1 + 10³ – 1 + 10⁴ – 1 + . . . . . . . . . . . . .
= 10 + 10² + 10³ + 10⁴ – 1 × 100
=10(10¹⁰⁰-1)/(10-1)-100
=(10¹⁰¹-910)/9
Q2. 5 + 55 + 555 + . . . . . . . . t₁₀₀ , Find the Sum? 
5 + 55 + 555 + . . . . . . . . t₁₀₀ , योग ज्ञात कीजिए?
5/9( (10¹⁰¹-900)/9)
5/9( (10¹⁰¹-910)/9)
5/9( (10¹⁰¹-920)/9)
5/9( (10¹⁰¹-930)/9)
Solution:
S2. Ans.(b) Sol. 5 + 55 + 555 + . . . . . . . . . . . . . + upto 100 terms
= 5 [ 1 + 11+ 111 + . . . . . . . . . . . . . ] = 5/9 [9 + 99 + 999 + . . . . . . . . . . . . . ]
=5/9 [(10¹⁰¹-910)/9]
Q3. The 7th term of G.P is 8 times the 4th term. What will be the 1st term if its 5th term is 48?
 ज्यामितीय प्रगति का सातवाँ पद चौथे पद का 8 गुना है. पहला पद क्या होगा यदि उसका पांचवां पद 48 है?
5
6
3
7
Solution:
S3. Ans.(c) Sol. T₇ = 8 × T₄
ar⁶ = 8 × ar³
r³ = 8 r = 2
T₅ = 48
ar⁴ = 48
a16 = 48
a = 3
Q4. Find the value of 25^[1/3+1/9+1/27+ . . . . . . . ∞] ? 
25^[1/3+1/9+1/27+ . . . . . . . ∞] का मान ज्ञात कीजिए?
5
6
7
9
Solution:
S4. Ans.(a) Sol. 〖25〗^[1/3+1/9+1/27+ . . . . . . . ∞]
S_∞=a/(1-r)=(1/3)/(1-1/3) =(1/3)/(2/3)=1/2 ⇒〖25〗^(1/2)=5
Q5. A ball is thrown from a height of 5000 meter on a ground the ball is bounced 4/5 times of its every last bounce then calculate the total distance covered by the ball till it stopped? एक गेंद को 5000 मीटर की ऊंचाई से जमीन पर फेंका जाता है,गेंद उसके प्रत्येक अंतिम उछाल का 4/5 गुना उछलती है तो गेंद जब तक वह रुक नहीं जाती तो उसके द्वारा तय की गयी कुल दूरी की गणना कीजिए?
45000
50000
40000
35000
Solution:
S5. Ans.(a) Sol. 1st drop height = 5000
2nd Bounce height = 5000 × 4/5 = 4000
3rd Bounce height = 4000 × 4/5 = 3200
4th Bounce height = 32000 × 4/5 = 25600
= 5000 + 4000 + 4000 + 3200 + 3200 + 2560
= 9000 + 7200 . . . . . . . . . . . . . ∞
S_∞=a/(1-r)
r=7200/9000=4/5
S_∞=9000/(1-4/5) = 9000 × 5 = 45000 m
Q6. A ball is thrown from a height of 540 m on a ground the ball is bounced. 2/3 times of its every last bounce then calculate the total distance covered by the ball till it stopped? 
एक गेंद को 540 मीटर की ऊंचाई से जमीन पर फेंका जाता है,गेंद उसके प्रत्येक अंतिम उछाल का 2/3 गुना उछलती है तो गेंद जब तक वह रुक नहीं जाती तो उसके द्वारा तय की गयी कुल दूरी की गणना कीजिए?
2700
2800
2900
3000
Solution:
S6. Ans.(a) Sol. 540 + 360 + 360 + 240 + 240 + 160 =
900 + 600 + 400
S_∞=a/(1-r)
a = 900, r=600/900=2/3
S_∞=900/(1-2/3)=2700
Q7. The side of a square is 16 cm infinite no. of squares are made by joining mid points of each side of the squares. Then calculate the total area of all the infinite no. of squares? 
एक वर्ग की भु215कर असंख्य संख्या में वर्ग बनाए जाते हैं.सभी असंख्य संख्या के वर्गों का कुल क्षेत्रफल ज्ञात कीजिए?
512
513
514
516
Solution:
S7. Ans.(a) Sol. Area of larger Square = (16)² = 256
Area of square made by joining mid points of the square = 256/2 = 128
Sum of Areas of Square = 256 + 128 + . . . . . . . . . . . ∞ =256/(1-1/2) = 256 × 2 = 512 cm²
Q8. The side of a right-angled triangle are 6, 8 & 10 cm respectively. A new right-angled triangle is made by joining the mid-points of all each sides of triangle. This process continues for infinite times then calculate the area of all infinite triangles? 
एक समकोणीय त्रिभुज की भुजा क्रमशः 6, 8 और 10 सेमी हैं. एक नया समकोणीय त्रिभुज, त्रिभुज की प्रत्येक सभी भुजाओं के मध्य बिन्दुओं को जोड़कर बनाया जाता है. इस प्रक्रिया को अनगिनत बार जारी रखा जाता है तो सभी असंख्य त्रिभुजों का क्षेत्रफल कितना होगा?
32
33
34
35
Solution:
S8. Ans.(a) Sol. Area of ∆ABC = 1/2 × 8 × 6 = 24
Area of ∆ after joining mid points of ∆ABC = 1/4 × 24 = 6
Sum of Area of triangle = 24/(1-1/4) = 32 cm²
Q9. If A = 1/4+1/16+1/64+ . . . . . . . n^term , and B = 1+1/2+1/4+ . . . . . . 2n^term , then find A/B = ?
यदि A = 1/4+1/16+1/64+ . . . . . . . n^term , और B = 1+1/2+1/4+ . . . . . . 2n^term , है तो A/B = ज्ञात कीजिए ?
1/6
1/5
1/7
1/8
Solution:
S9. Ans.(a) Sol. A=1/4+1/16+1/64+ . . . . . . . . n^th
B=1+1/2+1/4+. . . . . . . . 2^(n th) terms
A/B=((1/4 [1-(1/4)^n ])/(1-1/4))/(1[1-(1/2)^2n ]/(1-1/2)) =(1/3 [1-(1/4)^n ])/2[1-(1/4)^n ] =1/6
Q10. Find the sum n terms of the series 11 + 103 + 1005 + . . . . . n? श्रृंखला 11 + 103 + 1005 + . . . . . n के n पदों का योग ज्ञात कीजिए?
10/9 (〖10〗^n-1)+n²
10/9 (〖10〗^n-1)+n
10/9 (〖10〗^n-1)+n³
10/9 (〖10〗^n-1)+n⁴
Solution:
S10. Ans.(a) Sol. 11 + 103 + 1005 + . . . . . . . . . . . n ⇒
 10 + 1 + 100 + 3 + 1000 + 5 + . . . . . . . . . . . n terms.
=10(〖10〗^n-1)/9+n²

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