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Monday, 2 December 2019

Quantitative Aptitude Quiz For IBPS CLERK PRELIMS | 02 -12-19

Durgesh Mahendras

Quantitative Aptitude Quiz For IBPS CLERK PRELIMS | 02 -12-19

Dear All,
As you all know that IBPS Clerk Prelims exams dates are aproaching and it is high time for you to start your preparation for the exam. In order to make you prepared for the examination, Mahendras has started new quizzes of Maths, English & Reasoning section of the IBPS Clerk Prelims, these quizzes will be very beneficial for the exam. You can practice these quizzes on daily basis.
INSTRUCTION:-
There are five different fighter plane and there marked up percent and discount percent are given below. Study the data carefully and answer the question below.

निर्देश :- पांच विभिन्न प्रकार के लड़ाकू विमान दिए गए है जिनके बढे हुए मूल्यों का प्रतिशत और छुट प्रतिशत दिया गया है. इसको ध्यान पूर्वक पढ़े और निचे दिए गए प्रश्नों का उत्तर दें.




FIGHTER PLANE COST PRICE (IN CARORE RUPEES) SELLING PRICE (IN CARORE RUPEES)
F – 22 RAPTOR 1500 -------
SUKHOI Su – 57 -------- 2106
CHENGDU J – 20 -------- 2900
EUROFIGHTER TYPHOON 4500 ----
DASSAULT RAFALE -------- 3060


Q.1 what is the ratio of selling price of F – 22 RAPTOR to the cost price of DASSAULT RAFALE.

F – 22 RAPTOR के विक्रय मूल्य और DASSAULT RAFALE के क्रय मूल्य का अनुपात क्या होगा ?



  1.  111 : 211
  2.  Can’t be determined/ ज्ञात नहीं कर सकते
  3.  13 :14
  4.  14 :25
  5.  None of these/इनमे से कोई नहीं

Q.2:- If manufacturer of Eurofighter TYFOON sell 3  Eurofighter TYFOON to India first at given usual selling price second one is 80% of the selling price and for 3rd one  ratio of discount percent to profit percent is 3 : 5.  Find overall profit percent of the manufacturer?

यदि Eurofighter TYFOON के निर्माता ने 3  Eurofighter TYFOON इंडिया को बेचे . जिसमे पहला उसके वास्तविक विक्रय मूल्य पर, दुसरा विक्रय मूल्य के 80% पर और तीसरे के छूट प्रतिशत और उसके लाभ प्रतिशत का अनुपात 3 : 5 है तो निर्माता का कुल लाभ कितना है?



  1. 10 IBPS-CLERK-MATHS-QUIZ-02-12-19 %
  2.  13IBPS-CLERK-MATHS-QUIZ-02-12-19 %
  3.  Insufficient data/ अपर्याप्त आंकड़े
  4.  None of these/इनमे से कोई नहीं
  5.  13%


Q.3 what is the average of cost price of SUKHOI Su – 57, CHENGDU J – 20 , DASSAULT RAFALE and selling price of F – 22 RAPTOR and EUROFIGHTER TYPHOON.


SUKHOI Su – 57, CHENGDU J – 20, DASSAULT RAFALE के क्रय मूल्य और F – 22 RAPTOR, EUROFIGHTER TYPHOON के विक्रय मूल्य का औसत क्या होगा?



  1.  Rs.2745
  2.  Rs.2674
  3.  DATA INSUFFICIENT/ अपर्याप्त आकंडे
  4.  Rs.2741
  5.  NONE OF THESE/ इनमे से कोई नहीं


Q.4. what is the ratio of cost price of CHENGDU J – 20 to that the cost price of DASSAULT RAFAL?

CHENGDU J – 20 और DASSAULT RAFAL के क्रय मूल्यों का अनुपात क्या होगा?



  1.  15 :17
  2.  11 :15
  3.  5 : 6
  4.  13 : 11
  5.  113 : 115

Q.5:- when the cost price of F – 22 RAPTOR is increased by 100% and the profit is increased by (200 + x) % then its profit percent is become IBPS-CLERK-MATHS-QUIZ-02-12-19%. What is possible value of x ?


यदि F – 22 RAPTOR के क्रय मूल्य को 100 % बढ़ा दिया जाता है और लाभ को (200 + x)% बढ़ा दिया जाता है तो इसका लाभ IBPS-CLERK-MATHS-QUIZ-02-12-19% हो  जाता है ? तो x का संभावित  का मान ज्ञात करें ?



  1.  None of these/इनमे से कोई नहीं
  2. 100
  3.  200
  4.  120
  5.  Insufficient data/अपर्याप्त आंकड़े

Q.6:- In the following number series find the missing term?

निम्नलिखित संख्या श्रृंखला में लुप्त पद ज्ञात कीजिए। 


6250      7500       9000       10800        ?

  1.  12960
  2.  12600
  3.  12700
  4.  12860
  5.  12660

Q.7 In the following questions two equations numbered I and II are given. You have to solve both the equations and give answer

निम्नलिखित प्रश्नों में दो समीकरण I और II दी गयी हैं । आपको दोनों समीकरणों को हल करना होगा और उत्तर दीजिये


I. p^2 – 4 = 0
II. q^2 + 6q + 9 = 0


(1) p > q
(2) p < q
(3) p IBPS-CLERK-MATHS-QUIZ-02-12-19 p
(4) pIBPS-CLERK-MATHS-QUIZ-02-12-19 q
(5) If p = q or relation can’t be established/ सम्बन्ध स्थापित नहीं किया जा सकता


(8) In the following questions two equations numbered I and II are given. You have to solve both the equations and give answer

निम्नलिखित प्रश्नों में दो समीकरण I और II दी गयी हैं । आपको दोनों समीकरणों को हल करना होगा और उत्तर दीजिये

I.  p^2 – 0.7p + 0.12 = 0
II. q^2 + 0.1 × q – 0.12 = 0


(1) p > q
(2) p < q
(3) p IBPS-CLERK-MATHS-QUIZ-02-12-19 q
(4) pIBPS-CLERK-MATHS-QUIZ-02-12-19 q
(5) If p = q or relation can’t be established/ सम्बन्ध स्थापित नहीं किया जा सकता


(9) In the following questions two equations numbered I and II are given. You have to solve both the equations and give answer

निम्नलिखित प्रश्नों में दो समीकरण I और II दी गयी हैं । आपको दोनों समीकरणों को हल करना होगा और उत्तर दीजिये


I. p^4 – 227 = 398
II. q^2 + 321 = 346


1) p > q
(2) p < q
(3) q IBPS-CLERK-MATHS-QUIZ-02-12-19 p
(4) qIBPS-CLERK-MATHS-QUIZ-02-12-19 p
(5) If p = q or relation can’t be established/ सम्बन्ध स्थापित नहीं किया जा सकता

(10) In the following questions two equations numbered I and II are given. You have to solve both the equations and give answer

निम्नलिखित प्रश्नों में दो समीकरण I और II दी गयी हैं । आपको दोनों समीकरणों को हल करना होगा और उत्तर दीजिये

I. 2p^2 + 11p + 14 = 0
II. 4q^2 + 12q + 9 = 0

1) p > q
(2) p < q
(3) q IBPS-CLERK-MATHS-QUIZ-02-12-19 p
(4) qIBPS-CLERK-MATHS-QUIZ-02-12-19 p
(5) If p = q or relation can’t be established/ सम्बन्ध स्थापित नहीं किया जा सकता



Answer:-

(1)SOLUTION:- 

Selling price of F – 22 RAPTOR
Marked price = 1500 IBPS-CLERK-MATHS-QUIZ-02-12-19 IBPS-CLERK-MATHS-QUIZ-02-12-19 = Rs.2100
Selling price = 2100 IBPS-CLERK-MATHS-QUIZ-02-12-19 IBPS-CLERK-MATHS-QUIZ-02-12-19 = Rs.1680
Cost price of DASSAULT RAFALE
Marked price = 3060 IBPS-CLERK-MATHS-QUIZ-02-12-19 = Rs.3600
Cost price = 3600 IBPS-CLERK-MATHS-QUIZ-02-12-19 = 3000
Required ratio = 1680 : 3000
                       = 14 :25
हल :- F – 22 RAPTOR का विक्रय मूल्य
अंकित मूल्य = 1500 IBPS-CLERK-MATHS-QUIZ-02-12-19 IBPS-CLERK-MATHS-QUIZ-02-12-19 = Rs.2100
विक्रय मूल्य = 2100 IBPS-CLERK-MATHS-QUIZ-02-12-19 IBPS-CLERK-MATHS-QUIZ-02-12-19 = Rs.1680
DASSAULT RAFALE का क्रय मूल्य =
अंकित मूल्य = 3060 IBPS-CLERK-MATHS-QUIZ-02-12-19 = Rs.3600
क्रय मूल्य  = 3600 IBPS-CLERK-MATHS-QUIZ-02-12-19 = 3000
अभीष्ट अनुपात  = 1680: 3000
                      = 14:25
2. Solution:- 
Selling price of first one = 4500 IBPS-CLERK-MATHS-QUIZ-02-12-19 1.5 IBPS-CLERK-MATHS-QUIZ-02-12-19 0.7 = Rs.4725
Selling price of second one = 4725 IBPS-CLERK-MATHS-QUIZ-02-12-19 0.8 = Rs.3780
Selling price of 3rd one = 4500 IBPS-CLERK-MATHS-QUIZ-02-12-19 1.5 = Rs.6750
Required profit % = IBPS-CLERK-MATHS-QUIZ-02-12-19 IBPS-CLERK-MATHS-QUIZ-02-12-19 100 = 13%
3.SOLUTION:-

REQUIRED AVERAGE = IBPS-CLERK-MATHS-QUIZ-02-12-19 = IBPS-CLERK-MATHS-QUIZ-02-12-19 = 2741
अभीष्ट औशत :- IBPS-CLERK-MATHS-QUIZ-02-12-19 = IBPS-CLERK-MATHS-QUIZ-02-12-19 = 2741
4.Solution:- 
C.P of DASSAULT RAFAL = 3060 IBPS-CLERK-MATHS-QUIZ-02-12-19  IBPS-CLERK-MATHS-QUIZ-02-12-19 IBPS-CLERK-MATHS-QUIZ-02-12-19 IBPS-CLERK-MATHS-QUIZ-02-12-19 = Rs.3000
CP of CHENGDU J – 20 = 2900 IBPS-CLERK-MATHS-QUIZ-02-12-19  IBPS-CLERK-MATHS-QUIZ-02-12-19 IBPS-CLERK-MATHS-QUIZ-02-12-19 IBPS-CLERK-MATHS-QUIZ-02-12-19 = Rs.2500
Required average = 2500: 3000
                            = 5: 6

5.Solution:- 


CP = Rs.1500                                                              New CP = Rs.3000

Marked price = 1500 IBPS-CLERK-MATHS-QUIZ-02-12-19 IBPS-CLERK-MATHS-QUIZ-02-12-19 = Rs.2100                         
Selling price = 2100 IBPS-CLERK-MATHS-QUIZ-02-12-19 IBPS-CLERK-MATHS-QUIZ-02-12-19 = Rs.1680
Profit = Rs.180                                                       
New profit = 180 + IBPS-CLERK-MATHS-QUIZ-02-12-19 IBPS-CLERK-MATHS-QUIZ-02-12-19 180 = IBPS-CLERK-MATHS-QUIZ-02-12-19
New profit percent = IBPS-CLERK-MATHS-QUIZ-02-12-19
ATQ,
IBPS-CLERK-MATHS-QUIZ-02-12-19 IBPS-CLERK-MATHS-QUIZ-02-12-19 IBPS-CLERK-MATHS-QUIZ-02-12-19 100
IBPS-CLERK-MATHS-QUIZ-02-12-19  = IBPS-CLERK-MATHS-QUIZ-02-12-19 
IBPS-CLERK-MATHS-QUIZ-02-12-19
1.8x = 180
X = 100
Q.6 Sol:- (2) 12900

Logic:- IBPS-CLERK-MATHS-QUIZ-02-12-19

(7) SOLUTION:- 

.p = +2 , - 2 
q  = -3 , - 3  hence p > q

(8) Solution:- I.  p2 – 0.7p + 0.12 = 0
    p2 – 0.3p – 0.4p + 0.12 = 0
    p (p – 0.3) – 0.4 (p – 0.3) = 0
    (p – 0.3) (p – 0.4) = 0
    p = 0.3, 0.4
II. q2 + 0.1 × q – 0.12 = 0
      q2 + 0.4q – 0.3q – 0.12 = 0
      q (q + 0.4) – 0.3 (q + 0.4) = 0
      (q + 0.4) (q – 0.3) = 0
      q = - 0.4, 0.3
      Hence, q ≤ p
(9) Solution:-
I. p4 – 227 = 398
    p4 = 625
    p = 5
II. q2 + 321 = 346
    q2 = 346 – 321
    q2 = 25
    q = 5, - 5
    Hence, p ≥ q
(10) Solution:-

I. 2p^2 + 11p + 14 = 0

     2p^2 + 7p + 4p + 14 = 0

     p (2p + 7) + 2 (2p + 7) = 0

     (2p + 7) (p + 2) = 0

     p = - 2, - 7/2

II. 4q^2 + 12q + 9 = 0

     4q^22 + 6q + 6q + 9 = 0

     2q (2q + 3) + 3(2q + 3) = 0

     (2q + 3) (2q + 3) = 0

     q = - 3/2, - 3/2

      Hence, p < q

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