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SSC CPO : Quantitative Aptitude Quiz | 01 - 05 - 18

Mahendra Guru
SSC CPO : Quantitative Aptitude Quiz | 28 - 04 - 18

In SSC exam, quantitative Aptitude section is more scoring and easy, if you know the shorts tricks and formulas of all the topics. So, it is important to know the basic concepts of all the topics so you can apply the short tricks and solve the question with the new concept sin lesser time while giving the quiz. It will help you to score more marks from this section in less time period. Quantitative Aptitude section basically measures your mathematical and calculation approach of solving the question. SSC Quiz of quantitative Aptitude section helps you to analyse your preparation level for upcoming SSC examination. Mahendra Guru provides you Quantitative Aptitude Quiz for SSC examination based on the latest pattern so that you can practice on regular basis. It will definitely help you to score good marks in the exam. It is the most important section for all the govt exams like Insurance, SSC-MTS, SSC CPO, CGL, CHSL, State Level, and other Competitive exams.

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Q.1. If 40 men or 60 women or 80 children can do a piece of work in 6 months, then 10 men, 10 women and 10 children together do half of the work in

यदि 40 पुरूष या 60 महिलाएँ या 80 बच्चे किसी कार्य को 6 माह में कर सकते हैं, तो 10 पुरूष, 10 महिलाएँ और 10 बच्चे मिलकर आधे कार्य को कितने माह में पूरा करेंगे?
(A) months/माह 
(B) 6 months/माह
(C) months/माह 
(D) months/माह

Q.2. If, then the value of is

यदि, तोका मान है 
(A) 1448
(B) 1442
(C) 1444
(D) 1446

Q.3. 10 men can complete a piece of work in 15 days. They started work and after five days 10 women joined them and the work has finished in next 5 days. In how many days can 25 women finish the same piece of work?

10 व्यक्ति किसी कार्य को 15 दिन में पूरा करते है। उन्होंने कार्य शुरू किया और 5 दिन बाद 10 महिलायें उनके साथ शामिल हो गई जिसके परिणामस्वरूप कार्य अगले 5 दिनों में पूर्ण हो गया। यदि 25 महिलायें उस कार्य को करती तो कार्य कितने दिनों में पूर्ण हो जाता?
(A) 4 days/दिन 
(B) 5 days/दिन
(C) 6 days/दिन
(D) 8 days/दिन
Q.4. x = a secα cosβ, y = b secα sinβ, z = c tanα, then the value of  is
x = a secα cosβ, y = b secα sinβ, z = c tanα तो का मान है
(A) 2
(B) 0
(C) 1
(D) – 1 

Q.5. If tan2θ = 1– e2, then the value of secθ + tan3θ cosecθ is
        यदि tan2θ = 1 – e2, तो secθ + tan3θcosecθ का मान है 
(A) (2 + e2)3/2
(B) (2 – e2)1/2
(C) (2 + e2)1/2
(D) (2 – e2)3/2

Q.6. If x sin60o tan30o – tan245o = cosec60o cot30o – sec2 45o, then what will be x =?
        यदि x sin60o tan30o – tan2 45o = cosec60o cot30o – sec2 45o, तो x = क्या होगा ?
(A) 2
(B) – 2
(C) 6
(D) – 4
Q.7. If, then the value of is

यदि तोका मान है 
(A) 15
(B) 10
(C) 20
(D) 5

Q.8. I walk a certain distance and ride back taking a total time of 37 minutes. I could walk both ways in 55 minutes. How long would it take me to ride both ways?

मैं कुछ दूरी तक पैदल चलता हूँ, और वापस सवारी से आने में कुल 37 मिनट लगते हैं दोनों ओर से पैदल चलने में मुझे 55 मिनट लगते हैं दोनों ओर से सवारी से आने-जाने में कुल कितना समय लगेगा?
(A) 20 minutes 
(B) 19 minutes
(C) 15 minutes 
(D) 25 minutes

Q.9. If (1 – x– 2) is divided by (x– 1 + 1), the result is –
        यदि (1 – x– 2) को (x– 1 + 1) से विभाजित किया गया है, परिणाम है
(A)
(B)
(C)
(D)
Q.10. The simplification of gives
का सरलीकृत परिणाम है
(A)
(B)
(C)
(D)

ANSWER KEY

Q1. (C) 40M = 60W = 80C
2M = 3W = 4C
Now, 10M + 10W + 10C = 10M + M + 5M
= M
Required number of days/अभीष्ट दिनों की संख्या = months/माह

Q2. (B),

Q3. M1 × d1 = M2 × d2
        10M × 15 = (10M × 5) + (10M + 10W) × 5
        150 M = 50 M + 50 M + 50W
        50 M = 50W
        1M = 1W [10 M = 10W)
        Again/पुनः, M1 × d1 = M2 × d2
        10W × 15 = 25W × d2
        d2 = 6 days
/दिन

Q4. (C) x = a secα.cosβ, y = b secα.sinβ, z = c tanα 
sec2α.cos2β + sec2α.sin2 β - tan2α

= sec2α - tan2α = 1
Q5. (D) secθ + tan3θ cosecθ =
 = =
   tan2θ = 1 - e

Q6. (A) x sin600. tan300 - tan2 450 =cosec 600 cot300 - sec2 450
x = 2 
Q7. (D)
2x2 + 1 + 5x = 15x
2x2 + 1 = 10x

Q8. (B) x + y = 37 … (I)
x + x = 55 … (II)
x = 27.5
y = 9.5
Req. time/अभीष्ट समय = 9.5 + 9.5 = 19 minute/मिनट
Q9. (B) (1 – x– 2) ÷ (x– 1 + 1) =
=
Q10. (D) =
=
=
=
=

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