SSC CPO : Quantitative Aptitude Quiz

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 concepts in 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 x⁵ – 9x² + 12x – 14 is divided by (x – 3), what is the remainder?

यदि x⁵ – 9x² + 12x – 14 को (x – 3) से भाग दिया जाता है तो शेषफल क्या है?

(A) 0

(B) 1

(D) 184

Q.2. The distance between the centres of two circles having radii 4.5 cm and 3.5 cm respectively is 10 cm. What is the length of the transverse common tangent of these circle?

क्रमशः 4.5 सेमी. और 3.5 सेमी. त्रिज्या वाले दो वृत्तों के केंद्र की बीच की दूरी 10 सेमी.है। इन वृत्तों की अनुप्रस्थ उभयनिष्ठ स्पर्श रेखा की लम्बाई क्या है?

(A) 8 cm/सेमी.

(B) 7 cm/सेमी.

(D) 5 cm/सेमी.

Q.3. A certain amount of money has to be divided between two persons P and Q in the ratio 3 : 5 .But it was divided in the ratio of 2 : 3 and thereby Q loses Rs. 10 .What was the amount ?

एक निश्चित धनराशि दो व्यक्तियों P और Q के मध्य 3 : 5 के अनुपात में बांटी जानी है। लेकिन यह 2 : 3 में बांटी गई और Q को 10 रू. की हानि हुयी। धनराशि क्या थी?

(A) Rs.250

(B) Rs.300

(D) Rs.400

Q.4. The curved surface of a cylinder is 1000 cm² .A wire of diameters 5 mm. is wound around it ,so as to cover it completely .What is the length of the wire used?

एक बेलन का वक्रपृष्ठ 1000 सेमी² है। एक 5 मिमी. व्यास वाले तार को इसके चारों ओर लपेटा जाता है। जिसमें यह इसकी पूरी तरीके से ढ़क ले। प्रयोग हुये तार की लम्बाई क्या है?

(A) 22 m.

(B) 20 m.

(D) 34 m.

Q.5. 'X' is twice as old as 'Y' 3 Years ago, when 'X' was as old 'Y' today. If the difference between their ages at present is 3 years ,how old is 'X' at present?

'X' की उम्र उस उम्र से दोगुनी है जो 'Y' की तीन वर्ष पूर्व थी, जब X की उम्र 'Y' की आज की उम्र के बराबर थी। यदि वर्तमान में उनकी उम्रों का अंतर 3 वर्ष है, तो 'Y' की उम्र वर्तमान में कितनी है?

(A) 18 years

(B) 12 years

(D) 8 years

Q.6. x⁴ + 4y⁴ is divisible by -

x⁴ + 4y⁴ भाज्य है -

(A) (x² + 2xy + 2y²)

(B) (x² + 2y²)

(D) (x² - 2y² + 2xy)

Q.7. When 0 < q < 90⁰, then the value of [(1 - sin²q) sec²q + tan²q] (cos²q + 1) is-

जब 0 < q < 90⁰ हो, तो [(1 - sin²q) sec²q + tan²q] (cos²q + 1) का मान क्या है-

(A) 2

(B) > 2

(D) < 2

Q.8. If the root of equation lx² + nx + n = 0 in the ratio of p: q, then the value of is -

यदि समीकरण lx² + nx + n = 0 के मूल p: q के अनुपात में है, तो का मान है-

(A) 0

(B) – 1

(D) 3

Q.9. If a = then is equals to -

यदि a = हो, तो बराबर है-

(A)

(B)

(C)

(D)

Q.10. If (3.7)^x = (0 .037)^y = 10000 then what is the value of ?

यदि (3.7)^x = (0 .037)^y = 10000 हो, तो का मान क्या है?

(A) 2

(B) – 2

(C)

(D)

Answer Key:

Q1. (D) f (x) = x⁵ – qx² + 12x – 14

                  f (3) = (3)⁵ – 9(3)² + 12(3) – 14
                          = 243 – 9 × 9 + 36 – 14
                          = 184

Q2. (C) l

Q3. (D) The amount Q should have received/वह राशि जो Q को प्राप्त होनी चाहिए = 5x

But the amount he actually get/लेकिन वह राशि जो उसे वास्तविकता में मिली

Difference/अंतर = 5x – =

= 10

x = 50

Total amount/कुल राशि = 8x = Rs.400

Q4. (B)

Let the wire be wounded n times around the cylinder, then

माना तार की बेलन के चारों ओर n बार मोड़ा जाता है, तो

= height of the cylinder/बेलन की ऊँचाई

Length of the wire/तार की लम्बाई = = 2000 cm./सेमी. = 20 m. /मी.

Q5. (B) X Y

3 years ago/3 वर्ष पहले y + 3 y

Present age/वर्तमान आयु 2y y + 3

The difference/अन्तर = 2y – (y + 3) = 3

y = 6

Present age of X/X की वर्तमान आयु = 6 × 2 = 12 years/वर्ष

Q6. (A) x⁴ + 4y⁴ = x⁴ + 4y⁴ + 4x²y² – 4x²y²

                               = (x + 2y)² – (2xy)²
                               = (x² + 2y² – 2xy) (x² + 2y² + 2xy)
                  That is divisible by (x² + 2y² + 2xy)/जोकि (x² + 2y² + 2xy) से विभाज्य है ।