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SSC CPO : Quantitative Aptitude Quiz | 13- 09 - 18

Mahendra Guru
SSC CPO : Quantitative Aptitude Quiz | 13- 09 - 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 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.


Mahendra Guru also provides you important notes and study material for all subject and test through its website, Mahendra Guru App and YouTube channel apart from it Speed Test Portal. Most of these preparation products are also available for purchase on my shop. You can also visit Mahendras.org to get more information about our endeavour for your success. You can also study in details through our E-Mahendras Facebook and Mahendra Guru YouTube channel of Quantitative Aptitude.

Q.1 – ABCD is a cyclic quadrilateral. A tangent PQ is drawn on the point B of the circle. If ∠DBP = 650 then find ∠BCD-

ABCD рдПрдХ рдЪрдХ्рд░ीрдп рдЪрддुрд░्рднुрдЬ рд╣ै। рдПрдХ рд╕्рдкрд░्рд╢ рд░ेрдЦा PQ рд╡ृрдд्рдд рдХे рдмिंрджु B рдкрд░ рдЦींрдЪी рдЧрдИ рд╣ै। рдпрджि ∠ DBP = 6500рддो ∠ BCD рдЬ्рдЮाрдд рдХीрдЬिрдпे-
(A) 1050
(B) 1150
(C) 1250
(D) 950

Q.2 – Two chords AB and CD of a circle intersect at E such that AE = 2.4 cm, BE = 3.2 cm, and CE = 1.6 cm. The length of DE is-

рдПрдХ рд╡ृрдд्рдд рдХी рджो рдЬीрд╡ा AB рдФрд░ CD, E рдкрд░ рдЗрд╕ рдк्рд░рдХाрд░ рдк्рд░рддिрдЪ्рдЫेрджिрдд рдХрд░рддी рд╣ै рдХि AE = 2.4 рд╕ेрдоी, BE = 3.2 рд╕ेрдоी рдФрд░ CE = 1.6 рд╕ेрдоी. DE рдХी рд▓рдо्рдмाрдИ рдХ्рдпा рд╣ै?
(A) 1.6 cm/рд╕ेрдоी.
(B) 3.2 cm/рд╕ेрдоी
(C) 4.8 cm/рд╕ेрдоी
(D) 6.4 cm/рд╕ेрдоी

Q.3 – The areas of two similar triangle are 12 cm2 and 48 cm2 . If the height of the smaller one is 2.1 cm, then the corresponding height of the bigger one is-

рджो рд╕рдорд░ूрдк рдд्рд░िрднुрдЬों рдХे рдХ्рд╖ेрдд्рд░рдлрд▓ 12 рд╕ेрдоी2 рдФрд░ 48 рд╕ेрдоी2 рд╣ै рдпрджि рдЫोрдЯे рдд्рд░िрднुрдЬ рдХी рдКँрдЪाрдИ 2.1 рд╕ेрдоी рд╣ै рддो рдмреЬे рдд्рд░िрднुрдЬ рдХी рд╕ंрдЧрдд рдКँрдЪाрдИ рдХ्рдпा рд╣ै?
(A) 4.8 cm/рд╕ेрдоी
(B) 8.4 cm/рд╕ेрдоी
(C) 4.2 cm/рд╕ेрдоी
(D) 0.525 cm/рд╕ेрдоी

Q.4 – 10 years ago Ram was 5 times as old as Shyam but 20 year later from now he will be only twice as old as Shyam. How many years old is Shyam?

10 рд╡рд░्рд╖ рдкрд╣рд▓े рд░ाрдо, рд╢्рдпाрдо рд╕े рдЖрдпु рдоें 5 рдЧुрдиा рдеा, рд▓ेрдХिрди рдЕрдм рд╕े 20 рд╡рд░्рд╖ рдмाрдж рд╡рд╣, рд╢्рдпाрдо рдХी рдЖрдпु рдХा рдХेрд╡рд▓ рджोрдЧुрдиा рд╣ोрдЧा। рд╢्рдпाрдо рдХी рдЖрдпु рдХ्рдпा рд╣ै?
(A) 20 years/рд╡рд░्рд╖
(B) 30 years/рд╡рд░्рд╖
(C) 40 years/рд╡рд░्рд╖
(D) 50 years/рд╡рд░्рд╖

Q.5 – When x40 + 2 is divided by x4 +1, what is the remainder?

рдЬрдм x40 + 2 рдХो x4 +1 рд╕े рднрдЧ рджिрдпा рдЬाрддा рд╣ै рддो рд╢ेрд╖рдлрд▓ рдХ्рдпा рд╣ै?
(A) 1
(B) 2
(C) 3
(D) 4

Q-6 If x = 2015, y = 2014 and z = 2013, then value of x2+y2 +z2 – xy – yz – zx is 

рдпрджि x = 2015, y = 2014 рдФрд░ z = 2013 рд╣ो рддो x2+y2 +z2 – xy – yz – zx рдХा рдоाрди рд╣ै-
(A) 7.5
(B) 5
(C) 4.5
(D) 3 

Q-7 If a+b+c = 0, then the value of (a+b – c)2 + (b+c – a)2 + (c+a – b)2 is -

рдпрджि a+b+c = 0 рд╣ो рддो (a+b – c)2 + (b+c – a)2 + (c+a – b)2 рдХा рдоाрди рд╣ै-
(A) (a2 + b2 + c2)
(B) 2 (a2 + b2 + c2)
(C) 8 (a2 + b2 + c2)
(D) 4 (a2 + b2 + c2)

Q-8 DE is a tangent to the circum-circle of ╬ФABC at the vertex A such that DE||BC. If AB = 17 cm., then the length of AC is equal to -

╬ФABC рдХे рдкрд░िрд╡ृрдд्рдд рдоें DE рд░ेрдЦा A рд╢ीрд░्рд╖ рдмिрди्рджु рдкрд░ рдЗрд╕ рдк्рд░рдХाрд░ рд╕्рдкрд░्рд╢ рдХрд░рддी рд╣ै рдХि DE||BC рдпрджि AB = 17 рд╕ेрдоी. рд╣ै рддो AC рдХी рд▓рдо्рдмाрдИ рдмрд░ाрдмрд░ рд╣ै-
(A) 20
(B) 18
(C) 15
(D) 17

Q-9 The distance between the centres of two circles with radii 9 cm. and 16 cm. is 25 cm. Then find the length of the common transverse tangent.

рджो 9 рд╕ेрдоी. рдФрд░ 16 рд╕ेрдоी. рдд्рд░िрдЬ्рдпा рд╡ाрд▓े рд╡ृрдд्рддों рдХे рдХेрди्рдж्рд░ों рдХे рдмीрдЪ рдХी рджूрд░ी 25 рд╕ेрдоी. рд╣ै рддो рдЙрдирдХे рдЙрднрдпрдиिрд╖्рда рдЕрдиुрд╕्рдкрд░्рд╢ рд░ेрдЦा рдЦрдг्рдб рдХी рд▓рдо्рдмाрдИ рдЬ्рдЮाрдд рдХीрдЬिрдП। 
(A) 27
(B) 24
(C) 25
(D) 29

Q-10 The sum of two numbers is 232 and their HCF is 29. What is the numbers of such pairs of numbers satisfied the above condition.

рджो рд╕ंрдЦ्рдпाрдУं рдХा рдпोрдЧ 232 рдФрд░ рдЙрдирдХा рдо.рд╕.рдк. 29 рд╣ै। рдРрд╕ी рд╕ंрдЦ्рдпाрдУं рдХे рдХिрддрдиे рдЬोреЬे рд╕ंрднрд╡ рд╣ै рдЬो рдЙрдкрд░ोрдХ्рдд рд╢рд░्рдд рдХो рд╕ंрддुрд╖्рдЯ рдХрд░ेंрдЧे।
(A) 0
(B) 1
(C) 2
(D) 3

Answer Key- 
Q1. (B) 
∠DBQ = 180o – ∠DBP = 180o – 65o = 115o
∠DBP = ∠DCB = 115o (alternate angle/рдПрдХाрди्рддрд░ рдХोрдг)
∠BCD = 115o 

Q2. (C) 
AE × BE = CE × DE
2.4 × 3.2 = 1.6 × x
x = = 4.8 cm. /рд╕ेрдоी 

Q3. (C) 

Q4. (A) 
Let the present age of Ram and Shyam be x and y years./рдоाрдиा рд░ाрдо рдФрд░ рд╢्рдпाрдо рдХी рд╡рд░्рддрдоाрди рдЖрдпु x рдФрд░ y рд╡рд░्рд╖ рд╣ैं |
(x – 10) = 5 (y – 10)
x – 5y = – 40 ________(I)
and/рдФрд░ (x+20) = 2(y+20)
x – 2y = 20 _________(II)
On solving/рд╣рд▓ рдХрд░рдиे рдкрд░,
y = 20 years/рд╡рд░्рд╖ 

Q5. (C) 
Let/рдоाрдиा f(x) = x40 + 2
Put/рд░рдЦिрдпे x4 = –1,
f(x) = (–1)10 + 2 = 3 

Q6. (D)

Q7. (D) 
a + b + c = 0
a + b = - c, b + c = - a, c + a = - b
(a + b - c)2 + (b + c - a)2 + (c + a - b)2
= 4c2 + 4a2 + 4b2 = 4 (a2 + b2 + c2 ) 

Q8. (D)
OA = OB = OC
AB = BC = AC
AC = 17 cm

Q9. (A) Length of tangent /рд╕्рдкрд░्рд╢ рд░ेрдЦा рдХी рд▓рдо्рдмाрдИ 

Q10. (C) Let two number is 29a and 29b 
рдоाрдиा рджो рд╕ंрдЦ्рдпाрдПं 29a рддрдеा 29b рд╣ैं 
29a + 29b = 232
a + b = 232/29 = 8
(a,b)= (1,7) (3,5)
The pair is / рдпुрдЧ्рдо рд╣ैं (87, 145) (29,203)

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