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SSC CGL & CHSL Quiz : Quantitative Aptitude | 10 -01-2020

Durgesh Mahendras
SSC CGL & CHSL Quiz : Quantitative Aptitude | 10 -01-2020


As SSC CGL notification is out and candidates have started their preparation for this exam. Mahendras also has started special quizzes for this examination. This series of the quizzes is based on the latest pattern of the SSC CGL examination. Regular practice of the questions included in the quizzes will boost up your preparations and it will be very helpful in scoring good marks in the examination.



Q.1:- The ratio of two number are 3 : 5, if 8 is added to first and 7 to second the ratio become 2 : 3. What will be the ratio when 6 will be added to each?

दो संख्याओं का अनुपात 3 : 5 है, यदि 8 पहले में और 7 दुसरे में जोड़ा जाए तो अनुपात 2 : 3. यदि 6 दोनों में जोड़ा जाये तो अनुपात क्या होगा?

(1)  8 : 13
(2) 13 : 8
(3) 14 : 9
(4) 9 : 14

Q.2:- A 84 cm long cylinder made of iron whose internal and external diameter are x and 8 cm respectively. If weight of cylinder is 1848 SSC-CGL-10-12-20 find the internal radius?

एक 84 सेमी लम्बा बेलन लोहे का बना है जिसकी आतंरिक और बाह्य व्यास x और 8 सेमी है यदि सिलिंडर का भार 1848SSC-CGL-10-12-20है तो आतंरिक त्रिज्या ज्ञात करिए.

(1) 3 cm
(2) 6 cm
(3) 3.5 cm
(4) 2.5 cm

Q.3:- If MNOP is a rectangle X is a point on MN such that MX = 4XN and area of triangle is SSC-CGL-10-12-20 then what is the area of rectangle in SSC-CGL-10-12-20?

यदि MNOP एक आयत है X , भुजा MN पर एक बिंदु इस प्रकार है कि MN  = 4XN और त्रिभुज का क्षेत्रफल SSC-CGL-10-12-20 तो आयत का क्षेत्रफल ज्ञात करें?

(1) SSC-CGL-10-12-20
(2) SSC-CGL-10-12-20
(3) SSC-CGL-10-12-20
(4) SSC-CGL-10-12-20

Q.4:-if/ यदि  SSC-CGL-10-12-20, SSC-CGL-10-12-20

(1)  2
(2) 1
(3) 0
(4) SSC-CGL-10-12-20

Q.5 A pole is standing at a road which goes from  East to west two mile stone are present at East side named M and N, MN = a and angle of elevation from point M and N to the top of pole is SSC-CGL-10-12-20 find the height of tower?

एक पोल एक सड़क के बीच में खड़ा है जो पूर्व से पश्चिम दिशा में जा रही है दो मील के पत्थर जिनके नाम M और N इस प्रकार है की MN  = a है  और बिंदु M और N से पोल के शीर्ष का उन्नयन कोण क्रमशः SSC-CGL-10-12-20 है तो पोल की लम्बाई ज्ञात करें?

(1) SSC-CGL-10-12-20 
(2)  SSC-CGL-10-12-20 
(3)SSC-CGL-10-12-20 
(4) SSC-CGL-10-12-20 

Q.6:-  What is the value of/ का मान क्या होगा  SSC-CGL-10-12-20 

(1) 0.2
(2) 0.4
(3) 0.8
(4) 0.02

Q.7:- An article is sold at Rs. 98496 after giving three successive discount of 10%, 5% and 4% . What is the marked price of Article?

एक सामान को 98496 रुपये में तीन क्रमागत 10% , 5%  और 4 % छूटों के बाद बेचा जाता है. सामान का अंकित मूल्य ज्ञात करें?

(1)  120000
(2) 12000
(3) 112000
(4) 11200

Q.8:- If (320 + 342 + 530 + 915) SSC-CGL-10-12-20 (20 + 22 – x + 18 )  = 43 then find the value of / तो का मान ज्ञात करें x?

(1) 11
(2) 7
(3) 12
(4) 19

Q.9 . The LCM of two numbers is 132. The numbers are in the ratio 2: 3. The sum of the numbers is-

दो संख्याओं का ल.स.प.
132 है। संख्याएँ 2: 3के अनुपात में हैं| संख्याओं का योग है-

(1) 112
(2) 100
(3) 120
(4) 110

Q10. SSC-CGL-10-12-20

(1) 2
(2) 1
(3) SSC-CGL-10-12-20
(4) SSC-CGL-10-12-20









Answer:-

1.Sol:- (4)

Let number be 3x and 5x
SSC-CGL-10-12-20 
After solving
= x = 10
Numbers are = 30 and 50
(30 + 6 ) : (50 + 6)
36 : 56
9 : 14

2.Sol:- (1)

Volume of hollow cylinder = SSC-CGL-10-12-20
Where, SSC-CGL-10-12-20 = external radius ,  SSC-CGL-10-12-20 internal Radius
SSC-CGL-10-12-20  = SSC-CGL-10-12-20 = 1848
X = 3 cm

3.Sol:- (3)

SSC-CGL-10-12-20
Area of triangle = SSC-CGL-10-12-20
SSC-CGL-10-12-20 = SSC-CGL-10-12-20
Area of rectangle = SSC-CGL-10-12-20

4.Sol:- (1)

SSC-CGL-10-12-20

5.Sol:-  (4)

SSC-CGL-10-12-20
Angle AMN = SSC-CGL-10-12-20 and angle ANB =  SSC-CGL-10-12-20
SSC-CGL-10-12-20 SSC-CGL-10-12-20
AM = SSC-CGL-10-12-20, SSC-CGL-10-12-20
SSC-CGL-10-12-20

6.Sol:- (1)

SSC-CGL-10-12-20  = 0.2


7.Sol:- (1)

Marked price/ अंकित मूल्य  = 98496 SSC-CGL-10-12-20 = 120000

8.Sol:- (1)

SSC-CGL-10-12-20
2107 = 2580 – 43x
43x = 473
X = 11

9.Sol:- (4)

2 × 3 = 132
6 = 132
1 = 22
2 = 44
3 = 66
66 + 44 = 110

10.Sol:- (2)

SSC-CGL-10-12-20
1 + 1 – 1 = 2 – 1 = 1

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