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SSC CGL Mains Quantitative aptitude Questions : 6th August

Dear aspirants,

As you all know, the upcoming months are lined up with various important exams like SSC CGL Mains 2018, so we are here to help you with the subject that is common to all of the given exams. We are providing daily quantitative aptitude quizzes, practice which will help you to score good marks in this section. We aim to provide the best study material to our readers with exam level questions to help them get used to the recent pattern. Attempt this quiz and check your preparation.

Q1. In the given figure, find the length of QS./ दी गयी आकृति में , QS की लम्बाई ज्ञात कीजिये

SSC CGL Mains Quantitative aptitude Questions : 6th August_50.1
(A) √31
(B) √43
(C) √61
(D) √73
S1. C
The length of PR=3+5=8. Applying the Pythagorean theorem to triangle PRS, we get 
8²+(PS)²=10²
(PS)²=100-64=36 
∴PS=√36=6  
Now, applying the Pythagorean theorem to triangle PQS, we get
(QS)²=5²+6²=25+36=61 
∴QS=√61 


Q2. For the given figure, which of the following best describes the value of y?/ दी गयी आकृति में , निम्न में से कौन सा y को सही तरीके से निरुपित करता है
∠POQ=70° and x >15

SSC CGL Mains Quantitative aptitude Questions : 6th August_60.1

(A) y<35

(B) y>35

(C) y<55

(D) y>55

 S2.Ans (A)
Sol.
Since ∠POQ=70° we get x+y+20=70 Solving this equation for y yields y=50-x. Now, we are given that x>15. Hence, the expression (50-x) must be less than 35. Hence, y<35.
 

Q3. In given figure, if L∥K, then find the value of y.
दी गयी आकृति में , यदि L∥K है तो y का मान ज्ञात कीजिये.

 SSC CGL Mains Quantitative aptitude Questions : 6th August_70.1
(A) 55
(B) 60
(C) 70
(D) 75
S3. Ans. (D) 
Sol. Since lines L and K are parallel, we know that the corresponding angles are equal. Hence,
 y=2y-75 
 y=75 
Q4. In the given figure, both triangles are right triangles. Find the area of the shaded region.
दी गयी आकृति में , दोनों त्रिभुज समकोण त्रिभुज हैं. छायांकित भाग का क्षेत्रफल ज्ञात कीजिये.
 SSC CGL Mains Quantitative aptitude Questions : 6th August_80.1
(A) 3/5
(B) 7/8
(C) 7/9
(D) 3/7
S4. Ans. (B)
Sol. Since the height and base of the larger triangle are the same, the slope of the hypotenuse is 45°. Hence, the base of the smaller triangle is the same as its height, 3/2. Thus, the area of the shaded region
=(area of the larger triangle)-(area of the smaller triangle) 
=1/2×2×2-1/2×3/2×3/2 
=2-9/8=(16-9)/8=7/8 
 
Q5. In the given figure, the radius of the larger circle is twice that of the smaller circle. If the circles are concentric, find the ratio of the shaded region’s area to the area of the smaller circle?
दी गयी आकृति में , बड़े वृत्त की त्रिज्या छोटे वृत्त की तुलना में दोगुनी है। यदि वृत्त संकेंद्रित हैं, तो छोटे वृत्त के क्षेत्र में छायांकित क्षेत्र के क्षेत्रफल का अनुपात ज्ञात कीजिए?
 SSC CGL Mains Quantitative aptitude Questions : 6th August_90.1
(A) 3 : 1
(B) 3 : 5
(C) 7 : 5
(D) 5 : 7
S5. Ans. (A)
Sol. Let us assume that the radius of the larger circle is 2 and the radius of the smaller circle is 1. Then, the area of the larger circle is πr²=π(2)²=4π, and the area of the smaller circle is πr²=π(1)²=π. Hence, the area of the shaded region is 4π-π=3π.
∴ (Area of shaded region)/(Area of smaller circle)=3π/π=3/1
 
Q6. In the given figure, ∆PST is an isosceles right triangle, and PS=2. Find the area of the shaded region URST.
दी गयी आकृति में , ∆PST एक समद्विबाहु समकोण त्रिभुज है, और PS = 2। छायांकित क्षेत्र URST का क्षेत्रफल ज्ञात कीजिए।
 SSC CGL Mains Quantitative aptitude Questions : 6th August_100.1
(A) 2/3
(B) 5/7
(C) 1/2
(D) 7/9
S6.Ans. (C)
Sol. Let, TP=TS=y. Applying the Pythagorean theorem to the right triangle PST, we get 
 TP²+TS²=PS²
y²+y²=2²
2y²=4⇒y=√2 
Now, area of the shaded region
=area of ∆PST-area of ∆PRU 
=1/2×√2×√2-1/2×1×1=1/2 
 
Q7. In the given figure, the area of ∆PQR is 40. Find the area of ∆QRS.
दी गयी आकृति में , दिया गया है कि त्रिभुज PQR का क्षेत्रफल 40 है। ∆QRS का क्षेत्रफल ज्ञात कीजिए।
 SSC CGL Mains Quantitative aptitude Questions : 6th August_110.1
(A) 25 
(B) 28
(C) 32
(D) 36
S7. Ans. (A)
Sol. Area of ∆PQS=1/2×5×6=15
Area of ∆QRS=Area of ∆PQR-Area of ∆PQS
=40-15=25 
Q8. In the given figure, PQRS is a square and M and N are midpoints of their respective sides. Find the area of the quadrilateral PMRN.
दी गयी आकृति में , PQRS एक वर्ग है और M और N उनके संबंधित भुजाओं के मध्यबिंदु हैं। चतुर्भुज PMRN का क्षेत्रफल ज्ञात कीजिए।
 SSC CGL Mains Quantitative aptitude Questions : 6th August_120.1
(A) 26
(B) 24
(C) 16
(D) 8
S8. Ans. (D)
Sol. Since M is the midpoint of side PQ, the length of MQ is 2. Hence, the are of ∆MQR=1/2×2×4=4. A similar analysis shows that the area of ∆NSR=4. Thus, the unshaded area of the figure=4+4=8
Hence, the area of quadrilateral PMRN
=Area of the square PQRS-The unshaded area of the figure 
=16-8=8 Q9. In the given figure, what is the greatest number of regions into which two straight lines with divide the shaded region?
दी गयी आकृति में , उन क्षेत्रों की सबसे बड़ी संख्या क्या है जिनमें दो सीधी रेखाएं छायांकित क्षेत्र को विभाजित करती हैं?
 SSC CGL Mains Quantitative aptitude Questions : 6th August_130.1
(A) 6
(B) 5
(C) 4
(D) 3
S9.Ans. (B)
Sol. Let us draw the two straight lines as follows.The lines must intersect in the shaded region.
Hence, the two straight lines will divide the shaded region into 5 parts.
 

Q10. In the given figure, O is the centre of the circle. If the area of the circle is 9π, then find the perimeter of the sector PRQO.
दी गयी आकृति में , O वृत्त का केंद्र है। यदि वृत्त का क्षेत्र 9π है, तो सेक्टर PRQO की परिधि का ज्ञात कीजिये।

 
 SSC CGL Mains Quantitative aptitude Questions : 6th August_140.1
(A) 1/2 π+6
(B) 1/2 π-6
(C) 2/3 π+6
(D) 2/3 π-6
S10. Ans. (A)
Sol. Since the area of the circle is 9π, we have
πr²=9π⇒r²=9⇒r=3  
The circumference of the circle is 
C=2πr=2π×3=6π 
Since the central angle is 30°, the length of arc PRQ
=30/360×C=1/12×6π=1/2 π 
Hence, the perimeter of the sector PRQO=1/2 π+3+3=1/2 π+6

 



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