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# Previous Year Quant Questions for SSC CGL/MTS/CPO Exams

Dear maths lovers, Let your practice begins in minuteness but ends in magnificence.It is impossible to study maths properly by just reading and listening. So, practise, practise & more practise. For that, we are providing here Quant Quiz of 15 questions on Time and Work, in accordance with the syllabus of SSC CGL.We have also provided Study Notes and quizzes on all the topics.

Directions (1-4): Study the bar diagram and answer questions based on it.
Persons killed in industrial accidents
Person killed in coal mines

Q1. The number of persons killed in coal mines in 2006 was what percent of those killed in industrial accidents in that year?
(a) 4
(b) 25
(c) 36
(d) 300

Q2. In which year, minimum number of persons killed in industrial accidents and coal mines together?
(a) 2006
(b) 2007
(c) 2008
(d) 2009

Q3. In which year, maximum number of persons were killed in industrial accidents other than those killed in coal mines?
(a) 2006
(b) 2007
(c) 2008
(d) 2009

Q4. In which year, minimum number of persons were killed in coal mines other than those killed in industrial accidents?
(a) 2006
(b) 2007
(c) 2008
(d) 2009

Q5. In the figure given below, the perimeter of the circle is 220 cm. What is the area of the shaded portion in cm^2?

(a) 2542(7/9)
(b) 2584(1/3)
(c) 2447(1/9)
(d) 2352(7/9)

Q6. If the incentre of an equilateral triangle lies inside the triangle and its radius is 3 cm, then the side of the equilateral triangle is :
(a)9√3 cm
(b) 6√3 cm
(c)3√3 cm
(d) 6 cm

Q7. D is a point on the side BC of a triangle ABC such that AD⊥BC.E is a point on AD for which AE : ED = 5 : 1. If ∠BAD=30° and  tan (∠ACB)= 6 tan (∠DBE), then find ∠ACB.
(a) 30 °
(b) 45 °
(c) 60 °
(d) 15°

Q8. The tangents drawn at P and Q on the circumference of a circle intersect at A. If ∠PAQ=68°, then the measure of the ∠APQ is
(a) 56°
(b) 68°
(c) 28°
(d) 34°

Q9. The external bisector of ABC of ∆ABC intersects the straight line through A and parallel to BC at the point D. If ∠ABC=50°,  then measure of ∠ADB is :
(a) 65°
(b) 55°
(c) 40°
(d) 20°

Q10. AB is a diameter of a circle with centre at O. DC is a chord of it such that DC || AB. If ∠BAC=20°,  then ∠ADC is equal to
(a) 120°
(b) 110°
(c) 115°
(d) 100°

Q11. Suppose ∆ABC be a right-angled triangle where ∠A=90° and  AD⊥BC. If Area (∆ABC) = 40 cm2, Area (∆ACD)= 10 cm^2 and AC = 9 cm, then the length of BC is :
(a) 12 cm
(b) 18 cm
(c) 4 cm
(d) 6 cm

Q12. Two circles touch each other externally at P. AB is a direct common tangent to the two circles, A and B are points of contact and ∠PAB=35°. Then ∠ABP is :
(a) 35°
(b) 55°
(c) 65°
(d) 75°

Q13. X and Y are centres of circles of radii 9 cm and 2 cm respectively, XY = 17 cm. Z is the centre of a circle of radius r cm which touches the above circles externally. Given that ∠XZY=90°, the value of r is:
(a) 13 cm
(b) 6 cm
(c) 9 cm
(d) 8 cm

Q14. I is the incentre of a triangle ABC. If ∠ABC=65° and ∠ACB=55°, then the value of ∠BIC is :
(a) 130°
(b) 120°
(c) 140°
(d) 110°

Q15. The angle of a triangle are in Arithmetic Progression. The ratio of the least angle in degrees to the number of radians in the greatest angle is 60∶π. The angles in degrees are :
(a) 30°, 60°, 90°
(b) 35°, 55°, 90°
(c) 40°, 50°, 90°
(d) 40°, 55°, 85°

Solutions

S1. Ans.(b)
Sol. Required percentage = 300/1200 × 100% = 25%

S2. Ans.(d)
Sol. Minimum number of persons were killed = 1000 (In 2009)

S3. Ans.(a)
Sol. Maximum number of persons were killed in industrial accidents = 1200 (in 2006)

S4. Ans.(b)
Sol. Minimum number of persons were killed in coalmines = 150 (in 2007)

S6. Ans.(b)
Sol. From the formula:
⇒3=side/(2√3)⇒  side=3×2√3 cm = 6√3 cm.

S7. Ans.(c)
Sol.

∴6 (BD/DC)=6  ⇒BD=DC
∴∠ACB=60°
∴∆ ABC is an equilateral triangle.

S8. Ans.(a)
Sol.

AP=AQ

∴∠APQ=∠AQP
∴∠APQ+∠AQP=180°-68°=112°
∴∠APQ=112/2=56°

S9. Ans.(a)
Sol.

BD is external bisector of ∠ABC.
∠ABC=50°
∴∠DAB=50° (alternate interior angle)
∠ABE=180°-50°=130°
∴∠DBA=(130°)/2=65°

S10. Ans.(b)
Sol.

∴∠BAC=∠BDC=20°
(On the same are BC)

S11. Ans.(b)
Sol. In ∆ACD and ∆BCA,
∠CDA= ∠CAB=90°
∠C  is common.
∴∆ACD ~∆BCA

∴(ar(∆ACD))/(ar(∆BCA))=(AC^2)/(BC^2 )
⇒10/40=9^2/(BC^2 )  ⇒BC^2=4×9^2
∴BC=(2×9)= 18 cm

S12. Ans.(b)
Sol.

OA=OP
∴∠PAB=OPA=35°
∴∠AOP=110°⇒∠POB=70°
∴∠ABP=(180°-70°)/2=110/2=55°

S13. Ans.(b)
Sol.

YZ=r+2 , XZ = r  + 9
∴XY^2=XZ^2+ ZY^2
⇒17^2=(r+9)^2+ (r+2)^2
⇒289=r^2+18r+81+r^2+4r+4
⇒2r^2+22r+85-289=0
⇒2r^2+22r-204=0
⇒r^2+11r-102=0
⇒r^2+17r-6r-102=0
⇒r(r+17)-6(r+17)=0
⇒r=6 cm

S14. Ans.(b)
Sol.

∠IBC=1/2∠ABC=65/2=32.5°
∠IBC=1/2∠ACB=55/2=27.5°
∴∠BIC=180°-32.5°-27.5°=120°
S15. Ans.(a)
Sol.  Angles of triangle = (a-d)°,a°,(a+d)°
∴a-d+a+a+d=180°
⇒3a=180°⇒a=60°
∴(a-d)/(a+d)=60/π=60/180=1/3
⇒(60-d)/(60+d)=1/3  ⇒180-3d=60+d
4d=120°            ⇒d=30°
∴ Angles of triangle :
∴a-d=60°-30°=30°
a=60°
a+d=60+30=90°

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