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# SSC CGL TIER-1 9th March 2020 shift -3 Quantitative aptitude questions with detailed solutions.

SSC CGL Previous Year Quant Paper: AÂ  study plan for cracking the SSC CGLÂ  Exam must include the daily task of practising Previous Year Question Papers. Students can attempt all four sections (General Intelligence & Reasoning, General Awareness & GK, Quantitative Aptitude and English Language & Comprehension) and build an innovative preparation strategy to clear this exam with a high score. Here you can download SSC CGL TIER-1: 9th March 2020 Shift -3 Quantitative aptitude questions with detailed solutions.

Watch Video Explanation :

Question 1:Â The value of $\frac{3(1 – 2 \sin^{2}x)}{\cos^{2}x – \sin^{2}x}$ is:

a)Â 4

b)Â 3

c)Â 1

d)Â 2

Question 2:Â The perimeter of a square plot is the same as that of a rectangular plot with sides 35 m and 15 m. The side of the square plot is:

a)Â 25 m

b)Â 20 m

c)Â 100 m

d)Â 50 m

Question 3:Â The value of $515\times485$ is:

a)Â 249775

b)Â 250225

c)Â 20825

d)Â 200825

Question 4:Â The following table shows the number of students enrolled in different streams in a particular college.

The ratio of the number of girls studying Arts to the number of girls studying in all other streams is:

a)Â 1 : 3

b)Â 2 : 1

c)Â 1 : 2

d)Â 3 : 1

Question 5:Â The compound interest on a certain sum at the end of two years is Rs.408. The simple interest on the same sum for the same time is Rs.400. The rate of interest per annum is:

a)Â 2%

b)Â 80%

c)Â 40%

d)Â 8%

Question 6:Â A car dealer purchased an old car for Rs.1,08,500 and spent some amount on its maintenance. He sold it for Rs.1,56,250, there by earning a profit of 25%. How much money did he spend on the maintenance of the car?

a)Â Rs. 16,500

b)Â Rs. 20,625

c)Â Rs. 8,687.5

d)Â Rs. 47,750

Question 7:Â The following table gives the frequency of vowels used in the page of a book.

As per the given data, which vowels occur less than 80 times?

a)Â a, e, i

b)Â a, i, o, u

c)Â a, o, u

d)Â a, e

Question 8:Â Eight persons can finish a work in 20 days. After 5 days they were requested to complete the work in the next 8 days. How many more persons should join the group to fulfil the requirement?

a)Â 23

b)Â 15

c)Â 7

d)Â 12

Question 9:Â The following table shows the income(in rupees) for a particular month, together with their source, in respect of 5 employees (A, B, C, D and E).

For employee D, the income from overtime is what percentage of his total income?

a)Â 5%

b)Â 7.5%

c)Â 6.25%

d)Â 22.5%

Question 10:Â Ina class, the average score of thirty students on a test is 69. Later on it was found that the score of one student was wrongly read as 88 instead of 58. The actual average score is:

a)Â 58

b)Â 88

c)Â 69

d)Â 68

Question 11:Â PAQ is a tangent to circle with centre O, at a point A on it. AB is a chord such that $\angle BAQ = x^\circ(x < 90)$. C is a point on the major arc AB such that $\angle ACB = y^\circ$. If $\angle ABO = 32^\circ$, then the value of $x + y$ is:

a)Â 98

b)Â 112

c)Â 110

d)Â 116

Question 12:Â The smallest number which may replace * in the number 1190*6 to make the number divisible by 9 is:

a)Â 1

b)Â 9

c)Â 3

d)Â 0

Question 13:Â The value of $(26-13\times2)\div2+1$ is:

a)Â $0$

b)Â $1$

c)Â $\frac{26}{3}$

d)Â $14$

Question 14:Â The value of $\cos 10^\circ \cos 30^\circ \cos 50^\circ \cos 70^\circ \cos 90^\circ$ is:

a)Â 3

b)Â 0

c)Â 5

d)Â 1

Question 15:Â A cell phone was marked at 20% above the cost price and a discount of 10% was given on its marked price. What is the profit percentage?

a)Â 10%

b)Â 8%

c)Â 11%

d)Â 9%

Question 16:Â The value of $(\cosec 30^\circÂ – \tan 45^\circ) \cot 60^\circ \tan 30^\circ$ is:

a)Â $\frac{1}{3}$

b)Â $1$

c)Â $3$

d)Â $\frac{1}{\sqrt{3}}$

Question 17:Â If $x+3y+2=0$ then value of $x^{3}+27y^{3}+8-18xy$ is:

a)Â -2

b)Â 2

c)Â 1

d)Â 0

Question 18:Â As per data shown in the following table, what is the percentage of students who got less than 50 marks?

a)Â 74%

b)Â 48%

c)Â 72%

d)Â 38%

Question 19:Â In the triangle, If $AB = AC$ and $\angle ABC = 72^\circ$, then $\angle BAC$ is:

a)Â $36^\circ$

b)Â $30^\circ$

c)Â $54^\circ$

d)Â $18^\circ$

Question 20:Â In a circle, AB is a the diameter and CD is a chord. AB and CD produced meet at a point P, outside the circle. If PD = 15.3 cm, CD = 11.9 cm and AP = 30.6 cm,then the radius of the circle is is:

a)Â 8 cm

b)Â 8.5 cm

c)Â 9cm

d)Â 7.5 cm

Question 21:Â From an external point P, a tangent PQ is drawn to a circle, with the centre O, touching the circle at Q. If the distance of P from the centre is 13 cm and length of the tangent PQ is 12 cm, then the radius of the circle is:

a)Â 3cm

b)Â 5 cm

c)Â 10 cm

d)Â 12.5 cm

Question 22:Â If $p+q=7$ and $pq=5$, then the value of $p^{3}+q^{3}$ is:

a)Â 34

b)Â 238

c)Â 448

d)Â 64

Question 23:Â If an amount of Rs.990 is divided among A, B and C in the ratio of 3 : 4 : 2, then B will get:

a)Â Rs. 247.5

b)Â Rs. 440

c)Â Rs. 110

d)Â Rs. 350

Question 24:Â The price of cooking oil increased by 25%. Find by how much percentage a family must reduce its consumption in order to maintain the same budget?

a)Â 70%

b)Â 80%

c)Â 30%

d)Â 20%

Question 25:Â The distance between two stations A and B is 575 km. A train starts from station â€˜Aâ€™ at 3:00 p.m. and moves towards station â€˜Bâ€™ at an average speed of 50 km/h. Another train starts from station â€˜Bâ€™ at 3:30 p.m. and moves towards station â€˜Aâ€™ at an average speed of 60 km/h. How far from station â€˜Aâ€™ will the trains meet ?

a)Â 275 km

b)Â 325 km

c)Â 300 km

d)Â 225 km

$\frac{3(1-2sin^{2}x)}{cos^{2}x-sin^{2}x}$

=Â $\frac{3(1-2sin^{2}x)}{1-sin^{2}x-sin^{2}x}$
($\because sin^{2}x +Â cos^{2}x = 1$)
= $\frac{3(1-2sin^{2}x)}{1-2sin^{2}x}$ = 3

The perimeter ofÂ  a rectangular plot = 2(length + breadth) = 2(35 + 15) = 100 m

The perimeter of a square plot = perimeterÂ of a rectangular plot

The perimeter of a square plot = 100 m

100 = 4 $\times$ side

side = 100/4 = 25 m

$515\times485$

=Â $(500 + 15)\times(500 – 15)$

= $(500)^2 – (15)^2$

($\because (a – b)(a + b) = a^2 – b^2$)

= 250000 = 225 = 249775

Number of girls studying Arts = 45

Number of girls studying in all other streams = 18 + 42 + 30 = 90

The ratio of the number of girls studying Arts to the number of girls studying in all other streams = 45 :Â 90 = 1 :Â 2

Difference of interest = 408 – 400 = 8

In the 2nd year, extra interest counts on the interest of the 1st year so.

Interest = $\frac{prt}{100} 8 =$\frac{400 \times r \times 1}{100}$r = 2% 6)Â AnswerÂ (A) Total cost price of old car =Â 1,08,500 + maintenance money Selling price =Â 1,56,250 Profit = 25% Total cost price =$\frac{156250}{125} \times 100 = 125000

1,08,500 + maintenance money = 125000

Maintenance money = 125000 – 108500 = Rs. 16,500

Vowels occur less than 80 times = a, o, u

Total work = number of persons $\times time = 8 \times 20 = 160$

Work done in 5 days = 8 $times 5 = 40$

Remaining Work = 160 – 40 = 120

According to question,

Number of man $\times$Â time = 120

Number of man $\times$ 8 = 120

Number of man = 120/8 = 15

Number ofÂ persons should join the group = 15 – 8 = 7

Total income of employee D = 40000

Income from overtimeÂ of employee D = 2500

Required percentage = $\frac{2500}{40000} \times 100 = 6.25% 10)Â AnswerÂ (D) When data read wronglyÂ thenÂ totalÂ score of thirty students on a test = 69$\times$30 = 2070 (Average = sum of all terms/number of terms) Difference in data = 88 – 58 = 30 Actual score = 2070 – 30 = 2040 Actual average score = 2040/30 = 68 11)Â AnswerÂ (D) Download SSC CGL Previous Papers PDF Download SSC CHSL Previous Papers PDF 12)Â AnswerÂ (A) By the divisibility rule, if sum ofÂ all digits ofÂ a number is divisible by 9 the number will be divisible by 9. Number =1190*6 Sum of the digits = 1 + 1+ 9 + 0 + * + 6 = 17 + * For the smallest number, we took the minimum value of the ‘*’. so, value of * = 1 Sum of all digit = 17 + 1Â = 18$\therefore$the value of * will be 1. 13)Â AnswerÂ (B)$(26-13\times2)\div2+1$=Â$(26- 26)\div2+1$= 0 + 1 =Â 1 14)Â AnswerÂ (B)$cos10^{0}cos30^{0}cos50^{0}cos70^{0}cos90^{0}$=$cos10^{0}cos30^{0}cos50^{0}cos70^{0}\times 0$($\because cos90^{0} = 0$) = 0 15)Â AnswerÂ (B) Let the cost price be 100. Marked price = 100$\times$120/100 = 120 Discount = 10% Selling price = 120$\times 90/100 = 108

Profit = 108 – 100 = 8

Profit percentage = $\frac{8}{100} \times$ 100 = 8%

$(\cosec 30^\circÂ – \tan 45^\circ) \cot 60^\circ \tan 30^\circ$

On put the value,

= $(2 – 1)\frac{1}{\sqrt3} \times \frac{1}{\sqrt3}$

= 1/3

$x+3y+2=0$

x + 3y = -2

Taking cube both sides,

$(x + 3y)^3 = -8$

$x^3 + 27y^3 + 3x.3y(x +Â 3y) = -8$

$x^3 + 27y^3 + 9xy(-2) = -8$

$x^{3}+27y^{3} -18xy = -8$

$x^{3}+27y^{3}+8-18xy$ = 0

NumberÂ of students who got less than 50 marks = 5 + 11 + 18 + 38 = 72

Total number of students = 5 + 11 + 18 + 38 +Â 39 + 24 + 15 = 150

Percentage of students who got less than 50 marks = $\frac{72}{150} \times 100$ = 48%

By triangle properties,

When AB = AC then $\angleABC = \angleACB$ so,

In$\triangle$ ABC,

$\Rightarrow$ $\angle ABC +Â \angle ACB +Â \angle BAC = 180^{0}$

$\Rightarrow \angle ABC +Â \angle ABCÂ + \angle BAC = 180^{0}$

$\Rightarrow 72^{0} +Â 72^{0} +Â \angle BAC = 180^{0}$

$\Rightarrow \angle BAC =Â 180^{0} –Â 144^{0} = 36^{0}$

From the property,

PA $\times PB = PC \times PD$

30.6 $\times PB = (PD + CD) \times 15.3$

30.6 $\times PB = (15.3 +Â 11.9) \times 15.3$

30.6 $\times PB = 27,2 \times 15.3$

PB =Â 416.16/30.6 = 13.6 cm

Diameter (AB) =Â PA – PB = 30.6 – 13.6 = 17 cm

Radius = AB/2 = 17/2 = 8.5 cm

$\triangle$ OPQ is a right angle triangle because $\angle Q = 90\degree$,

By Pythagoras,

$(OQ)^2 + (PQ)^2 = (OP)^2$

$(OQ)^2 = (13)^2 – (12)^2$

$(OQ)^2 = 169 – 144$

$(OQ)^2 = 25$

$OQ = 5$

$\therefore$Â The radius of the circle is 5 cm.

$p^{3}+q^{3} = (p + q)^3 – 3pq(pÂ + q)$

=$7^3 – 3 \times 5(7)$

= 343 – 105 = 238

Ratio of the A, B and C = 3 : 4 : 2

Amount = 990

Share of B = $\frac{4}{3 +Â 4 + 2} \times 990Â = \frac{4}{9} \times 990 = Rs.440 24)Â AnswerÂ (D) By the formula, Decrements in the consumption in order to maintain the same budgetÂ =$\frac{increment in the rate}{100 + increment in the rate} \times 100$=$\frac{25}{100 + 25} \times 100 = \frac{25}{125} \times 100 = 20%

Distance between stations = 575 km

Speed of train A = 50 km/hr

Distance covered by train A in (30 min = 1/2 hr) = time $\times speed = 50 \times 1/2 = 25 km$

Trains are running in opposite direction so,

Relative speed = 50 + 60 = 110 km/hr

Distance covered by both train = 575 – 25 = 550 km

Time taken by both trainsÂ to meet = 550/110 = 5 hr

Distance covered by train A in 5 hr =Â Â 5 $\times 50 = 250 km$

Distance covered by train A from station A =Â distance covered by train A in 30 minÂ +Â distanceÂ covered by train A in 5 = 250 + 25 = 275 km

$\therefore$275 kmÂ far from station â€˜Aâ€™ will the trains meet.