sscexamforum...16.The value of k for which the graph of kx + 3y – (k – 3) = 0 and 12x + ky – k...

5/17/2016

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MATHS PRACTICE SET 17 SOLVED [CombinedGraduate Level Exam (CGLE)]

MAXIMUM TIME: 20 minutes

1. 61 + 62 + 63 + ... + 100 = ?

A) 2525B) 2975C) 3220D) 3775

2. Six bells commence tolling together and toll at intervals of 2, 4, 6, 8, 10 and 12seconds respectively. In 30 minutes, how many times do they toll together?

A) 4B) 10C) 15D) 16

3. 497/25. Find remainder.

A) 1B) 4C) 7D) 9

4. If x²+2 = 2x, then the value of x4 x3 + x2 + 2

A) 0B) 1C) 1D) √2

5. If A+ B+ C = 1 & AB + BC + CA = 1/3; then A : B : C is?

A) 1 : 2 : 2B) 2 : 1 : 2C) 1 : 1 : 1D) 1 : 2 : 1

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6. The ratio of water and spirit in the mixture is 1 : 3. If he volume of the solutionis increased by 25% by adding spirit only; what is the resultant ratio of water andspirit?

A) 2 : 3B) 4 : 1C) 1 : 4D) 2 : 5

7. Average of marks scored by some students in an exam was found to be 60. Afterthe error having been corrected, the average of 100 students became 30 and for allthe students the average came down to 45. Find the number of students appearedin that exam.

7. A) 150B) 175C) 200D) 250

8. Find the third proportional to (x/y + y/x) and √(x² + y²)

A) xyB) √(xy)C) ∛(xy)D) ∜(xy)

9. A, B, C subscribe Rs 50,000 for a business. A subscribes Rs 4000 more than Band B Rs 5000 more than C. Out of a total profit of Rs 35,000, A receives?

A) Rs 8400B) Rs 11,900C) Rs 13,600D) Rs 14,700

10. In an election between two candidates, 75% of voters cast their votes, out ofwhich 2% of votes declared invalid. A candidate got 9261 votes which were 75% ofthe total valid votes. The total number of voters enrolled were?

A) 16000B) 16400C) 16800

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D) 18000

11. The cash down value of an article is 12500. It can be brought at credit 14300payable one year hence at 10% per annum. The trader gets a profit of?

A) 300B) 400C) 500D) 250

12. The salary of an employee of a company increases every month by 4%. If hissalary in August was Rs 6300, what would be his approximate salary in the monthof October in the same year?

A) Rs 6600B) Rs 6814C) Rs 6875D) Rs 7000

13. A refrigerator of MRP Rs 5000 is available in Rs 1000 cash down payment and9 monthly equal installment of Rs 500 each. Find rate of interest per annum.

A) 100/3%B) 50/3%C) 100/7%D) 33%

14. A can finish a job in 12 days, B in 20 days and C in 30 days. A, B & C start worktogether. After 3 days, A left. B+C worked for 2 days, then B left and A joined. A+Ccompleted the remaining work. In how many days was the work finished?

A) 55/7 daysB) 20/7 daysC) 9 daysD) 20 days

15. Two bullets are fired at a platform in the interval of 13 seconds. A passengerwho is sitting in the train coming towards the platform heard the sound in theinterval of 12.5 seconds. If speed of sound is 330 m/s, train's speed is?

A) 47 13∕25 km/h

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B) 50 km/hC) 66/5 m/hD) 45 km/h

16. The value of k for which the graph of kx + 3y – (k – 3) = 0 and 12x + ky – k =0 has infinitely many solutions?

A) 6B) 6C) 0D) ±6

17. The radii of the circular ends of a frustum of height 6 cm are 14 cm and 6cm respectively. Find the lateral surface area and total surface area ofthe frustum.

A) 620.57 cm², 1300 cm²B) 628.57 cm², 1357.71 cm²C) 528.56 cm², 1300.75 cm²D) 600.50 cm², 1400.15 cm²

18. Adjacent sides of a parallelogram are 18 and 16. If one of the diagonal is 20find the other. All units are in cm.

A) 2√180B) 180C) 2√190D) 2√203

19. Let C be a point on a straight line AB. Circles are drawn with diameters AC andAB. Let P be any point on the circumference of the circle with diameter AB. If APmeets the other circle at Q, then?

A) QC || PBB) QC is never parallel to PBC) QC = 1/2 PBD) QC || PB and QC = 1/2 PB

20. If 2 cos²θ = 3sinθ*cosθ; sinθ is not equal to cosθ, then tanθ will be?

A) 1

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B) 1/3C) 1/2D) 2/3

SOLUTION

1. 61 + 62 + 63 + ... + 100 = (1 + 2 + 3 + ... + 100) (1 + 2 + 3 + ... + 60)

[using n = n∕2(n+1)]= 100∕2(100 + 1) 60∕2(60 + 1)

= (50 x 101) (30 x 61)

= (5050 183o)

= 3220(option 'C')

2. Obviously all the bells will toll together in the time equal to the LCM of the time of theirintervals.

Now the LCM of the times of their intervals = LCM of 2, 4, 6, 8, 10, 12 i.e. 120 seconds= 2 minutes

But they will also toll together once in the beginning

Thus the number of times they will toll together in 30 minutes = 30/2 + 1 = 16 (option 'D')

3. 497

= (43)32)*4

= (6432)*4

Remainder obtained after dividing 64 by 25 is 14 [so when (1432)*4 is divided by 25 will give

us the same remainder when (6432)*4 is divided by 25]

Now (1432)*4

= (142)16)*4

= (19616)*4


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Now (2116)*4

= (212)8)*4

= (4418)*4



Now (168)*4

= (162)4)*4

= (2564)*4

Remainder obtained after dividing 256 by 25 is 6 [so when (64)*4 is divided by 25 will give us

the same remainder when (2564)*4 is divided by 25]

Now (64)*4

= (62)2)*4

= (362)*4

Remainder obtained after dividing 36 by 25 is 11 [so when (112)*4 is divided by 25 will give us

the same remainder when (362)*4 is divided by 25]

Now (112)*4=121*4= 484

You see when 484 is divided by 25 the remainder is 9 (option 'D')

4. Given x² + 2 = 2x=> x² = 2x 2

Now the find expression

= x4 x3 + x² + 2= (x²)² x²*x + x² +2

(by substituting x² = 2x 2 in the above)= (2x 2)² (2x 2)*x + 2x 2 + 2

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= (2x 2)² (2x 2)*x + 2x= 2x² 4x + 4 (on solving)= 2(x² 2x + 2)

Again by substituting x² = 2x 2= 2(2x 2 2x + 2)= 2*0= 0 (option 'A')

Another Method

x4 x3 + x² + 2

= x²(x² + 2) x3 + 2 x²

Putting x² + 2 = 2x

x²(2x) x3 + 2 x²

= x3 x² + 2

=> x(x² + 2) x² + 2 2x

Again putting x² + 2 = 2xx(2x) x² + 2 2x

=> x² + 2 2x = 0 (option 'A') [x² + 2 2x = 0 is equal to the given equation x²+ 2 = 2x]

5. To satisfy both A+ B+ C = 1 & AB + BC + CA = 1/3, it's a must that A, B & C each equal to 1/3So the ratio is 1 : 1 : 1 (option 'C')

6. According to the ratio the solution is = 1+3 = 4 unitsAfter the increase of the solution = 4 + 25% of 4 i.e. 5 units; means 1 more unit than earlier

But only spirit is added; therefore it is 1 + (3+1) = 1 + 4

So the ration now = 1 : 4 (option 'C')

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7. Let the total number of students appeared = xThus, incorrect total of marks = 60xAnd correct total of marks = 45x

Total marks of 100 students according to the incorrect average = 100*60 = 6,000Their correct total = 100*30 = 3,000Total of 100 students that was wrongly included = 6,000 – 3,000 = 3,000

Now wrong total of marks = correct total of marks + wrongly included marks60x = 45x + 3,000=> x = 200

Therefore the total students appeared = 200 (option 'C')

NOTE: This question can speedily be done by answer options.

8. First let one know what the third proportional isIf a : b = b : c, then 'c' is called the third proportional to 'a' and 'b'

Obviously in the question given above a = x/y + y/x = (x² + y²)/xyAnd b = √(x² + y²)

Therefore the third proportional = b²/a = [√(x² + y²)]²/[(x² + y²)/xy]

= (x² + y²)*xy/(x² + y²) = xy (option 'A')

9. Let C subscribes = Rs x

Then, B subscribes = x + 5000and A subscribes = x + 5000 + 4000 = x + 9000.

Hence, x + (x + 5000) + (x + 9000) = 50000=> x = 12000

Therefore ratio of their investments i.e. A : B : C = (12000 + 9000) : (12000 + 5000) : 12000 =21000 : 17000 : 12000 = 21 : 17 : 12.

Thus A's share in the profit = 35000 x 21/50 = Rs 14,700 (option 'D')

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10. Let the total number of enrolled voters = 200Therefore the number who cast votes = 75% of 200 = 150And invalid votes = 2% of 150 = 3So valid votes = 150 – 3 = 147

Number of votes got by the candidate = 75% of 147 = (3/4)*147 = 441/4

But the actual votes got by the candidate = 9261

Now, if the votes got is 441/4 the total number of voters enrolled = 200If the votes got is 9261 the total number of voters enrolled = (200*4*9261)/441 = 16800 (option‘C’)

11. If 10% for 1 year on the amount (principal + interest) 14300; then the principal =(100/110)*14300 = 13000Means the sum of Rs 12500 was given on simple interest considering it as Rs 13000So the profit of the trader = 13000 12500 = 500 (option 'C')

NOTE: You must remember that the interest got is not a part of profit, rather it's compensationfor the time that one waits to get his principal back.

12. This is a case as we see in compound interest. But this is just a 2 month affair, we need notapply the formula meant for compound interest here as it's a little time consuming. So better ifit's done applying simple approach. Look how:

Salary in the month of September = 6300 + 4% of 6300 = 6300 + 252 = 6552

Salary in October: 6552 + 4% of 6552 = 6552 + 262 .08 = 6814.08 = 6814 (approx) [option 'B']

13. When it's a monthly installment it's better to find the one month equivalent principal.

Now the principal got as loan = MRP Cash Down Payment = 5,000 1,000 = 4,000

As the sum of Rs 4,000 has be returned in 9 monthly equal installments, the payment of theprincipal to be made will end on the completion of 8 months, Rs 500 being for every month.

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And the principal for the first month will be Rs 4,000, for the second month Rs 3,500, for thethird month Rs 3,000, and so on.

Thus the one month equivalent sum = 4,000 + 3,500 + 3,000 + 2,500 + 2,000 + 1,500 + 1,000+ 500 = 18,000

As this sum of Rs 4,000 has to be paid in 9 monthly installments of Re 500 each, the amount(principal + interest) will be Rs 4,500 and the interest to be paid = 4,500 4,000 = 500

Now the principal (equivalent to one month) = 18,000;interest = 500time = 1 month = 1/12 years (as the principal has been converted equivalent to one month)

So rate = (100*500)/(18,000*1/12) (rate = 100*interest/principal*time)=> 50,000*12/18,000=> 100/3 (option 'A')

14. CaseI when A, B and C all worked together for 3 daysTherefore their 3 days work = (1/12 + 1/20 + 1/30)*3 = (10/60)*3 = 1/2

CaseII when B and C worked together for 2 daysTherefore their 2 days work = (1/20 + 1/30)*2 = (5/60)*2 = 1/6

So total worked finished in these 5 days = 1/2 + 1/6 = 4/6 = 2/3Now the remaining work that has to be done by A and C = 1 2/3 = 1/3

Now 1 day work of A and C if done together = 1/12 + 1/30 = 7/60

Time taken by them to complete 7/60 of work = 1 dayTherefore time to be taken by them to complete the remaining 1/3 of work = (60/7)*(1/3) =20/7 days [By unitary method)

Hence the total time taken to finish the work = 5 + 20/7 = 55/7 days (option 'A')

15. Let the speed of the train be x m/sObviously, the distance traveled by train in 12.5 seconds = distance traveled by sound in 13seconds 12,5 seconds i.e. 0.5 seconds

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therefore according to above; x*12.5 = 330*0.5

or x = 66/5 m/s = 47 13∕25 km/h (option 'A')

16. For a pair of linear equations to have infinitely many solutions:a1/a2 = b1/b2 = c1/c2

Here, a1/a2 = k/12 b1/b2 = 3/k c1/c2 = (k 3)/k

So, k/12 = 3/k and 3/k = (k 3)/k=> k² = 36 and k² = 6k=> k = ±6 and k = 0 or k = 6

Therefore, the value of k, that satisfies both the conditions, is k = 6. For this value, the pair oflinear equations has infinitely many solutions (option 'A')

17. Here, R = 14 cm, r = 6 cm and h = 6 cmNow, let l be the slant height of the frustum;then, l = √[(R r)² + h²]=> l = √(64 + 36)=> l = 10 cm

Now, lateral surface area of the frustum = π(R + r)l= π(14 + 6)10= 628.57 cm² (option 'B')

And, total surface area of the frustum = π[(R + r)l + R² + r²]= π(200 + 196 + 36)= π432= 1357.71 cm² (option 'B')

18. In a parallelogram sum of the squares of the diagonals = 2(a² + b² ), where a and b are thelengths of the sides of the parallelogram.

Now let the other diagonal be x cmTherefore x² + 20² = 2(18² + 16² )

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=> x² = 2(324 + 256) 400

=> x = √(1160 400) = √760

=> x = 2√190 (option 'C')

19.

∠APB = ∠AQC = 90 (both are angles of semi circles)Now in ΔAPB and ΔAQC∠A = ∠PAB = ∠QAC .......same angles∠APB = ∠AQC .......proved above

Therefore both triangles are similar with ∠APB and ∠AQC being corresponding.Hence QC ll PB

QC cannot necessarily be 1/2 PB as 'C' must be half of AB then, which cannot be proved

Hence QC ll PB (option 'A')

20. 2 cos²θ = 3sinθcosθ

Divide both sides by cos²θ

2sec²θ 1 = 3tanθ

=> 2(1 + tan²θ) 1 = 3tanθ

=> 2 + 2tan²θ 1 = 3tanθ

=> 2tan²θ 3tanθ + 1 = 0

=> 2tan²θ 2tanθ tanθ + 1 = 0

=> 2tanθ(tanθ 1) 1(tanθ 1) = 0

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=> (2tanθ 1)(tanθ 1) = 0

=> tanθ = 1/2 or tanθ = 1

But tanθ = 1 is rejected as sinθ≠cosθ

Therefore tanθ = 1/2 (option 'C')

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sscexamforum...16.The value of k for which the graph of kx + 3y – (k – 3) = 0 and 12x + ky – k...

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