ssc_previousyear_questions_solved_examsbook.pdf

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EExamplebook

SSC CombinedGraduate Level-Tier-

I,II

2013

Shiv

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1/20/2013


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SSC UPSC previous years solved questions with answers

Arithmetical Ability

1. The product of 2 numbers is 1575 and their quotient is 9/7. Then thesum of the numbers is

a. 74

b. 78

c. 80

d. 90

Ans : Let the numbers be x and y .

xy = 1575

And x/y = 9/7

xy/x/y = 1575/9/7

y2 = 1225

y = 35 and x = 45

The sum of the numbers = 45+35

= 80

2. The value of (81)3.6 * (9)2.7/ (81)4.2 * (3) is __

a. 3

b. 6

c. 9

d. 8.2


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Ans : (81)3.6 * (9)2.7/(81)4.2 * (3) = (3)14.4 * (3)5.4/(3)16.8 * (3)

= 314.4+5.4-16.8-1

= 32

= 9

3. 6+6+6+ is equal to

a. 2

b. 5

c. 4

d. 3

Ans : Let 6+6+6+. be x.

x =6+x

x2 = 6+x

x2 + x 6 = 0

(x-3) (x+2) = 0

x = 3

4. The sum of the squares of two natural consecutive odd numbers is 394.The sum of the numbers is

a. 24


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b. 32

c. 40

d. 28

Ans : Let the consecutive odd numbers be x and (x+2)

x2 + (x + 2)2 = 394

x2 + x2 + 4x + 4 = 394

2 x2 +4x 390 = 0

x2 + 2x 195 = 0

(x +15) (x-13) = 0

x = 13

Required sum = 13 +15 =28

5. When (6767 +67) is divided by 68, the remainder is-

a. 1

b. 63

c. 66

d. 67

Ans: (6767+ 67) = 67(6766 + 166)

As 66 is an even number


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6766 is an even number

(6766 + 1) is perfectly divisible by (67 + 1)

i.e. 68

The remainder = 67

6. In a division sum, the divisor is 4 times the quotient and twice theremainder. If a and b are respectively the divisor and the dividend then-

a. 4b-a2 /a =3

b. 4b-2a /a2 =2

c. (a+1)2 = 4b

d. A(a+2)/b = 4

Ans : As divisor is a, and dividend is b.

Quotient = a/4

And Remainder = a/2

b = a * a/4 + a/2

4b = a2 + 2a

a(a+2)/b = 4

7. If 738 A6A is divisible by 11, then the value of A is-

a. 6


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b. 3

c. 9

d. 1

Ans : As 738A6 A is divisible by 11.

A + A + 3 = 6 + 8 + 7

A = 9

8. The east positive integer that should be subtracted from 3011 * 3012 sothat the difference is perfect square is-

a. 3009

b. 3010

c. 3011

d. 3012

Ans : 3011 * 3012 = 3011 (3011 + 1)

= (3011)2 + 3011

Required least number = 3011

9. P, Q, R are employed to do a work for Rs. 5750. P and Q together

finished 19/23 of work and Q and R together finished 8/23 of work. Wage of

Q, in rupees, is-

a. 2850

b. 3750


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c. 2750

d. 1000

Ans : Work done by Q = 19/23 + 8/23 1

4/23

Wage of Q = 4/23 * 5750

= Rs. 1000

10. A can do a piece of work in 24 day, B in 32 days and C in 64 days. Allbegin to do it together, but A leaves after 6 days and B leaves 6 days before

the completion of the work. How many days did the work last?

a. 15

b. 20

c. 18

d. 30

Ans : Work done by A = 6/24 =

Work done by B = (x-6)/32

(where x is no. of days in which work is competed)

+ x 6/32 + x/64 = 1

16 + 24 12 + x /64 = 1

3x + 4 = 64


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x = 60/3 = 20 days

11. The square root of (0.75)3 /1-0.75 + [0.75 + 90.75)2 +1] is-

a. 1

b. 2

c. 3

d. 4

Ans : The square root of

[(0.75)3/1 0.75 + {0.75 + (0.75)2 + 1}]

= 1.6875 + 2.3125

=4

= 2

12. Given that 4096 = 64, the value of 4096 + 40.96 +0.004096 is-

a. 70.4

b. 70.464

c. 71.104

d. 71.4

Ans : Given Exp. = 4096 + 40.96 +0.004096

= 64 + 6.4 + 0.064


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= 70.464

13. By selling an article at 3/4th of the marked price, there is a gain of

25%. The ratio of the marked price and the cost price is-

a. 5 : 3

b. 3 : 5

c. 3 : 4

d. 4 : 3

Ans : Let of M.P. be Rs. x.

S.P. = Rs.3x/4

and C.P. = 3x/4 * 100/125

= Rs.3x/5

required ratio = x: 3x/5

= 5:3

14. A and B earn in the ratio 2:1. They spend in the ratio 5:3 and save in

the ratio 4:1. If the total monthly savings of both A and B are Rs.5,000, themonthly income of B is-

a. Rs.7,000

b. Rs.14,000

c. Rs.5,000

d. Rs.10,000


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Ans : Let the monthly income of B be Rs. x.

Monthly income of A = Rs. 2x and

Saving of A =5000 * 4/(4 + 1)

= Rs. 4000

Saving of B = Rs.1000

2x 4000/x-1000 = 5/3

6x 12000 = 5x 5000

x = Rs.7000

15. The ratio of the sum of two numbers and their difference is 5:1. Theratio of the greater number to the smaller number is-

a. 2 : 3

b. 3 : 2

c. 5 : 1

d. 1 : 5

Ans : Let the numbers be x and y.

x + y/ x y = 5/1

5x 5y = x + y

4x = 6y

x/y = 6/4


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x : y = 3 : 2

16. A cistern has 3 pipes A, B and C. A and B can fill it in 3 and 4 hours

respectively, and C can empty it in 1 hour. If the pipes are opened at 3p.m., 4 p.m. and 5 p.m. respectively on the same day, the cistern will be

empty at-

a. 7.12 p.m.

b. 7.15 p.m.

c. 7.10 p.m.

d. 7.18 p.m.

Ans : Let the cistern be emptied at x p.m.

x -3/3 + x 4/4 = x - 5/1

4x 12 + 3x 12/12 = x 5/1

7x 24 = 12x 60

5x = 36

x = 36/5

= 7 hr. + 12 min.

= 7.12 p.m.

17. If A works alone, he would take 4 days more to complete the job than if

both A and B worked together. If B worked alone, he would take 16 daysmore to complete the job than if A and B work together. How many days

would they take to complete the work if both of them worked together?


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a. 10 days

b. 12 days

c. 6 days

d. 8 days

Let x days are taken when they work together.

Time taken by A to complete the work = (x + 4) days

And Time taken by B to complete the work = (x + 16) days

1/x = 1/x + 4 + 1/x + 16

1/x = x + 16 + x + 4/(x + 4) (x +16)

2x2 + 20x = x2 + 20x +64

x2 = 64 (8)2

x = 8 days

18. 250 men can finish a work in 20 days working 5 hours a day. To finish

the work within 10 days working 8 hours a day, the minimum number ofmen required is-

a. 310

b. 300

c. 313

d. 312

Ans : Required no. of men = 250 * 5 * 20/8 * 10


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c. 16 2/3%

d. 16 1/3%

Ans : Let the M.P. be Rs.100.

S.P = (100-40) = Rs.60

and C. P. = 60 * 100/(100-30) = Rs.600/7

Reqd. % profit = 100-600/7 /600/7 * 100%

= (700 600)/7 * 600 * 100 * 7%

= 16 2/3%

21. Successive discount of 10% , 20% and 50% will be equivalent to a

single discount of-

a. 36%

b. 64%

c. 80%

d. 56%

Ans : Equivalent to a single discount

= [100 - (100 10) (100 20) (100 50)/100*100]%

= [100 - 90*80*50/10000]%

= [100-36]%

= 64%


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22. A retailer offers the following discount schemes for buyers on an article-

I. Two successive discounts of 10%

II. A discount of 12% followed by a discount of 8%.

III. Successive discounts of 15% and 5%

IV. A discount of 20%

The selling price will be minimum under the scheme-

a. I

b. II

c. III

d. IV

Ans : From (i) single discount = [10 + 10 10*10/100] % = 19%

From (ii) single discount = [12 + 8 -12*8/100] % = 19.04%

From (iii) single discount = [15 + 5 15*5/100] % = 19.25%

From (iv) single discount = 20%

The S.P. will be minimum under the scheme IV.

23. Of three numbers, the second is thrice the first and the third number is

three-fourth of the first. If the average of the three numbers is 114, thelargest number is


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a. 72

b. 216

c. 354

d. 726

Ans : Let the first number be x.

Second number = 3x and Third number = 3x/4

x + 3x + 3x/4 = 3*114

19x/4 = 342

x = 342*4/19 = 72

The largest number = 3*72

= 216

24. A car covers 1/5 of the distance from A to B at the speed of 8 km/hour,

1/10 of the distance at 25 km per hour and the remaining at the speed of 20km per hour. Find the average speed of the whole journey-

a. 12.625 km/hr

b. 13.625 km/hr

c. 14.625 km/hr

d. 15.625 km/hr


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Ans : If the whole journey be x km. The total time taken

= (x/5/8 + x/10/25 + 7x/10/20) hrs

= (x/40 + x/250 + 7x/200) hrs

= 25x + 4x + 35x/1000

= 64x/1000 hrs

Average speed = x/64x/1000

= 15.625 km/hr

25. The average of 3 numbers is 154. The first number is twice the second

and the second number is twice the third. The first number is-

a. 264

b. 132

c. 88

d. 66


Second number = x/2 and Third number = x/4

x + x/2 + x/4 = 3 x 154

7x/4 = 462

x = 462*4/7

=264


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26. The average salary of all the staff in an office of a corporate house is

Rs. 5,000. The average salary of the officers is Rs. 14,000 and that of therest is Rs. 4,000. If the total number of staff is 500, the number of officers is

a. 10

b. 15

c. 25

d. 50

Ans : Let the number of officers be x.

5000*500 = 14000x + 4000(500-x)

2500000 =14000x + 2000000 4000x

x = 500000/10000

= 50

27. The average marks of 40 students in an English exam are 72. Later it is

found that three marks 64, 62 and 84 were wrongly entered as 60, 65 and73. The average after mistakes were rectified is-

a. 70

b. 72

c. 71.9


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d. 72.1

Ans : Correct average

= 40*72 + (64 + 62 +84) 68 65 73/40

= 2880 + 210 206/40 = 2884/40

= 72.1

28. A and B are two alloys of gold and copper prepared by mixing metals in

the ratio 7:2 and 7:11 respectively. If equal quantities of the alloys are

melted to form a third alloy C, the ratio of gold and copper in C will be-

a. 5 : 7

b. 5 : 9

c. 7 : 5

d. 9 : 5

Ans : Quantity of gold in A = 7/9* wt. of A

Quantity of gold in B = 7/18* wt. of B

If 1 kg of each A and B are mixed to form third alloys C.

Then quantity of gold in 2 kg C = 7/9 + 7/18

= 7/6 kg

And Quantity of copper in 2 kg C = 2 7/6

= 5/6 kg

Required ratio = 7/6 : 5/6 = 7 : 5


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29. In a laboratory, two bottles contain mixture of acid and water in theratio 2 : 5 in the first bottle and 7 : 3 in the second. The ratio in which thecontents of these two bottles be mixed such that the new mixture has acid

and water in the ratio 2 : 3 is-

a. 4 : 15

b. 9 : 8

c. 21 : 8

d. 1 : 2

Ans : Quantity of acid in first bottle = 2/7 x mix.

and Quantity of acid in second bottle = 7/10 x mix.

If x and 1 volumes are taken from I and II bottle respectively to form new

mixture.

Then, (2/7 x + 7/10 * 1)/(5x/7 +3/10 * 1) = 2/3

6x/7 21/10 = 10x/7 +6/10

4x/7 = 15/10

x = 15/10*7/4 = 21/8

Required ratio = x : 1

= 21 : 8

30. A mixture contains 80% acid and rest water. Part of the mixture thatshould be removed and replaced by same amount of water to make the ratio

of acid and water 4 : 3 is-


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a. 1/3 rd

b. 3/7 th

c. 2/3 rd

d. 2/7 th

Ans : Let the initial wt. of mixture be 1 kg and x kg of mixture is taken out

and replaced by same amount of water.

Amt. of acid/Amt. of water = 0.8 0.8x/ (0.2 0.2x + x) = 4/3

2.4 2.4x = 0.8 + 3.2x

5.6x = 1.6

x = 1.6/5.6 = 2/7th part

31. An employer reduces the number of his employees in the ratio 9 : 8 and

increases their wages in the ratio 14 : 15. If the original wage bill was Rs.

189,900, find the ratio in which the wage bill is decreased-

a. 20 : 21

b. 21 : 20

c. 20 : 19

d. 19 : 21

Ans : Let the initial number of employees be 9x and the employer gives Rs.

14y as wage to each.

9x * 14y =18900

xy = 150 and The later bill = 8x*15y = 120xy


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= 120*150 = 18000

Required ratio = 18000 : 18900

= 20 : 21

32. The batting average for 40 innings of a cricketer is 50 runs. His highest

score exceeds his lowest score by 172 runs. If these two innings areexcluded, the average of the remaining 38 innings is 48 runs. The highestscore of the player is-

a. 165

b. 170

c. 172

d. 174

Ans : Let the max. number of runs be x.

The lowest score = (x-172)

40*50 = 38*48 + x + (x-172)

2000 = 1824 + 2x 172

x= 174

33. Walking at 3 km per hour, Pintu reaches his school 5 minutes late. If hewalks at 4 km per hour he will be 5 minutes early. The distance of Pintusschool from his house is-

a. 1 Km


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b. 2 Km

c. 2 Km

d. 5 Km

Ans : Distance of the school from the house

= 4*3/(4 3) * 5 +5/60 km

= 12*1/6

=2 km

34. Nitin bought some oranges at Rs. 40 a dozen and an equal number atRs.30 a dozen. He sold them at Rs. 45 a dozen and made a profit of Rs.480.

The number of oranges, he bought, was-

a. 48 dozen

b. 60 dozen

c. 72 dozen

d. 84 dozen

Ans : Let the number of oranges bought be x.

45x/12 70x/24 = 480

(45-35)/12 = 480

x = 480*12/10 = 576

= 48*12 = 48 dozen


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35. A man buys two chairs for a total cost of Rs.900. By selling one for 4/5

of its cost and the other for 5/4 of its cost, he makes a profit of Rs.90 on thewhole transaction. The cost of the lower priced chair is-

a. Rs.360

b. Rs.400

c. Rs.420

d. Rs.300

Ans : Let the cost price of 1 chair be Rs. x.

C.P. of other chair = Rs. (900-x)

4/5 x + 5/4 (900-x) = 900 + 90

4/5 x +1125 -5x/4 = 990

9x/20 = 135

x = 135*20/9

= Rs. 300

C.P. of the lower priced chair is Rs. 300.

36. By selling 100 oranges, a vendor gains the selling price of 20 oranges.

His gain per cent is-

a. 20

b. 25

c. 30

d. 32


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Ans : Let the S.P. of 100 Oranges be Rs. x.

S.P. of 20 oranges = x/100 *20 = Rs. x/5

C.P. of 100 oranges = x - x/5

= Rs. 4x/5

Reqd. Profit % = x/5 * 100*5/4x %

= 25%

37. 60% of the cost price of an article is equal to 50% of its selling price.Then the percentage of profit or loss on the cost price is-

a. 20% loss

b. 16 2/3% profit

c. 20% profit

d. 10% loss

Ans : Let the cost price be Rs. 100.

S.P * 50/100 = 100*16/100

S.P = 60*100/50

= Rs. 120

Reqd. % profit = (120 100)% = 20%

38. Maninder bought two horses at Rs.40,000 each. He sold one horse at

15% gain, but had to sell the second horse at a loss. If he hand suffered a


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loss of Rs.3,600 on the whole transaction, then the selling price of the

second horse is-

a. Rs.30,000

b. Rs.30,200

c. Rs.30,300

d. Rs.30,400

Ans: C.P. of two horses = 2*40000 = Rs. 80000

and S.P. of two horse = 80000 3600 = Rs. 76400

S.P. of the other horse = 76400 46000 = 30400

39. A fruit-seller buys x guavas for Rs.y and sells y guavas for Rs. x. If x>y,then he made-

a. x2 y2 / xy % loss

b. x2 y2 / xy % gain

c. x2 y2 / y2 % loss

d. x2 y2 / y2 * 100% gain

Ans : C.P. of 1 guava = Rs. y/x [x>y]

and S.P. of 1 guava = Rs. x/y

Reqd. Gain% = x/y y/x/ y/x * 100%

= x2 - y2 /y2 *100%


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40. A jar contain 10 red marbles and 30 green ones. How many red marblesmust be added to the jar so that 60% of the marbles will be red?

a. 25

b. 30

c. 35

d. 40

Ans : Let after adding x red marbles, the red marbles with be 60% of the

total.

(10+x)/ (10 + x) + 30 = 60/100

10 + x/40 + x = 3/5

50 + 5x = 120 + 3x

x = 70/2 = 35

41. If a number multiplied by 25% of itself gives a number which is 200%more than the number, then the number is

a. 12

b. 16

c. 35

d. 24

Ans : Let the number be x.


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x*25x/100 = x + 200x/100

x2/4 = 3x

x2 - 12x = 0

x 12 = 0

x = 12

42. The value of an article depreciates every year at the rate of 10% of its

value. If the present value of the article is Rs.729, then its worth 3 yearsago was-

a. Rs.1250

b. Rs.1000

c. Rs.1125

d. Rs.1200

Ans : Let the worth 3 years ago be Rs. x.

729 = x (1 10/100)3

729 = x*9*9*9/10*10*10

x = Rs. 1000

43. The price of onions has been increased by 50%. In order to keep the

expenditure on onions the same the percentage of reduction in consumptionhas to be-

a. 50%


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b. 33 1/3%

c. 33%

d. 30%

Ans : Reqd. Percentage of reduction

= 50*100/(100 + 50) %

= 5000/150 %

= 33 1/3%

44. A took two loans altogether of Rs.1200 from B and C. B claimed 14%simple interest per annum, while C claimed 15% per annum. The total

interest paid by A in one year was Rs.172. Then, A borrowed-

a. Rs.800 from C

b. Rs.625 from C

c. Rs.400 from B

d. Rs.800 from B

Ans : If A borrowed Rs. x from B. and A borrowed Rs. Rs. (1200 x) from C.

(1200 x)*15*1/100 + x*14*1/100

18000 15x + 14x = 172*100

x = Rs. 800

45. If a regular polygon has each of its angles equal to 3/5 times of tworight angles, then the number of side is-


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a. 3

b. 5

c. 6

d. 8

Ans : If the number of sides a regular polygon be n.

Then (2n-4)/n = 2*3/5

(2n 4)*5 = 6n

n = 5

46. A square is of area 200 sq. m. A new square is formed in such a waythat the length of its diagonal is 2 times of the diagonal of the givensquare. The the area of the new square formed is-

a. 2002 sq.m

b. 4002 sq.m

c. 400 sq.m

d. 800 sq.m

Ans : Length of the diagonal of Ist square

= 2*200

= 20 m

Length of the diagonal of new square = 202m


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91/(10 + x) + 91/(10 x) = 20

91(10 x + 10 + x)(10 + x) (10 x) = 20

91*20 = 20(100 x2)

x2 = 9 = (3)2

x = 3 km/hr

49. A man driving at 3/4th of his original speed reaches his destination 20

minutes later than the usual time. Then the usual time is-

a. 45 minutes

b. 60 minutes

c. 75 minutes

d. 120 minutes

Ans : Let the original speed be x km/hr and the usual time be y hours.

x * y = x(y+1/3)

4y = 3y + 1

y = 1 hr = 60 minutes

50. A motor-boat, travelling at the same speed, can cover 25 km upstreamand 39 km downstream in 8 hours. At the same speed, it can travel 35 km

upstream and 52 km downstream in 11 hours. The speed of the stream is

a. 2 km/hr


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b. 3 km/hr

c. 4 km/hr

d. 5 km/hr

Ans : Let the speeds of motor boat and the stream be x and y km/hrrespectively.

39/x + y + 25/x y = 8 (1)

and 52/x + y + 35/x y = 11 (2)

Solving equations (1) and (2), we get-

100 - 105/x y = 32 33

x y = 5

and x + y = 13

y = 4 km/hr

51. If a sum of money placed at compound interest, compounded annually,doubles itself in 5 years, then the same amount of money will be 8 times ofitself in-

a. 25 years

b. 20 years

c. 15 years

d. 10 years

Ans : Required time = 5log 8/log 2


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= 5*3log 2/log 2

= 15 years

52. A person has left an amount of Rs.1,20,000 to be divided between his 2son aged 14 years and 12 years such that they get equal amounts when

each attains 18 years of age. If the amount gets a simple interest of 5% per

annum, the younger sons share at present is-

a. Rs.48,800

b. Rs.57,600

c. Rs.62,400

d. Rs.84,400

Ans : Let the present share of the younger son be Rs. x.

The share of the elder son = Rs. (120000 x)

x + x*6*5/100

= (120000 x) + (120000 x)*4*5/100

130x/100 = (120000 x)*120/100

13x = 1440000 12x

25x = 1440000

x = Rs. 57600

53. If the simple interest on Rs.x at a rate of a% for m years is same asthat on Rs. y at a rate of a2% for m2 years, then x : y is equal to-


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a. m : a

b. am : 1

c. 1/m : 1/a

d. 1/am : 1

x*a*m/100 = y*a2*m2/100

x/y = am/1

x : y = am : 1

54. Base of a right prism is an equilateral triangle of side 6 cm. If the

volume of the prism is 1083 cc, its height is-

a. 9 cm

b. 10 cm

c. 11 cm

d. 12 cm

Ans : Height of the prism = 1083*4/3*(6)2 = 12cm

55. If a + 1/a + 2 = 0, then the value of (a37 1/a100) is-

a. 0


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b. -2

c. 1

d. 2

Ans : a + 1/a + 2 = 0

a2 + 1 +2a = 0

(a + 1)2 = 0

a +1 = 0

a = -1

a37 -1/a100 = (-1) (1) = -2

56. The value of k for which the graphs of (k-1) x+y-2 = 0 and (2-k) x -3y+ 1 = 0 are parallel is-

a.

b. -1/2

c. 2

d. -2

Ans : (k 1) x + y 2 = 0

y = (1 k) x + 2 .(1)

and (2 k) x 3y 1 = 0

3y = (2 k) x +1

Y = 2 k/3 x + 1/3 .(2)


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m1 = m2

1 k = 2 k/3

3 3k = 2 - k

k = 1/2

57. If a2 + b2 + c2 = 2 (a-b-c) 3, then the value of (a b + c) is-

a. -1

b. 3

c. 1

d. -2

Ans : a2 + b2 + c2 = 2(a b c) 3

a2 - 2a + 1 + b2 + 2b + 1 + c2 + 2c + 1 = 0

(a 1)2 + (b + 1)2 + (c + 1)2 = 0

a 1 = 0, b + 1 = 0, c + 1 = 0

a = 1, b = -1, c = -1

a + b c = 1 1 + 1 =1

58. If x2 + 3x + 1 = 0, then the value of x3 + 1/x3 is-

a. -18

b. 18


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c. 36

d. -36

Ans : x2 + 3x + 1 = 0

x + 3 + 1/x = 0

x + 1/x = -3

(x + 1/x)3 = (-3)3

x3 + 1/x3 + 3 (-3) = -27

x3 + 1/ x3 = -18

59. If xa, xb, xc = 1, then the value of a3 + b3 + c3 is

a. 9

b. abc

c. a + b + c

d. 3abc

Ans : xa. xb. xc = 1

xa + b + c = x0

a +b + c =0

a3 + b3 + c3 = 3abc


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60. Base of a right pyramid is a square, length of diagonal of the base is

242 m. If the volume of the pyramid is 1728 cu.m, its height is-

a. 7 m

b. 8 m

c. 9 m

d. 10 m

Ans : Area of the base of the pyramid

= (242)2 = 576m2

Height of pyramid + 1728*3/576 = 9m

61. The height of a right circular cone and the radius of its circular base arerespectively 9 cm and 3 cm. The cone is cut by a plane parallel to its base so

as to divide it into two parts. The volume of the frustum (i.e., the lower part)

of the cone is 44 cubic cm. The radius of the upper circular surface of thefrustum (taking = 22-7) is-

a. 312 cm

b. 313 cm

c. 36 cm

d. 320 cm

Ans : Let the radius of the upper circular part of the frustum be r cm.


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(Picture)

Then r/3 = x/9 = AC/AD

x = 3r (where AC = x cm)

/3h(r12 + r1r2+r22 = 44

/3(9 x)*(9 + 3r + r2) = 44

(9 x)*(9 + 3r + r2) = 44*3*7/22 = 42

81 + 27r + 9r2 9x 3rx - r2x = 42 On putting x = 3r,

81 + 27r + 9r2 - 9r2 - 3r3 27r = 42

3r3 = 39

r = 313 cm

62. The ratio of radii of two right circular cylinder is 2 : 3 and their heightsare in the ratio 5 : 4. The ratio of their curved surface area is-

a. 5 : 6

b. 3 : 4

c. 4 : 5

d. 2 : 3

Ans : Required ratio = 2*2r*5h/2*3r*4h = 5 : 6


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63. A solid cylinder has total surface area of 462 sq.cm. Curved surfacearea is 1/3rd of its total surface area. The volume of the cylinder is-

a. 530 cm3

b. 536 cm3

c. 539 cm3

d. 545 cm3

Ans : 2r(r + h) = 462

and 2rh = 1/3*462 = 154

r + h/h = 462/154 =3

r + h = 3h

r = 2h

2*2h2 = 154

h2 = 154*7/22*4 = 49/4

= (7/2)2

h = 7/2 cm

and r = 7 cm

Volume of the cylinder

= 22/7*49*7/2

= 539 cm3


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64. A cylinder and a cone have equal radii of their bases and equal heights.

If their curved surface areas are in the ratio 8 : 5, the ratio of their radiusand height is-

a. 1 : 2

b. 1 : 3

c. 2 : 3

d. 3 : 4

Ans : Let the radius and height of each are r and h respectively.

2rh/rh2 + r2 = 8/5

10h = 8r2 + h2

100h2 =64r2 + 64h2

h2 = 64 r2/36 = (4/3 r)2

h = 4/3r

r : h = 3 : 4

65. A solid is hemispherical at the bottom and conical above. If the surface

areas of the two parts are equal, then the ratio of radius and height of itsconical part is-

a. 1 : 3

b. 1 :1

c. 3 : 1

d. 1 : 3


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Ans :

rl = 2r2

l = 2r

h2 + r2 = 2r

h2 = 3r2

r : h = r/h

= 1 : 3

66. If O is the circumcentre of ABC and OBC = 350, then the BAC isequal to-

a. 550

b. 1100

c. 700

d. 350
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Ans : BOC = 1800 (350 + 350) = 1100

BAC = 1/2 * 1100 = 550

67. If I is the incentre of ABC and BIC = 1350, then ABC is-

a. Acute angled

b. Equilateral

c. Right angled

d. Obtuse angled

Ans :


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BIC = 1350

B/2 + C/2 = 1800 1350 = 450

B + C = 900

A = 1800 (B + C) = 900

i.e., ABC is a right angled.

68. If sin2 (+ /2) is-

a. 1

b. -1

c. 0

d. 0.5

sin2 + sin2 = 2

1 cos2 1 - cos2 = 2

cos2 cos2 = 0


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cos = 0

and cos = 0

= /2 and = /2

cos (+ /2) = cos [/2 + /2 / 2]

= cos /2 = 0

69. The length of a shadow of a vertical tower is 1/3 times its height. Theangle of elevation of the Sun is-

a. 300

b. 450

c. 600

d. 900

Ans :

tan = h/13 h = 3 = tan 600
http://placement.freshersworld.com/placement-papers/sites/default/files/SSC69.png


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(4x 2)2 = 3

4x = 2 + 3

64x3 = (23)3

=8 + 33 + 63 (2 + 3)

= 26 + 153

64x3 + 1/64x3 = (26 + 153) + 1/ (26 + 153)

= (26 + 153) + 26 - 153/676 -675

=52

72. If a, b, c, are three non-zero real numbers such that a + b + c = 0, andb2 ca, then the value of a2 + b2 + c2/ b2 ca is-

a. 3

b. 2

c. 0

d. 1

a + b + c = 0

a + c = -b

a2 + c2 = b2 -2ac

a2 + b2 + c2 = 2b2 2ac

a2 + b2 + c2/ b2 ac = 2


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73. If a4 + a2 b2 + b4 = 8 and a2 + ab + b2 = 4, then the value of ab is-

a. -1

b. 0

c. 2

d. 1

a4 + a2 b2 + b4/ a2 + ab + b2 = 8/4

(a2 + b2)2 (ab) 2/ (a2 + b2 + ab) = 2

a2 ab + b2 = 2 .(1)

and a2 + ab + b2 = 4 ..(2)

2ab = 2

ab = 1

74. If a = 25, b = 15, c = -10, then the value of a3 + b3 + c3 3abc/ (a-b)2 + (b-c)2 + (c-a)2 is-

a. 30

b. -15

c. -30

d. 15

a3 + b3 + c3 3abc


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= (25)3 + (15)3 + (-10)3 3*25*15*(-10)

=15625 + 3375 1000 + 11250 = 29250

and (a - b)2 + (b c)2 + (c a)2

= (10)2 + (25)2 + (-35)2

= (10)2 + 625 + 1225

= 1950

Required value = 29250/1950 =15

75. A, B, C are three points on a circle. The tangent at A meets BCproduced at T, BTA = 400, CAT = 440. The angle subtended by BC at the

centre of the circle is-

a. 840

b. 920

c. 960

d. 1040

Ans : (Picture)

ACB = 400 + 440 = 840

ACO = 900 - 440 = 460 = OAC

OCB = ACB - ACO

= 840 - 460 = 380 = OBC


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BOC = 1800 (OCB + OBC)

= 1800 (380 + 380) = 1040

76. If the length of a chord of a circle at a distance of 12 cm from theCentre is 10 cm, then the diameter of the circle is-

a. 13 cm

b. 15 cm

c. 26 cm

d. 30 cm

Ans : (Picture)

OA = OM2 + AM2

= 122 + 52 = 13

Diameter of the circle = 2*OA

= 2*13 = 26cm

77. In ABC, P and Q are the middle points of the sides AB and ACrespectively. R is a point on the segment PQ such that PR : RQ = 1 : 2. If PR= 2 cm, then BC =


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a. 4 cm

b. 2 cm

c. 12 cm

d. 6 cm

PR/RQ =

But, PR = 2cm

RQ = 2*PR

= 4cm

(Picture)

PQ = PR + RQ

= 2 + 4 = 6 cm

BC = 2*PQ = 12CM

78. If tan tan 2 = 1, then the value of sin2 2 +tan2 is equal to

a.

b. 10/3

c. 3

d. 3


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=1

80. If sin + cos = 17/13, 0


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1. The product of 2 numbers is 1575 and their quotient is 9/7. Then the

sum of the numbers is

a. 74

b. 78

c. 80

d. 90


xy = 1575

And x/y = 9/7

xy/x/y = 1575/9/7

y2 = 1225

y = 35 and x = 45


= 80

2. The value of (81)3.6 * (9)2.7/ (81)4.2 * (3) is __

a. 3

b. 6

c. 9

d. 8.2

Ans : (81)3.6 * (9)2.7/(81)4.2 * (3) = (3)14.4 * (3)5.4/(3)16.8 * (3)

= 314.4+5.4-16.8-1


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= 32

= 9


a. 2

b. 5

c. 4

d. 3

Ans : Let 6+6+6+. be x.

x =6+x

x2 = 6+x

x2 + x 6 = 0

(x-3) (x+2) = 0

x = 3

4. The sum of the squares of two natural consecutive odd numbers is 394.The sum of the numbers is

a. 24


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b. 32

c. 40

d. 28


x2 + (x + 2)2 = 394

x2 + x2 + 4x + 4 = 394

2 x2 +4x 390 = 0

x2 + 2x 195 = 0

(x +15) (x-13) = 0

x = 13



a. 1

b. 63

c. 66

d. 67

Ans: (6767+ 67) = 67(6766 + 166)

As 66 is an even number


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6766 is an even number

(6766 + 1) is perfectly divisible by (67 + 1)

i.e. 68

The remainder = 67

6. In a division sum, the divisor is 4 times the quotient and twice theremainder. If a and b are respectively the divisor and the dividend then-

a. 4b-a2 /a =3

b. 4b-2a /a2 =2

c. (a+1)2 = 4b

d. A(a+2)/b = 4

Ans : As divisor is a, and dividend is b.

Quotient = a/4

And Remainder = a/2

b = a * a/4 + a/2

4b = a2 + 2a

a(a+2)/b = 4

7. If 738 A6A is divisible by 11, then the value of A is-

a. 6


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b. 3

c. 9

d. 1

Ans : As 738A6 A is divisible by 11.

A + A + 3 = 6 + 8 + 7

A = 9

8. The east positive integer that should be subtracted from 3011 * 3012 sothat the difference is perfect square is-

a. 3009

b. 3010

c. 3011

d. 3012

Ans : 3011 * 3012 = 3011 (3011 + 1)

= (3011)2 + 3011

Required least number = 3011

9. P, Q, R are employed to do a work for Rs. 5750. P and Q together

finished 19/23 of work and Q and R together finished 8/23 of work. Wage of

Q, in rupees, is-

a. 2850

b. 3750


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c. 2750

d. 1000

Ans : Work done by Q = 19/23 + 8/23 1

4/23

Wage of Q = 4/23 * 5750

= Rs. 1000



a. 15

b. 20

c. 18

d. 30




+ x 6/32 + x/64 = 1

16 + 24 12 + x /64 = 1

3x + 4 = 64


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x = 60/3 = 20 days


a. 1

b. 2

c. 3

d. 4


[(0.75)3/1 0.75 + {0.75 + (0.75)2 + 1}]

= 1.6875 + 2.3125

=4

= 2


a. 70.4

b. 70.464

c. 71.104

d. 71.4

Ans : Given Exp. = 4096 + 40.96 +0.004096

= 64 + 6.4 + 0.064


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= 70.464

13. By selling an article at 3/4th of the marked price, there is a gain of

25%. The ratio of the marked price and the cost price is-

a. 5 : 3

b. 3 : 5

c. 3 : 4

d. 4 : 3


S.P. = Rs.3x/4

and C.P. = 3x/4 * 100/125

= Rs.3x/5


= 5:3



a. Rs.7,000

b. Rs.14,000

c. Rs.5,000

d. Rs.10,000


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x : y = 3 : 2

16. A cistern has 3 pipes A, B and C. A and B can fill it in 3 and 4 hours

respectively, and C can empty it in 1 hour. If the pipes are opened at 3p.m., 4 p.m. and 5 p.m. respectively on the same day, the cistern will be

empty at-

a. 7.12 p.m.

b. 7.15 p.m.

c. 7.10 p.m.

d. 7.18 p.m.


x -3/3 + x 4/4 = x - 5/1

4x 12 + 3x 12/12 = x 5/1

7x 24 = 12x 60

5x = 36

x = 36/5

= 7 hr. + 12 min.

= 7.12 p.m.

17. If A works alone, he would take 4 days more to complete the job than if

both A and B worked together. If B worked alone, he would take 16 daysmore to complete the job than if A and B work together. How many days

would they take to complete the work if both of them worked together?


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a. 10 days

b. 12 days

c. 6 days

d. 8 days




1/x = 1/x + 4 + 1/x + 16

1/x = x + 16 + x + 4/(x + 4) (x +16)

2x2 + 20x = x2 + 20x +64

x2 = 64 (8)2

x = 8 days



a. 310

b. 300

c. 313

d. 312



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c. 16 2/3%

d. 16 1/3%


S.P = (100-40) = Rs.60

and C. P. = 60 * 100/(100-30) = Rs.600/7

Reqd. % profit = 100-600/7 /600/7 * 100%

= (700 600)/7 * 600 * 100 * 7%

= 16 2/3%

21. Successive discount of 10% , 20% and 50% will be equivalent to a

single discount of-

a. 36%

b. 64%

c. 80%

d. 56%


= [100 - (100 10) (100 20) (100 50)/100*100]%

= [100 - 90*80*50/10000]%

= [100-36]%

= 64%


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a. I

b. II

c. III

d. IV









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a. 72

b. 216

c. 354

d. 726



x + 3x + 3x/4 = 3*114

19x/4 = 342

x = 342*4/19 = 72


= 216



a. 12.625 km/hr

b. 13.625 km/hr

c. 14.625 km/hr

d. 15.625 km/hr


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= (x/5/8 + x/10/25 + 7x/10/20) hrs

= (x/40 + x/250 + 7x/200) hrs

= 25x + 4x + 35x/1000

= 64x/1000 hrs


= 15.625 km/hr



a. 264

b. 132

c. 88

d. 66



x + x/2 + x/4 = 3 x 154

7x/4 = 462

x = 462*4/7

=264


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a. 10

b. 15

c. 25

d. 50


5000*500 = 14000x + 4000(500-x)

2500000 =14000x + 2000000 4000x

x = 500000/10000

= 50

27. The average marks of 40 students in an English exam are 72. Later it is

found that three marks 64, 62 and 84 were wrongly entered as 60, 65 and73. The average after mistakes were rectified is-

a. 70

b. 72

c. 71.9


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29. In a laboratory, two bottles contain mixture of acid and water in theratio 2 : 5 in the first bottle and 7 : 3 in the second. The ratio in which thecontents of these two bottles be mixed such that the new mixture has acid

and water in the ratio 2 : 3 is-

a. 4 : 15

b. 9 : 8

c. 21 : 8

d. 1 : 2




mixture.

Then, (2/7 x + 7/10 * 1)/(5x/7 +3/10 * 1) = 2/3

6x/7 21/10 = 10x/7 +6/10

4x/7 = 15/10

x = 15/10*7/4 = 21/8


= 21 : 8

30. A mixture contains 80% acid and rest water. Part of the mixture thatshould be removed and replaced by same amount of water to make the ratio

of acid and water 4 : 3 is-


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a. 1/3 rd

b. 3/7 th

c. 2/3 rd

d. 2/7 th

Ans : Let the initial wt. of mixture be 1 kg and x kg of mixture is taken out

and replaced by same amount of water.


2.4 2.4x = 0.8 + 3.2x

5.6x = 1.6

x = 1.6/5.6 = 2/7th part


increases their wages in the ratio 14 : 15. If the original wage bill was Rs.

189,900, find the ratio in which the wage bill is decreased-

a. 20 : 21

b. 21 : 20

c. 20 : 19

d. 19 : 21



9x * 14y =18900



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= 120*150 = 18000


= 20 : 21



a. 165

b. 170

c. 172

d. 174



40*50 = 38*48 + x + (x-172)

2000 = 1824 + 2x 172

x= 174

33. Walking at 3 km per hour, Pintu reaches his school 5 minutes late. If hewalks at 4 km per hour he will be 5 minutes early. The distance of Pintusschool from his house is-

a. 1 Km


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Ans : Let the S.P. of 100 Oranges be Rs. x.

S.P. of 20 oranges = x/100 *20 = Rs. x/5

C.P. of 100 oranges = x - x/5

= Rs. 4x/5

Reqd. Profit % = x/5 * 100*5/4x %

= 25%

37. 60% of the cost price of an article is equal to 50% of its selling price.Then the percentage of profit or loss on the cost price is-

a. 20% loss

b. 16 2/3% profit

c. 20% profit

d. 10% loss

Ans : Let the cost price be Rs. 100.

S.P * 50/100 = 100*16/100

S.P = 60*100/50

= Rs. 120

Reqd. % profit = (120 100)% = 20%

38. Maninder bought two horses at Rs.40,000 each. He sold one horse at

15% gain, but had to sell the second horse at a loss. If he hand suffered a


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loss of Rs.3,600 on the whole transaction, then the selling price of the

second horse is-

a. Rs.30,000

b. Rs.30,200

c. Rs.30,300

d. Rs.30,400








d. x2 y2 / y2 * 100% gain




= x2 - y2 /y2 *100%


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40. A jar contain 10 red marbles and 30 green ones. How many red marblesmust be added to the jar so that 60% of the marbles will be red?

a. 25

b. 30

c. 35

d. 40


total.

(10+x)/ (10 + x) + 30 = 60/100

10 + x/40 + x = 3/5

50 + 5x = 120 + 3x

x = 70/2 = 35


a. 12

b. 16

c. 35

d. 24



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b. 33 1/3%

c. 33%

d. 30%


= 50*100/(100 + 50) %

= 5000/150 %

= 33 1/3%

44. A took two loans altogether of Rs.1200 from B and C. B claimed 14%simple interest per annum, while C claimed 15% per annum. The total

interest paid by A in one year was Rs.172. Then, A borrowed-

a. Rs.800 from C

b. Rs.625 from C

c. Rs.400 from B

d. Rs.800 from B


(1200 x)*15*1/100 + x*14*1/100

18000 15x + 14x = 172*100

x = Rs. 800

45. If a regular polygon has each of its angles equal to 3/5 times of tworight angles, then the number of side is-


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a. 3

b. 5

c. 6

d. 8


Then (2n-4)/n = 2*3/5

(2n 4)*5 = 6n

n = 5

46. A square is of area 200 sq. m. A new square is formed in such a waythat the length of its diagonal is 2 times of the diagonal of the givensquare. The the area of the new square formed is-

a. 2002 sq.m

b. 4002 sq.m

c. 400 sq.m

d. 800 sq.m


= 2*200

= 20 m



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Area of the new square = *(20.2)2 = 400 sq. m

47. The heights of a cone, cylinder and hemisphere are equal. If their radii

are in the ratio 2 : 3 : 1, then the ratio of their volumes is-

a. 2 : 9 : 2

b. 4 : 9 : 1

c. 4 : 27 : 2

d. 2 : 3 : 1

Ans : Ratio of their volumes [cone : cylinder : hemisphere]

= 1/3(2)2h : (3)2*h : 2/3(1)2.h

= 4/3 : 9 : 2/3 = 4 : 27 : 2

48. A motor-boat can travel at 10 km/hr in still water. It travelled 91 km

downstream in a river and then returned to the same place, takingaltogether 2 hours. Find the rate of flow of river-

a. 3 km/hr

b. 4 km/hr

c. 2 km/hr

d. 5 km/hr

Ans : Let the rate of flow of river be x km/hr.


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b. 3 km/hr

c. 4 km/hr

d. 5 km/hr

Ans : Let the speeds of motor boat and the stream be x and y km/hrrespectively.

39/x + y + 25/x y = 8 (1)

and 52/x + y + 35/x y = 11 (2)

Solving equations (1) and (2), we get-

100 - 105/x y = 32 33

x y = 5

and x + y = 13

y = 4 km/hr

51. If a sum of money placed at compound interest, compounded annually,doubles itself in 5 years, then the same amount of money will be 8 times ofitself in-

a. 25 years

b. 20 years

c. 15 years

d. 10 years

Ans : Required time = 5log 8/log 2


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= 5*3log 2/log 2

= 15 years

52. A person has left an amount of Rs.1,20,000 to be divided between his 2son aged 14 years and 12 years such that they get equal amounts when

each attains 18 years of age. If the amount gets a simple interest of 5% per

annum, the younger sons share at present is-

a. Rs.48,800

b. Rs.57,600

c. Rs.62,400

d. Rs.84,400

Ans : Let the present share of the younger son be Rs. x.

The share of the elder son = Rs. (120000 x)

x + x*6*5/100

= (120000 x) + (120000 x)*4*5/100

130x/100 = (120000 x)*120/100

13x = 1440000 12x

25x = 1440000

x = Rs. 57600

53. If the simple interest on Rs.x at a rate of a% for m years is same asthat on Rs. y at a rate of a2% for m2 years, then x : y is equal to-


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a. m : a

b. am : 1

c. 1/m : 1/a

d. 1/am : 1

x*a*m/100 = y*a2*m2/100

x/y = am/1

x : y = am : 1

54. Base of a right prism is an equilateral triangle of side 6 cm. If the

volume of the prism is 1083 cc, its height is-

a. 9 cm

b. 10 cm

c. 11 cm

d. 12 cm

Ans : Height of the prism = 1083*4/3*(6)2 = 12cm

55. If a + 1/a + 2 = 0, then the value of (a37 1/a100) is-

a. 0


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b. -2

c. 1

d. 2

Ans : a + 1/a + 2 = 0

a2 + 1 +2a = 0

(a + 1)2 = 0

a +1 = 0

a = -1

a37 -1/a100 = (-1) (1) = -2

56. The value of k for which the graphs of (k-1) x+y-2 = 0 and (2-k) x -3y+ 1 = 0 are parallel is-

a.

b. -1/2

c. 2

d. -2

Ans : (k 1) x + y 2 = 0

y = (1 k) x + 2 .(1)

and (2 k) x 3y 1 = 0

3y = (2 k) x +1

Y = 2 k/3 x + 1/3 .(2)


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m1 = m2

1 k = 2 k/3

3 3k = 2 - k

k = 1/2

57. If a2 + b2 + c2 = 2 (a-b-c) 3, then the value of (a b + c) is-

a. -1

b. 3

c. 1

d. -2

Ans : a2 + b2 + c2 = 2(a b c) 3

a2 - 2a + 1 + b2 + 2b + 1 + c2 + 2c + 1 = 0

(a 1)2 + (b + 1)2 + (c + 1)2 = 0

a 1 = 0, b + 1 = 0, c + 1 = 0

a = 1, b = -1, c = -1

a + b c = 1 1 + 1 =1

58. If x2 + 3x + 1 = 0, then the value of x3 + 1/x3 is-

a. -18

b. 18


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60. Base of a right pyramid is a square, length of diagonal of the base is

242 m. If the volume of the pyramid is 1728 cu.m, its height is-

a. 7 m

b. 8 m

c. 9 m

d. 10 m

Ans : Area of the base of the pyramid

= (242)2 = 576m2

Height of pyramid + 1728*3/576 = 9m

61. The height of a right circular cone and the radius of its circular base arerespectively 9 cm and 3 cm. The cone is cut by a plane parallel to its base so

as to divide it into two parts. The volume of the frustum (i.e., the lower part)

of the cone is 44 cubic cm. The radius of the upper circular surface of thefrustum (taking = 22-7) is-

a. 312 cm

b. 313 cm

c. 36 cm

d. 320 cm

Ans : Let the radius of the upper circular part of the frustum be r cm.


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(Picture)

Then r/3 = x/9 = AC/AD

x = 3r (where AC = x cm)

/3h(r12 + r1r2+r22 = 44

/3(9 x)*(9 + 3r + r2) = 44

(9 x)*(9 + 3r + r2) = 44*3*7/22 = 42

81 + 27r + 9r2 9x 3rx - r2x = 42 On putting x = 3r,

81 + 27r + 9r2 - 9r2 - 3r3 27r = 42

3r3 = 39

r = 313 cm

62. The ratio of radii of two right circular cylinder is 2 : 3 and their heightsare in the ratio 5 : 4. The ratio of their curved surface area is-

a. 5 : 6

b. 3 : 4

c. 4 : 5

d. 2 : 3

Ans : Required ratio = 2*2r*5h/2*3r*4h = 5 : 6


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63. A solid cylinder has total surface area of 462 sq.cm. Curved surfacearea is 1/3rd of its total surface area. The volume of the cylinder is-

a. 530 cm3

b. 536 cm3

c. 539 cm3

d. 545 cm3

Ans : 2r(r + h) = 462

and 2rh = 1/3*462 = 154

r + h/h = 462/154 =3

r + h = 3h

r = 2h

2*2h2 = 154

h2 = 154*7/22*4 = 49/4

= (7/2)2

h = 7/2 cm

and r = 7 cm

Volume of the cylinder

= 22/7*49*7/2

= 539 cm3


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64. A cylinder and a cone have equal radii of their bases and equal heights.

If their curved surface areas are in the ratio 8 : 5, the ratio of their radiusand height is-

a. 1 : 2

b. 1 : 3

c. 2 : 3

d. 3 : 4

Ans : Let the radius and height of each are r and h respectively.

2rh/rh2 + r2 = 8/5

10h = 8r2 + h2

100h2 =64r2 + 64h2

h2 = 64 r2/36 = (4/3 r)2

h = 4/3r

r : h = 3 : 4

65. A solid is hemispherical at the bottom and conical above. If the surface

areas of the two parts are equal, then the ratio of radius and height of itsconical part is-

a. 1 : 3

b. 1 :1

c. 3 : 1

d. 1 : 3


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BAC = 1/2 * 1100 = 550

67. If I is the incentre of ABC and BIC = 1350, then ABC is-

a. Acute angled

b. Equilateral

c. Right angled

d. Obtuse angled

Ans : (Picture)

BIC = 1350

B/2 + C/2 = 1800 1350 = 450

B + C = 900

A = 1800 (B + C) = 900

i.e., ABC is a right angled.

68. If sin2 (+ /2) is-

a. 1

b. -1


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c. 0

d. 0.5

sin2 + sin2 = 2

1 cos2 1 - cos2 = 2

cos2 cos2 = 0

cos = 0

and cos = 0

= /2 and = /2

cos (+ /2) = cos [/2 + /2 / 2]

= cos /2 = 0

69. The length of a shadow of a vertical tower is 1/3 times its height. Theangle of elevation of the Sun is-

a. 300

b. 450

c. 600

d. 900

Ans : (Picture)


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tan = h/13 h = 3 = tan 600

70. The graphs of x +2y =3 and 3x-2y = 1 meet the Y-axis at two points

having distance-

a. 8/3 units

b. 4/3 units

c. 1 unit

d. 2 units

Ans : When the graphs meet the Y-axis at two points.

Then, [x + 2y = 3] at x = 0 [0, y1 =3/2]

[3x - 2y = 1] at x = 0

and i.e., [0, y2 = -1/2]

Required distance = (y1 y2)

= 3/2 (- ) = 2 units

71. If x+1/16x = 1, then the value of 64x3 + 1/64x3 is-

a. 4

b. 52

c. 64

d. 76

x + 1/16x = 1


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16x2 16x + 1 =0

16x2 16x + 4 = 3

(4x 2)2 = 3

4x = 2 + 3

64x3 = (23)3

=8 + 33 + 63 (2 + 3)

= 26 + 153

64x3 + 1/64x3 = (26 + 153) + 1/ (26 + 153)

= (26 + 153) + 26 - 153/676 -675

=52

72. If a, b, c, are three non-zero real numbers such that a + b + c = 0, and

b2 ca, then the value of a2 + b2 + c2/ b2 ca is-

a. 3

b. 2

c. 0

d. 1

a + b + c = 0

a + c = -b

a2 + c2 = b2 -2ac

a2 + b2 + c2 = 2b2 2ac


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a2 + b2 + c2/ b2 ac = 2

73. If a4 + a2 b2 + b4 = 8 and a2 + ab + b2 = 4, then the value of ab is-

a. -1

b. 0

c. 2

d. 1

a4 + a2 b2 + b4/ a2 + ab + b2 = 8/4

(a2 + b2)2 (ab) 2/ (a2 + b2 + ab) = 2

a2 ab + b2 = 2 .(1)

and a2 + ab + b2 = 4 ..(2)

2ab = 2

ab = 1

74. If a = 25, b = 15, c = -10, then the value of a3 + b3 + c3 3abc/ (a-b)2 + (b-c)2 + (c-a)2 is-

a. 30

b. -15

c. -30

d. 15


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a3 + b3 + c3 3abc

= (25)3 + (15)3 + (-10)3 3*25*15*(-10)

=15625 + 3375 1000 + 11250 = 29250

and (a - b)2 + (b c)2 + (c a)2

= (10)2 + (25)2 + (-35)2

= (10)2 + 625 + 1225

= 1950

Required value = 29250/1950 =15

75. A, B, C are three points on a circle. The tangent at A meets BC

produced at T, BTA = 400, CAT = 440. The angle subtended by BC at the

centre of the circle is-

a. 840

b. 920

c. 960

d. 1040

Ans : (Picture)

ACB = 400 + 440 = 840

ACO = 900 - 440 = 460 = OAC


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OCB = ACB - ACO

= 840 - 460 = 380 = OBC

BOC = 1800 (OCB + OBC)

= 1800 (380 + 380) = 1040

76. If the length of a chord of a circle at a distance of 12 cm from theCentre is 10 cm, then the diameter of the circle is-

a. 13 cm

b. 15 cm

c. 26 cm

d. 30 cm

Ans : (Picture)

OA = OM2 + AM2

= 122 + 52 = 13

Diameter of the circle = 2*OA

= 2*13 = 26cm


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77. In ABC, P and Q are the middle points of the sides AB andACrespectively. R is a point on the segment PQ such that PR : RQ = 1 : 2. If PR= 2 cm, then BC =

a. 4 cm

b. 2 cm

c. 12 cm

d. 6 cm

PR/RQ =

But, PR = 2cm

RQ = 2*PR

= 4cm

(Picture)

PQ = PR + RQ

= 2 + 4 = 6 cm

BC = 2*PQ = 12CM

78. If tan tan 2 = 1, then the value of sin2 2 + tan2 is equal to

a.

b. 10/3


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c. 3

d. 3

tan * tan 2= 1

tan * 2 tan /1 tan2 = 1

2 tan2 = 1 - tan2

3 tan2 = 1

tan = 1/3 = tan 300

= 300

sin2 2+ tan2 2= sin2 600 + tan2 600

= + 3

=3

79. The value of cot

/20 cot 3

/20 cot 5

/20 cot 7

/20 cot 9

/20 is-

a. -1

b.

c. 0

d. 1

Ans : Given Exp.

= cot /20. cot 3/20. cot 5/20. cot 7/20. cot 9/20

= cot 90. cot 270. cot 450. cot 630. cot 810


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Arithmetical Ability

1. The product of 2 numbers is 1575 and their quotient is 9/7. Then the

sum of the numbers is

a. 74

b. 78

c. 80

d. 90


xy = 1575

And x/y = 9/7

xy/x/y = 1575/9/7

y2 = 1225

y = 35 and x = 45


= 80

2. The value of (81)3.6 * (9)2.7/ (81)4.2 * (3) is __

a. 3

b. 6

c. 9

d. 8.2

Ans : (81)3.6 * (9)2.7/(81)4.2 * (3) = (3)14.4 * (3)5.4/(3)16.8 * (3)


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= 314.4+5.4-16.8-1

= 32

= 9


a. 2

b. 5

c. 4

d. 3

Ans : Let 6+6+6+. be x.

x =6+x

x2 = 6+x

x2 + x 6 = 0

(x-3) (x+2) = 0

x = 3

4. The sum of the squares of two natural consecutive odd numbers is 394.

The sum of the numbers is


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a. 24

b. 32

c. 40

d. 28


x2 + (x + 2)2 = 394

x2 + x2 + 4x + 4 = 394

2 x2 +4x 390 = 0

x2 + 2x 195 = 0

(x +15) (x-13) = 0

x = 13



a. 1

b. 63

c. 66

d. 67

Ans: (6767+ 67) = 67(6766 + 166)


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b. 3750

c. 2750

d. 1000

Ans : Work done by Q = 19/23 + 8/23 1

4/23

Wage of Q = 4/23 * 5750

= Rs. 1000



a. 15

b. 20

c. 18

d. 30




+ x 6/32 + x/64 = 1

16 + 24 12 + x /64 = 1


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3x + 4 = 64

x = 60/3 = 20 days


a. 1

b. 2

c. 3

d. 4


[(0.75)3/1 0.75 + {0.75 + (0.75)2 + 1}]

= 1.6875 + 2.3125

=4

= 2


a. 70.4

b. 70.464

c. 71.104

d. 71.4

Ans : Given Exp. = 4096 + 40.96 +0.004096


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= 64 + 6.4 + 0.064

= 70.464

13. By selling an article at 3/4th of the marked price, there is a gain of25%. The ratio of the marked price and the cost price is-

a. 5 : 3

b. 3 : 5

c. 3 : 4

d. 4 : 3


S.P. = Rs.3x/4

and C.P. = 3x/4 * 100/125

= Rs.3x/5


= 5:3



a. Rs.7,000

b. Rs.14,000

c. Rs.5,000


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d. Rs.10,000

Ans : Let the monthly income of B be Rs. x.

Monthly income of A = Rs. 2x and

Saving of A =5000 * 4/(4 + 1)

= Rs. 4000

Saving of B = Rs.1000

2x 4000/x-1000 = 5/3

6x 12000 = 5x 5000

x = Rs.7000

15. The ratio of the sum of two numbers and their difference is 5:1. Theratio of the greater number to the smaller number is-

a. 2 : 3

b. 3 : 2

c. 5 : 1

d. 1 : 5

Ans : Let the numbers be x and y.

x + y/ x y = 5/1

5x 5y = x + y

4x = 6y


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x/y = 6/4

x : y = 3 : 2

16. A cistern has 3 pipes A, B and C. A and B can fill it in 3 and 4 hoursrespectively, and C can empty it in 1 hour. If the pipes are opened at 3

p.m., 4 p.m. and 5 p.m. respectively on the same day, the cistern will beempty at-

a. 7.12 p.m.

b. 7.15 p.m.

c. 7.10 p.m.

d. 7.18 p.m.


x -3/3 + x 4/4 = x - 5/1

4x 12 + 3x 12/12 = x 5/1

7x 24 = 12x 60

5x = 36

x = 36/5

= 7 hr. + 12 min.

= 7.12 p.m.

17. If A works alone, he would take 4 days more to complete the job than ifboth A and B worked together. If B worked alone, he would take 16 days

more to complete the job than if A and B work together. How many dayswould they take to complete the work if both of them worked together?


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a. 10 days

b. 12 days

c. 6 days

d. 8 days




1/x = 1/x + 4 + 1/x + 16

1/x = x + 16 + x + 4/(x + 4) (x +16)

2x2 + 20x = x2 + 20x +64

x2 = 64 (8)2

x = 8 days



a. 310

b. 300

c. 313

d. 312


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=312.5

313

19. 2 men and 5 women can do a work in 12 days. 5 men 2 women can dothat work in 9 days. Only 3 women can finish the same work in-

a. 36 days

b. 21 days

c. 30 days

d. 42 days

(2m + 5w) * 12 = (5m + 2w) * 9

24m + 60w = 45m + 18w

21m = 42w

1m = 2w

2m + 5w = 9w

required no. of days = 9 * 12/3

= 36

20. While selling, a businessman allows 40% discount on the marked priceand there is a loss of 30%. If it is sold at the marked price, profit per cent

will be

a. 10%


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b. 20%

c. 16 2/3%

d. 16 1/3%


S.P = (100-40) = Rs.60

and C. P. = 60 * 100/(100-30) = Rs.600/7

Reqd. % profit = 100-600/7 /600/7 * 100%

= (700 600)/7 * 600 * 100 * 7%

= 16 2/3%

21. Successive discount of 10% , 20% and 50% will be equivalent to asingle discount of-

a. 36%

b. 64%

c. 80%

d. 56%


= [100 - (100 10) (100 20) (100 50)/100*100]%

= [100 - 90*80*50/10000]%

= [100-36]%


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d. 15.625 km/hr


= (x/5/8 + x/10/25 + 7x/10/20) hrs

= (x/40 + x/250 + 7x/200) hrs

= 25x + 4x + 35x/1000

= 64x/1000 hrs


= 15.625 km/hr



a. 264

b. 132

c. 88

d. 66



x + x/2 + x/4 = 3 x 154

7x/4 = 462


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x = 462*4/7

=264



a. 10

b. 15

c. 25

d. 50


5000*500 = 14000x + 4000(500-x)

2500000 =14000x + 2000000 4000x

x = 500000/10000

= 50

27. The average marks of 40 students in an English exam are 72. Later it isfound that three marks 64, 62 and 84 were wrongly entered as 60, 65 and

73. The average after mistakes were rectified is-

a. 70


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b. 72

c. 71.9

d. 72.1

Ans : Correct average

= 40*72 + (64 + 62 +84) 68 65 73/40

= 2880 + 210 206/40 = 2884/40

= 72.1

28. A and B are two alloys of gold and copper prepared by mixing metals inthe ratio 7:2 and 7:11 respectively. If equal quantities of the alloys are

melted to form a third alloy C, the ratio of gold and copper in C will be-

a. 5 : 7

b. 5 : 9

c. 7 : 5

d. 9 : 5

Ans : Quantity of gold in A = 7/9* wt. of A

Quantity of gold in B = 7/18* wt. of B

If 1 kg of each A and B are mixed to form third alloys C.

Then quantity of gold in 2 kg C = 7/9 + 7/18

= 7/6 kg

And Quantity of copper in 2 kg C = 2 7/6


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= 5/6 kg

Required ratio = 7/6 : 5/6 = 7 : 5

29. In a laboratory, two bottles contain mixture of acid and water in theratio 2 : 5 in the first bottle and 7 : 3 in the second. The ratio in which the

contents of these two bottles be mixed such that the new mixture has acidand water in the ratio 2 : 3 is-

a. 4 : 15

b. 9 : 8

c. 21 : 8

d. 1 : 2




mixture.

Then, (2/7 x + 7/10 * 1)/(5x/7 +3/10 * 1) = 2/3

6x/7 21/10 = 10x/7 +6/10

4x/7 = 15/10

x = 15/10*7/4 = 21/8


= 21 : 8


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30. A mixture contains 80% acid and rest water. Part of the mixture that

should be removed and replaced by same amount of water to make the ratioof acid and water 4 : 3 is-

a. 1/3 rd

b. 3/7 th

c. 2/3 rd

d. 2/7 th

Ans : Let the initial wt. of mixture be 1 kg and x kg of mixture is taken outand replaced by same amount of water.


2.4 2.4x = 0.8 + 3.2x

5.6x = 1.6

x = 1.6/5.6 = 2/7th part


increases their wages in the ratio 14 : 15. If the original wage bill was Rs.189,900, find the ratio in which the wage bill is decreased-

a. 20 : 21

b. 21 : 20

c. 20 : 19

d. 19 : 21


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33. Walking at 3 km per hour, Pintu reaches his school 5 minutes late. If he

walks at 4 km per hour he will be 5 minutes early. The distance of Pintusschool from his house is-

a. 1 Km

b. 2 Km

c. 2 Km

d. 5 Km

Ans : Distance of the school from the house

= 4*3/(4 3) * 5 +5/60 km

= 12*1/6

=2 km

34. Nitin bought some oranges at Rs. 40 a dozen and an equal number atRs.30 a dozen. He sold them at Rs. 45 a dozen and made a profit of Rs.480.The number of oranges, he bought, was-

a. 48 dozen

b. 60 dozen

c. 72 dozen

d. 84 dozen

Ans : Let the number of oranges bought be x.

45x/12 70x/24 = 480


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(45-35)/12 = 480

x = 480*12/10 = 576

= 48*12 = 48 dozen

35. A man buys two chairs for a total cost of Rs.900. By selling one for 4/5of its cost and the other for 5/4 of its cost, he makes a profit of Rs.90 on thewhole transaction. The cost of the lower priced chair is-

a. Rs.360

b. Rs.400

c. Rs.420

d. Rs.300

Ans : Let the cost price of 1 chair be Rs. x.

C.P. of other chair = Rs. (900-x)

4/5 x + 5/4 (900-x) = 900 + 90

4/5 x +1125 -5x/4 = 990

9x/20 = 135

x = 135*20/9

= Rs. 300

C.P. of the lower priced chair is Rs. 300.

36. By selling 100 oranges, a vendor gains the selling price of 20 oranges.His gain per cent is-


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= Rs. 120

Reqd. % profit = (120 100)% = 20%

38. Maninder bought two horses at Rs.40,000 each. He sold one horse at15% gain, but had to sell the second horse at a loss. If he hand suffered a

loss of Rs.3,600 on the whole transaction, then the selling price of thesecond horse is-

a. Rs.30,000

b. Rs.30,200

c. Rs.30,300

d. Rs.30,400








d. x2 y2 / y2 * 100% gain


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= x2 - y2 /y2 *100%

40. A jar contain 10 red marbles and 30 green ones. How many red marbles

must be added to the jar so that 60% of the marbles will be red?

a. 25

b. 30

c. 35

d. 40


total.

(10+x)/ (10 + x) + 30 = 60/100

10 + x/40 + x = 3/5

50 + 5x = 120 + 3x

x = 70/2 = 35


a. 12


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b. 16

c. 35

d. 24


x*25x/100 = x + 200x/100

x2/4 = 3x

x2 - 12x = 0

x 12 = 0

x = 12

42. The value of an article depreciates every year at the rate of 10% of its

value. If the present value of the article is Rs.729, then its worth 3 yearsago was-

a. Rs.1250

b. Rs.1000

c. Rs.1125

d. Rs.1200

Ans : Let the worth 3 years ago be Rs. x.

729 = x (1 10/100)3

729 = x*9*9*9/10*10*10

x = Rs. 1000


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43. The price of onions has been increased by 50%. In order to keep the

expenditure on onions the same the percentage of reduction in consumptionhas to be-

a. 50%

b. 33 1/3%

c. 33%

d. 30%


= 50*100/(100 + 50) %

= 5000/150 %

= 33 1/3%

44. A took two loans altogether of Rs.1200 from B and C. B claimed 14%simple interest per annum, while C claimed 15% per annum. The totalinterest paid by A in one year was Rs.172. Then, A borrowed-

a. Rs.800 from C

b. Rs.625 from C

c. Rs.400 from B

d. Rs.800 from B


(1200 x)*15*1/100 + x*14*1/100

18000 15x + 14x = 172*100


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x = Rs. 800

45. If a regular polygon has each of its angles equal to 3/5 times of two

right angles, then the number of side is-

a. 3

b. 5

c. 6

d. 8


Then (2n-4)/n = 2*3/5

(2n 4)*5 = 6n

n = 5

46. A square is of area 200 sq. m. A new square is formed in such a way

that the length of its diagonal is 2 times of the diagonal of the givensquare. The the area of the new square formed is-

a. 2002 sq.m

b. 4002 sq.m

c. 400 sq.m

d. 800 sq.m


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= 2*200

= 20 m


Area of the new square = *(20.2)2 = 400 sq. m

47. The heights of a cone, cylinder and hemisphere are equal. If their radiiare in the ratio 2 : 3 : 1, then the ratio of their volumes is-

a. 2 : 9 : 2

b. 4 : 9 : 1

c. 4 : 27 : 2

d. 2 : 3 : 1

Ans : Ratio of their volumes [cone : cylinder : hemisphere]

= 1/3(2)2h : (3)2*h : 2/3(1)2.h

= 4/3 : 9 : 2/3 = 4 : 27 : 2

48. A motor-boat can travel at 10 km/hr in still water. It travelled 91 km

downstream in a river and then returned to the same place, taking

altogether 2 hours. Find the rate of flow of river-

a. 3 km/hr

b. 4 km/hr


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c. 2 km/hr

d. 5 km/hr

Ans : Let the rate of flow of river be x km/hr.

91/(10 + x) + 91/(10 x) = 20

91(10 x + 10 + x)(10 + x) (10 x) = 20

91*20 = 20(100 x2)

x2 = 9 = (3)2

x = 3 km/hr

49. A man driving at 3/4th of his original speed reaches his destination 20

minutes later than the usual time. Then the usual time is-

a. 45 minutes

b. 60 minutes

c. 75 minutes

d. 120 minutes

Ans : Let the original speed be x km/hr and the usual time be y hours.

x * y = x(y+1/3)

4y = 3y + 1

y = 1 hr = 60 minutes


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50. A motor-boat, travelling at the same speed, can cover 25 km upstream

and 39 km downstream in 8 hours. At the same speed, it can travel 35 kmupstream and 52 km downstream in 11 hours. The speed of the stream is

a. 2 km/hr

b. 3 km/hr

c. 4 km/hr

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