## SAMPLE PAPER FOR SYMBIOSIS NATIONAL APTITUDE TEST

1—————–168/15

15/24———-168/15*15/24=7 days..

8.A trader makes a profit equal to the selling price of 75 articles when he sold 100 articles.What % of profit did he make?

(a)33.33 (b)75 (c)300 (d)150

Ans.Let the c.p of each article be Rs.1

Then profit made =100-75=25

% of profit=75/25*100=300

9.In a100m race,A gives B a start of 200m,then A wins by 5 second.Alternatively,if A gives B a start of 40m,then the race end in a dead heat.How long does A take to run 200m?

Ans.B’s speed=20/5=4

A’s speed/B’s speed=100/60

A’s speed/4=100/60

A’s speed=20/3

Time taken by A’s 200*20/3=30s

10.A 4cm cube is cut into 1cm cubes.What is the percentage increase in the surface area after such cutting?

Ans.% change in surface area=

(6*64-6*16)/6*16)*100=300%

11.A number G236G0 can be divided by 36 if G is

(a)8 (b)6 (c)1 (d)0

Ans.Factors of 36=4×9

For a number to be divisible by 36,it has to be divisible by both 4 and 9

For a number to be divisible by 4,its last digit must me divisible by 4.As the last digit is 0 ,it is divisible by 4.

Further for a number to be divisible by 9,when the sum of the digit is divisible by 9.

So,G+2+3+6+G+0=11+2G

Putting G=8,we have the sum of digit as 27 which is divisible by 9,so the value of G is 8.

12.Amit can do a work in 12 days and Sagar in 15 days.If they work on it together for 4 days,then the fraction of work left?

(a)3/20 (b)3/5 (c)2/5 (d)2/20

Ans.Amit’s 1 day work=1/12

Sagar’s 1 day work=1/15

(Amit+Sagar)’s 1 day work=1/12+1/15=3/20

(Amit+Sagar)’s 4 day work=4*3/20=3/5

So,work left=1-3/5=2/5

13.A rectangular park 60m long and 40 m wide has two concrete crossroads running in the middle of the park and the rest of the park has been used as a lawn.If the area of the lawn is 2109sqm,then what is the width of the road?

(a)2.91m (b)3m (c)5.82 (d)none

Ans.Area of the park=60*40=2400

Area of the cross road=(60*m)+(40*m)-(m*m)

=m(100-m)=2109

M=3

14.A bag contains 5 white and 3 black balls,another bag contains 4 white and 5 black balls.From any one of these two bags a single draw of 2 ball is made.Find the probability that one of them would be white and other black ball.

(a)275/504 (b)5/18 (c)5/9 (d)None

Ans.Required probability=1/2((5c1*3c1)/8c2)+(4c1*5c1)/9c2))

=1/2(15/28+20/36)

=275/504

15.A solid sphere is melted and recast into right circular cone with base radius equal to to the radius of the sphere.What is the ration of the height and radius of the cone so formed.

Ans.Area of solid sphere=4/3*pi*r^3

Area of right Circular cone=1/3*pi*r^2*h

Given,4/3*pi*r^3=1/3*pi*r^2*h

So,h/r=4/1

So the required ratio of h:r=4:1

16.A reduction of 20% in the price of sugar enables a person to purchase 6kg more of Rs.240.What is the original price per kg of sugar?

Ans.Reduction in price=20%=1/5,So the increase in amount of sugar=1/4

Since,25%=6kg,it means the original amount=24kg.

Therefore the original price of the sugar is 240/24=10

17.The radius of a wire is decreased to one-third and its volume remains the same.The new length is how many times the original length?

(a)2 times (b)4 times (c)5 times (d)9 times.

Ans.(a)

18.A,B,C and D purchased a hotel for Rs.56 lakh.The contribution of B,C and D together is 460% that of A,alone. The contribution of A,C and D together is 366.66% that of B’s Contribution and C’s contribution is 40% that of A,B and D together.The amount contributed by D is

(a)10 lakhs (b)12 lakhs (c)16 lakhs (d)18 lakhs

Ans.Given

A:(B+C+D)=100:460

So A contributed Rs.10 lakhs

C:(A+B+D)=40:100

So C contributed Rs 16 lakh

Thus the contribution of D=56-(10+12+16)

=18 lakhs

19.The probability of a leap year selected at random contains either 53 Sundays or 53 Mondays.

(a)17/53 (b)1/53 (c)3/7 (d)None

Ans.3/7

20.If ncx=56 and npx=336,find n and x.

(a)7,3 (b)8,4 (c)8,3 (d)9,6

Ans.Go through options Only (c) satisfies the given conditions.

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