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Question Booklet Series: B
Group A : General Mental Ability Marks : 30 x 1.5= 45
1. The difference between simple interest and compound interest on a sum of money for 2 years at 5% is Rs.25. The sum is :-C) Rs. 10000
Solution :
CI – SI = P. R^2/100^2 = 25
Or, P. 5*5/10000 =25
Or, P= 25 *10000/25 = 10000
2. A bag contains 50 paise, 25 paise and 10 paise coin in the ratio 1:2:5 amounting to Rs. 132. The number of 10 paise coins is:
c) 440
Solution :
Let, Number coins are x, 2x and 5x.
Now, 2 fifty paise coins make Rs. 1, 4 twenty paise coins make Rs.1 and 10 ten paise coins make Rs. 1.
Now , x/2 + 2x/4 + 5x/10 = 132
Or, x/2 + x/2 + x/2 = 132
Or, 3x/2=132
Or, x= 132 * 2/3 = 88
Then 5x = 5*88=440
3. The incorrect statement of the following :
a) 115:138>1:2 b) 115:138<2:3
c) 3:4<115:138 d) 115:138=5:6
d) Ans:- d) 115:138=5:6
Solution : 115/138 = 0.8333, ½ = 0.5 , 2/3=0.67 , ¾=0.75 , 5/6=0.8333. Then, 115:138=5:6
4. The sides of a triangle are 12 cm, 16 cm and 20 cm. The area of the triangle is : Ans.- C) 96 sq. cm
Solution : s= (a+b+c)/2= (12+16+20)/2= 24
Area= sqrt(s(s-a)(s-b)(s-c))=sqrt(24*12*8*4)=sqrt(24*24*4*4)=24*4=96.
5. H.C.F and L.C.M of x and y are 3 and 105 respectively. If x+y=36, value of 1/x +1/y will be –
Ans- a) 4/35
Solution :
xy = H.C.F x L.C.M= 3*105=315
1/x + 1/y = (y+x)/xy = 36/315 = 4/35
6. A train running at speed 54 km/hr. crosses a passenger standing on a platform of length 200 meters in 10 seconds. Length of the train in meters is :
Ans:- D) 150
Solution : Train passed the man standing on the platform not the whole platform, so train passed own length in 10 seconds.
54km/hr= 54 *5/18 = 15 m/s
In 10 seconds train goes 15*10=150meters.
7. The ratio of a rod and its shadow is 1:v3. Angle of elevation of sun is :
Ans- c)30 degree
Solution : Let, length of road = x and length of shadow is v3x .
tan ѳ = x/v3x = 1/v3 = tan 30
or, ѳ=30 degree
8. After starting from a point , a man walks 4 k.m towards west, the turning to his right he moves 4 km. After this he again turns right and moves 4 km. Which choice given below indicates the correct direction in which he is from his starting point ?
a) North
Solution:
Starting point A and end point B.
9. If the profit is 20% on sell price, profit on cost price will be –
b) 25 %
Solution : - SP= CP+ Profit
According to question, Profit %= Profit/ SP
Or, 20/100 = Profit / CP+Profit
Or, 1/5= profit / CP+Profit
Or, CP+Profit = 5 * Profit
Or, CP = 4*Profit
Now, Profit % on CP= Profit *100/ CP = Profit *100/4*Profit [:. CP= 4*Profit]
= 100/4=25%
10. Out of four numbers, the average of first three is 15 and of the last three is 20. If the last number is 17, the first number is –
Ans- A) 2
Solution :
Let the four number is a,b,c,d.
Now , a+b+c=15*3=45 ----- (i)
And b+c+d= 60 -------- (ii)
We get from equation (ii) – (i) and putting d= 17
d-a= 15 or, 17-a=15 or a= 2
11. The length of the largest possible square tiles that can be used in flooring a room of length 5.44 m and of breadth 3.7 m is :-
a) 36 cm b) 34 cm c) 32 cm d) 30 cm
Ans- I don’t know.
12. A and B invested in a business in the ratio :2. If 9% of total profit goes to charity and A’s share is Rs. 650, the total profit is –
Ans- A) Rs. 1000
Solution : Let the total profit is x.
After pay to the charity profit left = x- x*9/100= 91x/100
A’s share = 91x/100 * 5/7 = 65x/100
Or, 650 = 65x/100
Or, x= 650*100/65= 1000
13. Rupees 12100 is divided into three parts such that they become equal after 1,2 and 3 years respectively. The rate of simple interest being 7% pa in all case. The smallest part is –
Ans- B) Rs. 2200
Solution :
Let, three parts are x,y and z.
We know that, SI= P*R*T/100.
Accordingly, x*7*1/100= y*7*2/100= z*7*3/100 =k
Or, x=2y=3z =k
X=k, y=k/2, z=k/3. X:y:z= k:k/2:k/3 = 6:3:2
Then the small part = 12100 *2/11 = 1100*2=2200
14. A cloth merchant claims to sell cloth at cost price. However his meter, scale is 96 cm long. His gain is –
Ans- c) 4 1/6%
Solution :-
Gain% = (Correct measurement – Defective measurement)*100 / Defective measurement
= (100-96)*100/96 = 4*100/96 = 25/6 = 4 1/6
15. I a certain language , ROAD is written as ORDA, how will DATE be written in the same code ?
Ans- b) ADET
16. The area of a circle is 100pi square cm. The side of the largest square inscribed in the circle is –
Ans- b) 10v2
Solution :
Let the radius of the circle = r
Pi r*r= 100 pi
Or, r=10
Then Diameter = 2r= 20= Diagonal of the Biggest Square
Or, v2 * side of the square = Diagonal
Or, v2 * side of the square = 20
Or, side of the square = 20/ v2 = 10v2
17. In a regular polygon every interior angle is 8 times of its exterior angle. The number of sides of the regular polygon is-
Ans- a) 18
Solution :
Interior angle of a regular polygon is = (n-2)*180/n
And Exterior angle = 360/ n
Accordingly, (n-2)*180/n=8 * 360/n
Or, n-2= 8*360/180=16
Or, n= 18
18. A wire 25 cm long bent into the form of a rectangle of area 25 sq. cm. The length of the longer side is –
Ans- C) 10 cm
Solution :
Solution of this question is based on trial and error methods.
If you take the longer side 25 then other side should be zero because total length is 25 cm, if you take longer side 20 then length other sides should be 5 cm but it is not possible, same way it is not possible if one side is 5 cm.
Total length > 25 Total length >25 Total length=25 area=25
Total length <25
19. a_c b_ _ b c _ c a b c b_
Ans- c) bcabc
Solution:- The pattern is -abcbc | abcbc | abcbc
20. A person is walking at 3 km/hr due northeast and his friend is walking at (data not given here) km/hr. due southeast. After two hours the distance between them will be-
Ans- No answer
Solution:-
His friend speed not given here. If it was given then solution will be –
Let take his friend speed x.
Then distance between them will be – sqrt(3^2 + x^2)
Let x=4, then distance between them will be = sqrt(3^2+4^2)= sqrt(9+16)=sqrt(25)=5
21. The number to be subtracted from each of the terms of the ratio 5:7 to make it 1:2 is –
Ans- C) 3
Solution :
5-x/7-x= ½ or, 10-2x=7-x or, x=3
22. The decimal 0.2576 is equivalent to –
Ans- c) 1287/4995
Solution: 1278/4995 = 0.2576576….
23. Starting fro place at a speed of x km/hr. a motor cyclist reaches a destination an returns to the same place at a speed of y km/hr. His average speed in the whole journey (in km/hr) is –
Ans- C) 2xy/(x+y)
Solution: Let the distance between A and B is r km.
Time taken to reach A to B = r/x and from B to A = r/y
Total distance traveled = r+r=2r
Total time taken = r/x+r/y = r(y+x)/xy.
Average speed = Total distance covered / total time taken = 2r / r(x+y)/xy = 2xy/(x+y)
24. The sides of a triangle are in the ratio 3:2:5. If perimeter is 75 cm, the length of the smallest side is –
Ans- D) 15 cm
Solution - Let, Sides are 3x, 2x and 5x cm. Perimeter = 3x+2x+5x= 10x cm
Accordingly, 10x=75 or, x= 75/10=7.5. Length of the smallest side = 7.5*2=15 cm
25. From a container having pure milk 20% is replaced by water and the process is repeated thrice. At the end of the third operation purity of the milk is –
Ans- c) 51.2%
Solution : Let total milk was 100 liters.
At the end of three successive process : 100(1-20/100)(1-20/100)(1-20/100) = 100 * 4/5 *4/5*4/5 = 4*4*4*4/5 = 256/5=51.2.
Now purity of milk is 51.2 % in the mixture.
26. From the top of a 60 meter high tower, the angle of depression of the top and the bottom of a building are observed to be 30 degree and 60 degree respectively. The height of the building is –
Ans- C) 40
Solution :-
Tan30= (60-h)/x or, x= (60-h)/tan30
Tan60= 60/x or, x= 60/tan60
Now, 60/tan60 = (60-h)/tan30
Or, 60/v3= (60-h)/1/v3
Or, 60/v3= v3(60-h) Or, 60= 3(60-h)
Or, 60/3= 60-h or, h= 60-20=40 meters
27. 3,16,42,81,__
Ans- b) 133
Solution :
16-3=13 =13*1
42-16=26 =13*2
81-42=39=13*3
x-81=13*4 or, x= 52+81=133
28. A cylindrical vessel of diameter 12 cm is filled with eater. A spherical ball is dropped in it and the water level raised by 1 cm. The diameter of the ball is –
Ans- a) 6 cm
Solution:
Volume of the 1 cm height water in the vessel = volume of the ball
Let radius of the vessel= R = 12/2=6 cm
Radius of the Ball=r cm
Pi * R*R*h= 4/3 pi*r*r*r
Or, 6*6*1= 4/3 * r^3
Or, r^3=36*3/4=27
Or, r= 3
Diameter= 2r=2*3=6 cm
29. A started at 8:30 am from a place P to go a place Q. 30 minutes later B left P for Q and caught up with A at 11 am. B reached Q at 12 noon. A will reach Q at –
Ans- B) 12.15 pm
Solution-
Let Speed of A is A km/h and Speed of B is B km/h.
Distance between P and Q is x km.
After 2 hr 30 minutes(for A) A and B (2 hr for B) meat each other.
A’s speed * 2 ½ = B’s speed * 2
A*5/2=2B
Or, A=4B/5
X= B* (12 noon- 9.00 am)= 3 B
A takes reach P to Q = x/A = 3 B / A = 3B/4B/5 = 15/4 = 3 hr 45 minutes
A reach to Q at = 8.30 A.M + 3 hr 45minutes= 12.15 PM
30. In triangle ABC , the point equidistant from AB and AC I the –
Ans- C) In center
Solution:- Center of the in-circle is called in-center.
------ End of GROUP- A- General Mental Ability -------------
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