Q1. What is the maximum volume of a right circular cone which can be cut from a solid cuboid of dimensions 12*10*15? (Ï€ is PI) a.100...
Q1. What is the maximum volume of a right circular cone which can be cut from a solid cuboid of dimensions 12*10*15? (Ï€ is PI)
a.100Ï€
b.120Ï€
c.125Ï€
d.225Ï€
Following three scenarios are possible:
Case 1:
In this case the cone is constructed using side wall as the base of the cone.
It is clear from the picture that the diameter of the circle is equal to the breadth of the cuboid  10 and so radius is 5 and the height of cone is equal to the length of the cuboid – 12.
So in this case the volume of the cone will be: Ï€ * 52 * 12/3 = 100Ï€
Case 2:
In this case the cone is constructed using floor/ceiling as the base of the cone.
It is clear from the picture that the diameter of the circle is equal to the breadth of the cuboid  10 and so radius is 5 and the height of cone is equal to the height of the cuboid – 15.
So in this case the volume of the cone will be: Ï€ * 52 * 15/3 = 125Ï€
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Case 3:
In this case the cone is constructed front wall as the base of the cone.
It is clear from the picture that the diameter of the circle is equal to the length of the cuboid  12 and so radius is 12 and the height of cone is equal to the breadth of the cuboid – 10.
So in this case the volume of the cone will be: Ï€ * 62 * 10/3 = 120Ï€
So from the above three cases it is clear that maximum volume can be 125Ï€.
Q2. How many solutions exists for √(x+5) = x√(x+5)
a.1
b.2
c.3
d.4
Solution:
Given, √(x+5) = x√(x+5)
Squaring both sides we get:
(x + 5) = x^{2}(x + 5)
=> x^{2}(x + 5)  (x + 5) = 0
=> (x^{2 } 1)(x + 5) = 0
=> x = 1, 1, 5
Note: Verify the solution before answering.
When x = 1:
√(x+5) = x√(x+5)
=> √(1+5) = 1 * √(1+5)
=> √6 = √6 (which is true)
When x = 5:
√(x+5) = x√(x+5)
=> √(5+5) = 5 * √(5+5)
=> √0 = 5 * √0
=> 0 = 0 (which is true)
When x = 1:
√(x+5) = x√(x+5)
=> √(1+5) = 1 * √(1+5)
=> √4 = √4
=> 2 = 2 (which is not true)
So for the given equation we can have two values of x.
Q3. When we perform a ‘digit slide’ on a number we move its unit digit’s to the front of the number. For example, the result of a ‘digit slide’ on 6471 is 1647. Let ‘z’ be the smallest positive integer with 5 as its unit’s digit such that the result of a ‘digit slide’ on number equals 4 times the number. How many digits will ‘z’ have?
a. 7
b. 6
c. 4
d. 3
a. Let the result of a ‘digit slide’ on z be x.
Given that unit digit of z is 5.
So z will be of the form:
z = *********5, number of stars in z do not indicate the number of digits in z
*

*

*

*

*

*

*

*

*

5

Also, given that x is 4 times of z so the unit digit of x will be 0
Hence x will be of the form:
x = *********0, number of stars in x do not indicate the number of digits in x
*

*

*

*

*

*

*

*

*

0

Since x is a ‘digit slide’ of z, the unit place digit of z is the leading digit of x.
Hence x will be of the form:
x = 5********0
5

*

*

*

*

*

*

*

*

0

Notice that x is a ‘digit slide’ of z, so the unit place digit of x will be the ten place digit of z.
So now z is of the form:
z = **********05
*

*

*

*

*

*

*

*

0

5

One important point: Since x is 4 times of z and the leading digit in x is 5. This implies that the leading digit in z must have been 1. This is because, after multiplying z by 4 we are getting just 5 at the leading position which has been a result of multiplication by 4 and adding the carry 1 from the previous one.
So now z is of the form:
1

*

*

*

*

*

*

*

0

5

Again, x is 4 times that of z and 4 times of z produces a result whose ten place digit is 2, so x will have 2 as ten place digit.
x = 5*******20
So now x is of the form:
5

1

*

*

*

*

*

*

2

0

Ten place digit of x will be hundred place digit of z
So now z is of the form:
z = 1********205
1

*

*

*

*

*

*

2

0

5

4 times of z now gives hundred place digit as 8.
So now x is of the form:
5

1

*

*

*

*

*

8

2

0

Hundred place digit of x will be thousand place digit of z
So now z is of the form:
1

*

*

*

*

*

8

2

0

5

4 times of z now gives thousand place digit as 2.
So now x is of the form:
5

1

*

*

*

*

2

8

2

0

Thousand place digit of x will be ten thousand’s place digit.
So now z is of the form:
1

*

*

*

*

2

8

2

0

5

At this stage z is 128205 and x is 512820.
Here x is 4 times of z.
If we proceed further this way, we might get another such number but our aim was to find first such number and that number is 128205 which is a 6 digit number.
Q4. Q4. O and O’ are centers of the two circles with radii 7cm and 9cm respectively. The distance between the centers is 20cm. If PQ is the transverse common tangent to the circles which cuts OO’ at X, what is the length of O’X in cm?
a. 35/4
b. 45/4
c. 10
d. 11
The given scenario is depicted in the following figure:
As clear from the figure itself, triangle OQX and triangle O'PX are similar.
So,
OQ/O'P = OX/O'X
=> OX = (7/9)*O'X
And since, OX + O'X = 20
=> (7/9)*O'X + O'X = 20
=> O'X = 45/4
Q5. A cow was standing on a bridge, 5m away from the middle of the bridge. A train was coming towards the bridge from the end nearest to the cow. Seeing this, the cow ran towards the train and managed to escape when the train was 2m away from the bridge. If it had run in the opposite direction (i.e. in the opposite direction), it would have been hit by the train 2m before the end of the bridge. What is the length of the bridge in meter assuming speed of train is 4 times the speed of the cow?
a. 32
b. 36
c. 40
d. cannot be determined
The initial scenario is depicted in the following diagram:
Where,
dc is the distance of the cow from the end of the bridge
dt is the distance of the train from the end of the bridge
When the cow manages to escape, the scenario is as follows:
The time in which the cow runs dc distance with speed v, the train travels a distance dt with speed 4v
So,
dc/v = (dt2)/4v
Ã° 4*dc = dt – 2
Ã° dt = 4*dc + 2
if the cow was hit while running in the opposite direction, the scenario would have been:
Half right of the bridge is (5 + dc) and so left half will also be the same.
So distance covered by the cow is: 5 + (dc + 5) – 2 = dc + 8
And, the distance covered by train in the same time is: dt + dc + 5 + dc + 5 – 2 = dt + 2*dc + 8
Speed of cow is v and speed of train is 4v
So,
(dc+8)/v = (dt+2*dc+8)/4*v
Ã° 4*dc + 32 = dt + 2*dc + 8
Ã° 2*dc – dt + 24 = 0 (put the value of dt in the form of dc from the above got equation)
Ã° 2*dc – 4*dc – 2 + 24 = 0
Ã° dc = 11
So the length of the bridge is: 2*(dc + 5) = 2*(11+5) = 32
For questions 6 to 8
Mark a if question can be answered using one of the statement alone but cannot be answered using the other statement.
Mark b if question can be answered using either of the statement alone
Mark c if question can be answered using both the statement together but cannot be answered using either of the statement alone.
Mark d if question cannot be answered even by using both the statement together
Q6. Find the value of 2x + 3y
(1) 2x + 15y = 24
(2) 6x + 9y = 18
(a) (b) (c) (d)
From statement (2):
3(2x + 3y) = 18
=> 2x + 3y = 6
So the value of 2x + 3y can be calculated just with the help of statement 2.
If you use statement (1) and (2) together, you can solve for x and y and then find the value of 2x + 3y which will again come the same.
But statement (2) is sufficient to find the value of 2x + 3y
Q7. Given that (a+b)^{2} = 1 and (ab)^{2} = 25, find the value of a and b
(1) 'a' is a prime number
(2) a > 2 and 'a' is an integer.
(a+b)^{2} = 1 => (a + b) = 1, 1
and, (ab)^{2} = 25 => (a  b) = 5, 5
Following cases are possible:
case 1:
(a + b) = 1 and (a  b) = 5 => a = 3, b = 2
case 2:
(a + b) = 1 and (a  b) = 5 => a = 2, b = 3
case 3:
(a + b) = 1 and (a  b) = 5 => a = 2, b = 3
case 4:
(a + b) = 1 and (a  b) = 5 => a = 3, b = 2
In all the 4 cases above 'a' is prime. So just by using statement (1) we cannot answer the question.
In case 1 and case 4, a > 2. So just by using statement (2) we cannot answer the question.
Even if we combine the statement (1) and (2) together the answer can be either the result of case 1 or case 4.
So we are not able to answer one value for 'a' and 'b', by using either of the statement or using both the statement together.
Hence, (d) will be the answer.
Q8. A piece of wood is cut into three pieces A, B and C. Are they of equal length?
(1) The sum of length of A and B is 2/3 of the original length.
(2) B and C are of the same length.
Suppose the length of wood is l unit.
Let's solve it by using just statement (1):
A + B = (2/3)*l => C = l/3
The sum of the length of A and B are 2/3 of the original length but from this we cannot say that A and B will also be of the same length l/3 each
So just by using statement (1) we cannot solve it.
Let's solve it by using just statement (2):
B and C are of the same length.
For an example B and C can be 0.4*l each and hence A will be 0.2*l.
So A, B and C may not be same.
So just by using statement (2) we cannot solve it.
Let's solve it by using statement (1) and statement (2) together:
A + B = (2/3)*l and B = C
Since, A + B = (2/3)*l => C = (1/3)*l
and, B = C => B = (1/3)*l
so, A = (1/3)*l
So A, B and C are of same length.
So by using statement (1) and statement(2) together we can solve it.
Hence (c) is the correct answer.
Q9. What is the value of 1/1 + 1/3 + 1/6 + 1/10 + ...
a. 2
b. 3/2
c. 5/3
d. 3
1/1 + 1/3 + 1/6 + 1/10
= 2/1*2 + 2/2*3 + 2/3*4 + 2/4*5 + ....
= 2(11/2) + 2(1/2  1/3) + 2(1/3  1/4) + 2(1/4  1/5) + ....
= 2[1  1/2 + 1/2  1/3 + 1/3  1/4 + 1/4  1/5 + ....]
= 2(1)
= 2
Q10. A shop stores x kg of rice. The first customer buys half this amount plus half a kg of rice. The second customer buys half the remaining amount plus half a kg of rice. Then the third customer also buys half the remaining amount plus half a kg of rice. Thereafter no rice is left in the shop. Which of the following best describes the value of x?
a. 2<=x<=6
b. 5
c. 9<=x<=12
d. 11<=x<=14
Initial quantity of the rice = x
The first customer buys half the total rice and another half kg.
Rice purchased by first customer = x/2 + 1/2 = (x + 1)/2
So, rice left after the first customer = x  (x + 1)/2 = (x  1)/2
Now the second customer buys half the left rice and another half kg.
Rice purchased by second customer = (x  1)/4 + 1/2 = (x + 1)/4
So, rice left after the second customer = (x  1)/2  (x + 1)/4 = (x  3)/4
Now the third customer buys half the left rice and another half kg.
Rice purchased by third customer = (x  3)/8 + 1/2 = (x + 1)/8
So, rice left after the second customer = (x  3)/4  (x + 1)/8 = (x  7)/8
Given that (x  7)/8 = 0 => x = 7
So option (b) is correct.
Q11. Let f(x) = ax^{2 }+ bx + c, where a,b and c are certain constants and a#0. It is known that f(5) = 3f(2) and that 3 is a root of f(x) = 0. What is the other root of f(x) = 0
a. 7
b. 4
c. 2
d. 6
3 is a root of f(x) = 0
=> 9a + 3b + c = 0 (1)
and, f(5) = 3f(2)
=> 25a + 5b + c = 12a  6b 3c
=> 37a + 11b + 4c = 0 (2)
Solving equation (1) and (2) we get:
a = b
=> b/a = 1
Sum of roots is: (b/a)
=> 3 + n = 1 (where n is the other root of f(x) = 0)
=> n = 4
So other root is 4
Q12. Find the sum: √(1 + 1/1^{2} + 1/2^{2}) + √(1 + 1/2^{2} + 1/3^{2}) + ..... + √(1 + 1/2007^{2} + 1/2008^{2})
a. 2008  1/2008
b. 2007  1/2007
c. 2007  1/2008
d. 2008  1/2007
Take the first term:
√(1 + 1/1^{2} + 1/2^{2})
= √(1 + 1 + 1/4)
= √(9/4) = 3/2 =
2  1/2
Take the first two terms:
√(1 + 1/1^{2} + 1/2^{2}) + √(1 + 1/2^{2} + 1/3^{2})
= 3/2 + √(1 + 1/4 + 1/9)
= 3/2 + √(49/36)
= 3/2 + 7/6
= 8/3
3  1/3
Similarly:
√(1 + 1/1^{2} + 1/2^{2}) + √(1 + 1/2^{2} + 1/3^{2}) + ..... + √(1 + 1/2007^{2} + 1/2008^{2})
= 2008  1/2008
So option (a) is correct.
Q13. Two circles both of radii 1cm intersect each other such that the circumference of each one passes through the center of the circle of the other. What is the area(in sq cm) of the intersecting region?
a. Ï€/3  √3/4
b. 2Ï€/3  √3/2
c. 4Ï€/3  √3/2
d. 4Ï€/3 + √3/2
The given scenario is depicted in the following diagram.
We have to find the area of the region OPO'Q.
Triangle OPO' is an equilateral triangle since OP = PO' = OO' = 1cm.
We will do the following to accomplish our task:
Find the area of triangle OPO'. Let's say it A1.
Find the area of smaller segment of circle created by chord OP. Let's say it A2.
Required area is: 2*A1 + 4*A2
Let's find A1
Area of equilateral triangle is: (√3/4)*(side)^{2 }
So, A1 = √3/4
Let's find A2
Area of smaller segment of circle created by chord OP = Area of sector O'OP  Area of triangle O'OP
Area of sector O'OP will be (1/6)th of the area of the circle as angle OO'P is 60^{o}
Area of equilateral triangle is (√3/4)*(side)^{2}
So A2 = (1/6)*Ï€*1^{2 } (√3/4)*1^{2} = Ï€/6  √3/4
So the total required area is: 2*A1 + 4*A2 = 2*(√3/4) + 4*(Ï€/6  √3/4) = (2Ï€/3  √3/2)
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