Question 1] If a certain number when divided by the numbers 5,3,2 one
after other successively, leaves a remainder 0,2,1 Then what will be
the remainders if it is divided by the numbers 2,3,5 ?
A. 1 , 0 , 4
B. 1 , 0 , 6
C. 1 , 3, 4
D. 2 , 3 , 4
Explanation :
Let whatever be the number, it can be formulated as :-
Suppose X is the Quotient after having divided the number by 2 then
The original number = 5(3(2X + 1) + 2)) // this is formulated using remainder theorem.
Hence original number = 30X + 25.
This when divided by 2 will leave 1 as remainder, because 30X is divisible by 2 but 25 will leave remainder 1.
After being divided by 2 the number gets modified as 15X + 12.
Now, similarly divide this by 3, remainder will be 0 as both 15X and 12 are divisible by 3. then the number is modified as: 5X+4, which on dividing with 5 remains 4.
Therefore, 1,0,4 is the answer!
Question 2] How many two digit numbers are there such that the product of their digits after reducing it to the smallest form is a prime number? for example if we take 98 then 9*8=72, 72=7*2=14, 14=1*4=4. Consider only 4 prime no.s (2,3,5,7).
A. 16
B. 18
C. 17
D. 14
Explanation :
2 = 12 or 21 So 1×2, 2×1, 3×4, 4×3, 2×6, 6×2, 3×7, 7×3
3 = 13 or 31 So 1×3, 3×1
5 = 15, 51 So 1×5, 5×1, 3×5, 5×3, 7×5, 5×7
7 = 17 or 71 So 1×7, 7×1
15 = 3×5 = 5×3
So total 18 numbers = 12,13,15,17,21,26,31,34,35,37,43,51,53,57,62,71,73,75
so Ans is >>18.
Question 3] What is the remainder when 30^72^87 is divided by 11 ?
A. 5
B. 4
C. 1
D. 6
Explanation :
Rem [30^72^87 / 11] = Rem [(-3)^72^87 / 11] = Rem [3^72^87 / 11]
Now, we need to observe the pattern:-
3^1 when divided by 11, leaves a remainder of 3
3^2 when divided by 11, leaves a remainder of 9
3^3 when divided by 11, leaves a remainder of 5
3^4 when divided by 11, leaves a remainder of 4
3^5 when divided by 11, leaves a remainder of 1
And then the same cycle of 3, 9, 5, 4 and 1 will continue.
If a number is of the format of 3^(5k + 1), it will leave a remainder of 3
If a number is of the format of 3^(5k + 2), it will leave a remainder of 9
If a number is of the format of 3^(5k + 3), it will leave a remainder of 5
If a number is of the format of 3^(5k + 4), it will leave a remainder of 4
If a number is of the format of 3^(5k), it will leave a remainder of 1
The number given to us is 3^72^87
Let us find out Rem[Power / Cyclicity] t0 find out if it 3^(5k + what?)
Rem [72^87 / 5]
= Rem [2^87 / 5]
= Rem [2*4^43/5]
= Rem [2*(-1) / 5]
= -2
= 3
=> The number is of the format 3^(5k + 3).
=> Rem [3^72^87 / 11] = 5.
Question 4] If ABC is a three digit number such that no one number is similar to other than how many possible values of ( a + 4b + c ) will be divisible by 40 ?
A. 16
B. 15
C. 18
D. 14
Explanation :
a b c
4 9 0 -> 1 because if 0 comes in first it is not a three digit number
1 9 3 -> 2
1 8 7 -> 2
2 8 6 -> 2
3 8 5 -> 2
3 7 9 ->2
4 7 8 -> 2
7 6 9 -> 2
Hence, total such 15 values are possible.
Question 5] A right angled cone has its height 'h'. The cone is being cut parallel to its base at h/5 distance from its vertex. What will be the ratio of the cone and the cutting portion ?
A. 1:124
B. 1:125
C. 1:24
D. 1:25
Explanation :
volume of smaller cone / (volume of bigger cone - volume of smaller cone)
also by similarity r/R = h/H.
=> r/R=1/5.
Substituting and solving we get 1/124 ans.
Question 6 ] A square ABCD is inscribed into the circle. A triangle is formed inside the square with one side CD and other two sides meet at the mid point of AB. Find the ratio of Area of circle to the area of Triangle.
A. pi
B. pi/2
C. 2pi
D. pi/4
Explanation :
Let X be the side of square.
Diagonal will be X*2^1/2
Radius=Diagonal/2
Area of circle=pi*X^2/2
Area of the triangle = 1/2*X*X
So, the ratio is PI.
Question 7] There are two concentric circles, between them a running track is there, Milkha Singh can run 24m straight starting from one point on the outer circle to the other point on the outer circle touching only once the inner circle. What is the amount of synthetic material required to build the track.
A. 144pi
B. 162pi
C. 154pi
D. 192 pi
Explanation :
Draw a straight line from a point on outer circle touching inner circle & passing through other end of outer circle. it will be a tangent of length (24 cm) to smaller circle.
Draw a perpendicular to this tangent, it will divide tangent in two equal parts.
If 'r' be radius of inner circle & 'R' of outer circle then
=> R2 = (122 + r2).
Amount of synthetic material required = Pi(R2 - r2) = Pi*[(122 + r2) - r2] = 144pi.
Question 8] Consider the sequence of numbers a, a1, a2, a3 ....to infinity where a1 = 81.33 and a2 = -19. And , aj =aj-1 - a-2 for j ≥ 3. What is the sum of the first 6002 terms of this sequence ?
A. -100.33
B. -30.00
C. 62.33
D. 119.33
Explanation :
Given a1 = 81.33; a2 = –19.
Also, aj =aj-1 - a-2 for j ≥ 3.
=> a3 = a2 – a1 = –100.33.
=> a4 = a3 – a2 = –81.33.
=> a5 = a4 – a3 = 19
=> a6 = a5 – a4 = +100.33.
=> a7 = a6 – a5 = +81.33.
=> a8 = a7 – a6 = –19.
Clearly onwards there is a cycle of 6 and the sum of terms in every such cycle = 0. Therefore, when we add a1, a2, a3 ... upto a6002, we will eventually be left with a1 + a2 only i.e. 81.33 – 19 = 62. 33.
Question 9 ] Let n! = 1 x 2 x 3 ........n for integer n ≥ 1. if p =1! + (2 x 2!) + + (3 x 3!) + .......+ (10 x 10!), then p + 2 when divided by 11! leaves a remainder of
A. 10
B. 0
C. 7
D. 1
Explanation :
P = 1 + 2.2! + 3.3!+ ….10.10!
= (2 –1)1! + ( 3 – 1)2! + (4 – 1)3! + ….(11 – 1)10!
= 2! – 1! + 3! – 2! + ….. 11! –10!
= 1 + 11!
Hence the remainder is 1.
Question 10 ] There are ten pair of socks in a cupboard from which 4 individual socks are picked at random. The probability that there is at least one pair is
A. 195/323
B. 99/323
C. 198/323
D. 185/323
Explanation :
We will solve this question in a moment. First let us understand one concept which is bit confusing.
Suppose there are 4 objects and I ask you to pick up 2 from them. In how many ways can you do that?
The answer is simple: 4C2 = 6
Can I do like this:
Out of 4 given objects first I will pick one object(4C1). Then again I will pick one more object in the next draw from the remaining 3 objects(3C1).
This way I have selected two objects from fours objects and number of ways to do this is: 4C1* 3C1 = 12
Do you see in both the way I am selecting two objects from four objects but in first way of selection the result is 6 while in the other way it is 12. Why so ??
This is because in the second way when we are selecting two objects one by one, their order of selection is being counted in the result. Didn't understand ?
Suppose we mark all 4 objects from 1 to 4.
One way is: In the first draw we select object marked 1 and in the second draw we pick object marked 2.
The other way is: the first selected object is marked 2 and second object is marked 1.
We are selecting the same two objects but the number of way is 2 because their order of selection (1,2) and (2,1) made it happen so.
But our ultimate goal was to pick up two object and the order does not matter to us.
Similarly if you select 3 objects from these 4 objects you can have following orders for objects marked 1,2,3: (1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2), (3,2,1) = 6 ways= 3! ways. While it should be just 1 way and not 6 ways.
So we can conclude that we will have to divide the result by the factorial of the number of objects we are selecting.
Like in the first example when we are selecting 2 objects we will have to divide the result by 2! and in the second example when we are selecting 3 objects we will have to divide the result by 3!.
Now coming back to our given problem:
At least on pair of socks must be selected means either one pair or two pair.
So required probability will be: 1 - (no pair of socks are selected)
The way we can select no pair of socks is:
Select one sock from 20 sock in the first pick. In the second draw exclude the pair of the first selected sock and pick from the remaining 18 socks and so on.
= 20C1 * 18C1 * 16C1 * 14C1/4! (4! because of the above explained concept).
So required probability: 1 - ((20C1 * 18C1 * 16C1 * 14C1/4!)/20C4) = 99/323.
Question 11] In a cosmetics company , there are 5 engineers out of 20 officers. If 3 officers are picked at random, what is the probability that at least one is engineer.
A. 1/114
B. 91 / 228
C. 137 / 228
D. None of these
Explanation :
No. of engineer = 5.
No. of total non-engineer = 15.
Total no. of officers = 20 of which 3 officers are picked.
In order to find the probability of at leats one engineer, it will be easier to find the probability of the complementary event (i.e none is an engineer)
P(none is an engineer) =
=> 15C3 / 20C3.
=> 455 / 1140
=> 91 / 228.
P(at least one engineer) =
=> 1 - ( 91 / 228 ).
=> 137 / 228.
Hence, the required probability is 137 / 228.
Question 12] How much water must be added to 60 liters of milk at 1.5 liters for Rs. 20 So as to have a mixture worth Rs.32/3 a liter?
A. 10 liters
B. 12 liters
C. 15 liters
D. 14 liters
Explanation :
C.P. of 1 litre of milk = Rs. 20×2/3 = Rs. 40/3.
C.P. of 1 litre of water =Rs 0.
Mean price = Rs. 32/3.
By the rule of alligation, we have :-
C.P of 1 litre C.P of 1 litre
of water of milk
(0) (Rs. 40/3)
\ /
Mean Price
(Rs. 32/3)
/ \
40/3−32/3 32/(3−0)
8/3 32/3
The ratio of water and milk =8/3:32/3.
=>8:32=1:4.
Thus, Quantity of water to be added to 60 liters of milk:
=>(1/4)×60 litres.
=>15 litres.
Question 13] The price of a Maruti car rises by 30% while the sales of the car come down by 20%. What is the percentage change in the total revenue?
A. -4%
B. -2%
C. +4%
D. +2%
Explanation :
Let initial price of Maruti car be Rs. 100.
As price increases 30%, the price of car become 100 + 30% of 100 = Rs. 130.
Due to increase in price, sales is down by 20 %.
It means, it is going to make 20% less revenue as expected after increment of price.
So, new revenue = 130 - 20 % of 130 = Rs 104.
The initial revenue was Rs 100 which becomess Rs 104 at the end.
It means there is 4% increment in the total revenue.
Question 14] Mr.P,Mr.Q and Mr.R takes a project and they can complete in 36 hours , 54 hours and 72 hours respectively. Unfortunately Mr.P met an accident and left from the project 8 hours before the completion while Mr.Q left 12 hours before the completion. Then for how many hours did Mr.r worked?
A. 36
B. 30
C. 24
D. 28
Explanation :
Assume that the project has to be completed in X hours.
So R has worked for X hours.
Since P has left 8 hours before completion, he worked for X - 8 hours and Q had left 12 hours before, his participation in work is for X - 12 hours.
Therefore (X - 8) + (X - 12) + X = 1(completion of project).
P can complete a project in 36 hours.
In (X - 8) hours P can complete = (X - 8) / 36.
Similarly, Q can complete a project in 54 hours.
In (X - 12) hours P can complete = (X - 12) / 54.
In X hours R can complete = X / 72.
=> (X - 8) / 36 + (X - 12) / 54 + X / 72 = 1.
=> 6 (X - 8) + 4(X - 12) + 3X = 216.
=> 13X = 312.
=> X = 24.
Question 15] An escalator is descending at constant speed. A walks down and takes 50 steps to reach the bottom. B runs down and takes 90 steps in the same time as A takes 10 steps. How many steps are visible when the escalator is not operating.
A. 150
B. 100
C. 120
D. 200
Explanation :
Lets suppose that A walks down 1 step / min and escalator moves n steps/ min
It is given that A takes 50 steps to reach the bottom
In the same time escalator would have covered 50n steps
So total steps on escalator is 50+50n.
Again it is given that B takes 90 steps to reach the bottom and time
taken by him for this is equal to time taken by A to cover 10 steps i.e 10 minutes.
So in this 10 min escalator would have covered 10n steps.
So total steps on escalator is 90 + 10n
Again equating 50 + 50n = 90 +10n we get n = 1
Hence total no. of steps on escalator is 100.
Question 16] Ram and Shakil run a race of 2000m. First Ram gives Shakil a start of 200m and beats him by 1 minute.Next Ram gives shakil a start of 6 min and is beaten by 1000 meters . Find the time in minutes in which Ram and Shakil can run the race seperately.
A. 12 , 18
B. 8 , 10
C. 18 , 20
D. 20 , 25
Explanation :
Let speed of ram and shakil be x and y resp.
Case 1 : Ram gives a start of 200m and beats Shakil by 1 minute.
2000/x - 1800/y= -60.
2000/x= 1800/y -60 -----------------(1)
Case2: Ram gives Shakil a start of 6 min and is beaten by 1000 meters
distance travelled by shakil in 6 minutes= (6*60*y) meters = 360y meters
1000/x = (2000- 360y)/y -------------------(2)
Solve equations (1) and (2) to obtain x and y
Answer is x = 8 minutes, y = 10 minutes.
Question 17] If f(x) = log { (1+x) / (1-x) } , f(x) + f(y) is
A. f(x+y)
B. f { (x+ y) / (1 + xy) }
C. (x+ y) f { 1 /(1 + xy) }
D. ( f(x) + f(y) ) / ( 1 + xy )
Explanation :
f(x) + f(y)
=> log (1+x)/( 1-x) + log (1+ y) / (1– y).
=> log [ (1+x)(1+ y) / ( 1-x)(1– y) ] .
=> log [ ( 1+x+y+xy) / (1+xy -x-y) ] .
=> log [ ( 1+x+y+xy) / (1+xy -(x+y)) ] .
=> log [ (1+ ((x+y)/(1 + xy))) / (1- ((x+y)/(1 + xy))) ] .
=> f[(x+ y) /(1 + xy)] .
Question 18] Which of the following values of x do not satisfy the inequality x2 - 3x + 2 > 0 at all ?
A. 1 ≤ x ≤2
B. -1 ≥ x ≥ -2
C. 0 ≤ x ≤ 2
D. 0 ≥ x ≥ -2
Explanation :
On solving x2 - 3x + 2 > 0 , we get ( x - 1 )( x - 2 ) > 0.
That is the roots are x = 1 or x = 2.
The equation assumes a positive value for all values which belong to the interval less than 1 and greater than 2
For x = 1 and x = 2, it becomes zero and for any value of x between 1 and 2 it becomes negative.
Question 19] There are 10 white, 8 red and 6 green balls on a billiard board. Find the number of ways in which one or more balls can be put in the pocket.
A. 672
B. 692
C. 693
D. None of these
Explanation :
Here, number of balls to be put is not given, but the minimum number i.e 1 and the maximum number of balls (10 + 8 + 6 ) i.e 24 is known.
At least one ball is to be included in the required selection.
Hence, the required number of ways is :-
=> Selection without any restriction - none.
=> (10 + 1) x (8 + 1) x (6 + 1) - 1. [ Since,there is only one way when none is selected ]
=> 11 x 9 x 7 - 1.
=> 692.
Question 20] There are 5 copies of university physics, 8 copies of university chemistry, 3 different books on law and 2 different books on history in a library. Find the number of ways in which one or more than one book can be selected.
A. 1727
B. 1728
C. 1208
D. None of these
Explanation :
Here, the books on physics are identical, also the books on chemistry are identical, but the books on law and history are all different.
Hence, the required number of ways of selection is
=> (S1 + 1) (S2 + 1)2n - 1.
where, S1 =5 (identical physics books),
S2 = 8 (identical chemistry books)
and, n= 3 + 2 ( 3 different law book and 2 different history books).
=> (5+1)(8 + 1) 23+2 - 1.
=> 6 x 9 x 25 - 1.
=> 54 x 32 - 1.
=> 1727.
A. 1 , 0 , 4
B. 1 , 0 , 6
C. 1 , 3, 4
D. 2 , 3 , 4
Explanation :
Let whatever be the number, it can be formulated as :-
Suppose X is the Quotient after having divided the number by 2 then
The original number = 5(3(2X + 1) + 2)) // this is formulated using remainder theorem.
Hence original number = 30X + 25.
This when divided by 2 will leave 1 as remainder, because 30X is divisible by 2 but 25 will leave remainder 1.
After being divided by 2 the number gets modified as 15X + 12.
Now, similarly divide this by 3, remainder will be 0 as both 15X and 12 are divisible by 3. then the number is modified as: 5X+4, which on dividing with 5 remains 4.
Therefore, 1,0,4 is the answer!
Question 2] How many two digit numbers are there such that the product of their digits after reducing it to the smallest form is a prime number? for example if we take 98 then 9*8=72, 72=7*2=14, 14=1*4=4. Consider only 4 prime no.s (2,3,5,7).
A. 16
B. 18
C. 17
D. 14
Explanation :
2 = 12 or 21 So 1×2, 2×1, 3×4, 4×3, 2×6, 6×2, 3×7, 7×3
3 = 13 or 31 So 1×3, 3×1
5 = 15, 51 So 1×5, 5×1, 3×5, 5×3, 7×5, 5×7
7 = 17 or 71 So 1×7, 7×1
15 = 3×5 = 5×3
So total 18 numbers = 12,13,15,17,21,26,31,34,35,37,43,51,53,57,62,71,73,75
so Ans is >>18.
Question 3] What is the remainder when 30^72^87 is divided by 11 ?
A. 5
B. 4
C. 1
D. 6
Explanation :
Rem [30^72^87 / 11] = Rem [(-3)^72^87 / 11] = Rem [3^72^87 / 11]
Now, we need to observe the pattern:-
3^1 when divided by 11, leaves a remainder of 3
3^2 when divided by 11, leaves a remainder of 9
3^3 when divided by 11, leaves a remainder of 5
3^4 when divided by 11, leaves a remainder of 4
3^5 when divided by 11, leaves a remainder of 1
And then the same cycle of 3, 9, 5, 4 and 1 will continue.
If a number is of the format of 3^(5k + 1), it will leave a remainder of 3
If a number is of the format of 3^(5k + 2), it will leave a remainder of 9
If a number is of the format of 3^(5k + 3), it will leave a remainder of 5
If a number is of the format of 3^(5k + 4), it will leave a remainder of 4
If a number is of the format of 3^(5k), it will leave a remainder of 1
The number given to us is 3^72^87
Let us find out Rem[Power / Cyclicity] t0 find out if it 3^(5k + what?)
Rem [72^87 / 5]
= Rem [2^87 / 5]
= Rem [2*4^43/5]
= Rem [2*(-1) / 5]
= -2
= 3
=> The number is of the format 3^(5k + 3).
=> Rem [3^72^87 / 11] = 5.
Question 4] If ABC is a three digit number such that no one number is similar to other than how many possible values of ( a + 4b + c ) will be divisible by 40 ?
A. 16
B. 15
C. 18
D. 14
Explanation :
a b c
4 9 0 -> 1 because if 0 comes in first it is not a three digit number
1 9 3 -> 2
1 8 7 -> 2
2 8 6 -> 2
3 8 5 -> 2
3 7 9 ->2
4 7 8 -> 2
7 6 9 -> 2
Hence, total such 15 values are possible.
Question 5] A right angled cone has its height 'h'. The cone is being cut parallel to its base at h/5 distance from its vertex. What will be the ratio of the cone and the cutting portion ?
A. 1:124
B. 1:125
C. 1:24
D. 1:25
Explanation :
volume of smaller cone / (volume of bigger cone - volume of smaller cone)
also by similarity r/R = h/H.
=> r/R=1/5.
Substituting and solving we get 1/124 ans.
Question 6 ] A square ABCD is inscribed into the circle. A triangle is formed inside the square with one side CD and other two sides meet at the mid point of AB. Find the ratio of Area of circle to the area of Triangle.
A. pi
B. pi/2
C. 2pi
D. pi/4
Explanation :
Let X be the side of square.
Diagonal will be X*2^1/2
Radius=Diagonal/2
Area of circle=pi*X^2/2
Area of the triangle = 1/2*X*X
So, the ratio is PI.
Question 7] There are two concentric circles, between them a running track is there, Milkha Singh can run 24m straight starting from one point on the outer circle to the other point on the outer circle touching only once the inner circle. What is the amount of synthetic material required to build the track.
A. 144pi
B. 162pi
C. 154pi
D. 192 pi
Explanation :
Draw a straight line from a point on outer circle touching inner circle & passing through other end of outer circle. it will be a tangent of length (24 cm) to smaller circle.
Draw a perpendicular to this tangent, it will divide tangent in two equal parts.
If 'r' be radius of inner circle & 'R' of outer circle then
=> R2 = (122 + r2).
Amount of synthetic material required = Pi(R2 - r2) = Pi*[(122 + r2) - r2] = 144pi.
Question 8] Consider the sequence of numbers a, a1, a2, a3 ....to infinity where a1 = 81.33 and a2 = -19. And , aj =aj-1 - a-2 for j ≥ 3. What is the sum of the first 6002 terms of this sequence ?
A. -100.33
B. -30.00
C. 62.33
D. 119.33
Explanation :
Given a1 = 81.33; a2 = –19.
Also, aj =aj-1 - a-2 for j ≥ 3.
=> a3 = a2 – a1 = –100.33.
=> a4 = a3 – a2 = –81.33.
=> a5 = a4 – a3 = 19
=> a6 = a5 – a4 = +100.33.
=> a7 = a6 – a5 = +81.33.
=> a8 = a7 – a6 = –19.
Clearly onwards there is a cycle of 6 and the sum of terms in every such cycle = 0. Therefore, when we add a1, a2, a3 ... upto a6002, we will eventually be left with a1 + a2 only i.e. 81.33 – 19 = 62. 33.
Question 9 ] Let n! = 1 x 2 x 3 ........n for integer n ≥ 1. if p =1! + (2 x 2!) + + (3 x 3!) + .......+ (10 x 10!), then p + 2 when divided by 11! leaves a remainder of
A. 10
B. 0
C. 7
D. 1
Explanation :
P = 1 + 2.2! + 3.3!+ ….10.10!
= (2 –1)1! + ( 3 – 1)2! + (4 – 1)3! + ….(11 – 1)10!
= 2! – 1! + 3! – 2! + ….. 11! –10!
= 1 + 11!
Hence the remainder is 1.
Question 10 ] There are ten pair of socks in a cupboard from which 4 individual socks are picked at random. The probability that there is at least one pair is
A. 195/323
B. 99/323
C. 198/323
D. 185/323
Explanation :
We will solve this question in a moment. First let us understand one concept which is bit confusing.
Suppose there are 4 objects and I ask you to pick up 2 from them. In how many ways can you do that?
The answer is simple: 4C2 = 6
Can I do like this:
Out of 4 given objects first I will pick one object(4C1). Then again I will pick one more object in the next draw from the remaining 3 objects(3C1).
This way I have selected two objects from fours objects and number of ways to do this is: 4C1* 3C1 = 12
Do you see in both the way I am selecting two objects from four objects but in first way of selection the result is 6 while in the other way it is 12. Why so ??
This is because in the second way when we are selecting two objects one by one, their order of selection is being counted in the result. Didn't understand ?
Suppose we mark all 4 objects from 1 to 4.
One way is: In the first draw we select object marked 1 and in the second draw we pick object marked 2.
The other way is: the first selected object is marked 2 and second object is marked 1.
We are selecting the same two objects but the number of way is 2 because their order of selection (1,2) and (2,1) made it happen so.
But our ultimate goal was to pick up two object and the order does not matter to us.
Similarly if you select 3 objects from these 4 objects you can have following orders for objects marked 1,2,3: (1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2), (3,2,1) = 6 ways= 3! ways. While it should be just 1 way and not 6 ways.
So we can conclude that we will have to divide the result by the factorial of the number of objects we are selecting.
Like in the first example when we are selecting 2 objects we will have to divide the result by 2! and in the second example when we are selecting 3 objects we will have to divide the result by 3!.
Now coming back to our given problem:
At least on pair of socks must be selected means either one pair or two pair.
So required probability will be: 1 - (no pair of socks are selected)
The way we can select no pair of socks is:
Select one sock from 20 sock in the first pick. In the second draw exclude the pair of the first selected sock and pick from the remaining 18 socks and so on.
= 20C1 * 18C1 * 16C1 * 14C1/4! (4! because of the above explained concept).
So required probability: 1 - ((20C1 * 18C1 * 16C1 * 14C1/4!)/20C4) = 99/323.
Question 11] In a cosmetics company , there are 5 engineers out of 20 officers. If 3 officers are picked at random, what is the probability that at least one is engineer.
A. 1/114
B. 91 / 228
C. 137 / 228
D. None of these
Explanation :
No. of engineer = 5.
No. of total non-engineer = 15.
Total no. of officers = 20 of which 3 officers are picked.
In order to find the probability of at leats one engineer, it will be easier to find the probability of the complementary event (i.e none is an engineer)
P(none is an engineer) =
=> 15C3 / 20C3.
=> 455 / 1140
=> 91 / 228.
P(at least one engineer) =
=> 1 - ( 91 / 228 ).
=> 137 / 228.
Hence, the required probability is 137 / 228.
Question 12] How much water must be added to 60 liters of milk at 1.5 liters for Rs. 20 So as to have a mixture worth Rs.32/3 a liter?
A. 10 liters
B. 12 liters
C. 15 liters
D. 14 liters
Explanation :
C.P. of 1 litre of milk = Rs. 20×2/3 = Rs. 40/3.
C.P. of 1 litre of water =Rs 0.
Mean price = Rs. 32/3.
By the rule of alligation, we have :-
C.P of 1 litre C.P of 1 litre
of water of milk
(0) (Rs. 40/3)
\ /
Mean Price
(Rs. 32/3)
/ \
40/3−32/3 32/(3−0)
8/3 32/3
The ratio of water and milk =8/3:32/3.
=>8:32=1:4.
Thus, Quantity of water to be added to 60 liters of milk:
=>(1/4)×60 litres.
=>15 litres.
Question 13] The price of a Maruti car rises by 30% while the sales of the car come down by 20%. What is the percentage change in the total revenue?
A. -4%
B. -2%
C. +4%
D. +2%
Explanation :
Let initial price of Maruti car be Rs. 100.
As price increases 30%, the price of car become 100 + 30% of 100 = Rs. 130.
Due to increase in price, sales is down by 20 %.
It means, it is going to make 20% less revenue as expected after increment of price.
So, new revenue = 130 - 20 % of 130 = Rs 104.
The initial revenue was Rs 100 which becomess Rs 104 at the end.
It means there is 4% increment in the total revenue.
Question 14] Mr.P,Mr.Q and Mr.R takes a project and they can complete in 36 hours , 54 hours and 72 hours respectively. Unfortunately Mr.P met an accident and left from the project 8 hours before the completion while Mr.Q left 12 hours before the completion. Then for how many hours did Mr.r worked?
A. 36
B. 30
C. 24
D. 28
Explanation :
Assume that the project has to be completed in X hours.
So R has worked for X hours.
Since P has left 8 hours before completion, he worked for X - 8 hours and Q had left 12 hours before, his participation in work is for X - 12 hours.
Therefore (X - 8) + (X - 12) + X = 1(completion of project).
P can complete a project in 36 hours.
In (X - 8) hours P can complete = (X - 8) / 36.
Similarly, Q can complete a project in 54 hours.
In (X - 12) hours P can complete = (X - 12) / 54.
In X hours R can complete = X / 72.
=> (X - 8) / 36 + (X - 12) / 54 + X / 72 = 1.
=> 6 (X - 8) + 4(X - 12) + 3X = 216.
=> 13X = 312.
=> X = 24.
Question 15] An escalator is descending at constant speed. A walks down and takes 50 steps to reach the bottom. B runs down and takes 90 steps in the same time as A takes 10 steps. How many steps are visible when the escalator is not operating.
A. 150
B. 100
C. 120
D. 200
Explanation :
Lets suppose that A walks down 1 step / min and escalator moves n steps/ min
It is given that A takes 50 steps to reach the bottom
In the same time escalator would have covered 50n steps
So total steps on escalator is 50+50n.
Again it is given that B takes 90 steps to reach the bottom and time
taken by him for this is equal to time taken by A to cover 10 steps i.e 10 minutes.
So in this 10 min escalator would have covered 10n steps.
So total steps on escalator is 90 + 10n
Again equating 50 + 50n = 90 +10n we get n = 1
Hence total no. of steps on escalator is 100.
Question 16] Ram and Shakil run a race of 2000m. First Ram gives Shakil a start of 200m and beats him by 1 minute.Next Ram gives shakil a start of 6 min and is beaten by 1000 meters . Find the time in minutes in which Ram and Shakil can run the race seperately.
A. 12 , 18
B. 8 , 10
C. 18 , 20
D. 20 , 25
Explanation :
Let speed of ram and shakil be x and y resp.
Case 1 : Ram gives a start of 200m and beats Shakil by 1 minute.
2000/x - 1800/y= -60.
2000/x= 1800/y -60 -----------------(1)
Case2: Ram gives Shakil a start of 6 min and is beaten by 1000 meters
distance travelled by shakil in 6 minutes= (6*60*y) meters = 360y meters
1000/x = (2000- 360y)/y -------------------(2)
Solve equations (1) and (2) to obtain x and y
Answer is x = 8 minutes, y = 10 minutes.
Question 17] If f(x) = log { (1+x) / (1-x) } , f(x) + f(y) is
A. f(x+y)
B. f { (x+ y) / (1 + xy) }
C. (x+ y) f { 1 /(1 + xy) }
D. ( f(x) + f(y) ) / ( 1 + xy )
Explanation :
f(x) + f(y)
=> log (1+x)/( 1-x) + log (1+ y) / (1– y).
=> log [ (1+x)(1+ y) / ( 1-x)(1– y) ] .
=> log [ ( 1+x+y+xy) / (1+xy -x-y) ] .
=> log [ ( 1+x+y+xy) / (1+xy -(x+y)) ] .
=> log [ (1+ ((x+y)/(1 + xy))) / (1- ((x+y)/(1 + xy))) ] .
=> f[(x+ y) /(1 + xy)] .
Question 18] Which of the following values of x do not satisfy the inequality x2 - 3x + 2 > 0 at all ?
A. 1 ≤ x ≤2
B. -1 ≥ x ≥ -2
C. 0 ≤ x ≤ 2
D. 0 ≥ x ≥ -2
Explanation :
On solving x2 - 3x + 2 > 0 , we get ( x - 1 )( x - 2 ) > 0.
That is the roots are x = 1 or x = 2.
The equation assumes a positive value for all values which belong to the interval less than 1 and greater than 2
For x = 1 and x = 2, it becomes zero and for any value of x between 1 and 2 it becomes negative.
Question 19] There are 10 white, 8 red and 6 green balls on a billiard board. Find the number of ways in which one or more balls can be put in the pocket.
A. 672
B. 692
C. 693
D. None of these
Explanation :
Here, number of balls to be put is not given, but the minimum number i.e 1 and the maximum number of balls (10 + 8 + 6 ) i.e 24 is known.
At least one ball is to be included in the required selection.
Hence, the required number of ways is :-
=> Selection without any restriction - none.
=> (10 + 1) x (8 + 1) x (6 + 1) - 1. [ Since,there is only one way when none is selected ]
=> 11 x 9 x 7 - 1.
=> 692.
Question 20] There are 5 copies of university physics, 8 copies of university chemistry, 3 different books on law and 2 different books on history in a library. Find the number of ways in which one or more than one book can be selected.
A. 1727
B. 1728
C. 1208
D. None of these
Explanation :
Here, the books on physics are identical, also the books on chemistry are identical, but the books on law and history are all different.
Hence, the required number of ways of selection is
=> (S1 + 1) (S2 + 1)2n - 1.
where, S1 =5 (identical physics books),
S2 = 8 (identical chemistry books)
and, n= 3 + 2 ( 3 different law book and 2 different history books).
=> (5+1)(8 + 1) 23+2 - 1.
=> 6 x 9 x 25 - 1.
=> 54 x 32 - 1.
=> 1727.
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