Question 1] A number 10,10,101 when multiplied with N gives product
P, such that N is any no with only 1 as its digit (eg. N =
11,111,1111,.......) .How many different digit can be possible in the
product P.
A. 2
B. 3
C. 4
D. None of these
Explanation :
N*10*10*101=P
P=N*100*(100+1)
P=10000*N+100*N
Suppose if N=11 then P=111100.
if N=111 then P=1121100
if N=1111 then P=11221100
and N=11111 then P=112221100
an so P only contains digits such as 0,1,2 with repetition.
So answer is 3 (option b).
Question 2] If dP1, dP2, ........, dP24 are the difference between 24 prime numbers in series. 89 is the 24th Prime numbers Find the sum of dP1 + .... + dP24?
A. 95
B. 97
C. 85
D. 89
Explanation :
25 prime no.s between 1 & 100 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.
Difference between two primes i.e
dp1 =p2-p1= 3-2 =1
dp2 =p3-p2= 5-3 =2
...
dp24 = p25 - p24 = 97-89 = 8.
i.e 1, 2, 2, 4, 2, 4, 2, 4, 6, 2, 6, 4, 2, 4, 6, 6, 2, 6, 4, 2, 6, 4, 6, 8
Thus, Sum = dP1 + .... + dP24 = 95.
Question 3] A hollow cube of size 5cm is taken, with the thickness of 1cm. It is made of smaller cubes of size 1cm .If the outer surface of the cube is painted how many faces of the smaller cubes remain unpainted?
A. 450
B. 438
C. 550
D. 500
Explanation :
Volume of Big Cube considering it is not hollow = L3 = 5*5*5 = 125 cm3
Size of hollow cube (considering 1 cm thickness on two faces of large cube = 5 - 2 = 3cm
Volume of hollow cube = 3*3*3 = 27 cm3
So Total Volume filled up by smaller cubes = Volume of Larger Cube - Volume of hollow cube
= 125 - 27
= 98 cm3
Volume of 1 small cube = 1*1*1 = 1 cm3
Total number of small cubes in the larger cube = 98 / 1 = 98.
and Number of faces of 98 small cubes (6 faces each cube has) = 98*6 = 588 faces.
Total Surface area of 6 faces of larger cube painted = 6*L2 = 6*5*5 = 150cm2
Surface area of one face of small cube = 1*1 = 1cm2
Number of faces of small cube painted = 150/1 = 150 faces.
=> Hence number of faces of the smaller cubes remain unpainted = 588-150 => 438.
Hence (B) is the correct answer.
Question 4] Three pieces of cakes of weights 9/2lb, 27/4 lb and 36/5 lb respectively are to be divided into parts of equal weight. Further, each part must be as heavy as possible. If one such part is served to each guest, then what is the maximum number of guests that could be entertained?
A. 54
B. 72
C. 20
D. None of these
Explanation :
=> HCF( 9/2, 27/4, 36/5) = HCF(9,27,36) / LCM(2,4,5)
=> 9/20 lb i.e the weight of each piece.
Total weight is 18.45 lb.
Maximum number of guests ( 18.45 x 20 ) / 9 = 41.
Question 5] The internal bisector of an angle A in a triangle ABC meets the side BC at point D. AB = 4, AC =3 and ∠A = 600. Then what is the length of the bisector AD ?
A. 12 √3 /7
B. 12 √13 /7
C. 4√13 /7
D. 4√3 /7
Explanation :
Area of triangle ABC = Area of ADB + Area ADC
=> (1/2) * AB * AC * sinA = (1/2) * AB * AD * Sin(A/2) + (1/2) * AC * AD * Sin(A/2).
=> 12sin60 = 4ADsin30 + 3ADsin30.
=> 12√3/2= (7/2)*AD.
Hence, AD = 12√3/7.
Question 6] There is a common chord of 2 circles with radius 15 and 20. The distance between the two centers is 25. The length of the chord is
A. 48
B. 24
C. 36
D. 28
Explanation :
In the triangle OAO' we can see that angle A is 90 (because O'A is a tangent to circle with center O and the angle at point of tangency is 90).
So, in triangle OAO' we have :
1/2 * base * height = 1/2 * product of sides containing right angle = area
1/2 * 25 * h = 1/2 * 15 * 20
From here we get h = 12 = AC
So, length of common chord = 2AC = AC + BC = 24.
Question 7 ] Each side of a given polygon is parallel to either the X or the Y axis. A corner of such a polygon is said to be convex if the internal angle is 90° or concave if the internal angle is 270°. If the number of convex corners in such a polygon is 25, the number of concave corners must be
A. 20
B. 0
C. 21
D. 22
Explanation :
=> 90(25) + 270(n-25) = (n-2)*180.
=> 25 + 3(n-25) = (n-2)*2.
=> 3n - 50 = 2n-4
=> n = 46.
Hence, the required number of corners must be 46 - 25 = 21.
Question 8] The infinite sum 1 + (4/7) + (9/72) + (16/73) + (25/74) + ..... equals
A. 27/14
B. 21/13
C. 49/27
D. 256/147
Explanation :
Let S = 1 + (4/7) + (9/72) + (16/73) + (25/74) ...................(i)
=> (1/7)S = (1/7) + (4/72) + (9/73) + (16/74) . ....................(ii)
On subtracting (ii) from (i) gives,
=> S(1- 1/7) = 1+ (3/7) + (5/72) + (7/73) + (9 / 74) . ......................(iii)
=> (1/7) x S(1- 1/7) = (1/7) + (3/72) + (5/73) + (7/74) ......................(iv)
On subtracting (iv) from (iii) gives,
=> S(1- 1/7) -(1/7) x S(1- 1/7) = 1 + (2/7) + (2/72) + (2/73) + (2/74) + ......
=> S(1- 1/7) (1- 1/7) = 1+ (2/7) [ 1 + (1/7) + (1/72) + (1/73) + (1/74) + ...... ∞ ]
=> S(1- 1/7)2 = 1 + (2/7) x ( 1 / ( 1 - (1/7))).
=> S(6/7)2 = 1 + (2/7 x 7/6) .
=> S x (36/49) = 1 + (1/3).
=> S = (49/36) x (4/3).
=> S = 49/27.
Question 9] 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 10] 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 11] It takes six technicians a total of 10 hr to build a new server from Direct Computer, with each working at the same rate. If six technicians start to build the server at 11 am, and one technician per hour is added beginning at 5 pm, at what time will the server be completed?
A. 6.40 pm
B. 7 pm
C. 7.20 pm
D. 8 pm
Explanation :
Total amount of work = 6 men x 10 hr = 60 man-hour.
From 11am to 5 pm, 6 technicians = 36 man-hours.
From 5 pm to 6 pm, 7 technicians = 7 man-hours.
From 6 pm to 7 pm, 8 technicians = 8 man-hours.
From 7pm to 8 pm, 9 technicians = 9 man-hours.
Total of 60 man-hour and server will be completed at 8 pm.
Question 12] Find the number of ways you can fill a 3 x 3 grid(with four corners defined as a,b,c,d) if u have 3 white marbles and 6 black marbles.
A. 80
B. 84
C. 76
D. 90
Explanation :
Number of grids=9.
Number of ways filling 3 white marbles in 9 places= 9C3.
Number of filling remaining 6 places in grid with 6 marbles=6C6.
Total ways of simultaneous filling= 9C3*6C6 implies 9C3=84.
Question 13] A car after traveling 18 km from a point A developed some problem in the engine and speed became 4/5 of its original speed As a result, the car reached point B 45 minutes late. If the engine had developed the same problem after traveling 30 km from A, then it would have reached B only 36 minutes late. The original speed of the car (in km per hour) and the distance between the points A and B (in km.) is
A. 25
B. 30
C. 20
D. none of these
Explanation :
He moves with (4/5)S where S is his usual speed that means 1/5 decrease in speed lead to 1/4 increase in time.
Now, the main difference is in 12 km(30-18) and change in difference in time = (45-36) min = 9 min.
Thus,
=> 1/4 * T = 9 min where T is the time required to cover a distance of (30-18) i.e 12 km.
T = 36 min = 36 / 60 hours = 0.6 hours.
Hence, Speed of the car = 12/0.6 = 20 kmph.
Question 14] A man travels from A to B at a speed x km/hr. He then rests at B for x hours. He then travels from B to C at a speed 2x km/hr and rests for 2x hours. He moves further to D at a speed twice as that between B and C. He thus reaches D in 16 hr. If distances A-B, B-C and C-D are all equal to 12 km, the time for which he rested at B could be
A. 3 hr
B. 6 hr
C. 2 hr
D. 4 hr
Explanation :
Total time taken by the man to travel from A to D = 16 hr and total distance traveled = 36 km.
The time that he would have taken had he not rested in between will be (16 - x - 2x) = (16 - 3x).
But, this time should be equal to the addition of the times that he takes to travel individual segments. This is given as:
=> 12/x + 12/2x + 12/4x = 21/x.
Therefore, 21/x = 16 - 3x.
=> x = 3 or 7/3.
Question 15] What is the sum of 'n' terms in the series logm + log(m2 / n) + log(m3 / n2) + log(m4 / n3) + ..........?
A. log [ n(n-1)/m(n+1) ](n/2)
B. log [ mm/ nn ](n/2)
C. log [m(1-n) / n(1-m) ](n/2)
D. log [m(n+1) / n(n-1) ](n/2)
Explanation :
Sum of log m + log(m2 / n) + log(m3 / n2) + log(m4 / n3) + ......... n terms such problem must be solved by taking the value of number of terms.
Let’s say 2 and check the given option. If we look at the sum of 2 terms of the given series it comes out to be log m + log(m2 / n) .
=> log(m x m2) / n = log(m3 /n ).
Now look at the option and put number of terms as 2, only option (4) validates the above mentioned answer.
As log [m(n+1) / n(n-1) ](n/2) = log [m3 / n]1 = log [m3 / n] .
Question 16] Instead of a meter scale, a cloth merchant uses a 120 cm scale while buying, but uses an 80 cm scale while selling the same cloth. If he offers a discount of 20% on cash payment, what is his overall profit percentage?
A. 20%
B. 25%
C. 40%
D. 15%
Explanation :
Assume that the merchant while purchasing, purchases 120cm at Rs120 , that is 1cm at Rs1.
He sells 80 cm at Rs 120.
While the original CP = Rs 80 for 80 cm.
20% discount on 120 = 24,
So, actual SP = 120-24 = 96
Profit = 96-80 = 16
P%= (P/CP) x 100 = (16/80) x 100 = 20%.
Question 17] A milkman mixes 20 liters of water with 80 liters of milk. After selling one-fourth of this mixture, he adds water to replenish the quantity that he had sold. What is the current proportion of water to milk?
A. 2:3
B. 1:2
C. 1:3
D. 3:4
Explanation :
As it is clear from the diagram, removal of 25 litres at stage I will result in volume of milk reduced by 80% of 25 lit i.e 20 lit and volume of water being reduced by the remaining 5 lit. so, M = 60 lit and W = 15 lit.
Addition of 25 lit water will finally given M= 60 lit and W = 40 lit.
Hence, the ratio of W and M = 40 : 60 = 2 : 3.
Question 18] If log10x - log10√x = 2logx10, then the possible value of x is given by
A. 10
B. 1/100
C. 1/1000
D. None of these
Explanation :
=> log10x - log10√x = 2logx10.
=> log10 [ x/√n ] = logx100.
=> log10 √x = log10100 / log10x.
=> (1/2)log10x = 2 / log10x .
=> (log10x)2 = 4.
=> log10x = ∓ 2.
∴ log10x = -2 or log10 x = 2.
=> 102 = x or 10-2 = x.
x = 100 or x = 1/100.
Question 19] How many 7 digit number are there having the digit 3 three times & the digit 5 four times?
A. 7!/(3!*5!)
B. 33*55
C. 77
D. 35
Explanation :
Total Way we can form the number of 7 digit is 7!
but condition is given 3 repeat in 3 times and 5 repeats four time means(3,3,3) and (5,5,5)
(3,3,3) can arrange in 3 ways so 3! .
(5,5,5,5) can arrange in 4 ways so 4!.
Hence 7!/(3!*4!) = 35.
Question 20] Raju and Reenu are standing in a row there are 9 peoples. What is the probability that at least 3 people will stand between Raju and Reenu ?
A. 9!
B. 5/12
C. 15*7!*2!
D. None of these
Explanation :
There are at least 3 people between Raju and Reena... people can be 4,5,6,7.
when there are 3 people between them... in 5*7!*2!
when there are 4 people between them... in 4*7!*2!
when there are 5 people between them... in 3*7!*2!
when there are 6 people between them... in 2*7!*2!
when there are 7 people between them... in 1*7!*2!
so total 15*7!*2!....
Total no. of possibilities... 9!
probability will be 15*7!*2!/9! = 5/12.
A. 2
B. 3
C. 4
D. None of these
Explanation :
N*10*10*101=P
P=N*100*(100+1)
P=10000*N+100*N
Suppose if N=11 then P=111100.
if N=111 then P=1121100
if N=1111 then P=11221100
and N=11111 then P=112221100
an so P only contains digits such as 0,1,2 with repetition.
So answer is 3 (option b).
Question 2] If dP1, dP2, ........, dP24 are the difference between 24 prime numbers in series. 89 is the 24th Prime numbers Find the sum of dP1 + .... + dP24?
A. 95
B. 97
C. 85
D. 89
Explanation :
25 prime no.s between 1 & 100 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.
Difference between two primes i.e
dp1 =p2-p1= 3-2 =1
dp2 =p3-p2= 5-3 =2
...
dp24 = p25 - p24 = 97-89 = 8.
i.e 1, 2, 2, 4, 2, 4, 2, 4, 6, 2, 6, 4, 2, 4, 6, 6, 2, 6, 4, 2, 6, 4, 6, 8
Thus, Sum = dP1 + .... + dP24 = 95.
Question 3] A hollow cube of size 5cm is taken, with the thickness of 1cm. It is made of smaller cubes of size 1cm .If the outer surface of the cube is painted how many faces of the smaller cubes remain unpainted?
A. 450
B. 438
C. 550
D. 500
Explanation :
Volume of Big Cube considering it is not hollow = L3 = 5*5*5 = 125 cm3
Size of hollow cube (considering 1 cm thickness on two faces of large cube = 5 - 2 = 3cm
Volume of hollow cube = 3*3*3 = 27 cm3
So Total Volume filled up by smaller cubes = Volume of Larger Cube - Volume of hollow cube
= 125 - 27
= 98 cm3
Volume of 1 small cube = 1*1*1 = 1 cm3
Total number of small cubes in the larger cube = 98 / 1 = 98.
and Number of faces of 98 small cubes (6 faces each cube has) = 98*6 = 588 faces.
Total Surface area of 6 faces of larger cube painted = 6*L2 = 6*5*5 = 150cm2
Surface area of one face of small cube = 1*1 = 1cm2
Number of faces of small cube painted = 150/1 = 150 faces.
=> Hence number of faces of the smaller cubes remain unpainted = 588-150 => 438.
Hence (B) is the correct answer.
Question 4] Three pieces of cakes of weights 9/2lb, 27/4 lb and 36/5 lb respectively are to be divided into parts of equal weight. Further, each part must be as heavy as possible. If one such part is served to each guest, then what is the maximum number of guests that could be entertained?
A. 54
B. 72
C. 20
D. None of these
Explanation :
=> HCF( 9/2, 27/4, 36/5) = HCF(9,27,36) / LCM(2,4,5)
=> 9/20 lb i.e the weight of each piece.
Total weight is 18.45 lb.
Maximum number of guests ( 18.45 x 20 ) / 9 = 41.
Question 5] The internal bisector of an angle A in a triangle ABC meets the side BC at point D. AB = 4, AC =3 and ∠A = 600. Then what is the length of the bisector AD ?
A. 12 √3 /7
B. 12 √13 /7
C. 4√13 /7
D. 4√3 /7
Explanation :
Area of triangle ABC = Area of ADB + Area ADC
=> (1/2) * AB * AC * sinA = (1/2) * AB * AD * Sin(A/2) + (1/2) * AC * AD * Sin(A/2).
=> 12sin60 = 4ADsin30 + 3ADsin30.
=> 12√3/2= (7/2)*AD.
Hence, AD = 12√3/7.
Question 6] There is a common chord of 2 circles with radius 15 and 20. The distance between the two centers is 25. The length of the chord is
A. 48
B. 24
C. 36
D. 28
Explanation :
In the triangle OAO' we can see that angle A is 90 (because O'A is a tangent to circle with center O and the angle at point of tangency is 90).
So, in triangle OAO' we have :
1/2 * base * height = 1/2 * product of sides containing right angle = area
1/2 * 25 * h = 1/2 * 15 * 20
From here we get h = 12 = AC
So, length of common chord = 2AC = AC + BC = 24.
Question 7 ] Each side of a given polygon is parallel to either the X or the Y axis. A corner of such a polygon is said to be convex if the internal angle is 90° or concave if the internal angle is 270°. If the number of convex corners in such a polygon is 25, the number of concave corners must be
A. 20
B. 0
C. 21
D. 22
Explanation :
=> 90(25) + 270(n-25) = (n-2)*180.
=> 25 + 3(n-25) = (n-2)*2.
=> 3n - 50 = 2n-4
=> n = 46.
Hence, the required number of corners must be 46 - 25 = 21.
Question 8] The infinite sum 1 + (4/7) + (9/72) + (16/73) + (25/74) + ..... equals
A. 27/14
B. 21/13
C. 49/27
D. 256/147
Explanation :
Let S = 1 + (4/7) + (9/72) + (16/73) + (25/74) ...................(i)
=> (1/7)S = (1/7) + (4/72) + (9/73) + (16/74) . ....................(ii)
On subtracting (ii) from (i) gives,
=> S(1- 1/7) = 1+ (3/7) + (5/72) + (7/73) + (9 / 74) . ......................(iii)
=> (1/7) x S(1- 1/7) = (1/7) + (3/72) + (5/73) + (7/74) ......................(iv)
On subtracting (iv) from (iii) gives,
=> S(1- 1/7) -(1/7) x S(1- 1/7) = 1 + (2/7) + (2/72) + (2/73) + (2/74) + ......
=> S(1- 1/7) (1- 1/7) = 1+ (2/7) [ 1 + (1/7) + (1/72) + (1/73) + (1/74) + ...... ∞ ]
=> S(1- 1/7)2 = 1 + (2/7) x ( 1 / ( 1 - (1/7))).
=> S(6/7)2 = 1 + (2/7 x 7/6) .
=> S x (36/49) = 1 + (1/3).
=> S = (49/36) x (4/3).
=> S = 49/27.
Question 9] 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 10] 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 11] It takes six technicians a total of 10 hr to build a new server from Direct Computer, with each working at the same rate. If six technicians start to build the server at 11 am, and one technician per hour is added beginning at 5 pm, at what time will the server be completed?
A. 6.40 pm
B. 7 pm
C. 7.20 pm
D. 8 pm
Explanation :
Total amount of work = 6 men x 10 hr = 60 man-hour.
From 11am to 5 pm, 6 technicians = 36 man-hours.
From 5 pm to 6 pm, 7 technicians = 7 man-hours.
From 6 pm to 7 pm, 8 technicians = 8 man-hours.
From 7pm to 8 pm, 9 technicians = 9 man-hours.
Total of 60 man-hour and server will be completed at 8 pm.
Question 12] Find the number of ways you can fill a 3 x 3 grid(with four corners defined as a,b,c,d) if u have 3 white marbles and 6 black marbles.
A. 80
B. 84
C. 76
D. 90
Explanation :
Number of grids=9.
Number of ways filling 3 white marbles in 9 places= 9C3.
Number of filling remaining 6 places in grid with 6 marbles=6C6.
Total ways of simultaneous filling= 9C3*6C6 implies 9C3=84.
Question 13] A car after traveling 18 km from a point A developed some problem in the engine and speed became 4/5 of its original speed As a result, the car reached point B 45 minutes late. If the engine had developed the same problem after traveling 30 km from A, then it would have reached B only 36 minutes late. The original speed of the car (in km per hour) and the distance between the points A and B (in km.) is
A. 25
B. 30
C. 20
D. none of these
Explanation :
He moves with (4/5)S where S is his usual speed that means 1/5 decrease in speed lead to 1/4 increase in time.
Now, the main difference is in 12 km(30-18) and change in difference in time = (45-36) min = 9 min.
Thus,
=> 1/4 * T = 9 min where T is the time required to cover a distance of (30-18) i.e 12 km.
T = 36 min = 36 / 60 hours = 0.6 hours.
Hence, Speed of the car = 12/0.6 = 20 kmph.
Question 14] A man travels from A to B at a speed x km/hr. He then rests at B for x hours. He then travels from B to C at a speed 2x km/hr and rests for 2x hours. He moves further to D at a speed twice as that between B and C. He thus reaches D in 16 hr. If distances A-B, B-C and C-D are all equal to 12 km, the time for which he rested at B could be
A. 3 hr
B. 6 hr
C. 2 hr
D. 4 hr
Explanation :
Total time taken by the man to travel from A to D = 16 hr and total distance traveled = 36 km.
The time that he would have taken had he not rested in between will be (16 - x - 2x) = (16 - 3x).
But, this time should be equal to the addition of the times that he takes to travel individual segments. This is given as:
=> 12/x + 12/2x + 12/4x = 21/x.
Therefore, 21/x = 16 - 3x.
=> x = 3 or 7/3.
Question 15] What is the sum of 'n' terms in the series logm + log(m2 / n) + log(m3 / n2) + log(m4 / n3) + ..........?
A. log [ n(n-1)/m(n+1) ](n/2)
B. log [ mm/ nn ](n/2)
C. log [m(1-n) / n(1-m) ](n/2)
D. log [m(n+1) / n(n-1) ](n/2)
Explanation :
Sum of log m + log(m2 / n) + log(m3 / n2) + log(m4 / n3) + ......... n terms such problem must be solved by taking the value of number of terms.
Let’s say 2 and check the given option. If we look at the sum of 2 terms of the given series it comes out to be log m + log(m2 / n) .
=> log(m x m2) / n = log(m3 /n ).
Now look at the option and put number of terms as 2, only option (4) validates the above mentioned answer.
As log [m(n+1) / n(n-1) ](n/2) = log [m3 / n]1 = log [m3 / n] .
Question 16] Instead of a meter scale, a cloth merchant uses a 120 cm scale while buying, but uses an 80 cm scale while selling the same cloth. If he offers a discount of 20% on cash payment, what is his overall profit percentage?
A. 20%
B. 25%
C. 40%
D. 15%
Explanation :
Assume that the merchant while purchasing, purchases 120cm at Rs120 , that is 1cm at Rs1.
He sells 80 cm at Rs 120.
While the original CP = Rs 80 for 80 cm.
20% discount on 120 = 24,
So, actual SP = 120-24 = 96
Profit = 96-80 = 16
P%= (P/CP) x 100 = (16/80) x 100 = 20%.
Question 17] A milkman mixes 20 liters of water with 80 liters of milk. After selling one-fourth of this mixture, he adds water to replenish the quantity that he had sold. What is the current proportion of water to milk?
A. 2:3
B. 1:2
C. 1:3
D. 3:4
Explanation :
As it is clear from the diagram, removal of 25 litres at stage I will result in volume of milk reduced by 80% of 25 lit i.e 20 lit and volume of water being reduced by the remaining 5 lit. so, M = 60 lit and W = 15 lit.
Addition of 25 lit water will finally given M= 60 lit and W = 40 lit.
Hence, the ratio of W and M = 40 : 60 = 2 : 3.
Question 18] If log10x - log10√x = 2logx10, then the possible value of x is given by
A. 10
B. 1/100
C. 1/1000
D. None of these
Explanation :
=> log10x - log10√x = 2logx10.
=> log10 [ x/√n ] = logx100.
=> log10 √x = log10100 / log10x.
=> (1/2)log10x = 2 / log10x .
=> (log10x)2 = 4.
=> log10x = ∓ 2.
∴ log10x = -2 or log10 x = 2.
=> 102 = x or 10-2 = x.
x = 100 or x = 1/100.
Question 19] How many 7 digit number are there having the digit 3 three times & the digit 5 four times?
A. 7!/(3!*5!)
B. 33*55
C. 77
D. 35
Explanation :
Total Way we can form the number of 7 digit is 7!
but condition is given 3 repeat in 3 times and 5 repeats four time means(3,3,3) and (5,5,5)
(3,3,3) can arrange in 3 ways so 3! .
(5,5,5,5) can arrange in 4 ways so 4!.
Hence 7!/(3!*4!) = 35.
Question 20] Raju and Reenu are standing in a row there are 9 peoples. What is the probability that at least 3 people will stand between Raju and Reenu ?
A. 9!
B. 5/12
C. 15*7!*2!
D. None of these
Explanation :
There are at least 3 people between Raju and Reena... people can be 4,5,6,7.
when there are 3 people between them... in 5*7!*2!
when there are 4 people between them... in 4*7!*2!
when there are 5 people between them... in 3*7!*2!
when there are 6 people between them... in 2*7!*2!
when there are 7 people between them... in 1*7!*2!
so total 15*7!*2!....
Total no. of possibilities... 9!
probability will be 15*7!*2!/9! = 5/12.
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