Question 1 ] Find the values of n for which n+18 and n+90 is a perfect square ?
A. 6
B. 5
C. 4
D. 3
Explanation :
Let, n+18 = a2 and n+90 = b2
=> n+18 = a2
=> n = (a2 - 18) ------------------ (1)
=> n = (b2 - 90) ------------------ (2)
from (1) and (2):
=> (a2 - 18) = (b2 - 90)
=> b2 - a2 = 72.
=> (b+a)(b-a) = 72.
=> (b+a)(b-a) = 36 x 2 OR 24 x 3 OR 18 x 4 OR 12 x 6 OR 9 x 8
Case 1:
(b+a)(b-a) = 36 x 2.
Possible value of a and b are: 17 and 19
In this case, n = (a2 - 18) => n = 271
Case 2:
(b+a)(b-a) = 24 x 3.
Possible value of a and b are: 10.5 and 13.5
In this case, n = (a2 - 18) => n = 92.25
Case 3:
(b+a)(b-a) = 18 x 4.
Possible value of a and b are: 7 and 11
In this case, n =(a2 - 18) => n = 31
Case 4:
(b+a)(b-a) = 12 x 6.
Possible value of a and b are: 3 and 9
In this case, n = (a2 - 18) => n = -9
Case 5:
(b+a)(b-a) = 9 x 8.
No real values of a and b satisfy this equation.
So, there are total 4 possible values of n.
Question 2] If | r- 6 | = 11 and |2q - 12| = 8, what is the minimum possible value of q / r ?
A. -2/5
B. -2
C. 10/17
D. None of these
Explanation :
=> | r- 6 | = 11.
=>r- 6 =11.
=>r = 17.
or, –(r – 6) = 11, r = –5.
Similarly,
=> 2q – 12= 8 .
=> 2q –12 = 8.
=> q = 10.
or, 2q – 12 = –8 , q =2.
Hence, minimum value of q/r is 10/-5 i.e -2.
Question 3 ] If a1 = 1 and an+1 = 2an + 5, n=1, 2....... then a100 is equal to ?
A. 5 x 2^99 - 6
B. 5 x 2^99 + 6
C. 6 x 2^99 + 5
D. 6 x 2^99 - 5
Explanation :
a1 = 1, a2 = 7, a3 = 19, a4 = 43.
The difference between successive terms is in series 6, 12, 24, 48, ..., i.e. they are in GP.
Hence,
=> a100 = a1 + a [ (r^n- 1)/ (r-1) ] .
= 1 + 6[(2^n- 1)/ (2-1) ] .
= 6 x 2^99 - 5.
Question 4 ] Let R1 and R2 respectively denote the maximum and minimum possible remainders when (276)n is divided by 91 for any natural number n,n>= 144. Find R1+R2.
A. 90
B. 82
C. 84
D. 64
Explanation :
The remainder when 276 divided by 91 is 3 hence,
remainder of (276)n/ 91 = remainder 3n/91
Taking n = 1,2,3,4,5.
We get the possible remainders as 3, 9, 27, 81, 61, 1 respectively.
For n>5, the pattern repeats.
Hence, R1 = 81 and R2 = 1.
and, R1 + R2 = 81 + 1 = 82.
Question 5] If 2x+y = 10, 2y+z = 20 and 2x+z = 30. Where x, y,and z are real number. What is the value of 2x ?
A. 3/2
B. sqrt(15)
C. sqrt(6)/2
D. 15
Explanation :
Let, 2x = a, 2y = b and 2z = c
Then from the question:
2x+y = 2x * 2y = ab = 10
Similarly, bc = 20 , ac = 30
=> ab/bc = 1/2
=> a/c = 1/2
and ac = 30
(a/c)*ac = (1/2)*30 = 15 .
=>a = sqrt(15).
Question 6] Two dices are thrown simultaneously. What is the probability that the sum of the two number is 10 or the product of two numbers is >= 25 or both?
A. 5/27
B. 4/27
C. 5/36
D. None of these
Explanation :
Let, A be the event of obtaining a sum of the two number is 10 then, the score on two dice can be in order of (4,6), (5,5) and (6,4).
Hence, the number of favourable outcomes are 3 and the total number of possible outcomes T = 62 = 36.
Hence, the probability is :-
=> P(A)
=> 3/36.
Let, B be the event of obtaining product of numbers >=25, then, the score on two dice can be in order of (5,5), (5,6), (6,5) and (6,6).
Hence, the number of favorable outcomes are 4 and the total number of possible outcomes T = 62 = 36.
Hence, the probability is :-
=> P(B)
=> 4 / 36.
Hence , the required probability is :-
P(A+B) => P(A) + P(B) - P(A) x P(B).
=> (3/36) + (4/36) - (3/36) x (4/36).
=> 5/27.
Question 7 ] In how many ways can 10 engineers and 4 doctors be seated at a round table if all the 4 doctors do not sit together?
A. 13! - (10! × 4!)
B. 13! × 4!
C. 14!
D. 10! × 4!
Explanation :
Initially let's find out the total number of ways in which 10 engineers and 4 doctors can be seated at a round table.
In this case, n = total number of persons = 10 + 4 = 14
Hence, the total number of ways in which 10 engineers and 4 doctors can be seated at a round table
= (14-1)! = 13! -------------------------------(A)
Now let's find out the total number of ways in which 10 engineers and 4 doctors can be seated at a round table where all the 4 doctors sit together.
Since all the 4 doctors sit together, group them together and consider as a single doctor.
Hence, n = total number of persons = 10 + 1 = 11.
These 11 persons can be seated at a round table in (11-1)! = 10! ways ----------------------(B)
However these 4 doctors can be arranged among themselves in 4! Ways --------------------(C)
From (B) and (C), the total number of ways in which 10 engineers and 4 doctors can be seated at a round table where all the 4 doctors sit together = 10! 4! -----------------------(D)
From (A) and (D),
The total number of ways in which 10 engineers and 4 doctors can be seated at a round table if all the 4 doctors do not sit together = 13! - (10! x 4!)
Question 8 ] What is the approx. value of W, if W=(1.5)11, Given log2 = 0.301, log3 = 0.477.
A. 85
B. 86
C. 84
D. 89
Explanation :
W = (1.5)^11
Take log both sides :-
=> logW = 11 x log(1.5) = 11 x log(3/2).
=> logW = 11 x (log3 - log2) = 11 x 0.176 = 1.936.
=> W = 101.936.
=> W = 86 (approx).
Question 9 ] Excluding stoppages, the speed of a bus is 54 kmph and including stoppages, it is 45 kmph. For how many minutes does the bus stop per hour?
A. 9
B. 10
C. 11
D. 12
Explanation :
Speed of the bus excluding stoppages = 54 kmph
Speed of the bus including stoppages = 45 kmph
Loss in speed when including stoppages = 54 - 45 = 9kmph.
=> In 1 hour, bus covers 9 km less due to stoppages.
Hence, time that the bus stop per hour = time taken to cover 9 km.
=> distance / speed= 9 / 54 hour = 1/6 hour = 60/6 min=10 min.
Question 10 ] A boat M leaves shore A and at the same time boat B leaves shore B. They move across the river. They met at 500 yards away from A and after that they met 300 yards away from shore B without halting at shores. Find the distance between the shore A & B.
A. 1100 yards
B. 1200 yards
C. 1300 yards
D. 1400 yards
Explanation :
The boats first met after 500 yards from A.
Let the total distance between A and B be x.
Then,
500/a = x - 500 / b, ------------(1)
where, a = speed of A,
b = speed of B.
Assuming both boats are coming back to their original start place without halting at the shores,
Distance coverred by A = x + 300.
Distance covered by B = 2x - 300.
x+300/a = 2x-300/b. ------------(2)
From above,
500/x-500 = x+(300/2x)-300.
Solve, for x = 1200.
Question 11] Harish and Peter working separately can paint a building in 18 days and 24 days respectively. If they work for a day alternately, Harish beginning, in how many days, the painting work will be completed ?
A. 41/2 days.
B. 21/2 days
C. 63/4 days
D. 65/4 days
Explanation :
Let A = Harish, B = Peter.
(A+B)'s 2 day's work =(1/18 + 1/24) = 7/72.
work done in 10 pairs of days = (10*7/72) = 70/72.
remaining work = (1 - 70/72) = 2/72 = 1/36.
on 21st day, it is A's turn. 1/18 work is done by him in 1 days.
1/36 work id done by him in (18 * 1/36) = 1/2 day.
Hence, total time taken = (20 + 1/2) days = 41/2 days.
Question 12] A tank has two compartments I and II. Two taps X and Y, whose filling rates are in ratio of 2:1, are used to fill the tank. The ration fo time taken by tap X to fill compartment I and tap Y to fill compartment II is 16:25. Find the ratio of the times taken by tap X to fill compartment II and tap Y to fill compartment I.
A. 25:64
B. 16:25
C. 64:25
D. 4:5
Explanation :
Let the time (in minutes) taken by tap X to fill compartment I be 16x and that taken by tap Y to fill compartment II be 25x. As the filling rates of X and Y are in the ratio of 2:1, the ratio of times they take to fill the same volume is 1:2.
So tap Y will take (16x)2 i.e. 32x minutes to fill compartment I and tap X will take 25x/2 i.e. 12.5x minutes to fill the compartment II.
Hence the ratio 12.5x:32x i.e. 25:64.
Question 13] At the end of year 1998, Shepard bought nine dozen goats. Henceforth, every year he added p% of the goats at the beginning of the year and sold q% of the goats at the end of the year where p > 0 and q > 0. If Shepard had nine dozen goats at the end of year 2002, after making the sales for that year, which of the following is true?
A. p = q
B. p < q
C. p > q
D. p = q/2
Explanation :
The number of goats remain the same. If the percentage that is added every time is equal to the percentage that is sold, then there should be net decrease.
The same will be the case if the percentage added is less than the percentage sold.
The only way the number of goats remain the same if p>q.
Question 14 ] A real number x satisfying 1- (1/n) ≤ 3 + (1/n), for every positive integer n, is best described by ≤
A. 1 < x < 4
B. 1 < x ≤ 3
C. 0 < x ≤4
D. 1 ≤ x ≤ 3
Explanation :
Given,
1- (1/n) ≤ 3 + (1/n)
Put n =1 , we get 0 < x ≤4.
Question 15 ] Given that -1 ≤ v ≤ 1, -2 ≤ u ≤ -0.5, and -2 ≤ z ≤ -0.5 and w = vz/u, then which of the following is necessarily true ?
A. -0.5 ≤ w ≤ -2
B. -4 ≤ w ≤ 4
C. -4 ≤ w ≤ 2
D. -2 ≤ w ≤ -0.5
Explanation :
"u" is always negative. Hence, for us to have a minimum value of vz/u , vz should be positive.
Also for the least value, the numerator has to be the maximum positive value and the denominator has to be the smallest negative value. In other words, vz has to be 2 and u has to be –0.5.
Hence the minimum value of vz/u = 2−0.5 = –4.
For us to get the maximum value, vz has to be the smallest negative value and u has to be the highest negative value.
Thus, vz has to be –2 and u has to be –0.5. Hence the maximum value of vz/u = 2 / -0.5 = 4.
Question 16 ] A can complete a piece of work in 4 days. B takes double the time taken by A, C takes double that of B, and D takes double that of C to complete the same task. They are paired in groups of two each. One pair takes two-thirds the time needed by the second pair to complete the work. Which is the first pair?
A. A and B
B. A and C
C. B and C
D. A and D
Explanation :
Work done in one day by A,B,C and D are (1/4), (1/8), (1/16) and (1/32) respectively .
Using options, we note that
the pair B and C does 3/16 of work in one day.,
the pair A and D does (1/4) + (1/32) = (9/32) of work in one day.
Hence, A and D take 32/9 days.
B and C take 16/3 days.
Hence, the first pair must comprise A and D.
Question 17] Two men X and Y started working for a certain company at similar jobs on January 1, 1950. X asked for an initial salary of Rs. 300 with an annual increment of Rs. 30. Y asked for an initial salary of Rs. 200 with a rise of Rs. 15 every 6 months. Assume that the arrangements remained unaltered till December 31, 1959. Salary is paid on the last day of the month. What is the total amount paid to them as salary during the period?
A. Rs. 93,300
B. Rs. 93,200
C. Rs. 93,100
D. None of these
Explanation :
Amount of money given to X
= 12 x 300 + 12 x 330 + ....... + 12 x 570.
= 12[ 300 + 330 + .........+ 540 + 570 ].
= 12 x (10/2) [ 600 + 9 x 30 ].
= 52200.
Amount of money given to Y is
= 6 x 200 + 6 x 215 + 6 x 230 + 6 x 245 + ... upto 20 terms
= 6 [ 200 + 215 + 230 + .....485 ]
= 6 x (20/2)[ 400 + 19 x 15]
= 6 x 10 [ 400 + 285 ] .
= 60 x 685 = 41110
Hence, the total amount paid is 52200 + 41110 = Rs 93,300.
Question 18 ] A rectangular sheet of paper, when halved by folding it at the midpoint of its longer side, results in a rectangle, whose longer and shorter sides are in the same proportion as the longer and shorter sides of the original rectangle. If the shorter side of the original rectangle is 2, what is the area of the smaller rectangle?
A. 4√2
B. 2√2
C. √2
D. None of the above
Explanation :
Original rectangle: Shorter side = 2 Longer side = x
Second rectangle: Longer side = 2 Shorter side = x/2 (As it is folded into half at midpoint)
Ratio of 1st rectangle = x/2.
2nd rectangle =( 2/x)/2 = 4/x.
Ratio is same. x/2 = 4/x.
x2 = 8, x = 2√2.
So, area of smaller rectangle = (x/2) * 2 => x = 2√2.
Question 19] A cow is tethered at point A by a rope. Neither the rope nor the cow is allowed to enter ∆ABC.
∠BAC = 30°
I ( AB ) = I ( AC ) = 10 m
What is the area that can be grazed by the cow if the length of the rope is 8 m?
A. 134 π/3 sq. m
B. 121π sq. m
C. 132π sq. m
D. 176 π/3 sq. m
Explanation :
It can be seen that if the length of the rope is 8 m, then the cow will be able to graze an area equal to (the area of the circle with radius 8m) – (Area of the sector of the same circle with angle 30° ). This can further be expressed as
=> π (8)2 − (30/360)π (8)2
=> 64π - (1/12)(64π) = 176π/3 sq.m.
Shortcut:
Area grazed without restriction is 64π m2. it should be less than 64 π sq. m. with restriction.
Question 20 ] In the figure below, AB = BC = CD = DE = EF = FG = GA. Then ∠DAE is approximately?
A. 15°
B. 20°
C. 30°
D. 25°
Explanation :
Let us assume, ∠DAE = x
Triangle ABC is isosceles as AB = BC => ∠BCA = ∠CAB = x
Hence, ∠CBD = ∠CAB + ∠BCA = x + x = 2x .............. [External angle of triangle ABC]
Triangle BCD is isosceles as BC = CD => ∠CBD = ∠CDB = 2x
Hence, ∠DCE = ∠DAE + ∠CDA = x + 2x = 3x .............. [External angle of triangle ACD]
Triangle CDE is isosceles as CD = DE => ∠DCE = ∠DEC = ∠AED = 3x
Similarly, ∠ADE = ∠EFD = ∠AEF + ∠DAE = ∠EGF + ∠DAE = (∠DAE + ∠GFA) + ∠DAE = ∠DAE + ∠DAE + ∠DAE = 3x
Hence, in triangle ADE, ∠ADE + ∠DAE + ∠AED = 3x + x + 3x = 7x
Hence, 7x = 180 => x = 180/7 = 25.7.. ≈ 25
A. 6
B. 5
C. 4
D. 3
Explanation :
Let, n+18 = a2 and n+90 = b2
=> n+18 = a2
=> n = (a2 - 18) ------------------ (1)
=> n = (b2 - 90) ------------------ (2)
from (1) and (2):
=> (a2 - 18) = (b2 - 90)
=> b2 - a2 = 72.
=> (b+a)(b-a) = 72.
=> (b+a)(b-a) = 36 x 2 OR 24 x 3 OR 18 x 4 OR 12 x 6 OR 9 x 8
Case 1:
(b+a)(b-a) = 36 x 2.
Possible value of a and b are: 17 and 19
In this case, n = (a2 - 18) => n = 271
Case 2:
(b+a)(b-a) = 24 x 3.
Possible value of a and b are: 10.5 and 13.5
In this case, n = (a2 - 18) => n = 92.25
Case 3:
(b+a)(b-a) = 18 x 4.
Possible value of a and b are: 7 and 11
In this case, n =(a2 - 18) => n = 31
Case 4:
(b+a)(b-a) = 12 x 6.
Possible value of a and b are: 3 and 9
In this case, n = (a2 - 18) => n = -9
Case 5:
(b+a)(b-a) = 9 x 8.
No real values of a and b satisfy this equation.
So, there are total 4 possible values of n.
Question 2] If | r- 6 | = 11 and |2q - 12| = 8, what is the minimum possible value of q / r ?
A. -2/5
B. -2
C. 10/17
D. None of these
Explanation :
=> | r- 6 | = 11.
=>r- 6 =11.
=>r = 17.
or, –(r – 6) = 11, r = –5.
Similarly,
=> 2q – 12= 8 .
=> 2q –12 = 8.
=> q = 10.
or, 2q – 12 = –8 , q =2.
Hence, minimum value of q/r is 10/-5 i.e -2.
Question 3 ] If a1 = 1 and an+1 = 2an + 5, n=1, 2....... then a100 is equal to ?
A. 5 x 2^99 - 6
B. 5 x 2^99 + 6
C. 6 x 2^99 + 5
D. 6 x 2^99 - 5
Explanation :
a1 = 1, a2 = 7, a3 = 19, a4 = 43.
The difference between successive terms is in series 6, 12, 24, 48, ..., i.e. they are in GP.
Hence,
=> a100 = a1 + a [ (r^n- 1)/ (r-1) ] .
= 1 + 6[(2^n- 1)/ (2-1) ] .
= 6 x 2^99 - 5.
Question 4 ] Let R1 and R2 respectively denote the maximum and minimum possible remainders when (276)n is divided by 91 for any natural number n,n>= 144. Find R1+R2.
A. 90
B. 82
C. 84
D. 64
Explanation :
The remainder when 276 divided by 91 is 3 hence,
remainder of (276)n/ 91 = remainder 3n/91
Taking n = 1,2,3,4,5.
We get the possible remainders as 3, 9, 27, 81, 61, 1 respectively.
For n>5, the pattern repeats.
Hence, R1 = 81 and R2 = 1.
and, R1 + R2 = 81 + 1 = 82.
Question 5] If 2x+y = 10, 2y+z = 20 and 2x+z = 30. Where x, y,and z are real number. What is the value of 2x ?
A. 3/2
B. sqrt(15)
C. sqrt(6)/2
D. 15
Explanation :
Let, 2x = a, 2y = b and 2z = c
Then from the question:
2x+y = 2x * 2y = ab = 10
Similarly, bc = 20 , ac = 30
=> ab/bc = 1/2
=> a/c = 1/2
and ac = 30
(a/c)*ac = (1/2)*30 = 15 .
=>a = sqrt(15).
Question 6] Two dices are thrown simultaneously. What is the probability that the sum of the two number is 10 or the product of two numbers is >= 25 or both?
A. 5/27
B. 4/27
C. 5/36
D. None of these
Explanation :
Let, A be the event of obtaining a sum of the two number is 10 then, the score on two dice can be in order of (4,6), (5,5) and (6,4).
Hence, the number of favourable outcomes are 3 and the total number of possible outcomes T = 62 = 36.
Hence, the probability is :-
=> P(A)
=> 3/36.
Let, B be the event of obtaining product of numbers >=25, then, the score on two dice can be in order of (5,5), (5,6), (6,5) and (6,6).
Hence, the number of favorable outcomes are 4 and the total number of possible outcomes T = 62 = 36.
Hence, the probability is :-
=> P(B)
=> 4 / 36.
Hence , the required probability is :-
P(A+B) => P(A) + P(B) - P(A) x P(B).
=> (3/36) + (4/36) - (3/36) x (4/36).
=> 5/27.
Question 7 ] In how many ways can 10 engineers and 4 doctors be seated at a round table if all the 4 doctors do not sit together?
A. 13! - (10! × 4!)
B. 13! × 4!
C. 14!
D. 10! × 4!
Explanation :
Initially let's find out the total number of ways in which 10 engineers and 4 doctors can be seated at a round table.
In this case, n = total number of persons = 10 + 4 = 14
Hence, the total number of ways in which 10 engineers and 4 doctors can be seated at a round table
= (14-1)! = 13! -------------------------------(A)
Now let's find out the total number of ways in which 10 engineers and 4 doctors can be seated at a round table where all the 4 doctors sit together.
Since all the 4 doctors sit together, group them together and consider as a single doctor.
Hence, n = total number of persons = 10 + 1 = 11.
These 11 persons can be seated at a round table in (11-1)! = 10! ways ----------------------(B)
However these 4 doctors can be arranged among themselves in 4! Ways --------------------(C)
From (B) and (C), the total number of ways in which 10 engineers and 4 doctors can be seated at a round table where all the 4 doctors sit together = 10! 4! -----------------------(D)
From (A) and (D),
The total number of ways in which 10 engineers and 4 doctors can be seated at a round table if all the 4 doctors do not sit together = 13! - (10! x 4!)
Question 8 ] What is the approx. value of W, if W=(1.5)11, Given log2 = 0.301, log3 = 0.477.
A. 85
B. 86
C. 84
D. 89
Explanation :
W = (1.5)^11
Take log both sides :-
=> logW = 11 x log(1.5) = 11 x log(3/2).
=> logW = 11 x (log3 - log2) = 11 x 0.176 = 1.936.
=> W = 101.936.
=> W = 86 (approx).
Question 9 ] Excluding stoppages, the speed of a bus is 54 kmph and including stoppages, it is 45 kmph. For how many minutes does the bus stop per hour?
A. 9
B. 10
C. 11
D. 12
Explanation :
Speed of the bus excluding stoppages = 54 kmph
Speed of the bus including stoppages = 45 kmph
Loss in speed when including stoppages = 54 - 45 = 9kmph.
=> In 1 hour, bus covers 9 km less due to stoppages.
Hence, time that the bus stop per hour = time taken to cover 9 km.
=> distance / speed= 9 / 54 hour = 1/6 hour = 60/6 min=10 min.
Question 10 ] A boat M leaves shore A and at the same time boat B leaves shore B. They move across the river. They met at 500 yards away from A and after that they met 300 yards away from shore B without halting at shores. Find the distance between the shore A & B.
A. 1100 yards
B. 1200 yards
C. 1300 yards
D. 1400 yards
Explanation :
The boats first met after 500 yards from A.
Let the total distance between A and B be x.
Then,
500/a = x - 500 / b, ------------(1)
where, a = speed of A,
b = speed of B.
Assuming both boats are coming back to their original start place without halting at the shores,
Distance coverred by A = x + 300.
Distance covered by B = 2x - 300.
x+300/a = 2x-300/b. ------------(2)
From above,
500/x-500 = x+(300/2x)-300.
Solve, for x = 1200.
Question 11] Harish and Peter working separately can paint a building in 18 days and 24 days respectively. If they work for a day alternately, Harish beginning, in how many days, the painting work will be completed ?
A. 41/2 days.
B. 21/2 days
C. 63/4 days
D. 65/4 days
Explanation :
Let A = Harish, B = Peter.
(A+B)'s 2 day's work =(1/18 + 1/24) = 7/72.
work done in 10 pairs of days = (10*7/72) = 70/72.
remaining work = (1 - 70/72) = 2/72 = 1/36.
on 21st day, it is A's turn. 1/18 work is done by him in 1 days.
1/36 work id done by him in (18 * 1/36) = 1/2 day.
Hence, total time taken = (20 + 1/2) days = 41/2 days.
Question 12] A tank has two compartments I and II. Two taps X and Y, whose filling rates are in ratio of 2:1, are used to fill the tank. The ration fo time taken by tap X to fill compartment I and tap Y to fill compartment II is 16:25. Find the ratio of the times taken by tap X to fill compartment II and tap Y to fill compartment I.
A. 25:64
B. 16:25
C. 64:25
D. 4:5
Explanation :
Let the time (in minutes) taken by tap X to fill compartment I be 16x and that taken by tap Y to fill compartment II be 25x. As the filling rates of X and Y are in the ratio of 2:1, the ratio of times they take to fill the same volume is 1:2.
So tap Y will take (16x)2 i.e. 32x minutes to fill compartment I and tap X will take 25x/2 i.e. 12.5x minutes to fill the compartment II.
Hence the ratio 12.5x:32x i.e. 25:64.
Question 13] At the end of year 1998, Shepard bought nine dozen goats. Henceforth, every year he added p% of the goats at the beginning of the year and sold q% of the goats at the end of the year where p > 0 and q > 0. If Shepard had nine dozen goats at the end of year 2002, after making the sales for that year, which of the following is true?
A. p = q
B. p < q
C. p > q
D. p = q/2
Explanation :
The number of goats remain the same. If the percentage that is added every time is equal to the percentage that is sold, then there should be net decrease.
The same will be the case if the percentage added is less than the percentage sold.
The only way the number of goats remain the same if p>q.
Question 14 ] A real number x satisfying 1- (1/n) ≤ 3 + (1/n), for every positive integer n, is best described by ≤
A. 1 < x < 4
B. 1 < x ≤ 3
C. 0 < x ≤4
D. 1 ≤ x ≤ 3
Explanation :
Given,
1- (1/n) ≤ 3 + (1/n)
Put n =1 , we get 0 < x ≤4.
Question 15 ] Given that -1 ≤ v ≤ 1, -2 ≤ u ≤ -0.5, and -2 ≤ z ≤ -0.5 and w = vz/u, then which of the following is necessarily true ?
A. -0.5 ≤ w ≤ -2
B. -4 ≤ w ≤ 4
C. -4 ≤ w ≤ 2
D. -2 ≤ w ≤ -0.5
Explanation :
"u" is always negative. Hence, for us to have a minimum value of vz/u , vz should be positive.
Also for the least value, the numerator has to be the maximum positive value and the denominator has to be the smallest negative value. In other words, vz has to be 2 and u has to be –0.5.
Hence the minimum value of vz/u = 2−0.5 = –4.
For us to get the maximum value, vz has to be the smallest negative value and u has to be the highest negative value.
Thus, vz has to be –2 and u has to be –0.5. Hence the maximum value of vz/u = 2 / -0.5 = 4.
Question 16 ] A can complete a piece of work in 4 days. B takes double the time taken by A, C takes double that of B, and D takes double that of C to complete the same task. They are paired in groups of two each. One pair takes two-thirds the time needed by the second pair to complete the work. Which is the first pair?
A. A and B
B. A and C
C. B and C
D. A and D
Explanation :
Work done in one day by A,B,C and D are (1/4), (1/8), (1/16) and (1/32) respectively .
Using options, we note that
the pair B and C does 3/16 of work in one day.,
the pair A and D does (1/4) + (1/32) = (9/32) of work in one day.
Hence, A and D take 32/9 days.
B and C take 16/3 days.
Hence, the first pair must comprise A and D.
Question 17] Two men X and Y started working for a certain company at similar jobs on January 1, 1950. X asked for an initial salary of Rs. 300 with an annual increment of Rs. 30. Y asked for an initial salary of Rs. 200 with a rise of Rs. 15 every 6 months. Assume that the arrangements remained unaltered till December 31, 1959. Salary is paid on the last day of the month. What is the total amount paid to them as salary during the period?
A. Rs. 93,300
B. Rs. 93,200
C. Rs. 93,100
D. None of these
Explanation :
Amount of money given to X
= 12 x 300 + 12 x 330 + ....... + 12 x 570.
= 12[ 300 + 330 + .........+ 540 + 570 ].
= 12 x (10/2) [ 600 + 9 x 30 ].
= 52200.
Amount of money given to Y is
= 6 x 200 + 6 x 215 + 6 x 230 + 6 x 245 + ... upto 20 terms
= 6 [ 200 + 215 + 230 + .....485 ]
= 6 x (20/2)[ 400 + 19 x 15]
= 6 x 10 [ 400 + 285 ] .
= 60 x 685 = 41110
Hence, the total amount paid is 52200 + 41110 = Rs 93,300.
Question 18 ] A rectangular sheet of paper, when halved by folding it at the midpoint of its longer side, results in a rectangle, whose longer and shorter sides are in the same proportion as the longer and shorter sides of the original rectangle. If the shorter side of the original rectangle is 2, what is the area of the smaller rectangle?
A. 4√2
B. 2√2
C. √2
D. None of the above
Explanation :
Original rectangle: Shorter side = 2 Longer side = x
Second rectangle: Longer side = 2 Shorter side = x/2 (As it is folded into half at midpoint)
Ratio of 1st rectangle = x/2.
2nd rectangle =( 2/x)/2 = 4/x.
Ratio is same. x/2 = 4/x.
x2 = 8, x = 2√2.
So, area of smaller rectangle = (x/2) * 2 => x = 2√2.
Question 19] A cow is tethered at point A by a rope. Neither the rope nor the cow is allowed to enter ∆ABC.
∠BAC = 30°
I ( AB ) = I ( AC ) = 10 m
What is the area that can be grazed by the cow if the length of the rope is 8 m?
A. 134 π/3 sq. m
B. 121π sq. m
C. 132π sq. m
D. 176 π/3 sq. m
Explanation :
It can be seen that if the length of the rope is 8 m, then the cow will be able to graze an area equal to (the area of the circle with radius 8m) – (Area of the sector of the same circle with angle 30° ). This can further be expressed as
=> π (8)2 − (30/360)π (8)2
=> 64π - (1/12)(64π) = 176π/3 sq.m.
Shortcut:
Area grazed without restriction is 64π m2. it should be less than 64 π sq. m. with restriction.
Question 20 ] In the figure below, AB = BC = CD = DE = EF = FG = GA. Then ∠DAE is approximately?
A. 15°
B. 20°
C. 30°
D. 25°
Explanation :
Let us assume, ∠DAE = x
Triangle ABC is isosceles as AB = BC => ∠BCA = ∠CAB = x
Hence, ∠CBD = ∠CAB + ∠BCA = x + x = 2x .............. [External angle of triangle ABC]
Triangle BCD is isosceles as BC = CD => ∠CBD = ∠CDB = 2x
Hence, ∠DCE = ∠DAE + ∠CDA = x + 2x = 3x .............. [External angle of triangle ACD]
Triangle CDE is isosceles as CD = DE => ∠DCE = ∠DEC = ∠AED = 3x
Similarly, ∠ADE = ∠EFD = ∠AEF + ∠DAE = ∠EGF + ∠DAE = (∠DAE + ∠GFA) + ∠DAE = ∠DAE + ∠DAE + ∠DAE = 3x
Hence, in triangle ADE, ∠ADE + ∠DAE + ∠AED = 3x + x + 3x = 7x
Hence, 7x = 180 => x = 180/7 = 25.7.. ≈ 25
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