Question 1] Davji Shop sells samosas in boxes of different sizes. The
samosas are priced at Rs. 2 per samosa up to 200 samosas. For every
additional 20 samosas, the price of the whole lot goes down by 10 paise
per samosa. What should be the maximum size of the box that would
maximize the revenue?
A. 240
B. 300
C. 400
D. None of these
Explanation :
Number of samosas = 200 + 20n, where n is a natural number.
Price per samosa = Rs( 2 - 0.1n) .
Revenue = (200 + 20n)( 2 - 0.1n) = 400 + 20n -2n2.
For maxima 20 - 4n = 0, by differentiation n = 5.
=> Maximum revenue will be at (200 + 20 x 5) i.e 300 samosas.
Question 2 ] The number of positive integer valued pairs(x,y) satisfying 4x - 17y = 1 and x ≤ 1000 is
A. 55
B. 57
C. 58
D. 59
Explanation :
The difference between two integers will be 1, only if one is even and the other one is odd. 4x will always be even, so 17y has to be odd and hence y has to be odd.
Moreover, the number 17y should be such a number that is 1 less than a multiple of 4. In other words, we have to find all such multiples of 17, which are 1 less than a multiple of 4. The first such multiple is 51.
Here, as the multiples of 17 goes on increasing, the difference between it and its closest higher multiple of 4 is in the following pattern, 0, 3, 2, 1, e.g. 52 – 51 = 1, 68 – 68 = 0, 88 – 85 = 3, 104 – 102 = 2, 120 – 119 = 1, 136 – 136 = 0. So the multiples of 17 that we are interested in are 3,7, 11, 15 .
Now since, x ≤ 1000, 4x ≤ 4000 . The multiple of 17 closest and less than 4000 is 3995 (17 × 235). And incidentally, 3996 is a multiple of 4, i.e. the difference is 4. This means that in order to find the answer, we need to find the number of terms in the AP formed by 3, 7, 11, 15 … 235, where a = 3, d = 4.
Since, we know that Tn = a + (n – 1)d, so 235 = 3 + (n – 1) × 4.
Hence, n = 59.
Question 3 ] Total expenses of a boarding house are partly fixed and partly varying linearly with the number of boarders.The average expense per boarder is Rs. 700 when there are 25 boarders and Rs 600 when there are 50 boarders. What is the average expense per boarders when there are boarder ?
A. 550
B. 580
C. 540
D. 570
Explanation :
Let x be the fixed cost and y the variable cost
17500 = x + 25y … (i)
30000 = x + 50y … (ii)
Solving the equation (i) and (ii), we get :-
x = 5000, y = 500
Now if the average expense of 100 boarders be ‘A’.
Then,
=> 100 × A = 5000 + 500 × 100 .
∴ A = 550.
Question 4 ] Every ten years, the Indian government counts all the people living in the country. Suppose that the director of the census has reported the following data on two neighboring villages Chota Hazri and Mota Hazri :-
Chota Hazri has 4,522 fewer males than MotaHazri.
Mota Hazri has 4,020 more females than males.
Chota klazri has twice as many females as males.
Chota Hazri has 2,910 fewer females than Mota Hazri.
What is the total number of males in Chota Hazri?
A. 11,264
B. 14,174
C. 5,632
D. 10,154
Explanation :
Let ‘x’ be the number of males in Mota Hazri.
Chota Hazri Mota Hazri
Males x – 4522 x
Females 2(x – 4522) x + 4020
=> x + 4020 – 2(x – 4522) = 2910 .
=> x = 10154.
∴ Number of males in Chota Hazri = 10154 – 4522 = 5632.
Question 5 ] The rate of inflation was 1000%. Then what will be the cost of an article, which costs 6 units of currency now, 2 years from now?
A. 666
B. 626
C. 547
D. 726
Explanation :
Let x be initial cost .
1000 % increase => x + (1000/100)x = 11x.
1st yr = 6 * 11 = 66
2nd year = 66 * 11 = 726.
Question 6] The milk and water in two vessels A and B are in the ratio 4:3 and 2:3 respectively. In what ratio the liquids in both the vessels be mixed to obtain a new mixture in vessel c consisting half milk and half water?
A. 8:3
B. 7:5
C. 4:3
D. 2:3
Explanation :
Milk in 1 litre mixture of A = 4/7 litre.
Milk in 1 litre mixture of B = 2/5 litre.
Milk in 1 litre mixture of C = 1/2 litre.
By rule of alligation we have required ratio X:Y.
X : Y
4/7 2/5
\ /
(Mean ratio)
(1/2)
/ \
(1/2 – 2/5) : (4/7 – 1/2)
1/10 1/1 4
So, the required ratio = X : Y = 1/10 : 1/14 = 7:5.
Question 7] 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 8] The Tremors of earthquake were felt at intervals of 15 seconds. The first tremor was felt at 08:54:57 and last tremor was felt at 10:45:12. How many times were the tremors felt?
A. 460
B. 441
C. 450
D. 453
Explanation :
Considering that's 12 seconds
Total time gap in between =1 hr 50 min 15 seconds i.e 60*60+50*60+15 seconds .
=> 6615 seconds /15.
=> 440 tremors .
=> 440 + 1st one felt.
=> 441.
Question 9] A and B start from house at 10am. They travel on the MG road at 20kmph and 40 kmph. There is a Junction T on their path. A turns left at T junction at 12:00 noon, B reaches T earlier, and turns right. Both of them continue to travel till 2pm. What is the distance between A and B at 2 pm ?
A. 120 km
B. 140 km
C. 160 km
D. 200 km
Explanation :
At 12:00 noon A will travel=20*2 = 40km.
B will travel this 40 km in 40/40=1 hr i.e. by 11am.
After T junction for A- distance traveled = 2*20 = 40 km.
For B distance traveled =40*3=120.
So, the distance between A & B is=120+40=160km.
Question 10] Car A leaves city C at 5pm and is driven at a speed of 40 kmph. Two hours later another car leaves city C and is driven in same direction as car A. In how much time car B be 9 km ahead of car A if the speed of car B is 60 kmph.
A. 4 hours 27 minutes
B. 3 hours 27 minutes
C. 4 hours 08 minutes
D. 4 hours 00 minutes
Explanation :
Let after t time two cars will met.
So A will travel distance of 40t with 40kmph.
B will travel the distance of 60t with 60kmph.
And also A is ahead 80 km(40*2=80) from B
=> 60t - 40t = 80
=> t = 4hrs
Also time taken by B to cover 9kms more is 9/60 = 9 min.
Additional distance is 9 min
For additional time= (9/20)*60=27 min.
So correct answer = 4 hrs 27 min.
Question 11] Train X departs from station A at 11 a.m. for station B, which is 180 km so far. Train Y departs from station B at 11 a.m. for station A. Train X travels at an average speed of 70 km/hr and does not stop anywhere until it arrives at station B. Train Y travels at an average speed of 50 km/hr, but has to stop for 15 min at station C, which is 60 km away from station B en route to station A. Ignoring the lengths of the trains, what is the distance, to the nearest kilometer, from station A to the point where the trains cross each other?
A. 112 km
B. 118 km
C. 120 km
D. None of these
Explanation :
Total time taken by B to cover 60 km is 60/50 hr i.e 6/5 hr.
It stops at station C for 1/4 hr.
Now , in (6/5 + 1/4) hr , train X travels 70 x (29/20) = 101.5 km.
This means they do not cross each other by the time train Y finishes it stops at station C.
Let they meet after 't' hour.
Then, 70t + 50(t- 1/4) = 180.
t = 192.5/120 hr.
Distance from A will be 70 x 192.5/120 km i.e 112 km approximately.
Question 12] If logy x = a * log zy = b * log xz = a * b, then which of the following pairs of values for (a, b) is not possible?
A. -2, 1/2
B. 1,1
C. 0.4,2.5
D. π, 1/π
E. 2,2
Explanation :
Since,
log10100 = 2 (You have to know this)
log101000 = 3 (It is easy to remember such things, compared to the formulas)
This means, 102 = 100, 103 = 1000. That means logxy = a means xa = y.
Also, log x y = log y /log x (Both to the same base, 10 or e or anything)
Now,
We know, a*log z y = a * b. So, log z y = b.
We know, b*log x z = a * b. So, log x z = a.
We know, log y x = a * b.
Substituting equations 1 and 2 in equation 3, we get,
log y x = log z y * log x z .
log x / log y = log y / log z * log z / log x
log x / log y = log y / log x
log x2 = log y2
log x = log y or log x = -log y.
x = y or x = 1/y.
We know that log y x = a * b.
So a*b = log y y or log y 1/y (Substituting for x = 1/y, in the above equation)
So a * b = 1 or a * b = -1.
Hence, except option (E), all other options satisfies the above condition.
Question 13 ] Let S = 2x + 5x2 + 9x3 + 14x4 + 20x5 + ............∞. The coefficient of nth term is n(n+3) / 2.
The sum S is :-
A. x(2-x) / (1-x)3
B. (2-x) / (1-x)3
C. x(2-x) / (1-x)2
D. None of these
Explanation :
S = 2x + 5x2 + 9x3 + 14x4 + 20x5 + ..........∞. ..........(1)
xS = 2x2 + 5x3 + 9x4 + 14x5 + ..........∞. ..........(2)
On subtracting eq. 2 from eq 1,
S(1-x) = 2x + 3x2 + 4x3 + 5x4..... ..........(3)
Sx(1-x) = 2x2 + 3x3 + 4x4 + ..... ..........(4)
On subtracting eq. 4 from eq 3,
=> S[ (1- x) - x(1 - x)] = x + (x + x2 + x3 + x4 + ................... ).
=> S[ 1 - 2x + x2 ] = x + x/(1-x).
=> S[ 1-x ]2 = x(2-x)/(1-x).
=> S = x(2-x)/(1-x)3.
Question 14] A series S1 of five positive integers is such that the third term is half the first term and the fifth term is 20 more than the first term. In series S2, the nth term defined as the difference between the (n+1) term and the nth term of series S1, is an arithmetic progression with a common difference of 30.
First term of S1 is :-
A. 80
B. 90
C. 100
D. 120
Explanation :
According to the question,
S1:- 2a, b, a, c, 2a + 20
And,
S2:- b - 2a, a - b, c - a, 2a + 20 - c.
Also,
=> b - 2a + 30 = a - b i.e. 3a - 2b = 30 .
=> a - b + 30 = c - a .
=> c - a + 30 = 2a + 20 - c i.e. 3a - 2c = 10.
Solving them,
we get a = 50, b = 60, c = 70.
So series S1:- 100, 60, 50, 70, 120 .
and series S2: -40, -10, 20.
Question 15 ] If |b| ≥ 1 and x= -| a| b, then which one of the following is necessarily true?
A. a -xb < 0
B. a -xb ≥ 0
C. a -xb > 0
D. a -xb ≤ 0
Explanation :
x = –|a| b.
Now a – xb = a – (– |a| b) b.
= a + |a| b2
∴ a – xb = a + ab2 … a 0 ≥ OR a – xb.
= a - ab2 ….... a < 0
= a(1 + b2) = a(1 – b2).
Consider 1st case :-
As a ≥ 0 and |b| ≥ 1, therefore (1 + b2) is positive.
∴ a (1 + b2) ≥ 0
∴ a – xb ≥ 0
Consider 2nd case :-
As a < 0 and |b| ≥ 1, therefore (1 – b2) ≤ 0.
∴ a (1 – b2) ≥ 0. (Since –ve × -ve = +ve and 1 – b2 can be zero also), i.e. a – xb ≥ 0.
Therefore, in both cases a – xb ≥ 0.
Question 16] If 3y + x > 2 and x + 2y <= 3 , What can be said about the value of y?
A. y = 1
B. y >1
C. y > -2
D. y > -1
Explanation :
3y + x > 2 -------------------(1)
x + 2y < = 3
=> x + 2y + y < = 3 + y
=> x + 3y < = 3 + y -------------------(2)
From (1) & (2)
2 < 3y + x < 3 + y
=> 2 < 3 + y
=> -1 < y
=> y > -1.
Question 17 ] Consider the following two curves in the X - Y plane
y = x3 + x2 + 5.
y = x2 + x + 5.
Which of the following statements is true for −2 ≤ x ≤ 2?
A.The two curves intersect once
B.The two curves intersect twice
C.The two curves do not intersect
D.The two curves intersect thrice
Explanation :
Given,
y = x3 + x2 + 5.
y = x2 + x + 5.
For the two curves to intersect we need to find the value(s) of x for which the two equations are equal.
Equating the two we have,
x3 + x2 + 5 = x2 + x + 5.
On cancellation of common terms we have,
=> x3- x = 0.
=> x * (x2-1) = 0.
=> x * (x + 1)(x - 1) = 0.
=> x = 0, 1, -1.
So the points at which these curves meet are (0,5),(1,7) and (-1,5). They intersect at three points.
Question 18] The length, breadth and height of a room are in the ratio 3:2:1. If the breadth and height are halved while the length is doubled, then the total area of the four walls of the room will
A. remain the same
B. decrease by 13.64%
C. decrease by 15%
D. decrease by 18.75%
E. decrease by 30%
Explanation :
2 walls (facing each other) have length of the room and height of the room as their components. The other 2 walls (facing each other) have breadth of the room and height of the room as their components.
Area of 1 set of walls is l * h (rectangle area), and the other set of walls is b * h.
So, the total area of the 1st 2 walls are 2*l*h, while that of the next 2 walls are 2*b*h. (as there are 2 walls each).
Total area of 4 walls now are 2 * l * h + 2 * b * h = 2 * h * (l + b).
We know, l: b: h = 3:2:1. That means, l = 3x, b = 2x, h = x.
So, total area of 4 walls = 2 * x * (5 * x) = 10x2. (Applying the formula derived = 2*h*(l+b) ).
After the change,
l = 6x, (length is doubled) b = x, (breadth is halved) h = x/2. (height is halved)
So, total area of 4 walls = x *(7*x) = 7x2.
Total decrease = 3x2.
Percentage change = Change / Original * 100 = 3/10 * 100 = 30%.
Question 19] A punching machine is used to punch a circular hole of diameter two units from a square sheet of aluminum of width 2 units, as shown below. The hole is punched such that circular hole touches one corner P of the square sheet and the diameter of the hole originating at P is in line with a diagonal of the square.
Find the area of the part of the circle (round punch) falling outside the square sheet.
A. π /4
B. (π -1) /2
C. (π -1) /4
D. (π -2) /2
E. (π -2) /4
Explanation :
First, we have to expand the diagram to the following :-
Let PQRS be the square. Let A&B be the points of intersection of the circle on the square. Join AB. It is said that the diameter of the circle is along the same line as that of the diagonal of the square. So draw the diagonal PR of the square. Let it meet the circle at C. So PC is the diameter of the circle.
We know, ∟APB = 900 (PQRS is a square). Also, angle formed by a semicircle is 900. Hence the sector AB is semicircle, and AB is diameter.
Let the meeting point of AB and PC be O, both diameters meet at the centre of the circle, and hence O is the centre.
Now, since PC is the diameter of the circle, ∟PAC and ∟PBC is 900, and since AB is a diameter, ∟ACB is 900.
Now, in ACBP, we know all the angles are 900, and CP = AB (they are diameters of a circle). BP and AC are diagonals of the quadrilateral ACBP. Hence ACBP is a square.
Diagonals are perpendicular bisectors in a square and hence ∟POB = ∟BOC = ∟COA = ∟AOP = 900, and PO = CO = AO = BO = 1.
Now, the portion of the circle outside the square is :
Minor sector AP + Minor sector BP – Area of ∆AOP – Area of ∆BOP.
Area of sector AP = π/4 (circle is divided into 4 areas by the 2 diameters)
Area of sector AP = π/4 (circle is divided into 4 areas by the 2 diameters)
Area of ∆AOP = ½*b*h = ½*1*1 = ½.
Area of ∆BOP = ½*b*h = ½*1*1 = 1/2.
So, the portion outside the circle is (π/2) – 1 or (π – 2)/2.
Question 20] A bag has 1, 3, 5, 7, 9,11 and 13 cm sticks. Find the probability that they they will form a triangle if 3 sticks are drawn?
A. 13 / 35
B. 14 / 35
C. 1 / 2
D. 5 / 6
Explanation :
Total Cases: =7C3=35 .
Favorable Cases: (3,5,7),(3,7,9),(3,9,11),(3,11,13),(5,7,9),(5,7,11),(5,9,11),(5,9,13),(5,11,13),
(7,9,11), (7,9,13), (7,11,13), (9,11,13) = total 13.
[Others are rejected because to form a triangle it is necessary that the sum of 2 small sides must be greater than the third greatest side
A. 240
B. 300
C. 400
D. None of these
Explanation :
Number of samosas = 200 + 20n, where n is a natural number.
Price per samosa = Rs( 2 - 0.1n) .
Revenue = (200 + 20n)( 2 - 0.1n) = 400 + 20n -2n2.
For maxima 20 - 4n = 0, by differentiation n = 5.
=> Maximum revenue will be at (200 + 20 x 5) i.e 300 samosas.
Question 2 ] The number of positive integer valued pairs(x,y) satisfying 4x - 17y = 1 and x ≤ 1000 is
A. 55
B. 57
C. 58
D. 59
Explanation :
The difference between two integers will be 1, only if one is even and the other one is odd. 4x will always be even, so 17y has to be odd and hence y has to be odd.
Moreover, the number 17y should be such a number that is 1 less than a multiple of 4. In other words, we have to find all such multiples of 17, which are 1 less than a multiple of 4. The first such multiple is 51.
Here, as the multiples of 17 goes on increasing, the difference between it and its closest higher multiple of 4 is in the following pattern, 0, 3, 2, 1, e.g. 52 – 51 = 1, 68 – 68 = 0, 88 – 85 = 3, 104 – 102 = 2, 120 – 119 = 1, 136 – 136 = 0. So the multiples of 17 that we are interested in are 3,7, 11, 15 .
Now since, x ≤ 1000, 4x ≤ 4000 . The multiple of 17 closest and less than 4000 is 3995 (17 × 235). And incidentally, 3996 is a multiple of 4, i.e. the difference is 4. This means that in order to find the answer, we need to find the number of terms in the AP formed by 3, 7, 11, 15 … 235, where a = 3, d = 4.
Since, we know that Tn = a + (n – 1)d, so 235 = 3 + (n – 1) × 4.
Hence, n = 59.
Question 3 ] Total expenses of a boarding house are partly fixed and partly varying linearly with the number of boarders.The average expense per boarder is Rs. 700 when there are 25 boarders and Rs 600 when there are 50 boarders. What is the average expense per boarders when there are boarder ?
A. 550
B. 580
C. 540
D. 570
Explanation :
Let x be the fixed cost and y the variable cost
17500 = x + 25y … (i)
30000 = x + 50y … (ii)
Solving the equation (i) and (ii), we get :-
x = 5000, y = 500
Now if the average expense of 100 boarders be ‘A’.
Then,
=> 100 × A = 5000 + 500 × 100 .
∴ A = 550.
Question 4 ] Every ten years, the Indian government counts all the people living in the country. Suppose that the director of the census has reported the following data on two neighboring villages Chota Hazri and Mota Hazri :-
Chota Hazri has 4,522 fewer males than MotaHazri.
Mota Hazri has 4,020 more females than males.
Chota klazri has twice as many females as males.
Chota Hazri has 2,910 fewer females than Mota Hazri.
What is the total number of males in Chota Hazri?
A. 11,264
B. 14,174
C. 5,632
D. 10,154
Explanation :
Let ‘x’ be the number of males in Mota Hazri.
Chota Hazri Mota Hazri
Males x – 4522 x
Females 2(x – 4522) x + 4020
=> x + 4020 – 2(x – 4522) = 2910 .
=> x = 10154.
∴ Number of males in Chota Hazri = 10154 – 4522 = 5632.
Question 5 ] The rate of inflation was 1000%. Then what will be the cost of an article, which costs 6 units of currency now, 2 years from now?
A. 666
B. 626
C. 547
D. 726
Explanation :
Let x be initial cost .
1000 % increase => x + (1000/100)x = 11x.
1st yr = 6 * 11 = 66
2nd year = 66 * 11 = 726.
Question 6] The milk and water in two vessels A and B are in the ratio 4:3 and 2:3 respectively. In what ratio the liquids in both the vessels be mixed to obtain a new mixture in vessel c consisting half milk and half water?
A. 8:3
B. 7:5
C. 4:3
D. 2:3
Explanation :
Milk in 1 litre mixture of A = 4/7 litre.
Milk in 1 litre mixture of B = 2/5 litre.
Milk in 1 litre mixture of C = 1/2 litre.
By rule of alligation we have required ratio X:Y.
X : Y
4/7 2/5
\ /
(Mean ratio)
(1/2)
/ \
(1/2 – 2/5) : (4/7 – 1/2)
1/10 1/1 4
So, the required ratio = X : Y = 1/10 : 1/14 = 7:5.
Question 7] 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 8] The Tremors of earthquake were felt at intervals of 15 seconds. The first tremor was felt at 08:54:57 and last tremor was felt at 10:45:12. How many times were the tremors felt?
A. 460
B. 441
C. 450
D. 453
Explanation :
Considering that's 12 seconds
Total time gap in between =1 hr 50 min 15 seconds i.e 60*60+50*60+15 seconds .
=> 6615 seconds /15.
=> 440 tremors .
=> 440 + 1st one felt.
=> 441.
Question 9] A and B start from house at 10am. They travel on the MG road at 20kmph and 40 kmph. There is a Junction T on their path. A turns left at T junction at 12:00 noon, B reaches T earlier, and turns right. Both of them continue to travel till 2pm. What is the distance between A and B at 2 pm ?
A. 120 km
B. 140 km
C. 160 km
D. 200 km
Explanation :
At 12:00 noon A will travel=20*2 = 40km.
B will travel this 40 km in 40/40=1 hr i.e. by 11am.
After T junction for A- distance traveled = 2*20 = 40 km.
For B distance traveled =40*3=120.
So, the distance between A & B is=120+40=160km.
Question 10] Car A leaves city C at 5pm and is driven at a speed of 40 kmph. Two hours later another car leaves city C and is driven in same direction as car A. In how much time car B be 9 km ahead of car A if the speed of car B is 60 kmph.
A. 4 hours 27 minutes
B. 3 hours 27 minutes
C. 4 hours 08 minutes
D. 4 hours 00 minutes
Explanation :
Let after t time two cars will met.
So A will travel distance of 40t with 40kmph.
B will travel the distance of 60t with 60kmph.
And also A is ahead 80 km(40*2=80) from B
=> 60t - 40t = 80
=> t = 4hrs
Also time taken by B to cover 9kms more is 9/60 = 9 min.
Additional distance is 9 min
For additional time= (9/20)*60=27 min.
So correct answer = 4 hrs 27 min.
Question 11] Train X departs from station A at 11 a.m. for station B, which is 180 km so far. Train Y departs from station B at 11 a.m. for station A. Train X travels at an average speed of 70 km/hr and does not stop anywhere until it arrives at station B. Train Y travels at an average speed of 50 km/hr, but has to stop for 15 min at station C, which is 60 km away from station B en route to station A. Ignoring the lengths of the trains, what is the distance, to the nearest kilometer, from station A to the point where the trains cross each other?
A. 112 km
B. 118 km
C. 120 km
D. None of these
Explanation :
Total time taken by B to cover 60 km is 60/50 hr i.e 6/5 hr.
It stops at station C for 1/4 hr.
Now , in (6/5 + 1/4) hr , train X travels 70 x (29/20) = 101.5 km.
This means they do not cross each other by the time train Y finishes it stops at station C.
Let they meet after 't' hour.
Then, 70t + 50(t- 1/4) = 180.
t = 192.5/120 hr.
Distance from A will be 70 x 192.5/120 km i.e 112 km approximately.
Question 12] If logy x = a * log zy = b * log xz = a * b, then which of the following pairs of values for (a, b) is not possible?
A. -2, 1/2
B. 1,1
C. 0.4,2.5
D. π, 1/π
E. 2,2
Explanation :
Since,
log10100 = 2 (You have to know this)
log101000 = 3 (It is easy to remember such things, compared to the formulas)
This means, 102 = 100, 103 = 1000. That means logxy = a means xa = y.
Also, log x y = log y /log x (Both to the same base, 10 or e or anything)
Now,
We know, a*log z y = a * b. So, log z y = b.
We know, b*log x z = a * b. So, log x z = a.
We know, log y x = a * b.
Substituting equations 1 and 2 in equation 3, we get,
log y x = log z y * log x z .
log x / log y = log y / log z * log z / log x
log x / log y = log y / log x
log x2 = log y2
log x = log y or log x = -log y.
x = y or x = 1/y.
We know that log y x = a * b.
So a*b = log y y or log y 1/y (Substituting for x = 1/y, in the above equation)
So a * b = 1 or a * b = -1.
Hence, except option (E), all other options satisfies the above condition.
Question 13 ] Let S = 2x + 5x2 + 9x3 + 14x4 + 20x5 + ............∞. The coefficient of nth term is n(n+3) / 2.
The sum S is :-
A. x(2-x) / (1-x)3
B. (2-x) / (1-x)3
C. x(2-x) / (1-x)2
D. None of these
Explanation :
S = 2x + 5x2 + 9x3 + 14x4 + 20x5 + ..........∞. ..........(1)
xS = 2x2 + 5x3 + 9x4 + 14x5 + ..........∞. ..........(2)
On subtracting eq. 2 from eq 1,
S(1-x) = 2x + 3x2 + 4x3 + 5x4..... ..........(3)
Sx(1-x) = 2x2 + 3x3 + 4x4 + ..... ..........(4)
On subtracting eq. 4 from eq 3,
=> S[ (1- x) - x(1 - x)] = x + (x + x2 + x3 + x4 + ................... ).
=> S[ 1 - 2x + x2 ] = x + x/(1-x).
=> S[ 1-x ]2 = x(2-x)/(1-x).
=> S = x(2-x)/(1-x)3.
Question 14] A series S1 of five positive integers is such that the third term is half the first term and the fifth term is 20 more than the first term. In series S2, the nth term defined as the difference between the (n+1) term and the nth term of series S1, is an arithmetic progression with a common difference of 30.
First term of S1 is :-
A. 80
B. 90
C. 100
D. 120
Explanation :
According to the question,
S1:- 2a, b, a, c, 2a + 20
And,
S2:- b - 2a, a - b, c - a, 2a + 20 - c.
Also,
=> b - 2a + 30 = a - b i.e. 3a - 2b = 30 .
=> a - b + 30 = c - a .
=> c - a + 30 = 2a + 20 - c i.e. 3a - 2c = 10.
Solving them,
we get a = 50, b = 60, c = 70.
So series S1:- 100, 60, 50, 70, 120 .
and series S2: -40, -10, 20.
Question 15 ] If |b| ≥ 1 and x= -| a| b, then which one of the following is necessarily true?
A. a -xb < 0
B. a -xb ≥ 0
C. a -xb > 0
D. a -xb ≤ 0
Explanation :
x = –|a| b.
Now a – xb = a – (– |a| b) b.
= a + |a| b2
∴ a – xb = a + ab2 … a 0 ≥ OR a – xb.
= a - ab2 ….... a < 0
= a(1 + b2) = a(1 – b2).
Consider 1st case :-
As a ≥ 0 and |b| ≥ 1, therefore (1 + b2) is positive.
∴ a (1 + b2) ≥ 0
∴ a – xb ≥ 0
Consider 2nd case :-
As a < 0 and |b| ≥ 1, therefore (1 – b2) ≤ 0.
∴ a (1 – b2) ≥ 0. (Since –ve × -ve = +ve and 1 – b2 can be zero also), i.e. a – xb ≥ 0.
Therefore, in both cases a – xb ≥ 0.
Question 16] If 3y + x > 2 and x + 2y <= 3 , What can be said about the value of y?
A. y = 1
B. y >1
C. y > -2
D. y > -1
Explanation :
3y + x > 2 -------------------(1)
x + 2y < = 3
=> x + 2y + y < = 3 + y
=> x + 3y < = 3 + y -------------------(2)
From (1) & (2)
2 < 3y + x < 3 + y
=> 2 < 3 + y
=> -1 < y
=> y > -1.
Question 17 ] Consider the following two curves in the X - Y plane
y = x3 + x2 + 5.
y = x2 + x + 5.
Which of the following statements is true for −2 ≤ x ≤ 2?
A.The two curves intersect once
B.The two curves intersect twice
C.The two curves do not intersect
D.The two curves intersect thrice
Explanation :
Given,
y = x3 + x2 + 5.
y = x2 + x + 5.
For the two curves to intersect we need to find the value(s) of x for which the two equations are equal.
Equating the two we have,
x3 + x2 + 5 = x2 + x + 5.
On cancellation of common terms we have,
=> x3- x = 0.
=> x * (x2-1) = 0.
=> x * (x + 1)(x - 1) = 0.
=> x = 0, 1, -1.
So the points at which these curves meet are (0,5),(1,7) and (-1,5). They intersect at three points.
Question 18] The length, breadth and height of a room are in the ratio 3:2:1. If the breadth and height are halved while the length is doubled, then the total area of the four walls of the room will
A. remain the same
B. decrease by 13.64%
C. decrease by 15%
D. decrease by 18.75%
E. decrease by 30%
Explanation :
2 walls (facing each other) have length of the room and height of the room as their components. The other 2 walls (facing each other) have breadth of the room and height of the room as their components.
Area of 1 set of walls is l * h (rectangle area), and the other set of walls is b * h.
So, the total area of the 1st 2 walls are 2*l*h, while that of the next 2 walls are 2*b*h. (as there are 2 walls each).
Total area of 4 walls now are 2 * l * h + 2 * b * h = 2 * h * (l + b).
We know, l: b: h = 3:2:1. That means, l = 3x, b = 2x, h = x.
So, total area of 4 walls = 2 * x * (5 * x) = 10x2. (Applying the formula derived = 2*h*(l+b) ).
After the change,
l = 6x, (length is doubled) b = x, (breadth is halved) h = x/2. (height is halved)
So, total area of 4 walls = x *(7*x) = 7x2.
Total decrease = 3x2.
Percentage change = Change / Original * 100 = 3/10 * 100 = 30%.
Question 19] A punching machine is used to punch a circular hole of diameter two units from a square sheet of aluminum of width 2 units, as shown below. The hole is punched such that circular hole touches one corner P of the square sheet and the diameter of the hole originating at P is in line with a diagonal of the square.
Find the area of the part of the circle (round punch) falling outside the square sheet.
A. π /4
B. (π -1) /2
C. (π -1) /4
D. (π -2) /2
E. (π -2) /4
Explanation :
First, we have to expand the diagram to the following :-
Let PQRS be the square. Let A&B be the points of intersection of the circle on the square. Join AB. It is said that the diameter of the circle is along the same line as that of the diagonal of the square. So draw the diagonal PR of the square. Let it meet the circle at C. So PC is the diameter of the circle.
We know, ∟APB = 900 (PQRS is a square). Also, angle formed by a semicircle is 900. Hence the sector AB is semicircle, and AB is diameter.
Let the meeting point of AB and PC be O, both diameters meet at the centre of the circle, and hence O is the centre.
Now, since PC is the diameter of the circle, ∟PAC and ∟PBC is 900, and since AB is a diameter, ∟ACB is 900.
Now, in ACBP, we know all the angles are 900, and CP = AB (they are diameters of a circle). BP and AC are diagonals of the quadrilateral ACBP. Hence ACBP is a square.
Diagonals are perpendicular bisectors in a square and hence ∟POB = ∟BOC = ∟COA = ∟AOP = 900, and PO = CO = AO = BO = 1.
Now, the portion of the circle outside the square is :
Minor sector AP + Minor sector BP – Area of ∆AOP – Area of ∆BOP.
Area of sector AP = π/4 (circle is divided into 4 areas by the 2 diameters)
Area of sector AP = π/4 (circle is divided into 4 areas by the 2 diameters)
Area of ∆AOP = ½*b*h = ½*1*1 = ½.
Area of ∆BOP = ½*b*h = ½*1*1 = 1/2.
So, the portion outside the circle is (π/2) – 1 or (π – 2)/2.
Question 20] A bag has 1, 3, 5, 7, 9,11 and 13 cm sticks. Find the probability that they they will form a triangle if 3 sticks are drawn?
A. 13 / 35
B. 14 / 35
C. 1 / 2
D. 5 / 6
Explanation :
Total Cases: =7C3=35 .
Favorable Cases: (3,5,7),(3,7,9),(3,9,11),(3,11,13),(5,7,9),(5,7,11),(5,9,11),(5,9,13),(5,11,13),
(7,9,11), (7,9,13), (7,11,13), (9,11,13) = total 13.
[Others are rejected because to form a triangle it is necessary that the sum of 2 small sides must be greater than the third greatest side
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