Question 1] 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 2 ] A string of length 40 meters is divided into three parts of different lengths. The first part is three times the second part, and the last part is 23 meters smaller than the first part. Find the length of the largest part.
A. 27
B. 4
C. 5
D. 9
Explanation :
Let the 2nd piece length be x.
1st piece length is 3x.
3rd piece length is 3x – 23.
Sum of all the pieces is 40.
=>3x + x + 3x – 23 = 40.
=>7x = 63.
=>x = 9.
Lengths of the pieces are 27, 9, 4.
Question 3 ] Let x, y and z be distinct integers, x and y are odd and positive, and z is even and positive. Which one of the following statements cannot be true?
A. (x – z)^2 * y is even
B. (x – z)*y^2 is odd
C. (x – z)*y is odd
D. (x – y)^2* z is even
Explanation :
Let x = 5, y= 3, z = 2 (according to the given condition)
(x – z)2 * y :- ( 5 - 2)2 * 3 = 27
(x – z) * y2 :- (5 - 2) * 32 = 27
(x – z) * y :- (5 - 2) * 3 = 9.
(x – y)2 * z :- (5 - 3)2 * 2 = 8.
Question 4] What is the remainder when 787777 is divided by 100?
A. 43
B. 63
C. 67
D. 87
Explanation :
We have to find out the remainder of 787777 divided by 100.
This is the same as finding out the last two digits of 787777
Last two digits of the answer depend on the last two digits of the base.
=> We need to find out last two digits of 87777.
The key in questions like these is to reduce the number to something ending in 1.
=> 87777.
=> 87 * 87776.
=> 87 * (..69)388 {Just looking at last two digits of 872}
=> 87 * (...61)194 {Just looking at last two digits of 612}
=> 87 * (...41) {a number of the format ..a1..b will end in (a*b)1}
=> 67.
Question 5] In the X-Y plane, the area of the region bounded by the graph of |x + y| + |x – y| = 4 is ?
A. 8
B. 12
C. 16
D. 20
Explanation :
There are 4 cases here to be considered:
x & y positive
If |x| > |y|, then |x + y| = x + y & |x – y| = x - y. So, x + y + x - y = 4. x = 2.
If |x| < |y|, then |x + y| = x + y & |x – y| = y - x. So, x + y + y - x = 4. y = 2.
Since x and y is positive, this is the 1st quadrant - > (2,2)
x positive and y negative:
If |x| > |y|, then |x + y| = x + y & |x – y| = x - y. So, x + y + x - y = 4. x = 2.
If |x| < |y|, then |x + y| = -x - y & |x – y| = - y + x. So, -x - y - y + x = 4.y = -2.
Since x is positive and y is negative, this is the 3rd quadrant -> (2,-2)
x negative and y positive:
If |x| > |y|, then |x + y| = -x - y & |x – y| = -x + y. So, -x - y - x + y = 4. x = -2.
If |x| < |y|, then |x + y| = x + y & |x – y| = y - x. So, x + y + y - x = 4. y = 2.
Since x is negative and y is positive, this is the 2nd quadrant -> (-2,2)
x and y negative:
If |x| > |y|, then |x + y| = -x - y & |x – y| = -x + y. So, -x - y - x + y = 4. x = -2.
If |x| < |y|, then |x + y| = -x - y & |x – y| = -y + x. So, -x - y - y + x = 4. y = -2.
Since x and y is negative, this is the 4th quadrant -> (-2,-2)
So, now we have a square in the 4 quadrants passing through (2,2), (2,-2), (-2,2) and (-2,-2).
Hence, the area of square is 4 * 4 = 16 (Square of a side).
Question 6] A father and his son are waiting at a bus stop in the evening. There is a lamp post behind them. The lamp post, the father and his son stand on the same straight line. The father observes that the shadows of his head and his son’s head are incident at the same point on the ground. If the heights of the lamp post, the father and his son are 6 meters, 1.8 meters and 0.9 meters respectively, and the father is standing 2.1 meters away from the post, then how far (in meters) is the son standing from his father?
A. 0.9
B. 0.75
C. 0.6
D. 0.45
Explanation :
L is the lamp post position, F is father and S is son’s position. X is the point where the shadow falls.
LD = 0.9 = Son’s height , LB = 1.8 = Father’s height.
So AB = 6 – 1.8 = 4.2 Also BC = LF = 2.1
We observe that ∆ABC ~ ∆ADE (two triangles are similar).
Hence, the corresponding sides are proportional.
So, AB/AD = DC/DE.
4.2/5.1 = 2.1/DE.
DE = 5.1*2.1/4.2 = 5.1/2 = 2.55.
LS = DE = 2.55
FS = LS – LF = 2.55 – 2.1 = 0.45.
Question 7] In the adjoining figure, I and II are circles with centres P and Q respectively. The two circles touch each other and have a common tangent that touches them at points Rand S respectively. This common tangent meets the line joining P and Q at O. The diameters of I and II are in theratio 4 : 3. It is also known that the length of PO is 28 cm.
What is the ratio PQ:QO ?
A. 1:4
B. 1:3
C. 3:8
D. 3:4
Explanation :
We could solve this using the options .
The ratio of diameters given is 4:3 i.e. the radius for circle II will be multiple of 3.
Thus, the radius of the circle II is 3cm.
we get PQ=7 and PO=28(given)
Then, QO = 28 - 7 =21.
Hence, the required ratio i.e PQ:QO = 7 : 21 i.e 1:3.
Question 8 ] 2n boys are randomly divided into two sub-groups containing n boys each. Find the probability that two TOP RANK-HOLDER are in differed groups.
A. (n-1) / 2n
B. (n-1) / (2n-1)
C. n / (2n-1)
D. None of these
Explanation :
Without restriction
No. of ways to divide 2n boys into 2 equal groups having n boys each = 2nCn x nCn.
With restriction,
No. of ways to have two top ranks-holder in different group = 2n-2Cn-1 x n-1Cn-1 x 2!.
( i.e separate two top rankers and then make two groups out of (2n - 2) boys.
The required probability
= Selections with restriction / Selections without restriction .
= (2n-2Cn-1 x n-1Cn-1 x 2! ) / (2nCn x nCn ).
= { (2n-2)! / [ (n-1)! x (n-1)! ] } x 2! / (2n! / ( n! x n! ) ) .
= (n-1) / 2n - 1.
Hence, the probability is (n-1) / (2n-1).
Question 9] There are two men aged 30 and 36 years. The probability to live 35 years more is 0.67 for 30 -years old man and 0.60 for 36 year old man. Find the probability that at least one of these men will be alive 35 years hence.
A. 0.35
B. 0.868
C. 0.43
D. 0.44
Explanation :
Let, A ≡ event that 30 year-old man will live 35 years more.
B ≡ event that 36 year-old man will live 35 years more.
[ A and B are independent events ]
To find at least one of the men will be alive it easier to calculate the complementary of the event (i.e none will be alive) first. Since
P(at least one) = 1 - P(none).
P(none will be alive)
= P(~A~B)
= P(~A) x P(~B)
= [1 - P(A)] x [1 - P(B)]
= (1- 0.67)(1-0.6)
= 0.33 x 0.40
= 0.132
P ( at least one of them will be alive)
= 1 - 0.132
= 0.868.
Question 10] In how many ways can 20 different computers be divided equally among 5 persons?
A. 20! / 5!
B. 20! / 5!^5.
C. 20! / (4!)
D. 20! / (4!)5.
Explanation :
Here,
(i) the objects are different
(ii) the number of distinct objects n =20.
(iii) the number of persons p =5.
(iv) the computers are divided equally among the persons.
Hence, the required number of ways :-
=> n! / [(n/p)!]p
=> 20! / [(20/5)!]5.
=> 20! / (4!)5.
Question 11 ] A person has 12 friends and he wants to invite 8 of them to a birthday party. Find how many times 3 particular friends will always attend the parties.
A. 126
B. 95
C. 130
D. 155
Explanation :
No. of times 3 particular friends will attend the party always.
Hence the number of ways of selection is :-
=> n-kCr-k, where n= 12, r=8 and k=3.
=> No. of combinations = 12-3C8-3 .
=> 9C5.
=> 126.
Hence, the number of combinations with 3 particular friends present in each combination (party) is 126.
Question 12] On a 20 km tunnel, connecting two cities A and B, there are three gutters (1, 2 and 3). The distance between gutters 1 and 2 is half the distance between gutters 2 and 3. The distance from city A to its nearest gutter, gutter 1, is equal to the distance of city B from gutter 3. On a particular day, the hospital in city A receives information that an accident has happened at gutter 3. The victim can be saved only if an operation is started within 40 min. An ambulance started from city A at 30 km/hr and crossed gutter 1 after 5 min. If the driver had doubled the speed after that, what is the maximum amount of time would the doctor get to attend the patient at the hospital. Assume 1 min is elapsed for taking the patient into and out of the ambulance?
A. 4 min
B. 2.5 min
C. 1.5 min
D. The patient died before reaching the hospital
Explanation :
AG1 in 5 min at 30 km/hr = 2.5 km
G1G3 = 15 Km
Time for AG1 = 5 min.
Time for
G1G3 + G3A = 32.5 min = total of 37.5 min.
1 min is taken for transferring the patient into and out of the ambulance.
Hence, (40 - 37.5 - 1) i.e 1.5 min is remaining.
Question 13] A transport company charges for its vehicles in the following manner If the driving is 5 hours or less, the company charges Rs. 60 per hour or Rs. 12 per km (which ever is larger) If driving is more than 5 hours, the company charges Rs. 50 per hour or Rs. 7.5 per km (which ever is larger). If Anand drove it for 30 km and paid a total of Rs. 300, then for how many hours does he drive?
A. 4
B. 5.5
C. 7
D. 6
Explanation :
If he drives for 5hr or less, then he would have to give 360Rs (30km*12Rs/km).
So, he definitely traveled more than 5hrs.
Consider the options :-
7hrs – Either 350Rs(time based) or 225rs (km bases). So, 350Rs.
5.5 hrs – 275 or 225. So, 275 Rs.
6 hrs – 300 or 225. So 300rs.
Hence, he traveled for 6hrs.
Question 14] Two workers A and B manufactured a batch of identical parts. A worked for 2 hours and B worked for 5 hours and they completed half the job. Then they worked together for another 3 hours and they had to do (1/20)th of the job. How many hours time does B take to complete the job, if he worked alone?
A. 24hr
B. 12 hrs
C. 15 hrs
D. 30 hrs
Explanation :
Let 'a' hours be the time worker A takes to complete the job and 'b' hours be the time that worker B takes to complete the job.
A works for 2 hours and B works for 5 hours half the job is done.Then,
(2/a)+(5/b)=1/2 -------(i)
When they work together for the next three hours, (1/20)th of the job is yet to be completed. They have completed half the job earlier and (1/20)th is still left.
So by working for 3 hours, they have completed (1/2)−(1/20)=9/20th of the work.
Therefore,
(3/a)+(3/b)=9/20 -------(ii)
Solving equations (i) and (ii), we get, b = 15 hours.
Question 15 ] The rate of increase of the price of sugar is observed to be two percent more than the inflation rate expressed in percentage. The price of sugar, on January 1, 1994, is Rs. 20 per kg. The inflation rate for the years 1994 and 1995 are expected to be 8% each. The expected price of sugar on January 1, 1996 would be
A. 23.60
B. 24.00
C. 24.20
D. 24.60
Explanation :
Increase in the price of sugar = (8+2)= 10%
Hence, price of the sugar on Jan 1, 1996
=> (20 * 110 * 110)/( 100 * 100 ) = Rs 24.20.
Question 16] Two vessels A and B contain spirit and water mixed in the ratio 5:2 and 7:6 respectively. Find the ratio in which these mixture be mixed to obtain a new mixture in vessel c containing spirit and water in the ratio 8:5?
A. 1:7
B. 2:9
C. 7:9
D. 3:8
Explanation :
Spirit in 1 liter mix of A = 5/7 liter.
Spirit in 1 liter mix of B = 7/13 liter.
Spirit in 1 liter mix of C = 8/13 liter.
By rule of allegation we have required ratio X:Y X : Y
5/7 7/13
\ /
(8/13)
/ \
(1/13) : (9/91)
7 9
Therefore, the required ratio = 1/13 : 9/91 .
= 7:9.
Question 17 ] The number of common terms in the two sequences 17, 21, 25, … , 417 and 16, 21, 26, … , 466 is
A. 78
B. 19
C. 77
D. 20
Explanation :
The common terms also form an AP with first term 21 and common difference equal to LCM of common differences of the given APs
nth term of AP, tn = a+(n-1)*d
Hence 21+(n-1)*20 <= 417 (Since the last common term of APs can be 417)
=> n <= 20.8
Therefore n=20
Question 18] Consider the sequence 1,-2,3,-4,5,-6........what is the average of first 200 terms of sequence?
A. 0.5
B. -1
C. -0.5
D. -2
Explanation :
We have,1,-2,3,-4,5,-6....
If we find the sum of 2 consecutive numbers then we will have,
1+(-2)=1-2=-1
-1+3=2
2+(-4)=2-4=-2
-2+5=3
3+(-6)=3-6=-3
-3+7=4
4+(-8)=-4-8=-4
.
.
.
.
=> So 100+(-200)=100-200=-100.
=> So sum of the first 200 terms= -100.
=> Then average of the first 200 terms= -(100/200) = -0.5.
Question 19] Prakash, Varun and Gajini had to paint three identical fences. On the first day only Prakash turned up for work and he completed the only on the first fence taking m hours. On the second day all three of them turned up for work and they completed the work only on second fence taking (m-4) hours. On the third day Varun and Gajini turned up and they completed the work on third fence taking (m+4) hours. What is the value of m?
A. 8
B. 6
C. 10
D. 9
Explanation :
Varun and Gajini do not work individually.
Both work on the second and third days and neither works on the first day.
We can treat them as one unit (VG). P takes m hours and VG takes (m+5) hours.
Working together P and (VG) take (m-4) hours. P alone takes 4 hours more.
VG alone takes 9 hours more.
Hence,
=>(m-4) = (4*9)1/2
=>(m-4) = 6.
=>m = 10.
Question 20 ] If a, b and c are forming increasing terms of G.P., r is the common ratio then find the minimum value of (c-b), given that (log a+log b+log c)/log 6 = 6.Note that r can be any real no.
A. 36
B. 24
C. 18
D. 12
Explanation :
a,b,c are in G.P. so let the first term of G.P. = a/r, and common ratio = r.
Therefore, a = a/r, b = a, c = ar
Given,( log a+log b+log c) / log6 = 6.
=> log abc / log6=6.
⇒log6abc = 6.
=> abc = 66.
put the value of a,b,c in GP format
=> (a/r) × a × ar = 66.
=> a3 = 6.
=> a = 36.
Now a = 36/r, b = 36, c = 36r.
We have to find the minimum value of c - b = 36r - 36.
r can be any number. So for r < 0, we get c - b negative.
When r = 1, c - b = 0
But none of the options are not representing it.
From the given options, r = 4/3, then c = 48.
So, option d satisfies this.
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 2 ] A string of length 40 meters is divided into three parts of different lengths. The first part is three times the second part, and the last part is 23 meters smaller than the first part. Find the length of the largest part.
A. 27
B. 4
C. 5
D. 9
Explanation :
Let the 2nd piece length be x.
1st piece length is 3x.
3rd piece length is 3x – 23.
Sum of all the pieces is 40.
=>3x + x + 3x – 23 = 40.
=>7x = 63.
=>x = 9.
Lengths of the pieces are 27, 9, 4.
Question 3 ] Let x, y and z be distinct integers, x and y are odd and positive, and z is even and positive. Which one of the following statements cannot be true?
A. (x – z)^2 * y is even
B. (x – z)*y^2 is odd
C. (x – z)*y is odd
D. (x – y)^2* z is even
Explanation :
Let x = 5, y= 3, z = 2 (according to the given condition)
(x – z)2 * y :- ( 5 - 2)2 * 3 = 27
(x – z) * y2 :- (5 - 2) * 32 = 27
(x – z) * y :- (5 - 2) * 3 = 9.
(x – y)2 * z :- (5 - 3)2 * 2 = 8.
Question 4] What is the remainder when 787777 is divided by 100?
A. 43
B. 63
C. 67
D. 87
Explanation :
We have to find out the remainder of 787777 divided by 100.
This is the same as finding out the last two digits of 787777
Last two digits of the answer depend on the last two digits of the base.
=> We need to find out last two digits of 87777.
The key in questions like these is to reduce the number to something ending in 1.
=> 87777.
=> 87 * 87776.
=> 87 * (..69)388 {Just looking at last two digits of 872}
=> 87 * (...61)194 {Just looking at last two digits of 612}
=> 87 * (...41) {a number of the format ..a1..b will end in (a*b)1}
=> 67.
Question 5] In the X-Y plane, the area of the region bounded by the graph of |x + y| + |x – y| = 4 is ?
A. 8
B. 12
C. 16
D. 20
Explanation :
There are 4 cases here to be considered:
x & y positive
If |x| > |y|, then |x + y| = x + y & |x – y| = x - y. So, x + y + x - y = 4. x = 2.
If |x| < |y|, then |x + y| = x + y & |x – y| = y - x. So, x + y + y - x = 4. y = 2.
Since x and y is positive, this is the 1st quadrant - > (2,2)
x positive and y negative:
If |x| > |y|, then |x + y| = x + y & |x – y| = x - y. So, x + y + x - y = 4. x = 2.
If |x| < |y|, then |x + y| = -x - y & |x – y| = - y + x. So, -x - y - y + x = 4.y = -2.
Since x is positive and y is negative, this is the 3rd quadrant -> (2,-2)
x negative and y positive:
If |x| > |y|, then |x + y| = -x - y & |x – y| = -x + y. So, -x - y - x + y = 4. x = -2.
If |x| < |y|, then |x + y| = x + y & |x – y| = y - x. So, x + y + y - x = 4. y = 2.
Since x is negative and y is positive, this is the 2nd quadrant -> (-2,2)
x and y negative:
If |x| > |y|, then |x + y| = -x - y & |x – y| = -x + y. So, -x - y - x + y = 4. x = -2.
If |x| < |y|, then |x + y| = -x - y & |x – y| = -y + x. So, -x - y - y + x = 4. y = -2.
Since x and y is negative, this is the 4th quadrant -> (-2,-2)
So, now we have a square in the 4 quadrants passing through (2,2), (2,-2), (-2,2) and (-2,-2).
Hence, the area of square is 4 * 4 = 16 (Square of a side).
Question 6] A father and his son are waiting at a bus stop in the evening. There is a lamp post behind them. The lamp post, the father and his son stand on the same straight line. The father observes that the shadows of his head and his son’s head are incident at the same point on the ground. If the heights of the lamp post, the father and his son are 6 meters, 1.8 meters and 0.9 meters respectively, and the father is standing 2.1 meters away from the post, then how far (in meters) is the son standing from his father?
A. 0.9
B. 0.75
C. 0.6
D. 0.45
Explanation :
L is the lamp post position, F is father and S is son’s position. X is the point where the shadow falls.
LD = 0.9 = Son’s height , LB = 1.8 = Father’s height.
So AB = 6 – 1.8 = 4.2 Also BC = LF = 2.1
We observe that ∆ABC ~ ∆ADE (two triangles are similar).
Hence, the corresponding sides are proportional.
So, AB/AD = DC/DE.
4.2/5.1 = 2.1/DE.
DE = 5.1*2.1/4.2 = 5.1/2 = 2.55.
LS = DE = 2.55
FS = LS – LF = 2.55 – 2.1 = 0.45.
Question 7] In the adjoining figure, I and II are circles with centres P and Q respectively. The two circles touch each other and have a common tangent that touches them at points Rand S respectively. This common tangent meets the line joining P and Q at O. The diameters of I and II are in theratio 4 : 3. It is also known that the length of PO is 28 cm.
What is the ratio PQ:QO ?
A. 1:4
B. 1:3
C. 3:8
D. 3:4
Explanation :
We could solve this using the options .
The ratio of diameters given is 4:3 i.e. the radius for circle II will be multiple of 3.
Thus, the radius of the circle II is 3cm.
we get PQ=7 and PO=28(given)
Then, QO = 28 - 7 =21.
Hence, the required ratio i.e PQ:QO = 7 : 21 i.e 1:3.
Question 8 ] 2n boys are randomly divided into two sub-groups containing n boys each. Find the probability that two TOP RANK-HOLDER are in differed groups.
A. (n-1) / 2n
B. (n-1) / (2n-1)
C. n / (2n-1)
D. None of these
Explanation :
Without restriction
No. of ways to divide 2n boys into 2 equal groups having n boys each = 2nCn x nCn.
With restriction,
No. of ways to have two top ranks-holder in different group = 2n-2Cn-1 x n-1Cn-1 x 2!.
( i.e separate two top rankers and then make two groups out of (2n - 2) boys.
The required probability
= Selections with restriction / Selections without restriction .
= (2n-2Cn-1 x n-1Cn-1 x 2! ) / (2nCn x nCn ).
= { (2n-2)! / [ (n-1)! x (n-1)! ] } x 2! / (2n! / ( n! x n! ) ) .
= (n-1) / 2n - 1.
Hence, the probability is (n-1) / (2n-1).
Question 9] There are two men aged 30 and 36 years. The probability to live 35 years more is 0.67 for 30 -years old man and 0.60 for 36 year old man. Find the probability that at least one of these men will be alive 35 years hence.
A. 0.35
B. 0.868
C. 0.43
D. 0.44
Explanation :
Let, A ≡ event that 30 year-old man will live 35 years more.
B ≡ event that 36 year-old man will live 35 years more.
[ A and B are independent events ]
To find at least one of the men will be alive it easier to calculate the complementary of the event (i.e none will be alive) first. Since
P(at least one) = 1 - P(none).
P(none will be alive)
= P(~A~B)
= P(~A) x P(~B)
= [1 - P(A)] x [1 - P(B)]
= (1- 0.67)(1-0.6)
= 0.33 x 0.40
= 0.132
P ( at least one of them will be alive)
= 1 - 0.132
= 0.868.
Question 10] In how many ways can 20 different computers be divided equally among 5 persons?
A. 20! / 5!
B. 20! / 5!^5.
C. 20! / (4!)
D. 20! / (4!)5.
Explanation :
Here,
(i) the objects are different
(ii) the number of distinct objects n =20.
(iii) the number of persons p =5.
(iv) the computers are divided equally among the persons.
Hence, the required number of ways :-
=> n! / [(n/p)!]p
=> 20! / [(20/5)!]5.
=> 20! / (4!)5.
Question 11 ] A person has 12 friends and he wants to invite 8 of them to a birthday party. Find how many times 3 particular friends will always attend the parties.
A. 126
B. 95
C. 130
D. 155
Explanation :
No. of times 3 particular friends will attend the party always.
Hence the number of ways of selection is :-
=> n-kCr-k, where n= 12, r=8 and k=3.
=> No. of combinations = 12-3C8-3 .
=> 9C5.
=> 126.
Hence, the number of combinations with 3 particular friends present in each combination (party) is 126.
Question 12] On a 20 km tunnel, connecting two cities A and B, there are three gutters (1, 2 and 3). The distance between gutters 1 and 2 is half the distance between gutters 2 and 3. The distance from city A to its nearest gutter, gutter 1, is equal to the distance of city B from gutter 3. On a particular day, the hospital in city A receives information that an accident has happened at gutter 3. The victim can be saved only if an operation is started within 40 min. An ambulance started from city A at 30 km/hr and crossed gutter 1 after 5 min. If the driver had doubled the speed after that, what is the maximum amount of time would the doctor get to attend the patient at the hospital. Assume 1 min is elapsed for taking the patient into and out of the ambulance?
A. 4 min
B. 2.5 min
C. 1.5 min
D. The patient died before reaching the hospital
Explanation :
AG1 in 5 min at 30 km/hr = 2.5 km
G1G3 = 15 Km
Time for AG1 = 5 min.
Time for
G1G3 + G3A = 32.5 min = total of 37.5 min.
1 min is taken for transferring the patient into and out of the ambulance.
Hence, (40 - 37.5 - 1) i.e 1.5 min is remaining.
Question 13] A transport company charges for its vehicles in the following manner If the driving is 5 hours or less, the company charges Rs. 60 per hour or Rs. 12 per km (which ever is larger) If driving is more than 5 hours, the company charges Rs. 50 per hour or Rs. 7.5 per km (which ever is larger). If Anand drove it for 30 km and paid a total of Rs. 300, then for how many hours does he drive?
A. 4
B. 5.5
C. 7
D. 6
Explanation :
If he drives for 5hr or less, then he would have to give 360Rs (30km*12Rs/km).
So, he definitely traveled more than 5hrs.
Consider the options :-
7hrs – Either 350Rs(time based) or 225rs (km bases). So, 350Rs.
5.5 hrs – 275 or 225. So, 275 Rs.
6 hrs – 300 or 225. So 300rs.
Hence, he traveled for 6hrs.
Question 14] Two workers A and B manufactured a batch of identical parts. A worked for 2 hours and B worked for 5 hours and they completed half the job. Then they worked together for another 3 hours and they had to do (1/20)th of the job. How many hours time does B take to complete the job, if he worked alone?
A. 24hr
B. 12 hrs
C. 15 hrs
D. 30 hrs
Explanation :
Let 'a' hours be the time worker A takes to complete the job and 'b' hours be the time that worker B takes to complete the job.
A works for 2 hours and B works for 5 hours half the job is done.Then,
(2/a)+(5/b)=1/2 -------(i)
When they work together for the next three hours, (1/20)th of the job is yet to be completed. They have completed half the job earlier and (1/20)th is still left.
So by working for 3 hours, they have completed (1/2)−(1/20)=9/20th of the work.
Therefore,
(3/a)+(3/b)=9/20 -------(ii)
Solving equations (i) and (ii), we get, b = 15 hours.
Question 15 ] The rate of increase of the price of sugar is observed to be two percent more than the inflation rate expressed in percentage. The price of sugar, on January 1, 1994, is Rs. 20 per kg. The inflation rate for the years 1994 and 1995 are expected to be 8% each. The expected price of sugar on January 1, 1996 would be
A. 23.60
B. 24.00
C. 24.20
D. 24.60
Explanation :
Increase in the price of sugar = (8+2)= 10%
Hence, price of the sugar on Jan 1, 1996
=> (20 * 110 * 110)/( 100 * 100 ) = Rs 24.20.
Question 16] Two vessels A and B contain spirit and water mixed in the ratio 5:2 and 7:6 respectively. Find the ratio in which these mixture be mixed to obtain a new mixture in vessel c containing spirit and water in the ratio 8:5?
A. 1:7
B. 2:9
C. 7:9
D. 3:8
Explanation :
Spirit in 1 liter mix of A = 5/7 liter.
Spirit in 1 liter mix of B = 7/13 liter.
Spirit in 1 liter mix of C = 8/13 liter.
By rule of allegation we have required ratio X:Y X : Y
5/7 7/13
\ /
(8/13)
/ \
(1/13) : (9/91)
7 9
Therefore, the required ratio = 1/13 : 9/91 .
= 7:9.
Question 17 ] The number of common terms in the two sequences 17, 21, 25, … , 417 and 16, 21, 26, … , 466 is
A. 78
B. 19
C. 77
D. 20
Explanation :
The common terms also form an AP with first term 21 and common difference equal to LCM of common differences of the given APs
nth term of AP, tn = a+(n-1)*d
Hence 21+(n-1)*20 <= 417 (Since the last common term of APs can be 417)
=> n <= 20.8
Therefore n=20
Question 18] Consider the sequence 1,-2,3,-4,5,-6........what is the average of first 200 terms of sequence?
A. 0.5
B. -1
C. -0.5
D. -2
Explanation :
We have,1,-2,3,-4,5,-6....
If we find the sum of 2 consecutive numbers then we will have,
1+(-2)=1-2=-1
-1+3=2
2+(-4)=2-4=-2
-2+5=3
3+(-6)=3-6=-3
-3+7=4
4+(-8)=-4-8=-4
.
.
.
.
=> So 100+(-200)=100-200=-100.
=> So sum of the first 200 terms= -100.
=> Then average of the first 200 terms= -(100/200) = -0.5.
Question 19] Prakash, Varun and Gajini had to paint three identical fences. On the first day only Prakash turned up for work and he completed the only on the first fence taking m hours. On the second day all three of them turned up for work and they completed the work only on second fence taking (m-4) hours. On the third day Varun and Gajini turned up and they completed the work on third fence taking (m+4) hours. What is the value of m?
A. 8
B. 6
C. 10
D. 9
Explanation :
Varun and Gajini do not work individually.
Both work on the second and third days and neither works on the first day.
We can treat them as one unit (VG). P takes m hours and VG takes (m+5) hours.
Working together P and (VG) take (m-4) hours. P alone takes 4 hours more.
VG alone takes 9 hours more.
Hence,
=>(m-4) = (4*9)1/2
=>(m-4) = 6.
=>m = 10.
Question 20 ] If a, b and c are forming increasing terms of G.P., r is the common ratio then find the minimum value of (c-b), given that (log a+log b+log c)/log 6 = 6.Note that r can be any real no.
A. 36
B. 24
C. 18
D. 12
Explanation :
a,b,c are in G.P. so let the first term of G.P. = a/r, and common ratio = r.
Therefore, a = a/r, b = a, c = ar
Given,( log a+log b+log c) / log6 = 6.
=> log abc / log6=6.
⇒log6abc = 6.
=> abc = 66.
put the value of a,b,c in GP format
=> (a/r) × a × ar = 66.
=> a3 = 6.
=> a = 36.
Now a = 36/r, b = 36, c = 36r.
We have to find the minimum value of c - b = 36r - 36.
r can be any number. So for r < 0, we get c - b negative.
When r = 1, c - b = 0
But none of the options are not representing it.
From the given options, r = 4/3, then c = 48.
So, option d satisfies this.
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