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Q1)
Fortune, the latest SUV by Toyota Motors, consumes diesel at the rate of 1/400{(1000/x)+x} litres per km, when driven at the speed of x km per hour. If the cost of diesel is Rs.35 per litre and the driver is paid at the rate of Rs.125 per hour then find the approximate optimal speed (in km per hour) of Fortuner that will minimize the total cost of the round trip of 800 kms.
1)
49
2)
55
3)
50
4)
53
5)
Correct Answer: 1
Your Answer: Not Attempted
Course: CAT Quantitative
This Question Is Correctly Answered By: 50% test takers
Answer Description
Substitute the options in the equation and compare the values.
Substitute the options in the equation and compare the values.
Q2)
Two motorists Anil and Sunil are practising wiyh two different sports cars: Ferrari and Maclaron,on the circular racing track , for the car racing tournament to be held next month.Both Anil and Sunil start from the same point on the circular track. Anil completes one round of the track in 1 minute and Sunil takes 2 minutes to complete a round.While Anil maintains same speed for the all rounds, Sunil halves his speed after the completion of each round. How many times Anil and Sunil will meet between the 6th round and 9th round of sunil (6th and 9th round is excluded)? Assume that the completion of that round.
1)
260
2)
347
3)
382
4)
None of the above
5)
Correct Answer: 3
Your Answer: Not Attempted
Course: CAT Quantitative
This Question Is Correctly Answered By: 36% test takers
Answer Description
Consider when A takes 1min and S takes 2min. They meet only once at the starting point when A completes his second round and S completes his first round.
Next A takes 1min and S takes 4min. They meet 3 times by the time S finishes his round. In A’s first round they don’t meet but in each of his subsequent rounds he meets S once.
Similarly it can be derived that if A takes x mins and S takes y mins, A and S meet (y-x) times. In this question we need to consider 7th and 8th rounds of S for which S takes 128mins and 256mins respectively while A takes a constant time of 1min.
Thus they meet 127 + 255 = 382 times
Consider when A takes 1min and S takes 2min. They meet only once at the starting point when A completes his second round and S completes his first round.
Next A takes 1min and S takes 4min. They meet 3 times by the time S finishes his round. In A’s first round they don’t meet but in each of his subsequent rounds he meets S once.
Similarly it can be derived that if A takes x mins and S takes y mins, A and S meet (y-x) times. In this question we need to consider 7th and 8th rounds of S for which S takes 128mins and 256mins respectively while A takes a constant time of 1min.
Thus they meet 127 + 255 = 382 times
Q3)
The sum of the series is:
1/1.2.3 + 1/3.4.5 +1/5.6.7 + --------------------------
1/1.2.3 + 1/3.4.5 +1/5.6.7 + --------------------------
1)
e2 -1
2)
loge2 -1
3)
2log102 -1
4)
None of above
5)
Correct Answer: 4
Your Answer: Not Attempted
Course: CAT Quantitative
This Question Is Correctly Answered By: 40% test takers
Answer Description
S = 1/ 1. 2. 3 + 1/3. 4. 5 + 1/5. 6. 7 + ----------------
Tn = 1/M(n+1) (n+z)
= {1/n - 1/(n+1)} 1/(n+2)
=1/n(n+2) - 1/(n+1) (n+2)
=1/2[1/n - 1/(n+2)] - {1/(n+1) - 1/(n+2)}
To find the sum,put n = 1,2,3 and add.
Sn = 1/2 [1-1/3] - {1/2 -1/3}
+1/2[1/2 -1/4] - {1/3 -1/4}
+1/2[1/3 -1/5] - {1/4 - 1/5}
The terms start can celling out so, the sum is
Sn = 1/2[1+1/2] - 1/2
=1/2+1/4-1/2
=1/4
S = 1/ 1. 2. 3 + 1/3. 4. 5 + 1/5. 6. 7 + ----------------
Tn = 1/M(n+1) (n+z)
= {1/n - 1/(n+1)} 1/(n+2)
=1/n(n+2) - 1/(n+1) (n+2)
=1/2[1/n - 1/(n+2)] - {1/(n+1) - 1/(n+2)}
To find the sum,put n = 1,2,3 and add.
Sn = 1/2 [1-1/3] - {1/2 -1/3}
+1/2[1/2 -1/4] - {1/3 -1/4}
+1/2[1/3 -1/5] - {1/4 - 1/5}
The terms start can celling out so, the sum is
Sn = 1/2[1+1/2] - 1/2
=1/2+1/4-1/2
=1/4
Q4)
If Log2x.Logx/642 = Logx/16 2. Then x is:
1)
2
2)
4
3)
16
4)
12
5)
Correct Answer: 2
Your Answer: Not Attempted
Course: CAT Quantitative
This Question Is Correctly Answered By: 45% test takers
Answer Description
Log2x . Log(x/64)2 = Log(x/16)2
=>Logx/Log2 . Log2/Log(x/64) = Log2/Log(x/16)
=>Logx/Log2 . Log2/Log x - 6Log2 = Log2/Log x - 4Log2
=>Log x . (Log x - 4 Log 2) = Log2(Log x - 6 Log2)
=>(Log x)2 - 4Log2 . (Log x) = Log 2 . (Log x) - 6(Log2)2
=>(Log x)2 - 5Log2(Log x) + 6(Log2)2 = 0
=>(Log x)2 - 3Log2(Log x) - 2Log2(Log x) + 6(Log2)2 = 0
=>(Log x - 3Log2) (Log x - 2 Log2) = 0
=> so Log x = Log8 or Log x = Log4
=>x = 8 or x = 4
Log2x . Log(x/64)2 = Log(x/16)2
=>Logx/Log2 . Log2/Log(x/64) = Log2/Log(x/16)
=>Logx/Log2 . Log2/Log x - 6Log2 = Log2/Log x - 4Log2
=>Log x . (Log x - 4 Log 2) = Log2(Log x - 6 Log2)
=>(Log x)2 - 4Log2 . (Log x) = Log 2 . (Log x) - 6(Log2)2
=>(Log x)2 - 5Log2(Log x) + 6(Log2)2 = 0
=>(Log x)2 - 3Log2(Log x) - 2Log2(Log x) + 6(Log2)2 = 0
=>(Log x - 3Log2) (Log x - 2 Log2) = 0
=> so Log x = Log8 or Log x = Log4
=>x = 8 or x = 4
Q5)
A right circular cone is enveloping a right circular cylinder such that the base of the cylinder rests on the base of the cone. If the radius and the height of the cone is 4 cm and 10 cm respectively, then the largest possible curved surface area of the cylinder of radius r is :
1)
20πr2
2)
5π(4-r)
3)
5π(r-4)
4)
5π(2-r)
5)
Correct Answer: 2
Your Answer: Not Attempted
Course: CAT Quantitative
This Question Is Correctly Answered By: 57% test takers
Answer Description
5pi*r(4-r)
The curved surface area of the cylinder is 2pi*r*h.
Triangles ABC and ADE are similar.
Which implies (4-r)/4 = h/10
h= 5(4-r)/2
Therefore CSA = 5pi*r(4-r)
5pi*r(4-r)
The curved surface area of the cylinder is 2pi*r*h.
Triangles ABC and ADE are similar.
Which implies (4-r)/4 = h/10
h= 5(4-r)/2
Therefore CSA = 5pi*r(4-r)
Q6)
Radius of a spherical ballon, of radii 30cm, increases at the rate of 2cm per second. Then its curved surface area increases by:
1)
120π
2)
480π
3)
600π
4)
None of the above
5)
Correct Answer: 2
Your Answer: Not Attempted
Course: CAT Quantitative
This Question Is Correctly Answered By: 50% test takers
Answer Description
V = 4πr2
dv/dt = 8πr . dr/dt
r = 30 cm, dr/dt = 2 cm
so dv/dt = 8π x 30 x 2
= 480π
V = 4πr2
dv/dt = 8πr . dr/dt
r = 30 cm, dr/dt = 2 cm
so dv/dt = 8π x 30 x 2
= 480π
Q7)
Mohan was playing with a square cardboard of side 2metres.While playing , he sliced off the corners of the cardboard in such a manner that a figure having all its sides equal was generated.The area of this eight sided figure is:
1)
4
/( 
+1)
/( +1)
2)
4/( 
+1)
+1)
3)
2
/( 
+1)
/( +1)
4)
8/( 
+1)
+1)
5)
Correct Answer: 4
Your Answer: Not Attempted
Course: CAT Quantitative
This Question Is Correctly Answered By: 30% test takers
Answer Description
t----------x----------t
Let the length of the side of the octagon be x. And the length remaining on the square on either side of the side of the octagon be t.
The 2t + x = 2
Also, t2+t2 = x2
Solving this we get t = 2/
( 
+ 1)
Area of the octagon = Area of the square
- 4 x Area of right iso sceles triangle of equal side t.
= (2)2 - 4 x 1/2 x t2
Putting the value of t we get :
Area = 8 / (
+ 1)
t----------x----------t
Let the length of the side of the octagon be x. And the length remaining on the square on either side of the side of the octagon be t.
The 2t + x = 2
Also, t2+t2 = x2
Solving this we get t = 2/
( + 1)Area of the octagon = Area of the square
- 4 x Area of right iso sceles triangle of equal side t.
= (2)2 - 4 x 1/2 x t2
Putting the value of t we get :
Area = 8 / (
+ 1)
Q8)
Because of economic slowdown, a multinational company curtailed some of the allowences of its emloyees.Rashid, the marketing manager of the company whose monthly salary has ben reduced to Rs. 42000 is unable to cut down his expenditure.He finds that there is deficit, because of inflationary pressure,will keep on increasing by Rs.500 every month.Rashid has a saving of Rs.60000 which will be used to fill this defict.After his savings get exhausted, Rashid would start borrowing from his friends.How soon will he start borrowing?
1)
10th month
2)
11th month
3)
12th month
4)
13th month
5)
Correct Answer: 4
Your Answer: Not Attempted
Course: CAT Quantitative
This Question Is Correctly Answered By: 50% test takers
Answer Description
Options start from the tenth month. After the tenth month he would have used up [2000 + (2000 + 9*500)] * 10/2 = 42500 of his saving and the extra expenditure for the 10th month will be 6500. Similarly after the 12th month he would’ve used up 42500 + 7000 + 7500 = 57000 of his savings. Next month, he would not be able to use his savings of 3000 to meet an excess expenditure of 8000 and thus would start borrowing from the 13th month.
Options start from the tenth month. After the tenth month he would have used up [2000 + (2000 + 9*500)] * 10/2 = 42500 of his saving and the extra expenditure for the 10th month will be 6500. Similarly after the 12th month he would’ve used up 42500 + 7000 + 7500 = 57000 of his savings. Next month, he would not be able to use his savings of 3000 to meet an excess expenditure of 8000 and thus would start borrowing from the 13th month.
Q9)
The number of distinct terms in the expansion of (x + y + z + w)30 are:
1)
4060
2)
5456
3)
27405
4)
46376
5)
Correct Answer: 2
Your Answer: Not Attempted
Course: CAT Quantitative
This Question Is Correctly Answered By: 88% test takers
Answer Description
Number of distinct terms = n+r -1
C
r - 1
= 30 + 4 - 1
C
3
Number of distinct terms = n+r -1
C
r - 1
= 30 + 4 - 1
C
3
Q10)
A card is drawn at random from a well shuffled pack of 52 cards.
X : The card drawn is black or a king
Y : The card drawn is a club or a heart or a jack
Z : The card drawn is an ace or a diamond or a queen.
Then which of the following is correct?
X : The card drawn is black or a king
Y : The card drawn is a club or a heart or a jack
Z : The card drawn is an ace or a diamond or a queen.
Then which of the following is correct?
1)
P(X)>P(Y)>P(Z)
2)
P(X)>=P(Y) =P(Z)
3)
P(X)=P(Y)>P(Z)
4)
P(X) =P(Y) =P(Z)
5)
Correct Answer: 3
Your Answer: Not Attempted
Course: CAT Quantitative
This Question Is Correctly Answered By: 67% test takers
Answer Description
N (X) = 26+2 = 28
P (X) = N (X) / N (total) = 28/52 = 7/13
N(Y) = 13+13+ 2 = 28
P(Y) = N(Y) / N (total) = 28/52 = 7/13
N(Z) = 4+12+3 = 19
P(Z) = N(2) / N(total) = 19/52
We have, P(X) = P(Y)
P(Z)
N (X) = 26+2 = 28
P (X) = N (X) / N (total) = 28/52 = 7/13
N(Y) = 13+13+ 2 = 28
P(Y) = N(Y) / N (total) = 28/52 = 7/13
N(Z) = 4+12+3 = 19
P(Z) = N(2) / N(total) = 19/52
We have, P(X) = P(Y)
P(Z) Showing 1 to 10 out of 119 Answers
