AMCAT Quantitative Ability Previous Papers-6
51. If the logx 16 = 0.8, then what is the value of x?
A. 4 B. 2 C. 32 D. 16
Solution:
x^0.8=16
x=(2^4)^(10/8)
x=2^5
x=32
52. If f(x) = ex and g(x) = loge x, then what is the value of {f of g(x)}?
A. X B. ex C. logex D. cannot be determined
Ans: The answer is not solved, please comment the answer and it would be updated for others help
53. The value of log864 –log644096 is
A. -1 B. 0 C. 1 D. 2
Ans: 0
54. What is the relationship between the fractions 14/15 and 37/40?
A.14/15=37/40 B.14/15>37/40 C.14/15<37/40 D. Cannot be determined
Option c)
Take LCM of both 15, 40 we get 120
(14*8)/(15*8) = 112/120
(37*3)/(40*3) = 111/120
So 14/15>37/40
55. If LCM and HCF of two numbers are equal and product of two numbers is 2916, find their LCM?
A. 54 B. 56 C.64 D. 66
LCM and HCF will be equal if and only if both the numbers are equal.
therefore, 2916=54*54
so LCM of both number =54
56. If the sum of squares of two numbers is 2754 and their HCF and LCM are 9 and 135 respectively, then the numbers are
A.27, 36 B. 27, 35 C. 28, 45 D. 27, 45
As we know product of two nos.=their H.C.F*L.C.M
So,x*y=135*9=1215
and x^2+y^2=2754
So,(x+y)^2=x^2+y^2+2*x*y
=2754+2*1215=5184
So,x+y=72
Now on solving,the nos. are 45 and 27
57. The number of ways in the letters of the word “RESULT” can be arranged without repetition is
A.720 B. 120 C. 60 D. 840
Ans: The answer is not solved, please comment the answer and it would be updated for others help
58. The HCF of two numbers is 4 and LCM is 48. If one of the numbers is 12, then one of the divisions of the numbers is:
A.3 B. 6 C. 8 D. 12
Ans: The answer is not solved, please comment the answer and it would be updated for others help
59. Which number should be added to 113257 so that it can be divided by 9?
A. 4 B. 6 C. 8 D. 10
Ans: The answer is not solved, please comment the answer and it would be updated for others help
60. ABCD is a square PQRS is a rhombus lying inside the square such that P, Q, R and S are the mid-points of AB, BC, CD and DA respectively. A point is selected at random in the square. Find the probability that lies in the rhombus
A.1/3 B. 2/3 C. 1/2 D. 1/4
AMCAT Quantitative Ability Previous Papers-7
61. 4.28 and -3.28 are two numbers on a real number line. If 1 is added to the numbers, then which of the following is true?
A. Distance between the two new numbers is 2 more than distance between 4.28 and -3.28
B. Distance between the two new numbers is 2 less than distance between 4.28 and -3.28
C. Distance between the two new numbers is 2 equal distance between 4.28 and -3.28
D. None of these
62. When a local train travels at a speed of 60 kmph. It reaches the destination on time . when the same train travels at speed of 50kmph. It reaches its destination15 minutes late. What is the length of journey?
A. 75km B.50km C. 60km D.85km
Solution:
distance is constant.
so let time taken be t when it travels with 60kmph
let time be t' when it travels with 50kmph
d=s*t
therfore
s*t=s'*t'
60*t=50*(t+15/60)(in hrs)
on solving
t=5/4 hrs
d=60*5/4
75km
63. In a shooting competition, the probability of hitting the target by P is 3/5. By Q is 1/3 and by R is 2/5. If all they fire independently at the same target calculate the probability that only one of them will hit the target.
A. 25/75 B. 32/75 C. 39/75 D. 43/75
A. 25/75 B. 32/75 C. 39/75 D. 43/75
Solution:
ans .32/75.
P(p)=3/5 P(p')=2/5 same way for rest and by using formula
P(p)*P(q')*P(r')+P(p')*P(q)*P(r')+P(p')*P(q')*P(r)
64. For what value of M is the number 7M42876M divisible by 11?
A. 0 B. 8 C.4 D. 9
Solution:
Here an easy way to test for divisibility by 11. Take the alternating sum of the digits in the number, read from left to right. If that is divisible by 11, so is the original number.
So, for instance, 2728 has alternating sum of digits 2-7+2-8 = -11. Since -11 is divisible by 11, so is 2728.
Similarly, for 31415, the alternating sum of digits is 3-1+4-1+5 = 10. This is not divisible by 11, so neither is 31415.
Coming to the given question,
7-M+4-2+8-7+6-M
16-2M
so if M=8 then 16-16=0 so total number is divisible by 11.
65. A group of women can finish a piece of work in 50 days. In how many days will one-third the number of women be able to finish two-third of the work?
A. 150 days B. 75 days C. 50 days D. 100 days
Solution:
x women takes 50 days to complete a piece of work.
1/3 x women takes 150 days to complete a piece of work.
1/3 x women takes 100 days to complete 2/3 rd work.
Solution:
x women takes 50 days to complete a piece of work.
1/3 x women takes 150 days to complete a piece of work.
1/3 x women takes 100 days to complete 2/3 rd work.
66. What is the value of log512 8?
A. 3 B. 1/3 C. -3 D. -1/3
Solution:
log512 8
= log512 (512)1/3
=1/3
67. If log10 3 = 0.477, then the value of log10 9 is:
A. 1 B. 0.477 C. 0.954 D. 0.523
Solution:
log10 9= 2 log 10 3 = 2 * 0.477 = 0.954
A. 1 B. 0.477 C. 0.954 D. 0.523
Solution:
log10 9= 2 log 10 3 = 2 * 0.477 = 0.954
68. When the price of a pair of shoes is decreased by 10%, the number of pairs sold increased by 20%. What is the net effect on scales?
A. 8% decrease B.10% decrease C.10% increase D. 8% increase
A. 8% decrease B.10% decrease C.10% increase D. 8% increase
ans: let cp be 100, when price decreased by 10% then sp=90. Now sales increased by 20%, ie, 90+20%of90=108.
so cp=100, sp=108, gain=8, gain%=gain*100/cp=8%increase
69. Replace the symbols * and # in 9586*4# so that it is divisible by both 8 and 5.
A. 0,0 B. 1,0 C. 0,5 D. 1,5
Solution:
Rule for Divisible by 5 : number should end with either 0 or 5
Rule for divisible by 8: Last 3 digits should be divisible by 8
Solve from options: If option (A) is correct then number becomes 9586040
It is divisible with both 5 and 8 so Option(A) is correct.
A. 0,0 B. 1,0 C. 0,5 D. 1,5
Solution:
Rule for Divisible by 5 : number should end with either 0 or 5
Rule for divisible by 8: Last 3 digits should be divisible by 8
Solve from options: If option (A) is correct then number becomes 9586040
It is divisible with both 5 and 8 so Option(A) is correct.
70. If a coin with both heads is tossed, then the probability of obtaining a tail is:
A. 0 B. ½ C. 1/3 D. 1
A. 0 B. ½ C. 1/3 D. 1
Solution: If a coin contains both heads then no way we get tail so probablity is 0.
AMCAT Quantitative Ability Previous Papers-2
a)1/2 b)1/3 c)3/5 d)2/5
Ans) A single letter drawn at random from the above given word is 5c1/10 = 1/2
13)Suparna needs to browse through 75pages of a novel before she gives her review to the class. She has 2.5 hrs before the lecture. What should be her reading speed in pages/hour?
a) 16 b)30 c) 20 d) 22
Ans) speed=distance/time
S=75/2.5
S=30
14) The value of log100.1 is :
a) 0 b) -1 c) -10 d) -100
Ans)log100.1 = log1010-1 = -1 log 1010 = -1 (log 1010 =1)
15) A written exam consists of 6 questions with the answer options as yes/no/none. In how many ways can the examinees select the answers?
a) 6 ways b) 6 ways c) 3 .3 .3 .3 .3 d) (3 6
Ans) In 36 ways the examiners select the answers.
16) What is the sum of the two consecutive numbers, If the difference of whose squares is 19?
a. 9 b. 10 c. 18 d. 19
Ans) (n+1)2-n2=19
We get 2n=18, n=917) P is an integer. P>883. If (p-7) is a multiple of 11, then the largest number that will divide (p+4) (p+15) is :
11 121 242 None of the above
Ans) 242
Given P is an integer>883.
P-7 is a multiple of 11=>there exist a positive integer a such that
P-7=11 a=>P=11 a+7
(P+4)(P+15)=(11 a+7+4)(11 a+7+15)
=(11 a+11)(11 a+22)
=121(a+1)(a+2)
As a is a positive integer therefore (a+1)(a+2) is divisible by 2.Hence (P+4)(P+15) is divisible by 121*2=242
11 121 242 None of the above
Ans) 242
Given P is an integer>883.
P-7 is a multiple of 11=>there exist a positive integer a such that
P-7=11 a=>P=11 a+7
(P+4)(P+15)=(11 a+7+4)(11 a+7+15)
=(11 a+11)(11 a+22)
=121(a+1)(a+2)
As a is a positive integer therefore (a+1)(a+2) is divisible by 2.Hence (P+4)(P+15) is divisible by 121*2=242
18) Find the least number which when divided by 5, 7 and 13 leaves the same remainder 3 in each case
398 453 458 463
Ans) By trial method we get the answer as 458.Why because if we divide the number with 5,7,13 it leaves the remainder as 3 in all the three cases.
398 453 458 463
Ans) By trial method we get the answer as 458.Why because if we divide the number with 5,7,13 it leaves the remainder as 3 in all the three cases.
19)Which number should be subtracted from 321 so that it becomes prime?
2 4 6 9
Ans) If we apply trial method 4 is the answer.If we subtract 317 from 4 the result will be 317hence this is the prime number
2 4 6 9
Ans) If we apply trial method 4 is the answer.If we subtract 317 from 4 the result will be 317hence this is the prime number
410 210 216 416
28 *22 =28+2 =210
AMCAT Quantitative Ability Previous Papers-3
21) What will be the value of the expression a8/3 * a-6/9 ?
a-2 a-1 a0 a1 a2
a-2 a-1 a0 a1 a2
a8/3-6/9
a18/2=a2 .
22) What is the square root of 576/9?
1.4 2.8 3.12 4.16
Ans.8
1.4 2.8 3.12 4.16
Ans.8
23) Which number is the fourth power of 7?
1. 2401 2. 2421 3. 2601 4. 2621
1. 2401 2. 2421 3. 2601 4. 2621
ans: 72*72
=49*49 = 2401
24) HCF of two numbers is 11 and their LCM is 693. If one number is 77, find the other number?
1. 7 2. 9 3. 63 4. 99
1. 7 2. 9 3. 63 4. 99
77*x =11*693
X=11*693/77
X=99
25) Recycling 900 kg of paper saves 17 trees . How many trees are saved when 1200 kg of paper are recycled?
1. 19 2. 25 3. 20 4. 22
1. 19 2. 25 3. 20 4. 22
1200kg papers=?
The trees to find is=X
X= 1200*17/900
=68/3
=22
26) How many different four letter words can be formed ( the words need not to be meaningful ) using the letters of the word PACIFIC such that the first letter is p and the last letter is F?
1. 8 2. 3 3. 6 4. 7!/5!
1. 8 2. 3 3. 6 4. 7!/5!
Total number of letters n=7
R=2
Npr =n!/(n-r)!
=7!/(7-2)!
=7!/5!
27) Mauli purchased a designer saree from Mumbai at 8/9th of its MRP. When she came back to Delhi, her neighbour coaxed mauli to sell the saree to her. She was even ready to pay 9% more than its MRP. What would Mauli’s gain percentage be, if she decides to sell the saree to her neighbour?
1. 15.59% 2. 16.61% 3. 20.36% 4. 22.65%
cost prize = 800/9
selling prize=(100*9/100)+100 =109
sp=(100+gain%/100)cp
109 =((100+gain%)/100)800/9
By solving above equation we get gain=181/8
=22.65%
28) A goods carriage of length 2km, headed to Srinagar from Punjab was running at a speed of 30 km/hr. It crosses a tunnel which is 58 km long with that speed. Find the time taken by the goods carriage to cross the tunnel?
1. 4 hours 2. 3 hours 3. 2 hours 4. 1 hour
1. 4 hours 2. 3 hours 3. 2 hours 4. 1 hour
time =? Speed=30km/hr
Time =distance /speed
=(2+56) /30
Time=60/30
=2hours
29) A lucky draw is organized as part of the first anniversary celebration of new Age Company. There are 25 chits in a bowl one for each employee and the chits are marked from 1-25. Sarika and Rajesh have chits marked with numbers that are multiples of 3 or 7. They want to know if there are chances of them being awarded the trip to Goa which is the first prize of the lucky draw. When one chit is drawn at random, what is the probability that the chit has a number which is a multiple of 3 or 7?1. 3/25 2. 2/11 3. 11/25 4. 10/25
sarika and ragesh chits aremultiples of 3 and 7
3 multiples up to 25 =8
7multiples up to 25 =2
Total multiples= 8+2 =10
=total multiples / total no of chits
=10/25
30) What is the loss percentage incurred by a company when it buys an asset for Rs. 1,50,000 and sells it for Rs. 75,500?
1. 49.67% 2. 49.34% 3. 98.68% 4. 98.34%
1. 49.67% 2. 49.34% 3. 98.68% 4. 98.34%
=1,50,000-75000
=74,500
Loss% = (loss/cp)*100
=(74000/150000)*100
AMCAT Quantitative Ability Previous Question Papers with solutions-1
1) The probability of getting at least one tail in 5 throws of a coin is?
a)1/32 b)31/32 c)1/5 d)None of the above
Ans. a coin throws 5 times is equal to the 5 coins throws at one time.
So total no of choices are 25=32
Chances that not getting at least one tails is 1 i.e., (h,h,h,h,h)
So 1-(1/32)=31/32…option (b) is correct.
2) When a local train travels at a speed of 60kmph, it reaches the destination on time. When the same train travels at speed of 50kmph, it reaches its destination 15 minutes late. What is the length of journey?
a)75km b)50km c)60km d)85km
Ans. let the time taken by a train to reach destination at 60kmph is x. Then it takes (x+15) at 50kmph.
So distance, 60x=50(x+15)
X=75…option (a) is correct.
3)Give the greatest pair of twin primes which are below 100?
a)71,73 b)93,95 c)97,99 d)87,89
Ans. in the given options only 71 and 73 are primes.
So option (A) is correct.
4) In an examination involving quantitative aptitude and logical reasoning, 65% examinees cleared quantitative aptitude test while 70% cleared logical reasoning test. If 50% examinees passed both the tests, then how many failed in both tests?
a)35% b)15% c)30% d)20%
Ans. no of students who passed in at least one subject:
AUB=65+70-50=85
If total students are 100 .no of students who failed in both subjects = total students-students who passed in at least one subjective. 100-85=15
So option (b) is correct
5) A sum of money triples itself at compound interest in3 years. In 9 years it will be
a)6 times the principal b)12 times the principal
c)18 times the principal d)27 times the principal
Ans. let the principal be x.
X(1+R/100)3=3x
(1+R/100)^3^3=3^3=27
So option (d) is correct.
7) The number which should be subtracted frm 5a2-3ab+7b2 to make it equal to a2+ab+b2 ,is:
a)4a2-4ab+6b2 b)4a2-4ab+5b2 c)4a2+4ab+6b2 d)4a2-3ab+6b2
e)None of the above
Ans.Option a is correct.
8)Given that the interest is only earned on principal, if an investment of Rs.1000.00 amount to Rs.1440.00 in two years, then what is the rate of interest earned?
a)20% b)22% c)21% d)11% e)44%
Ans. 1000(1+r/100)2=1440
So the answer is 20%.
Option (A) is correct.
9)If nC5=nC0, then find the value of n.
a)n=0 b)n =1 c)n=5 d)n=10
Ans. nC5=nCn-5=nC0
n-5=0
n=5..so option (c) is correct
10)A bag contains 5 oranges, 4 bananas and 3 apples. Rohit wants to eat a banana or an apple.He draws a fruit from the bag randomly. What is the probability that he will get a fruit of this choice?
a)3.5/12 b)7/12 c)5/12 d)None of the above
Ans. total fruits are 12.
Chances to select banana or apple is: 4+3=7
Probability =7/12
So option (b) is correct.