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QUANTITAITVE APTITUDE CLUBBED MCQs
Total Questions : 1394
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The number will be of the form (2x)×(3y)×(7x).
And 0≤x,y,z≤2 (Since N is not divisible by any perfect cube greater than 1).
Hence sum of all possible combination of numbers is: (1+2+4)×(1+3+9)×(1+7+49)−1=5187−1=5186
Answer: Option A. -> 6
:
A
A Radius of semicircles ACB, AC and BC are 2.5, 2 and 1.5 respectively.
Area of Semi circles (ACB), (AC) and (BC) are 6.25π2, 2π and 2.25π2 respectively. And area of
triangle ABC =12 ×4 × 3 = 6
Sum of the area of region A and B = (Area of semicircle (ACB) - Area of triangle (ABC)} = (6.25π2−6)
Area of Shaded region = (Sum of the area of semicircles AC and BC - Sum of the area of region A and B)
Area of Shaded region = (2π+2.25π2) - (6.25π2−6) = (6.25π2)- (6.25π2−6) = 6
Alternate Solution:
As the options are far apart we can easily arrive at the answer by approximation. The shaded
region is approximately half of the areas of semicircles AC and BC = 14(4π+2.25π) = 6(approximately)
:
A
A Radius of semicircles ACB, AC and BC are 2.5, 2 and 1.5 respectively.
Area of Semi circles (ACB), (AC) and (BC) are 6.25π2, 2π and 2.25π2 respectively. And area of
triangle ABC =12 ×4 × 3 = 6
Sum of the area of region A and B = (Area of semicircle (ACB) - Area of triangle (ABC)} = (6.25π2−6)
Area of Shaded region = (Sum of the area of semicircles AC and BC - Sum of the area of region A and B)
Area of Shaded region = (2π+2.25π2) - (6.25π2−6) = (6.25π2)- (6.25π2−6) = 6
Alternate Solution:
As the options are far apart we can easily arrive at the answer by approximation. The shaded
region is approximately half of the areas of semicircles AC and BC = 14(4π+2.25π) = 6(approximately)
Answer: Option B. -> 2012
:
B
Method 1- Conventional Approach
f(x) = x15- 2013{x14 - x15 + x12 - ......... + x2 - x} or
f(x) = x15- 2013{-x + x2 - x3 + ........ - x13 + x14}
f(x) = x15 - (2013x(x14−1)(x+1))
Now, Put x = 2012
f(2012) = (2012)15 - (2013.2012(201214−1)(2012+1))
f(2012) = (2012)15 - 2012(2012^{14} - 1)
f(2012) = 2012.
Method 2- Assumption
Assume 2012 = n and 2013 = (n+ 1). Put x=1. Now replace the options in terms of "n" as
a) (n-1) b) n c) (n + 1) d) (n + 2)
f(1)=1 - 2 + 2 - 2 + 2 - 2 + 2 - ......................-2 + 2
f(1) = 1. Options (a),(c) & (d) can be eliminated as only option (b) gives us 1.
:
B
Method 1- Conventional Approach
f(x) = x15- 2013{x14 - x15 + x12 - ......... + x2 - x} or
f(x) = x15- 2013{-x + x2 - x3 + ........ - x13 + x14}
f(x) = x15 - (2013x(x14−1)(x+1))
Now, Put x = 2012
f(2012) = (2012)15 - (2013.2012(201214−1)(2012+1))
f(2012) = (2012)15 - 2012(2012^{14} - 1)
f(2012) = 2012.
Method 2- Assumption
Assume 2012 = n and 2013 = (n+ 1). Put x=1. Now replace the options in terms of "n" as
a) (n-1) b) n c) (n + 1) d) (n + 2)
f(1)=1 - 2 + 2 - 2 + 2 - 2 + 2 - ......................-2 + 2
f(1) = 1. Options (a),(c) & (d) can be eliminated as only option (b) gives us 1.
Answer: Option C. -> (n + 1)! - 1
:
C
Put n = 2
Given expression = 5
Option (a) = 2
Option (b) = 1
Option (c) = 5
Option (d) = 6
Hence option (c)
:
C
Put n = 2
Given expression = 5
Option (a) = 2
Option (b) = 1
Option (c) = 5
Option (d) = 6
Hence option (c)
Question 65. Two rugby teams (thirty players) have been selected to play for their school. Five of the players speak Spanish, Afrikaans and English. Nine of them speak only Spanish and English. Twenty speak Afrikaans, of which twelve also speak Spanish. Eighteen speak English. No one speaks only Spanish. How many players speak only English?
Answer: Option D. -> 3
:
D
Let us see the sequence of the numbers:
Number Last term of the number
11
23
36
410
−n−n(n+1)2
We have to find the value of n for the 2000th term.
If n = 62, the last term that ends with n is (12×62×63) = 1953.
Therefore, the next 63 terms are 63. So the 2000th term is 63. So the remainder when divided by 4 is 3.
:
D
Let us see the sequence of the numbers:
Number Last term of the number
11
23
36
410
−n−n(n+1)2
We have to find the value of n for the 2000th term.
If n = 62, the last term that ends with n is (12×62×63) = 1953.
Therefore, the next 63 terms are 63. So the 2000th term is 63. So the remainder when divided by 4 is 3.
Answer: Option C. -> 10
:
C
To get a set of 9 numbers, none of which are divisible by 10, we need to consider the lowest valued number in the set to have a value 1 above the divisor of 10. 0 is a divisor, one above that will be 1 and the lowest valued number in the set is n-4.
So, n-4=1. Now subsequent sets will occur in an AP with a common difference of 10.
1,11,21,31,41.....91 as n≤100
No. of such numbers =91−110+1=9010+1=9+1=10
:
C
To get a set of 9 numbers, none of which are divisible by 10, we need to consider the lowest valued number in the set to have a value 1 above the divisor of 10. 0 is a divisor, one above that will be 1 and the lowest valued number in the set is n-4.
So, n-4=1. Now subsequent sets will occur in an AP with a common difference of 10.
1,11,21,31,41.....91 as n≤100
No. of such numbers =91−110+1=9010+1=9+1=10
Answer: Option D. -> 8
:
D
Suppose we take x =50-paisa coins, y = 25-paisa coins and z =10-paisa coins.
We have x + y + z = 85 and 50x + 25y + 10z = 1875, because x, y and z are all integers, z should be a multiple of 5.
Possible values are
xyz263205552584730........23755
So there are eight solutions.
You should notice here that an increase of 5 in z and 3 in x and decrease of 8 in y neither changes the number of coins nor the value.
:
D
Suppose we take x =50-paisa coins, y = 25-paisa coins and z =10-paisa coins.
We have x + y + z = 85 and 50x + 25y + 10z = 1875, because x, y and z are all integers, z should be a multiple of 5.
Possible values are
xyz263205552584730........23755
So there are eight solutions.
You should notice here that an increase of 5 in z and 3 in x and decrease of 8 in y neither changes the number of coins nor the value.
Question 69. Answer the questions based on the information given below.
A microorganism Xeta combines with another microorganism of its kind to form a single microorganism unit. If it does not find another Xeta every second, it dies. There are 2x−1 Xeta in the incubation flask initially. Xeta will find another Xeta provided it is there in the same chamber.
When will all the Xeta be dead?
A microorganism Xeta combines with another microorganism of its kind to form a single microorganism unit. If it does not find another Xeta every second, it dies. There are 2x−1 Xeta in the incubation flask initially. Xeta will find another Xeta provided it is there in the same chamber.
When will all the Xeta be dead?
Answer: Option A. -> x seconds
:
A
Assume no. of Xeta=1 (you can assume any value which satisfies 2x−1)
Then, x = 1.
If there is 1 Xeta, it will die after the 1st second.
Put x=1 in the options and check which option gives answer as 1.
Answer will be option (a) - x seconds.
:
A
Assume no. of Xeta=1 (you can assume any value which satisfies 2x−1)
Then, x = 1.
If there is 1 Xeta, it will die after the 1st second.
Put x=1 in the options and check which option gives answer as 1.
Answer will be option (a) - x seconds.
Answer: Option C. -> 392
:
C
Aàcows
Bàcalves
2 cows and 7 calves are needed to do a piece of work in 14 days. This means to complete the work we need 2 cows to graze for 14 days and 7 calves to graze for 14 days. So if we want to complete the work in one day we need: 2×(14)=28cows and 7×(14)=98calves ⇒ 28A + 98B.
Also, 3 cows and 8 calves take 11 days, so if they need to complete the work in 1 day, we need 3(11) =33 cows and 8 (11) =88 calves ⇒ 33A + 88B
We can now equate the two cases as 28A + 98B = 33A + 88B
Solving we get5A=10B or 1A=2B
i.e. one cow is equivalent to two calves.
We need to calculate for 8 cows & 6 calves or 8+3=11 cows
3 cows and 8 calves can graze the field in 11 days ⇒ 7 cows can graze the field in 11 days
⇒ 11 cows can graze the field in 7 days
Thus, 11 cows can graze 4 fields in 7 × 4=28 days. Option (d)
:
C
Aàcows
Bàcalves
2 cows and 7 calves are needed to do a piece of work in 14 days. This means to complete the work we need 2 cows to graze for 14 days and 7 calves to graze for 14 days. So if we want to complete the work in one day we need: 2×(14)=28cows and 7×(14)=98calves ⇒ 28A + 98B.
Also, 3 cows and 8 calves take 11 days, so if they need to complete the work in 1 day, we need 3(11) =33 cows and 8 (11) =88 calves ⇒ 33A + 88B
We can now equate the two cases as 28A + 98B = 33A + 88B
Solving we get5A=10B or 1A=2B
i.e. one cow is equivalent to two calves.
We need to calculate for 8 cows & 6 calves or 8+3=11 cows
3 cows and 8 calves can graze the field in 11 days ⇒ 7 cows can graze the field in 11 days
⇒ 11 cows can graze the field in 7 days
Thus, 11 cows can graze 4 fields in 7 × 4=28 days. Option (d)
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