Free Objective Test 01 Practice Test - 11th and 12th
Question 1
The value of 2n{1.3.5.....(2n−3)(2n−1)} is
(2n)!n!
(2n)!2n
n!(2n)!
(2n−1)!n!
SOLUTION
Solution : A
1.3.5......(2n−1)2n = 1.2.3.4.5.6....(2n−1)(2n)2n2.4.5.....2n
= (2n)!2n2n(1.2.3......n) = (2n)!n!
Question 2
A five digit number divisible by 3 has to be formed using the numerals 0, 1, 2, 3, 4 and 5 without repetition. The total number of ways in which this can be done is
216
240
600
3125
SOLUTION
Solution : A
We know that a five digit number is divisible by 3, if and only if sum of its digits (= 15) is divisible by 3, therefore we should not use 0 or 3 while forming the five digit numbers.
Now,
(i) In case we do not use 0 the five digit number can be formed (from the digit 1, 2, 3, 4, 5) in 5P5 ways.
(ii) In case we do not use 3, the five digit number can be formed (from the digit 0, 1, 2, 4, 5) in 5P5−4P4 = 5! - 4!= 120 - 24 = 96 ways.
(4P4= cases when 0 is at the first position)∴The total number of such 5 digit number = 5P5+(5P5−4P4)=120+96=216
Question 3
The number of times the digit 3 will be written when listing the integers from 1 to 1000 is
269
300
271
302
SOLUTION
Solution : B
To find the number of times 3 occurs in listing the integer from 1 to 999. (since 3 does not occur in 1000). Any number between 1 to 999 is a 3 digit number xyzz where the digit x,y,z are any digits from 0 to 9.
Now, we first count the numbers in which 3 occurs once only. Since 3 can occur at one place in 3C1 ways, there are 3C1.(9×9) = 3.92 such numbers.
Again, 3 can occur in exactly two places in 3C1(9) such numbers. Lastly 3 can occur in all the three digits in one such number only 333.
∴The number of times 3 occurs is equal to 1×(3×92)+2×(3×9)+3×1=300.
Question 4
The number of numbers of 4 digits which are not divisible by 5 are
7200
3600
14400
1800
SOLUTION
Solution : A
The total number of 4 digits are 9999-999=9000.
The numbers of 4 digits number divisible by 5 are 90×20=1800. Hence required number of ways are 9000-1800 =7200.
{Since there are 20 numbers in each hundred (1 to 100) divisible by 5 and from 999 to 9999 there are 90 hundreds, hence the results}.
Question 5
An n-digit number is a positive number with exactly n digits. Nine hundred distinct n-digit numbers are to be formed using only the three digits 2, 5 and 7. The smallest value of n for which this is possible is
6
7
8
9
SOLUTION
Solution : B
Since at any place, any of the digits 2, 5 and 7 can be used, total number of such positive n-digit numbers are 3n. Since we have to form 900 distinct numbers, hence 3n≤900⇒n = 7
Question 6
How many numbers lying between 10 and 1000 can be formed from the digits 1, 2, 3, 4, 5, 6, 7, 8, 9 (repetition is allowed)
1024
810
2346
729
SOLUTION
Solution : B
The total number of numbers between 10 and 1000 are 989 but we have to form the numbers by using numerals 1,2,........9, so the numbers containing any '0' would be excluded i.e., Required number of ways
= 989 - ⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩20,30,40,...........................100=9101,102,...........................200=19201,...............................300=19..................................................................................901,..............................900=18⎫⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎬⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎭
= 989−(9+18+19×8)
Aliter:Between 10 and 1000, the numbers are of 2 digits and 3 digits.
Since repetition is allowed, so each digit can be filled in 9 ways.
Therefore number of 2 digit numbers = 989−(9+18+19×8)
Aliter:Between 10 and 1000, the numbers are of 2 digits and 3 digits.
Since repetition is allowed, so each digit can be filled in 9 ways.
Therefore number of 2 digit numbers = 9×9 = 81
and number of 3 digit numbers 9×9×9 = 729
Hence total ways = 81 + 729 = 810.
Question 7
The number of ways in which the letters of the word TRIANGLE can be arranged such that two vowels do not occur together is
1200
2400
14400
1440
SOLUTION
Solution : C
∘T∘R∘N∘G∘L∘
Three vowels can be arrange at 6 places in 6P3 = 120 ways. Hence the required number of arrangements = 120×5! =14400.
Question 8
If 56Pr+6:54Pr+3 = 30800:1, then r =
31
41
51
40
SOLUTION
Solution : B
56!(50−r)!×(51−r)!54!
= 308001⇒56×55×(51−r) = 30800⇒r = 41
Question 9
The number of ways in which the letters of the word ARRANGE can be arranged such that both R do not come together is
360
900
1260
1620
SOLUTION
Solution : B
The word ARRANGE, has AA,RR, NGE letters. That is two A' s, two R's and N, G, E one each.
∴The total number of arrangements 7!2!2!1!1!1!=1260
But, the number of arrangements in which both RR are together as one unit = 6!2!1!1!1!1! = 360
∴The number of arrangements in which both RR do not come together = 1260 - 360 = 900.
Question 10
There are n straight lines in a plane, no two of which are parallel and no three pass through the same point. Their points of intersection are joined. Then the number of fresh lines thus obtained is
n(n−1)(n−2)8
n(n−1)(n−2)(n−3)6
n(n−1)(n−2)(n−3)8
n(n−1)(n−2)(n−3)4
SOLUTION
Solution : C
Since no two lines are parallel and no three are concurrent, therefore n straight lines intersect at nC2 = N (say) points. Since two points are required to determine a straight line, therefore the total number of lines obtained by joining N points NC2. But in this each old line has been counted n−1C2 times, since on each old line there will be n-1 points of intersection made by the remaining (n-1) lines.
Hence the required number of fresh lines is NC2−n.n−1C2 = N(N−1)2−n(n−1)(n−2)2
= nC2(nC2−1)2−n(n−1)(n−2)2 = n(n−1)(n−2)(n−3)8
Question 11
There are 16 points in a plane out of which 6 are collinear, then how many lines can be drawn by joining these points
106
105
60
55
SOLUTION
Solution : A
Required number of lines
= 16C2−6C2 +1 = 120 -15+1 = 106
Question 12
The number of triangles that can be formed by choosing the vertices from a set of 12 points, seven of which lie on the same straight line is
185
175
115
105
SOLUTION
Solution : A
Required number of ways = 12C3−7C3
= 220 - 35 = 185
Question 13
There are four balls of different colours and four boxes of colours same as those of the balls. The number of ways in which the balls, one in each box, could be placed such that a ball does not go to box of its own colour is
8
7
9
6
SOLUTION
Solution : C
Since number of derangements in such a problems is given by
n!{1−11!+12!−13!+14!−...............(−1)n1n!}
∴ Number of dearangements are
= 4!{12!−13!+14!} = 12 -4 + 1 = 9
Question 14
Number of ways of selection of 8 letters from 24 letters of which 8 are a, 8 are b and the rest unlike, is given by
27
8.28
10.27
8.27
SOLUTION
Solution : C
The number of selections = coefficient of x8 in
(1+x+x2+..........+x8)(1+x+x2+.........+x8).(1+x)8
= coefficient of x8 in (1−x9)2(1−x)2(1+x)8
= coefficient of x8 in (1+x)3(1−x)−2
= coefficient of x8 in
(8C0+8C1x+8C2x2+..........+8C8x8)×(1+2x+3x2+4x3+.........+9x8+.........)
= 9.8C0+8.8C1+7.8C2+.............+1.8C8
= C0+2C1x+3C2x2+......9C8x8 = (1+x)8+8x(1+x)7
Putting x = 1, we get C0+2C1+3C2+........+9C8
= 28+8.27 = 27.(1+8) = 10.27.
Question 15
In how many ways can Rs. 16 be divided into 4 person when none of them get less than Rs. 3
70
35
64
192
SOLUTION
Solution : B
Required number of ways
= coefficient of x16 in (x3+x4+x5+.......x7)4
= coefficient of x16 in x12(1+x+x2+.......x4)4
= coefficient of x16 in x12(1−x5)4(1−x)−4
= coefficient of x4 in (1−x5)4(1−x)−4
= coefficient of x4 in (1−4x5+.....)
[1+4x+....+(r+1)(r+2)(r+3)3!x]
= (4+1)(4+2)(4+3)3! = 35
Aliter: Remaining 4 rupees can be distributed in 4+4−1C4−1 i.e., 35 ways