Lakshya Education MCQs

Question: The sum of first 3 terms of a G.P is 16 and the sum of next 3 terms is 128. Find the sum of n terms of the G.P?
Options:
A.167(2n + 1)
B.167(2n - 1)
C.97(3n - 1)
D.167(3n + 1)
Answer: Option B
: B

Conventional Method From the data given in the question S3=16=a(r31)r1................................(1)
S6=16+128=144=a(r61)r1=a(r31)(r3+1)r1................................(2)
9 = (r3 + 1)
Hence r =2
Substituting in equation 1, we get
a = 167
We know that,
Sn=a(rn1)r1
Substituting the values of a and r found above, we get
Sn=167(2n1)
Thus, answer is option (b)
Shortcut-
Substitution The answerto this question is in the question itself!
At n=3, the sum should be 16
Hence, in the correct answer option, at n=3, we should get 16
Option (a)= 167(23+ 1)16
Option(b)=167(23-1) = 16
Option (c)=97(33-1)16
Option (d)=167(33 -1)16
Options (a), (c) and (d) can never be the answer. Answer is option (b)

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More Questions on This Topic :

Question 1. In how many ways can the number 100 be written as a sum of two or more consecutive positive integers?
  1.    2
  2.    3
  3.    5
  4.    6
Answer: Option A
: A

Conventional Approach To write a number as a sum of two or more consecutive positive integers, we need to form an AP with "n” terms with a common difference 1 Sum to "n” terms of an AP= n2(2a + (n - 1)d) where d=1 (as the numbers are consecutive) n(2a+(n-1))=200 thus, we need to write 200 as a product of 2 numbers, where one number is odd and the other is even this can be done in 2 ways 200= 5×40 200=25×8 Thus 100 can be expressed in 2 ways as a sum of two or more consecutive positive integers Shortcut:- The answer is just one step. The number of ways of writing any number as a sum of two or more consecutive positive integers= number of odd factors of that number-1 100 = 22×52 Number of odd factors=3 Number of ways of expressing 100 as a sum of two or more consecutive positive integers= 3 - 1=2
Question 2. Both the H.C.F and the difference of two numbers is 7. If the L.C.M of the two numbers is a three digit number, then what is the maximum possible value of the smaller number?
  1.    77
  2.    84
  3.    70
  4.    63
Answer: Option A
: A

Let the smaller number be 7x, then the larger number will be (7x+7) = 7(x+1) Since x and (x+1) are co-prime, the LCM would be 7×x×(x+1). Maximum value of x such that LCM is a three digit number is x=11. So, smaller number = 77.
Question 3. Find the remainder when 287 is divided by 100
  1.    56
  2.    28
  3.    76
  4.    01
Answer: Option B
: B

Shortcut of last two digits: When the question is about finding the remainder when you divide a number by 100, what we actually need to find out is the last two digits, for which we have a direct shortcut. For an indepth explanation, please refer the demo tutorial

287=(210)8×27 (210)evevpowerwillalwaysendin..76

27 ends in ...28

Therefore 76x...28=28
Question 4. 1 unit of x% alcohol is mixed with 3 units of y% alcohol to give 60% alcohol. If x>y , how many integer values can x take?
  1.    19
  2.    20
  3.    21
  4.    13
  5.    15
Answer: Option D
: D

We know that (x+3y)4=60 or x+3y =240 and x>y As 0<x<100, let's start with x=99. When x= 99, y=47. When x=96, y=48 and so on. x has to be greater than y. Now in the limiting case, when x=60, y=60. So when x=63, y=59. So x can take the values of 63,66,69...... till 99, a total of 13 values.
Question 5. A tank has two pipes A and B. Pipe A is for filling the tank and Pipe B is for emptying the tank. Pipe A can fill the tank in 20 hours and Pipe B can empty the tank in 25 hours. How many hours will it take to completely fill a one-fourth empty tank?
  1.    10 hours
  2.    25 hours
  3.    50 hours
  4.    100 hours
  5.    5 hours
Answer: Option B
: B

In one hour pipe A can fill 10020 = 5% of the tank In one hour pipe B can empty 10025 = 4% of the tank (A+B)'s work = 5-4 = 1% To fill a one fourth of empty tank, it will take = 251 = 25 hours
Question 6. The sum of all the three digit numbers which when divided by 8 give a remainder of 5 is___

: For this condition,
Series is 109,117,125,..........997
a= 109 ,d=8 , n=112
Sn=n2[2a+(n1)d]
=1122[2×109+(1121)8]
=61936.

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