Lakshya Education MCQs

Question: Find the greatest number of five digits, which is exactly divisible by 7, 10,14, 15, 21 and 28.
Options:
A.99840
B.99900
C.99960
D.99990
Answer: Option C
: C

The number should be exactly divisible by 14(7,2),15 (3, 5), 21 (3, 7), 28 (4, 7).
Hence, it is enough to check the divisibility for 3, 4, 5 and 7. Only (c) is divisible by all.

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More Questions on This Topic :

Question 1. What digit does 20122012 end in?
  1.    2
  2.    6
  3.    4
  4.    8
Answer: Option B
: B

Since the number in the unit's place in 2012 is 2, we can find the last digit of the expression by considering only the power to which it is raised.
Now, we can write the following: 20122012=___24k. Therefore the last digit = 6.
Question 2. Two circles centered at points A and B, respectively, intersect at points E and F as shown. These circles intersect segment AB at points C and D. If AC=1 and CD = DB = 2, determine EF.
  1.    2.4
  2.    4.8
  3.    2√5
  4.    2√7
Answer: Option B
: B


In triangle AFB,12×AB×FP=12×AF×FB
FP=125
Hence, EF= 245 = 4.8.
Question 3. The figure below is a regular octagon. What fraction of its area is shaded?

  1.    13
  2.    14
  3.    15
  4.    38
Answer: Option B
: B

The diagram shows how the octagon can be divided into 4 congruent rectangles and 8 congruent triangles. Let R represent the area of a rectangle and let T represent the area of a triangle. Then the ratio of the shaded region to the area of the entire octagon isR+2T4R+8T=14
Question 4. How many 5 digit numbers are divisible by 3 and contain the digit 6?
  1.    7499
  2.    8776
  3.    12503
  4.    None of these
Answer: Option D
: D

Consider first the number of 5-digit numbers divisible by 3. The smallest is 10002=3334×3, the largest is 99999=33333×3, so there are 30000 such numbers.
Now consider the numbers that do not contain the digit 6. There are 8 choices for the 1st digit, 9 choices for each of the 2nd, 3rd and 4th digits.
If the total of the first 4 digits is a multiple of 3, then the last digit must be 0, 3 or 9.
If it gives a remainder 1 when divided by 3, then the last digit must be 2, 5 or 8.
If it gives a remainder 2 when divided by 3, then the last digit must be 1, 4 or 7.
So in all cases there are 3 choices for the last digit. Hence the total number is 8×93×3=17496.
So the total no. of 5 digit numbers which are divisible by 3 and contains a 6 is 12504.
Question 5. When you reverse the digits of the number 13, the number increases by 18. How many other two digit numbers increase by 18 when their digits are reversed?

___

: Let 10x + y be a two digit number, where x and y are positive single digit integers and x>0.
Its reverse = 10y + x
Now, 10y + x - 10x - y = 18
9(y - x) = 18 y - x = 2
Thus y and x can be (1, 3), (2, 4), (3, 5), (4, 6), (5, 7), (6, 8) and (7, 9)

Other than 13, there are 6 such numbers.
Question 6. ΔABC has a circle centred on vertex C that passes through points A and E. If ABC DEC, AC = 1 and CD = x, what is the distance between E and B?
  1.    2x
  2.    1−xx
  3.    1 - x
  4.    1 + x
Answer: Option B
: B

Since DEC ABC, it follows that ΔDEC ΔABC.
So, 1x=BE+11
x(BE+1)=1
(BE+1)=1x
BE=1xx

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