## Lakshya Education MCQs

Question: Find the complete range of values of "p” for which (p27p+13)p<1?
Options:
 A. (−∞,+∞) B. 0,∞ C. (−∞,4) D. (−∞,0)∪(3,4) E. 8
Answer: Option D
: D

Use substitution method, Put p= 5. Then LHS>RHS. Thus, the inequality is not satisfied for this value. Answer can never be option (a) or (b) Put p=2. Then LHS>RHS. Thus, the inequality is not satisfied for this value as well. Answer is option (d).

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## More Questions on This Topic :

Question 1. ABCDEF is a regular hexagon, with O as the centre. There is another regular hexagon with 2 of the end points PQ inscribed in the above hexagon (again with centre O). Find the area of the unshaded region, given that FPQC lies on a straight line and FP=CQ=2 and FP = 13 FO.

1.    20√3
2.    24√3
3.    18√3
4.    28√3
5.    118
Answer: Option B
: B

The hexagon can be graphically divided as follows Hence, there are totally 54 equilateral triangles out of which 24 are unshaded. Each of the triangles have aside 2. Total unshaded area = 34 x 4 x 24 = 243
Question 2. A shopkeeper keeps 100 marbles in two boxes, A and B, such that Box A contains only red and Box B contains only green marbles. Barring an equal number of marbles from each box, he sold all the others and earned Rs. 25 and Rs. 15 from Box A and Box B respectively. Later, he sold half of the remaining marbles. His total earning from all the transactions accumulated to Rs. 45. What is the cost of each marble - red or green.?
1.    40 paise
2.    25 paise
3.    75 paise
4.    50 paise
Answer: Option D
: D

Let the number of red and green marbles be a and b respectively. Let the selling price of each marble be Rs.e. Let the equal number of marbles not sold the first time = c. Then, e(a - c) = 25 ... (i) e(b - c) = 15 ... (ii) The number of marbles that remained that remained with the shopkeeper = 2c e(a - c) + e(b - c) + ec = 45 ec = 5 ...(iii) (from (i) and (ii)) Also, (100 - c)e - 45

100e - 5 = 45 (from (iii)

100e = 50
e = 0.5 = 50 Paise. Hence, [d].
Question 3. A number when divided by 3,4,5,6,7 gives remainders 2,3,4,5,6 respectively. The smallest such 5 digit number is?
1.    10019
2.    10076
3.    11079
4.    10079
Answer: Option D
: D

Take the LCM of the numbers 3,4,5,6,7. First number to satisfy the given conditions will be LCM-1 = 420-1 = 419 Numbers will fall in an AP with a common difference = LCM 2nd number = 419+420 = 839 3rd= 839+420 = 1259 and so on. First 5 digit number = 10080-1 = 10079.
Question 4. The population of a town was 3600 three years back. It is 4800 right now. What is the rate of growth of population, if it has been constant over the years and has been compounding annually?
1.    15%
2.    5%
3.    12%
4.    10%
5.    8
Answer: Option D
: D

Final Value = Initial Value (1+r100)t
Thus, Rate =[(FVIV)1t1]×100
FVIV=4836=1.33

Going from answer options -
If 10 is the answer -
1.13=1.3311.33=FVIV
Therefore, rate is approximately 10%.
Question 5. The product "n” of three positive integers is 6 times their sum. If one of the numbers is the sum of the other two, then the sum of all possible value(s) of "n” is?
1.    336
2.    252
3.    240
4.    192
5.    118
Answer: Option A
: A

Let the numbers be x,y, x+y.

Therefore xy(x+y) = 12(x+y). Hence, xy= 12. So, (x,y) = (1,12), (2,6) or (3,4).

Hence "n" = 156, 96 or 84. Sum "s" (of all possible values of n) = 336.
Question 6. The number of positive integers which divide (25)! are -
1.    213.33.52
2.    28.32.52
3.    211.32.5
4.    28.33.53
Answer: Option A
: A

32!=231.314.57.74.112.132.17.19.23.29.31 (on prime factorization) So it has (31+1)(14+1)(7+1)(4+1)(2+1)(2+1)(1+1)(1+1)(1+1)(1+1)(1+1)=213.33.52 factors. Hence, choice (a) is the right answer.

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