| 1. A is a set of n distinct whole numbers such that there are at least 10 even numbers, at least 5 prime numbers and at least 6 composite numbers. Find the minimum value of n. | Easy |
| A. 11 B. 9 C. 14 D. 5 |
View Answer
Answer: Option C
Explanation:
Out of the 5 prime numbers, one has to be necessarily 2 (because minimum is required). All 6 composite numbers need to be even (other than 2).
Since we need a minimum of 10 even numbers and 7 have already been accounted for (1 prime and 6 composite), we need 3 more even numbers. Hence, the minimum value of n = 5+6+3 = 14
| 2. If k+1 represents an odd integer, then which of the following may be false? | Easy |
| A. k(k-2) must be even. B. (k-1)(k+1) must be odd. C. (k+1)k must be even. D. (k+2)/2 must be even. |
View Answer
Answer: Option D
Explanation:
(k+1) is odd -> k is even -> (k-2) is even -> k(k-2) is even. Hence A is always true.
(k-1) is odd à (k+1)(k-1) will be necessarily odd (product of 2 odd numbers). Hence B is true.
(k+1)k will also be necessarily even (product of an even and an odd number).
Now, (k+2)/2 can be odd or even, depending on whether (k+2) is divisible by only 2 (odd) or both 2 and 4 (even). For example, if k+1 is 5 then (k+2)/2 is divisible by 2 but not by 4, and will be odd; if k+1 is 7 then (k+2)/2 is divisible by both 2 and 4, and will be even. Hence option D may be true or false.
| 3. If P represents a number between 4 and 8 and Q represents a number between 18 and 100, the Q/P represents a number between: | Easy |
| A. 2.25 and 25 B. 5 and 20 C. 2 and 5 D. 3 and 4 |
View Answer
Answer: Option A
Explanation:
Maximum Q/P = (Maximum of Q)/(Minimum of P) = 100/4 = 25
Minimum Q/P = (Minimum of Q)/(Maximum of P) = 18/8 = 2.25
-> Q/P would lie between 2.25 and 25. Hence, option A.
Alternatively, look at the answers and eliminate.
| 4. kn are either -1 or 1 and k≥4k1k2k3k4 + k2k3k4k5 + k3k45k6 + …… kn-1 knk1k2 + knk1k2k3 = 0, then n is: | Medium |
| A. Odd B. Even C. Prime D. Can’t be determined |
View Answer
Answer: Option B
Explanation:
Since kn are either -1 or 1 and the sum of the given terms is zero, it means that there has to be an equal number of 1s and -1s, which is possible only when the number of terms is even, which rules out option A. The only possibility of n being prime and even is 2, which is ruled out as n≥4. Hence options C and D are also ruled out.
| 5. If x+1 is odd, then under what conditions is (x+2)/2 divisible by 3? | Easy |
| A. (x+2)/2 is never divisible by 3. B. (x+2)/2 is always divisible by 3. C. (x+2)/2 is divisible by 3 if x+2 is equal to 6k, where k is any positive natural number. D. Can’t be determined. |
View Answer
Answer: Option C
Explanation:
Take values to conclude that options A and B are not true. For example, if x+1 = 5, then (x+2)/2 = 3, which is divisible by 3. However, if x+1=7, then (x+2)/2 = 4, which is not divisible by 3.
Option C is correct because if x+2 = 6k = 2(3k), then (x+2)/2 = 3k, which is always divisible by 3.
| 6. Three students went out for a stroll and noticed that a car broke the traffic regulations. None of them noticed the 4-digit license plate number but recalled that the first two-digits were the same and the last two digits were also alike. They also remembered that the 4-digit number was a perfect square. What was the plate number? | Medium |
| A. 5588 B. 1122 C. 7744 D. 2299 |
View Answer
Answer: Option C
Explanation:
A square can never end in 2, 3, 7, 8
-> Options a and b are eliminated.
Also, if a square ends in on odd digit the penultimate digit has to be even.
-> Option d is also eliminated.
Hence, option C is the answer (7744 = 882).
| 7. The difference in the squares of any two odd numbers is definitely divisible by: | Medium |
| A. 12 B. 16 C. 24 D. 8 |
View Answer
Answer: Option D
Explanation:
Two odd numbers can be expressed as (2m+1) and (2n+1), where m and n are natural numbers.
Difference of the squares of these odd numbers is (2m+1)2 – (2n+1)2
=(4m2+1+4m) – (4n2+1+4n)
=4[m2-n2+m-n]
=4[(m-n)(m+n)+(m-n)]
=4(m-n)(m+n+1)
Case I: Both m and n are odd. In this case (m-n) is even and (m+n+1) is odd
-> 4(m-n)(m+n+1) can be written as 8k1 where, k1 is a natural number.
Case II: Both m and n are even. In this case (m-n) is even and (m+n+1) is odd
-> 4(m-n)(m+n+1) can be written as 8k2 where, k2 is a natural number.
Case III: One is odd, other is even. In this case (m-n) is odd and (m+n+1) is even.
-> 4(m-n)(m+n+1) can be written as 8k3 where, k3 is a natural number.
In all 3 cases, 4(m-n)(m+n+1) can be expressed as a multiple of 8.
| 8. Suppose k is an integer such that the sum of the digits of k is 2 and 1011 < k < 1012. What is the number of different values of k? | Difficult |
| A. 12 B. 11 C. 10 D. 13 |
View Answer
Answer: Option A
Explanation:
1011 has 12 digits and 1012 has 13 digits.
As 1011 < k < 1012, therefore k must have twelve digits. Also, the sum of digits is 2 which is only possible with 2 ones or 1 two.
Numbers with 2 ones will have 10 zeros, and all such numbers will start with 1, and the second 1 can be placed in 11 different ways, leading to 11 such numbers.
Total numbers with one 2 will have 11 zeros and it is only possible if this number starts with 2 i.e., the number is 200000000000
-> Total number of values that k can take = 11+1 = 12
| 9. Consider three natural numbers p, q and r. Out of these, p and q are odd, while r is even. Which of the following statements cannot be true? | Medium |
| A. (p-q+r)2r is even. B. (p-r)2q is even. C. (p+q)3r is even. D. (p-r)q2 is odd. |
View Answer
Answer: Option B
Explanation:
Check all the options, either by applying properties of numbers or by taking random values.
Option A: (p-q+r) is (odd – odd + even), which is even -> (p-q+r)2 is even -> (p-q+r)2r is even *even, which is even. Hence option A is true.
Option B: (p-r) is (odd – even), which is odd -> (p-r)2 is odd -> (p-r)2q is odd *odd, which is odd. Hence option B is not true.
Option C: (p+q) is even -> (p+q)3 is even -> (p+q)3r is even*even, which is even. Hence option C is true.
Option D: (p-r) is (odd – even), which is odd; q2 is odd2, which is odd -> (p-r)q2 is odd*odd, which is odd. Hence, option D is true.
| 10. D1, D2, D3, D4 are recurring decimals given by D1 = 0.x1x2x3x4x5x1x2x3x4x5……, D2 = 0.x1x2x3x4x5x2x3x4x5……., D3 = 0.x1x2x3x4x5x3x4x5……, D4 = 0.x1x2x3x4x5x4x5……., where x1, x2, x3, x4 and x5 are distinct decimal digits. Which of the following numbers when multiplied by at least one of the above decimals will always result in an integer? | Difficult |
| A. 18000 B. 19980 C. 998 D. None of these |
View Answer
Answer: Option B
Explanation:
Converting all recurring decimals into equivalent fractions, we get the following results:
D1 = x1x2x3x4x5/99999
D2 = (x1x2x3x4x5 – x1)/99990
D3 = (x1x2x3x4x5 – x1x2)/99900
D4 = (x1x2x3x4x5 – x1x2x3)/99000
Out of the above 4 numbers, only D3 has a denominator (99900), which is a factor of one of the numbers in the options i.e., 19980. Hence, only this number will always produce an integer when multiplied 19980.