1. If 2x = 8y+1 and 9y = 3x-9. What is the value of xy? Easy
A. 4412
B. 4413
C. 215
D. None of these

View Answer

Answer: Option B

Explanation:

2x = 23(y+1)

x = 3y+3 —– ❶

9y = 3x-9

32y = 3x-9

2y = x-9 —– ❷

x = 21, y = 6

Therefore, xy = 216 = (212)3 = 4413

2. If ap=bq=cr where x,y,z ≠ 0 and c = ab, then which of the following is true? Medium
A. r = pq/(p+q)
B. r = pq/(p-q)
C. r = (p+q)/pq
D. None of these

View Answer

Answer: Option A

Explanation:

Let ap=bq=cr=k

a = k1/p, b = k1/q, c = k1/r

Now, c = ab (given)

Therefore, k1/r = k1/p. k1/q

-> k1/r = k1/p+1/q

-> 1/r = 1/p + 1/q
-> r = pq/(p+q)

3. If 2x=4y=8z, what is the least value of (x+y+z), if x,y and z are natural numbers?
Difficult
A. 10
B. 11
C. 12
D. Can’t be determined

View Answer

Answer: Option B

Explanation:

Given, 2x = 4y = 8z
-> if x =1, then y =1/2 and z = 1/3
Now, the LCM of 2 and 3 (denominators of the above values of y and z) is 6 -> the first set of values of x, y and z if they are all natural numbers, will be 6, 3 and 2 respectively.
Thus, 26 = 43 = 82
-> x+y+z = 11

4. If x raised to the power x5 equals 5, then what is the value of x? Medium
A. 51/5
B. 21/2
C. 31/3
D. None of these

View Answer

Answer: Option A

Explanation:

Let x5 = y
Then x = y1/5
Therefore, (y1/5)y = 5
-> yy/5 = 5
-> yy = 55

-> y = 5
-> x5 = 5

-> x = 51/5

 

Alternatively

Check the value of x from the options, which satisfies the equation

5. If xpqr = xp.xq.xr where p, q, r and x are all positive integers, what is the value of p3+q3+r3 ? Medium
A. 48
B. 14
C. 36
D. None of these

View Answer

Answer: Option C

Explanation:

Given, xpqr = xp.xq.xr = xp+q+r

-> pqr = p+q+r

By trial and error, p, q, r are 1, 2, 3 (respective values are not important as each variable undergoes the same treatment).

-> 13+23+33 = 36

6. 293-292-291 is the same as: Easy
A. 292
B. 290
C. 291
D. None of these

View Answer

Answer: Option C

Explanation:

293-292-291 = 291[22-2-1]

= 291

7. Which of the following is true? Easy
A. Nine raised to the power 87 equals 98 raised to the power 7.
B. Nine raised to the power 87 is less than 98 raised to the power 7
C. Nine raised to the power 87 is more than 98 raised to the power 7
D. None of these

View Answer

Answer: Option C

Explanation:

Nine raised to the power 87 = 98*8*8…..7 times
98 raised to the power 7 = (98)7 = 956
A. Therefore, nine raised to the power 87 is more than 98 raised to the power 7.

8. If x, y, and z are natural numbers such that x<y≤z and 3x+4y+5z is the largest five-digit number that satisfies the given conditions, find x4+y3+z2.  Difficult
A. 608
B. 1688
C. 8168
D. Can’t be determined

View Answer

Answer: Option B

Explanation:

A five-digit value of 3x+4y+5z is only possible if z is 6 or 7.  This expression will not be a five-digit number for z values less than 6 and more than 7. However, since 3x+4y+5z is the largest five-digit number, we take the value of z as 7.

Again, for 3x+4y+5z to be maximum under the given conditions, x= 6 and y=7. For x=6, y=7, z=7, 3x+4y+5z is still a five-digit number. Thus, plug in these values to find x4+y3+z2.

Desired value = 64 + 73 + 72 = 1296 +343 +49 = 1688

9. Which of the following might be true if p = q2 and q = pq? Difficult
A. p2 +q2 = 5/16
B. p/q3 = 2
C. p = q = 1
D. All of the above

View Answer

Answer: Option D

Explanation:

p = q2 —– ❶

q = pq —– ❷

Substituting ❶ in ❷

q = (q2)q = q2q  -> q = 1 or ½

when q = 1, p = 1 -> Option C is possible

when q = ½, p = ¼

We check other options using the above values.

Option A is possible when p = ¼, and q = ½ as (1/4)2 + (1/2)2 = 5/16

Option B is possible when p = ¼ and q = ½ as (1/4)/(1/2)3 = 8/4 =2

Therefore, all of them might be true.

10. Let x to the power 2n4 equals y, and x to the power 4n equals z. If yn = z4, where x and n are positive integers, find the value of (2n+1)√n20.

A. 64
B. 32
C. 16
D. None of these

Difficult

View Answer

Answer: Option C

Explanation:

Given, yn = z4
-> x raised to 2n4+1 = x raised to 4n+1
-> 2n4+1 = 4n+1
-> 2n5 = 22n+2
-> 2n5 = 2.22n+1
-> n5 = 22n+1
By trial and error, we find that n = 2
-> (2n+1)√n20 = 5√220 = (220)1/5 = 24 =16