# Quantitative Aptitude Questions Daily Quiz Day – 16

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142 Dear Aspirants, Our SSC Crackers team is providing a new series of Quantitative Aptitude Questions for Upcoming Exam so the aspirants can practice it on a daily basis. These questions are framed by our skilled experts after understanding your needs thoroughly. Aspirants can practice these new series questions daily to familiarize with the exact exam pattern and make your preparation effective.

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1) Find the value of (x5 + 1/x5 + 2)1/3 if x+ 1/x = 3

a) 5

b) 6

c) 7

d) 4

2) If X : Y = 3 : 7, Y : Z = 4 : 5 and W : X = 2 : 9, then what is the value of W : Y : Z?

a) 2 : 7 : 5

b) 2 : 3 : 5

c) 8 : 84 : 105

d) 10 : 21 : 105

3) Find the value of x2 + y2/ x2 – y2 if the value of 3x + 2y/ 3x – 2y = 5/3

a) 99/55

b) 73/55

c) 67/58

d) 189/196

4) What is the unit digit of 3749 – 2525 + 4816?

a) 5

b) 0

c) 6

d) 8

5) a) 0

b) 1

c) √3

d) √2

6) For what value of m, would the expression 28x2 + 9m + 12mx be a perfect square?

a) 3

b) 7

c) 9

d) 11

7) If (x + y) = 2z, then find the value of [x/(x – z)] + [z/(y – z)]?

a) -1

b) -2

c) 1

d) 2

8) What is the simplified value of [(p-1/2 + p1/2)] / [(p1/2 / {(1 – p1/2) × (1 + p1/2)}] + p?

a) 1 / p

b) (1 – p) / p

c) p / (1 – p)

d) p2

9) What least value must be given to *such that the number 91876*2 is divisible by 8?

a) 5

b) 3

c) 7

d) 4

10) A sum of Rs. 5640 is divided among P, Q and R in the ratio 13: 16: 18. How much amount did R get more than P?

a) Rs. 600

b) Rs. 660

c) Rs. 720

d) Rs. 780

=> x+ 1/x = 3

=> x2 + 1/x2 + 2 = 9

=> x2 + 1/x2 = 7

Also x3 + 1/x3 + 3 (x+ 1/x) =(x+1/x)3

=> x3 + 1/x3 + 3*3 =27

=> X3 + 1/x3 = 18

=> And x5 + 1/x5 = (x2 + 1/x2) (x3 +1/x3) – (x + 1/x)

=> 7* 18 – 3 =123

=> According to question,

=> (x5 + 1/x5 + 2)1/3

=> =(123 + 2)1/3

=> =(125)1/3 = 5

If X : Y = 3 : 7 and W : X = 2 : 9, on multiplying both the ratios, we get-

W : Y = 2 : 21

Now, Y : Z = 4 : 5; on making the common part of the ratios equal, we can write-

W : Y = 8 : 84 and Y : Z = 84 : 105

Thus, we get W : Y : Z = 8 : 84 : 105

=> 3 (3x + 2y) = 5 (3x -2y)

=> 9x + 6y =15x – 10y

=> 6x = 16y

=> x/y = 16/6 = 8/3

=> x2 + y2/x2 – y2

=> 64+9 / 64-9

=> 73/55

Unit digit of 71 = 7, 72 = 9, 73 = 3, 74 = 1, 75 = 7 ,… and so on.

Unit digit of 3749 = 7

Unit digit of 51 = 5, 52 = 5, 53 = 5, 54 = 5, 55 = 5,  … and so on.

Unit digit of 2525 = 5

Unit digit of 81 = 8, 82 = 4, 83 = 2, 84 = 6, 85 = 8, … and so on.

Unit digit of 4816 = 6

Thus, unit digit of 3749 – 2525 + 4816 = 7 – 5 + 6 = 8 Among the given values in the given options, for only m = 7, the expression 28x2 + 9m + 12mx will be a perfect square.

28x2 + 9 × 7 + 12 × 7x = 28x2 + 84x + 63

7(4x2 + 12x + 9) = 7(2x + 3)2 = (2√7x + 3√7)2

(x + y) = 2z

(x + y) = z + z

(x – z) = (z – y) = k (Let)

= [x/(x – z)] + [z/(y – z)]

= [x/k] – [z/k]

= (x – z)/k

= (x – z)/(x – z)

= 1

The given expression can be simplified as follows-

[{(1/√p + √p)} / {√p / (1 – p)}] + p

[{(1 + p) / √p} / {√p/ (1 – p)}] + p

[(1 – p2) / p] + p

(1 – p2 + p2) / p

= 1 / p

The number formed by the last three digits should be divisible by 8. So * =3.

Let the amounts P. Q and R got be 13x, 16x and 18x respectively.

Given that, 13x + 16x + 18x = Rs. 5640

→ 47x = Rs. 5640

→ x = 5640/47 = Rs, 120

Therefore, amount that R got more than P

= 18(Rs.120) – 13(Rs.120)

= (5) (Rs.120) = Rs. 600