# Mixed Quantitative Aptitude Questions Set 102

Directions(1-5): Find the missing term of the following questions.

1. 520,513,494,457,396,?
320
333
305
314
325
Option C
+1^3-2^3
+2^3-3^3
+3^3-4^3
+4^3-5^3
+5^3-6^3
? = 305

2. 1812,1820,1804,1828,?,1836
1796
1700
1770
1754
1784
Option A
+8*1
-8*2
+8*3
-8*4
+8*5
? = 1796

3. 42,22,24,39,82,?
215
198
222
210
200
Option D
*0.5+1
*1+2
*1.5+3
*2+4
*2.5+5
? = 210

4. 74,74,111,222,?,1665
520
500
525
533
555
Option E
*1
*1.5
*2
*2.5
*3
? =555

5. 5,28,74,143,?,350
222
235
242
255
274
Option B
+23*1
+23*2
+23*3
+23*4
+23*5
? = 235

6. The average weight of (x+5) eggs is 10 grams. If 5 more eggs are added then the average gets doubled. Find the value of x if the 5 eggs which are added weigh 200 grams.
4
5
2
3
6
Option B
[10(x+5)+200]/[x+10] = 2*10
=> x = 5

7. Three candidates contested an election and received 1136, 7636 and 11628 votes respectively. What percentage of the total votes did the winning candidate get?
44%
40%
57%
50%
63%
Option C
Required percentage =11628/20400×100
=969/17=57%

8. Ayisha’s age is 1/6th of her father’s age. Ayisha’s father’s age will be twice Shankar’s age after 10 years. If Shankar’s eight birthdays was celebrated two years before, then what is Ayisha’s present age.
6
4
7
3
5
Option E
Let Ayisha’s present age =x
Then, her father’s age =6x
Shankar’s age after 10 years =1/2(6x+10) =3x+5
Therefore, Shankar’s age after 10 years =8+12=20
3x+5=20
=>x=15/3=5

9. A train of length 250 m can cross a platform of length x m in seconds. Also it can cross a train of half of its length running with speed of 35 m/s in opposite direction in 5 seconds. Find the value of x.
100
125
150
135
144
Option C
(250+x)= 10*s —-(1)
(250+0.5*250) = 5*(s+35) —–(2)
On solving (1) and (2), we get
s= 40 m/s x = 10*40 – 250
=> x = 150 m

10. A bag contains 6 apples, 8 bananas and (x+2) oranges. 2 fruits are drawn at random . Find the value of x if the probability that both fruits are oranges is 2/51.
2
4
1
5
3
Option A
Probability of selecting 2 oranges = (x+2)C2/(16+x)C2 Now, (x+2)C2/(16+x)C2 = 2/51
=> (x+2)(x+1)/(x+16)(x+15) = 2/51
=> 49x^2 + 91x – 378 = 0
=> (x-2)(7x +27) = 0
=>x = 2
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