# Mixed Quantitative Aptitude Questions Set 74

1. There are two alloys copper and zinc. One alloy is weighing 15gram of zinc and copper in the ratio of 2:3 by weight . If 10 gram of zinc is added then find what amount of copper has to be removed from the alloy such that the final alloy has zinc and copper in the ratio 4:1 by weight?
7 gram
4 gram
10 gram
5 gram
9 gram
Option D
In first alloy , the amount of zinc = (15*2)/5 = 6 and the amount of copper = (15*3)/5 = 9
Let the copper is to be removed be x.
Therefore, (6+10)/(9-x) = 4/1
=> x= 5 gram

2. In a container two gallons of a mixture of spirit and water contains 10% of water. They are added to four gallons of another mixture, containing 10% of water and half a gallon of water is then added to the whole. Find the percentage of water in the resulting mixture.
22%
17%
33%
20%
15%
Option B
Total mixture = 2 + 4 + ½ = 13/2 = 6.5 gallon
Amount of water in first mixture = (2*10)/100 = 0.2
Amount of water in second mixture = (4*10)/100 = 0.4
Total water = 0.2+0.4+0.5 = 1.1 gallon
Required water = (1.1 *100)/6.5 = 17%(approx.)

3. 4 years ago, the respective ratio between 1/2 of P’s age at that time and four times of Q’s age at that time was 5 : 12. Eight years hence 1/2 of P’s age at that time will be less than Q’s age at that time by 2 years. What is Q’s present age?
8 yrs.
12 yrs.
10 yrs.
15 yrs.
5 yrs.
Option C
Let present ages of P and Q be x and y resp.
Now, ½(x-4)/4(y-4) = 5/12
=> 3x – 10y = -28——-(1)
Also, ½(x+8) = (y+8) -2
=> x – 2y = 4 ———(2)
On solving (1) and (2) , we get x = 24 years and y = 10 years
Present age of Q = 10 years

4. The simple interest accrued on a sum of certain principal is Rs. 2000 in 5 years at the rate of 4 p.c.p.a. What would be the compound interest accrued on same principal at the same rate in two years ?
Rs.720
Rs.816
Rs.600
Rs.520
Rs.850
Option B
Principal = (2000*100)/(4*5) = Rs.10,000
Required C.I. = 10000{(1+4/100)^2 – 1} = (104^2 – 100^2) = Rs.816

5. Annu beats Binay by 20 m in a 100 m race and Binay beats Chandu by 20 m in a 100 m race. How much start should Annu give to Chandu in a 100 m race so that both of them reach the winning post at the same time ?
36 m
45 m
50 m
30 m
55 m
Option A
Ratio of speed Annu and Binay = 10:8
Ratio of speed Binay and Chandu = 10:8
Annu:Binay:Chandu = 100:80:64
Annu should give 36m heat start to Chandu.

6. A runner runs from point X to Y at a speed of 15 km/hr. A second runner simultaneously runs from point Y to X and back at a speed of 20 km/h. If they cross each other 12 minutes after the start, after how much time will they cross each other ?
35 mins.
25 mins.
32 mins.
20 mins.
48 mins.
Option E
Since,12 minutes which means the distance between them is 7 km.
Ratio = 15:20 = 3:4
So, they cross each other after 48minutes.

7. There are four boxes in a truck. The weight of the first box is 300 kg and the weight of the second box is 20% higher than the weight of the third box, whose weight is 25% higher than the first box’s weight. The fourth box at 400 kg is 20% lighter than the fifth box. Find the difference in the average weight of the four heaviest boxes and the four lightest boxes.
60
50
80
70
90
Option B
Weight of the first box = 300kg
Weight of the second box = 300*(125/100)*(120/100) = 450kg
Weight of the third box = 300*(125/100) = 375kg
Weight of the fourth box = 400kg
Weight of the fifth box = 400 *(100/80) = 500kg
Required Difference = (500+450+400+375)/4 – (300+375+400+450)/4 = 1725/4 – 1525/4 = 431.25 – 381.25
= 50

8. If the areas of the three adjacent faces of a cuboidal box are 35 cm square, 42 cm square and 30 cm square respectively, then find the volume of the box .
210 cu.cm
200 cu.cm
150 cu.cm
125 cu. cm
145 cu.cm
Option A
lb = 35 sq. cm
bh= 42 sq. cm
hl = 30 sq. cm
=> lb * bh * lh = 35 * 42 * 30
=> (lbh)^2 = 35 * 42 * 30
=> lbh = 210
Hence, volume of the box = 210 cm^3

9. One card is drawn at random from a pack of 52 cards. What is the probability that the card drawn is a face card ( Queen and King only)?
4/13
1/13
3/13
5/13
2/13
Option E
Required Probability = 8/52 = 2/13

10. How many numbers can be formed using all the digits 6, 3, 5, 3, 6, 5, 2, 4, 3, such that odd digits occupy the odd places?
100
80
120
90
110
Option C
There are five odd places and four even places , of which 3 occurs three times and 5 occurs five timesand 6 occurs two times .
The five odd places can be arranged = 5!/(2!*3!) = 10
The four even places can be arranged by = 4!/(2!) = 12
Therefore, total number of ways = 12*10 = 120