- 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 gram4 gram10 gram5 gram9 gramOption 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 - 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.) - 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 - 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.720Rs.816Rs.600Rs.520Rs.850Option B

Principal = (2000*100)/(4*5) = Rs.10,000

Required C.I. = 10000{(1+4/100)^2 – 1} = (104^2 – 100^2) = Rs.816 - 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 m45 m50 m30 m55 mOption 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. - 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. - 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.
6050807090Option 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 - 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.cm200 cu.cm150 cu.cm125 cu. cm145 cu.cmOption 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 - 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)?
4/131/133/135/132/13Option E

Required Probability = 8/52 = 2/13 - 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?
1008012090110Option 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