- If we consider four numbers, the average of first three is 18 and that of last three is 19. If the last number is 16, find the first number?
1315111914Option A

Let four numbers be x,y,z,16.

Average of first three numbers = (x+y+z)/3 = 18 => x+y+z = 54 ——-(1)

Average of last three numbers = (y+z+16)/3 = 19

=> y + z+16 = 57 ——–(2)

On solving (1) & (2), we get x = 13 - A completes 32% of a task in 8 days and then takes the help of B and C. B is 50% as efficient as A is and C is 50% as efficient as B is. In how many more days will they complete the work?
3(1/4) days5(2/7) days10(1/11) days9(5/7) days8(1/3) daysOption D

A completes 32% of work in 8days.

Given, A: B is 2:1 and B: C is 2:1

Now, A: B: C = 4 : 2 : 1

A’s work 4 × 8 (days) = 32%

Remaining (68%) = 68/ (4 + 2 + 1) = 9 (5/7) days - If the average marks of 1/5 th class is 80% and that of and that of 1/4 th class is 75% and the average marks of the rest class is 65%, then what is the average of the whole class is (for the given subjects)?
6070.5755880Option B

Required average

= [(x/5)*(80/100) + (x/4)*(75/100) + (11x/20)*(65/100)]*100/x

= 70.5 - A style of curtains emporium the trader measures 30% less for every metre of cloth also he marks up goods by 30%. What is the approximate profit percentage?
70%72%50%66%86%Option E

If CP = Rs.1

MP = Rs.1.3

Cloth costs Rs.0.7 instead of Rs.1.

Therefore, Profit% = {(1.3 – 0.7)*100}/0.7 = 86%(approx.) - The diluted alcohol contains only 7 litres of alcohol and the rest is water. A new mixture whose concentration is 40%, is to be formed by replacing alcohol. How many litres of mixture shall be replaced with pure alcohol if there was initially 28 litres of water in the mixture?
42/5 L33/7 L35/4 L47/8 L14/3 LOption C

Alcohol : Water

7L : 28L

1L : 4L

20% ==== 80% (original ratio)

40%=====60% (required ratio)

Ratio of the left quantity to the initial quantity = 3:4

Therefore, ¾ = (1 – x/35)

=> x = 35/4 litre - A man deposited a total amount of Rs. 25300 with a bank in two different schemes at 20% p.a. interest being compounded annually . As per the schemes, he gets the same amount after 4 years on the first deposit as he gets after 3 years on the second deposit. How much money did he deposit for 5 years ?
Rs.12,000Rs.15,450Rs.16,620Rs.14,440Rs.13,800Option E

P1(1+r1/100)^4 = P2(1+r2/100)^3

=> P1/P2 = 6/5 Required amount = (6*25300)/11 = Rs.13800 - The ratio of speeds of a goods train and passenger train is 11:13 in the same direction. If the passenger train crosses the goods train in 90 seconds, while a passenger in the passenger train observes that he crosses the goods train in 75 seconds. Find the ratio of length of goods train to passenger train?
6:74:95:18:113:7Option C

Let the Speed of Goods train = 11m/sec.

Speed of the Passenger train = 13m/sec

Relative speed in the same direction = 13 – 11 = 2 m/sec.

Total Length (goods + passenger) = 2 × 90 = 180 m

Now, For Goods train; Relative Speed = 13 – 11 = 2 m/sec.

Length of the Goods Train = 2 × 75 = 150 m

Therefore, Length of the Passenger Train = 180 – 150 = 30 m

Hence Required Ratio = 150: 30 = 5: 1 - The radius of a circular field is equal to the side of a square field. If the difference between the perimeter of the circular field and that of the square field is 32m, what is the perimeter of the square field?
56 m60 m35 m48 m66 mOption A

Let the radius of the circular field = the side of the square be x m .

Now, 2*pi*x – 4*x = 32

=> x = 14 m

Hence, the perimeter of the square = 4*14 = 56 m . - In an Almira there are 3 maroon caps, 5 grey caps and 7 brown caps. 2 caps are drawn one by one without replacement. If the first cap comes out to be of maroon colour, then 8 more maroon coloured caps are added to Almira. Find the probability that both the caps drawn are of maroon colour.
1/114/152/193/14None of theseOption A

Required probability

= (3/15)*(10/22) = 1/11 - When six fair coins are tossed simultaneously, in how many of the outcomes will at most three of the coins turn up as heads?
5035524233Option D

Total number of outcomes = 6C0+6C1+6C2+6C3

= 1+6+15+20 = 42 outcomes