- Average age of the class of 36 students is 17 years. Average age of the class is increased by two years if the age of the class teacher and the principal is included. Age of the principal is x% is more than the age of the class teacher. Find the value of x if the age of the principal is 10 years more than the age of the class teacher.
50%40%10%20%30%Option D

Sum of the ages of the students = 36*17

= 612 years

Sum of the ages of students, class teacher and principal = 38*19 = 722 years

Let the age of the class teacher = ‘y’ years

Age of principal = ‘y+10’ years

612+y+y+10 = 722

=> y = 50

Value of x = (60-50)/50*100 = 20% - Amit had certain number of toffees with him. He gave 2/15, 2/7 and 1/6 of the toffees to Abhishek, Vishal and Govind resp. Now Amit is left with only 522 toffees. Find the total number of toffees received by Abhishek.
160168135144150Option B

Let the total number of toffees Amit had be x.

Number of toffees left with Amit = x – [2x/15 – 2x/17 – x/6] = 522

=> x = 1260

Number of toffees left with Abhishek = 1260*2/15 = 168 - A,B and C started a business with initial investments in the ratio of 5:2:7 resp. After one year A, B and C made additional investments in the ratio of 6:11:17 resp. Find the profit share of C out of the total profit of Rs.3360 after two years.
Rs.1450Rs.1330Rs.1420Rs.1550Rs.1680Option E

Let the initial investments of A, B and C be 5x , 2x and 7x resp.

Ratio of their profits = 5x+5x+6y : 2x+2x+11y : 7x+7x+17y

= 10x+6y : 4x+11y : 14x+17y

Profit share of C = (14x+17y)/(28x+34y)*3360 = Rs.1680 - 85% of the population of Kangra is registered in voters list. 25% of the registered voters didn’t cast their votes. 10% of the cast voters are invalid. The winning candidate got 55% of the valid votes and won by 4131 votes. Find the population of Kangra.
72,00068,00054,00066,00070,000Option A

Let the population of Kangra be x.

Number of registered voters = 0.85x

Number of cast votes = 0.75*0.85 = 0.6375x

Number of valid votes = 0.90*0.6375x = 0.57375x 0.10* 0.57375x = 4131

=> x = 72,000 - Upstream speed of the boat is 20% less than the downstream speed of the boat. If the boat can cover 90 km in still water in 4 hours, find the time taken by the boat to cover 120 km upstream and 125 km downstream.
8 hours11 hours10 hours12 hours9 hoursOption B

Speed of the boat in still water = 90/4 = 22.5 km/hr.

Let the speed of the stream be x km/hr.

Now, 0.80*(22.5+x) = 22.5 – x

=> x = 2.5

Upstream speed of the boat = 22.5 – 2.5 = 20 km/hr.

Downstream speed of the boat = 22.5 + 2.5 = 25 km/hr.

Time taken by the boat = 120/20 + 125/25 = 11 hours - Sahil invested Rs. 84000 in a scheme offering simple interest for three years. Rate of interest for first, second and third year is 20%, 24% and 25% resp. Find the interest earned by him after three years.
Rs. 61260Rs. 52550Rs. 48540Rs. 57960Rs. 51250Option D

Interest earned by Sahil after three years

= 84000*(0.20+0.24+0.25) = Rs. 57960 - A shopkeeper marked an article an article 45% above the cost price and sold it after a discount of 20%. Had he sold the article after a discount of 15% he would gained Rs.899 more. Find the cost price of the article.
Rs.13500Rs.15000Rs.14300Rs.12400Rs.13100Option D

Let the CP of the article be x.

MP = 1.45x

SP of the article if sold after a discount of 20% = 1.45x * 0.80 = Rs.1.16x

SP of the article if sold after a discount of 15% = 1.45x * 0.85 = Rs.1.2325x

Now, 1.2325x – 1.16x = 899

=> x = 12400

CP of the article = Rs.12400 - A vessel contains mixture of 180 litres of milk and 30 litres of water. A certain quantity of mixture is taken out of the vessel and is replaced with 52 litres of water such that the ratio of milk to water in the vessel becomes is 5:3. Find the final quantity of milk in the vessel.
95 litres105 litres110 litres120 litres100 litresOption D

Initial ratio = 180:30 = 6:1

7x litres of mixture contains 6x litres of milk and x litres of water.

(180-6x)/(30 –x + 52) = 5/3

=> x = 10

Quantity of milk present in the vessel = 180 – 6*10 = 180 – 60 = 120 litres - A man goes to his office from his house at a speed of 3 km/hr and returns at a speed of 2 km/hr. If he takes 5 hours in going and coming, what is the distance between his house and office?
2 km4 km5 km6 km8 kmOption D

Average speed = (2×3×2)/(2+3) = 12/5 km/hr

Total time taken = 5 hours

Distance travelled = 12/5 × 5 = 12 km

Therefore, distance between his house and office = 12/2 = 6 km - How many 10 letter words with or without meaning can be formed by using all the letters of the word ‘TEMPTATION’.
565200486300604800588000611250Option C

Number of words formed = 10!/3! = 604800