**Aptitude Test 2 : **

The test given in this section can be taken without any login credentials. At the end of the test, you can check you score as well as detailed answer for each question.

Question No.1 Question No.2 Question No.3 Question No.4 Question No.5 Question No.6 Question No.7 Question No.8 Question No.9 Question No.10 |
From 6
^{10} × 7^{17} × 55^{27}, we have to write each base in terms of multiplication of its prime factors.
That is,=(2X3) ^{10}×(7)^{17}×(5X11)^{27}
=2 ^{10}X3^{10}X7^{17}X5^{27}X11^{27}
The no. of prime factors = sum of the exponents = 10+10+17+27+27 = 91 Hence, the number of prime factors is 91.
When the two train running in opposite direction, relative speed = 60+48 = 108 kmph = 108X5/18 m/sec = 30 m/sec
Sum of the lengths of the two trains = sum of the distances covered by the two trains in the above relative speed Sum of the lengths of the two trains = 30X15 = 450 m When the two trains running in the same direction, relative speed = 60-48 =12 kmph = 12X5/18 = 10/3 m/sec When the two trains running in the same direction, a person in the faster train observes that he crossed the slower train in 36 seconds. The distance he covered in 36 seconds in the relative speed is equal to the length of the slower train. Length of the slower train = 36X10/3 = 120 m Length of the faster train = 450-120 = 330 m Hence, the length of the two trains are 330m and 120m
We can apply L.C.M method to solve this problem
Total work = 90 units (L.C.M of 15,30,18,9) (A+B) can complete 6 units/day (90/15 = 6) (B+C) can complete 3 units/day (90/30 = 3) (A+C) can complete 5 units/day (90/18 = 5) By adding, we get 2(A+B+C) = 14 units/day (A+B+C) = 14/2 = 7 units/day work done by A = (A+B+C)-(B+C)=7-3=4 units/day work done by B = (A+B+C)-(A+C)=7-5=2 units/day work done by C = (A+B+C)-(A+B)=7-6=1 unit/day work done by (A+B+C)in 9 days = 9X7 = 63 units. Balance work = 90-63 = 27 units This 27 units of work to be completed B & C. Because A left after 9 days of work. No. of days taken by B & C to complete the balance 27 units of work = 27/3 = 9 days. (Because B&C can complete 3 units of work per day)
Average speed = 2pq/(p+q), here p=25 q=4
= (2X25X4)/(25+4) Therefore, average speed = 200/29 km/hr And 5 hour 48 min = 5 ^{48}⁄_{60}hrs = 29/5 hours
Distance = Speed X Time Distance = (200/29)X(29/5) = 40 km Distance covered in 5 hrs 48 min = 40 km Hence,distance of the post office from the village = 40/2 = 20km
Let "x" be the present age of the son.Then the present age of father is (3x+3)
3 years hence, father's age will be 10 years more than twice the age of the son (3x+3)+3 = 2(x+3)+10 3x+6 = 2x+16 x = 10 To find the present age of the father, plug x = 10 in (3x+3) Present age of the father = 3(10)+3 = 33 years
If "x' be the first even number, then the four consecutive even numbers are x, x+2, x+4, x+6
Average of the four consecutive even numbers = 27 (x+x+2+x+4+x+6)/4 = 27 (4x+12) = 108 4x = 96 ===> x = 24 Hence the largest even number = x+6 = 30
From the given information, we have
Speed downstream = (14+4) = 18 kmph Speed upstream = (14-4) = 10 kmph Let "x" be the distance between A and B x/18 + (x/2)/10 =19 (Hint: Time = Distance/Speed) x/18 + x/20 =19 (10x + 9x)/180 = 19 (L.C.M of 18,20 is 180) 19x/180 = 19 x = 180 Hence the distance between A and B is 180 km.
Original weight of John = 56.7 kg (given)
He is going to reduce his weight in the ratio 7:6 [Hint:If a quantity increases or decreases in the ratio a:b, then new quantity = 'b' of the original quantity divided by 'a'] His new weight = (6x56.7)/7 = 6x8.1 = 48.6 kg.
Let the cost price of 1 ltr pure milk be $1
Now we take some quantity of milk (less than 1 ltr),add some water and make it to be 1 ltr mixture Let "x" be the money we invest for milk in 1 ltr of milk-water mixture Since the gain is 20%, selling price = x + 20%of x Selling price = 120% of x = (120/100)x = (6/5)x But the mixture is sold at the cost price of pure milk So, we have (6/5)x = 1 x = 5/6 Therefore cost price of the milk in the mixture = $(5/6 Cost price of the water = $0 Rule to find the ratio for producing mixture = (d-m):(m-c) (d-m):(m-c) = 1-5/6:5/6-0 = 1/6:5/6 = 1:5 So, water and milk to be mixed in the ratio to gain 20% is 1:5
Let A,B and C be the incomes of A,B and C respectively
From the given information, we have C = $1600 10% of B = 30% of C (10/100)B = (30/100)X1600 B = $ 4800 15% of A = 25% of B (15/100)A = (25/100)X4800 A = $8000 A+B+C = 8000+4800+1600 = 14400 Hence, the total income of A,B and C is $14400 |

After having practiced answering the above questions, we hope that the students would have understood, how to solve quantitative problems easily.

Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here.

Widget is loading comments...

You can also visit our following web pages on different stuff in math.

**WORD PROBLEMS**

**Word problems on simple equations **

**Word problems on linear equations **

**Word problems on quadratic equations**

**Area and perimeter word problems**

**Word problems on direct variation and inverse variation **

**Word problems on comparing rates**

**Converting customary units word problems **

**Converting metric units word problems**

**Word problems on simple interest**

**Word problems on compound interest**

**Word problems on types of angles **

**Complementary and supplementary angles word problems**

**Markup and markdown word problems **

**Word problems on mixed fractrions**

**One step equation word problems**

**Linear inequalities word problems**

**Ratio and proportion word problems**

**Word problems on sets and venn diagrams**

**Pythagorean theorem word problems**

**Percent of a number word problems**

**Word problems on constant speed**

**Word problems on average speed **

**Word problems on sum of the angles of a triangle is 180 degree**

**OTHER TOPICS **

**Time, speed and distance shortcuts**

**Ratio and proportion shortcuts**

**Domain and range of rational functions**

**Domain and range of rational functions with holes**

**Graphing rational functions with holes**

**Converting repeating decimals in to fractions**

**Decimal representation of rational numbers**

**Finding square root using long division**

**L.C.M method to solve time and work problems**

**Translating the word problems in to algebraic expressions**

**Remainder when 2 power 256 is divided by 17**

**Remainder when 17 power 23 is divided by 16**

**Sum of all three digit numbers divisible by 6**

**Sum of all three digit numbers divisible by 7**

**Sum of all three digit numbers divisible by 8**

**Sum of all three digit numbers formed using 1, 3, 4**

**Sum of all three four digit numbers formed with non zero digits**