Quantitative Aptitude Challenge-1- Solution


Quantitative Aptitude Challenge Solution (Difficulty was Moderate):
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Q1. Sum of the ages of a boy and his father is 70 years. Before 20 years Father's age was 14 times the age of the boy. What was the age of father at the time of birth of the child?
Options:
1) 25
2) 26
3) 27
4) 28

Answer: 26
Detailed Explanation:
Let present age of father and the son are x and y then according to the given condition x+y = 70 
before 20 years  (x-20) = 14(y-20) 
so 70-y-20 = 14(y-20) 
or 50-y = 14y-280
or 15y = 330
or y = 22 and x = 48
so at the time of birth of child, age of father = 48-22 = 26


Q2. What will be the probability that k1*k2*k3 is an even number, if k1, k2, k3 
are 3 positive integers taken randomly. 
Options. 
1) 1/3
2) 1/8
3) 2/3
4) 7/8

Answer : 7/8
Detailed Explanation:
Each of the three numbers could either be even or odd. 
Total number of combinations of such numbers = 2^3 = 8
There is only one comination where multiple of 3 numbers would be odd is when K1, K2 and K3 all are Odd numbers.
For rest all 7 combinations the multiple of 3 number would always be even.
Hence probability = 7/8. 
Key Learning: 
When any number is multiplied by an even number, the result is always a even number.


Q3.What is the sum of all the four digit numbers formed using the digits 0, 1, 2 and 3. None of the two digits should be repeated.?
1) 36664
2) 34664
3) 37664
4) 38664

Answer: 38664
Detailed Explanation:
From these four digits '0' cannot be at first place to form four digit number 
So we have 1, 2, or 3 can be at thousand's place 
Keeping one digit at thousand place(example 1XXX or 2XXX or 3XXX ) we can have 6 different four digit number.For example if 1 is fixed at thousands place following 6 four digit numbers are possible = 1023, 1032,1203,1230,1302,1320 . Similarly for  2 and 3 fixed at thousand place.
So that means 1*1000*6 + 2*1000*6 + 3*1000* 6 = 36000

Now come to hundred's place we can have all four digit at this place , '0' will not contribute any thing to sum, so in all four digits number 1,2,3 will be at 4 times on same place which means: 100*4 + 200* 4 + 300 * 4= 2400

Now come to tenth's place ...similar as hundred 
So 10*4 + 20*4 + 30*4= 240

Similarly for unit place 
1*4 + 2*4 + 3*4= 24 
At last if you add whole the total sum of all possible four digit number formed is 38664
Key Learning:
1. While solving such questions, fix digits at each position and find out all combinations of numbers at other positions. 
2. To understand the logic above easily, you should try to solve & master a simpler question such as sum of all the three digit numbers formed by 1,2,3.
All possible 3 digit numbers are:
123,132,213,231,312,321. Now the slowest way to solve this question is to identiyf all such numbers and add all these numbers to get the answers :).
Smart way to solve this would be based on the fact that 123= 1*100+2*20+3*1
Step-1 :Fix nummbers at at each place (100th place, 10th place & unit place) one by one and then finding all different numbers which are possible.
Fixing a number at 100th place there are 2 different numbers which could be formed by arranging numbers at 10th & unti place. Similarly on fixing numbers at 10th place and unit place there are 2 different numbers which could be formed by arranging numbers at 100th & unit place and 100th & 10th place respectively. 
Step-2 :Find sum at each place of all digits (100th place, 10th place & unit place)
So for 100th place  = (1*100+2*100+3*100)*2 = 6*100*2 = 1200
Similarly for 10th place = (1*10+2*10+3*10)*2 = 6*10*2 = 120
Similarly for unit place = (1*1+2*1+3*1)*2 = 6*1*2 = 12
Step-3 : Do sum of all numbers = 1200 + 120 + 12 = 1332

  
Q4. There are 5 lines in a plane. Difference between maximum and minimum number of intersecting points. No two lines are parallel 
Options. 
1) 9
2) 8
3) 10
4) 11

Answer  9
Detailed Explanation:
For n non parallel lines, maximum number of intersection points = n*(n-1)/2
Here n=5
max points of intersection = 5*4/2=10
Minimum number of intersection pointts = 1 (all lines intersecting at common point)
Difference = 10-1=9
Key Learning: 
Maximum number of intersection points for n non parallel lines = n*(n-1)/2


Q5. How many words can be formed by using all the letters of the word MASTERCHEF at a time if you have to arrange these letters in the same manner as these letters are arrange in english alphabet?
(! represents factorial in options) 
Options:
1) 10!/2
2) 5!
3) 1!+0!
4) 0!

Answer 0!
Detailed Explanation:
Only 1 method is possible on the basis of conditions(letters are arranged in english alphabet) given this word is ACEEFHMRST
Key Learning: 
Remember 0!=1


Q6. 1,4,9,16,......
sum of this sequence upto 20 terms will be 
Options. 
1) 1980
2) 2050
3) 2870
4) 2980

Answer: 2870
Detailed Explanation:
= 1^2+2^2+3^2+4^2+..............
= sum of squares of first 20 integers
according Sum of squares of first n integers = (n*(n+1)*(2n+1))/6
On putting n = 20, answer will be obtained
Key Learning:
Sum of the first n integers = n(n+1)/2
Sum of squares of first n integers = (n*(n+1)*(2n+1))/6
Sum of cubes of first n integers = (n(n+1)/2)^2
Sum of cubes = (sum of integers)^2
1³ + 2³ + 3³ + ... + n³ = ( 1 + 2 + 3 + ... + n )² = (n(n+1)/2)^2


=> Back to Questions Quantitative Aptitude Challenge-1 - Test your Quantitative Aptitude for MBA Aspirants

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