#### Data Interpretation

• ( 1 ) The least positive integer which is a perfect square and also divisible by each of 6,12, 18 and 24?

• 1) 36
• 2) 100
• 3) 144
• 4) 196
• Discussion in forum
Solution : Perfect Square : A number when written as product of prime factors , if each prime factor has even number as its power , then the number is perfect square.
Writing the given number as product of prime factors
6 = 2 * 3
12 = 23 * 3
18 = 2 * 32
24 = 23 * 3
LCM = 23 * 32 = 72
But 72 is not the perfect square. But multiples of 72 are also exactly divisible by given numbers.
72 * 2 = 144 is the perfect square. So answer is 144

discussion

• ( 2 ) What is the side of the largest possible square brick which can be paved on the floor of a room 4m 96cm long and 4m 3cm broad?

• 1) 27
• 2) 28
• 3) 29
• 4) 31
• Discussion in forum
Solution : Converting given measurements into centimeters
4m 96 cm = 400 + 96 = 496 cm
4m 3cm = 400 + 3 = 403 cm
To find the side of the largest possible brick , the side of brick must divide length as well as breadth exactly.
Using division method
403)496(1
403
_____
93) 403 (4
370
_____
31) 93 (3
93
____
0
HCF = 31

discussion

• ( 3 ) Find the smallest number which when increased by 5 is divisible by 9, 21, 25 and 30.?

• 1) 3145
• 2) 6305
• 3) 3155
• 4) 1605
• Discussion in forum
Solution : LCM = 3 * 5 * 3 * 7 * 5 * 2 = 3150
By definition, LCM of a given set of numbers is their smallest common multiple. In other words, it is the smallest numbers which is divisible by all the given numbers.
So, to get the answer , we have to deduct 5 from the LCM.
3150 - 5 = 3145.

discussion

• ( 4 ) LCM of 0.12, 0.15, 0.2 and 0.54?

• 1) 5.4
• 2) 10.8
• 3) 16.2
• 4) 2.7
• Discussion in forum
Solution : Make the number of decimal places in all the given numbers the same i.e., 0.12, 0.15, 0.2 and 0.54
LCM of 12, 15, 20 and 54
2|12   15   20   54
2|6  15   10   27
3|3  15   5   27
5|1  5   5   9
1   1   1   9
LCM = 22 * 3 * 5 * 9 = 540
LCM = 540/10 = 5.4

discussion

• ( 5 ) In a school, the number of students in Sections A , B and C of 10th class are 70,98 and 126 respectively. Due to overload of student in each class , the administration wants to increase the number rooms . What is the minimum number of rooms required, if in each room the same number of students are to seated and all of them being in the same section?

• 1) 18
• 2) 20
• 3) 21
• 4) 23
• Discussion in forum
Solution : The number students in each room = HCF of 70,98 and 126 = 14
In each room, maximum 14 students can be seated.
Total number students = 70 + 98 +126 = 294
Number of rooms required = 294/14 = 21

discussion

• ( 6 ) A wine seller had three types of wine. 403 liters of 1st kind, 434 liters of 2nd kind and 465 liters of 3rd kind. Find the least possible number of casks of equal size in which different types of wine can be filled without mixing.

• 1) 41
• 2) 42
• 3) 43
• 4) 44
• Discussion in forum
Solution : For the least possible number of casks of equal size, the size of each cask must be of the greatest volume.
To get the greatest volume of each cask, we have to find the largest number which exactly divides 403, 434 and 465.That is nothing but the H.C.F of (403,434,465)
The H.C.F of (403,434,465) = 31 liters
Each cask must be of the volume 31 liters.
Req. No. of casks = (403/31) + (434/31) + (465/31) = 13 + 14 + 15 = 42
Hence, the least possible number of casks of equal size required is 42

discussion

• ( 7 ) Find the H.C.F of 0.63, 1.05, and 2.1

• 1) 0.21
• 2) 0.22
• 3) 0.23
• 4) 0.24
• Discussion in forum
Solution : We we look in to the given numbers 0.63, 1.05, and 2.1, maximum number of digits after the decimal is 2.
So, let us multiply each number by 100 to avoid decimal.
When the given numbers are multiplied by100, we get 63, 105, 210
63 = 32 * 7
105 = 5 * 3 * 7
210 = 2 * 5 * 3 *7
In the prime factors (63, 105, 210), we find 3 and 7 in common
Take 3 and 7 with minimum power. They are 31 and 71
Now, H.C.F of (63, 105, 210) = 31 * 71 = 21 To get H.C.F of (0.63, 1.05, 2.1), divide 21 by 100. 21/100 = 0.21 Therefore, H.C.F of (0.63, 1.05, 2.1) = 0.21

discussion

• ( 8 ) The HCF of two numbers is 11 and their LCM is 693. If one of the numbers is 77, find the other.

• 1) 99
• 2) 88
• 3) 77
• 4) 66
• Discussion in forum
Solution : Given: One of the numbers = 77, H.C.F = 11 and L.C.M = 693
Let "x" be the other number.
Product of two numbers = Product of their H.C.F and L.C.M
77x = 11 * 693
x = 99
Hence, the other number is 99

discussion

• ( 9 ) Find the H.C.F of 513, 1134, and 1215.

• 1) 27
• 2) 28
• 3) 29
• 4) 30
• Discussion in forum
Solution : Given numbers are 513, 1134, 1215
513 = 33 * 19
1134 = 2 * 34 * 7
1215 = 35 * 5
In the above prime factors of (513, 1134, 1215), we find 3 in common.So, take 3 with minimum power. That is 33
Now, H.C.F = 33 = 27

discussion

• ( 10 ) Find the highest common factor of 36 and 84

• 1) 6
• 2) 12
• 3) 14
• 4) 16
• Discussion in forum
Solution : 36 = 22 * 32
84 = 22 * 3 * 7
H.C.F. = 22 * 3 = 12.

discussion