# Math reasoning skills – exercise 1

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Math reasoning test 1

Duration 16 mins

1 / 8

If

• 0<x<4
• x can be 0 or 4.
• and y<12

Which of the following cannot be a value of  xy?

2 / 8

If x and y are prime numbers. Which of the following cannot be the sum of x and y?

3 / 8

If the quotient a/b is positive, which of the following must be true?

4 / 8

Which of the following is the product of two integers whose sum is 11?

5 / 8

The difference of two number is 2. If the sum of their reciprocals is 7/24. What is the sum of the two numbers?

6 / 8

If xy>0. Which of the following must be greater than 0?

i. x+y

ii. x2+y

iii. (x+y)2+1

iv.  (x-y)2

7 / 8

For which value of x is the following inequality true: x2 < 2x

8 / 8

The numbers a,b and c are all positive integers. a is a factor of c and b is also factor of c. if a<b , then which of the following must also be a positive integer factor of c?

0%

In-depth solutions:

Question 1 :

If xy>0. Which of the following must be greater than 0?

i.x+y          ii. x2+y          iii (x+y)2+1           iv (x-y)2

Solution:  let’s eliminate answer options by substituting values

Identify the conditions…… xy>0 and each answer option must be greater than 0. For any value of x and y the answer option >0

since xy>0.  You can substitute

1. x as +ve or y as + ve
2. x as -ve or y as – ve
3. x =y

x = -1 y = -6… xy>0 but x+y is not greater than 0. So rule out  option i and ii

x = -1 y = -6… xy>0 but x2+y  is not greater than 0. So rule out  option ii

x =1 y=1 xy>0 but (x-y)2 is not greater than 0. So rule out  option iv

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Question 2:

For which value of x is the following inequality true: x2 < 2x

1.x<0       2. 0<x<2        3.-2<x<2          4. x<2         5. x>2

Solution:

Tip:  When ever you get ranges.. substitute the lowest possible value and the highest possible value. You can eliminate options faster.

Let x =-1. Hence  x2 < 2x  .. (-1)2 < 2(-1)… not true..

Hence eliminate options 1/4/3

Lets take x = 100 Hence  x2 < 2x  .. (100)2 < 2(100)… not true..

Hence eliminate options 5

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Question 3:

The numbers a,b and c are all positive integers. a is a factor of c and b is also factor of c.if a<b , then which of the following must also be a positive integer factor of c?

1. a+b     2. a*b      3. c/a       4.c/(b*a)

Solution:

a is a factor of c and b is also factor of c

2 is a factor of 6 and 3 is also factor of 6

2+3 is not a factor of 6. Hence eliminate option 1

Note- take big numbers too.. don’t stick to single digit numbers

Lets take c as 100. a is 20 and b is 50

Then 100/( 20*50)  = 0.1 is not a integer factor of 100. Hence eliminate option 4

Lets take c as 24. a is 8  and b is 12

8*12 =96 is not a factor of 24. Hence eliminate option 2

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Question 4:

The difference of two number is 2. If the sum of their reciprocals is 7/24. What is the sum of the two numbers?

1.20          2.14               3.15

Solution:

Let the two numbers be x and y.                 x-y=2               (1/x) + (1/y) = (7/24)

We can solve by quadratic equation. But that’s cumbersome. Lets solve by substituting numbers

(1/x) + (1/y) = (7/24)

Simplifying the fractions:  (y+x)/(xy) = 7/24

Lets break the fraction.    y+x =7              yx=24       and            x-y=2

Now lets substitute numbers .

None of the pairs satisfy all the 3 equations

Lets try again

(y+x)/(xy) = 7/24

Since this is a ratio

Lets try the next iteration of the ratio.. Multiply the numerator and denominator with 2

(y+x)/(xy) = 14/48

Lets break the fraction

y+x =14        yx=48       and      x-y=2

yx=48. Then possibilities are (24,2) (8,6),(12,4)…

8,6 satisfies all the conditions.

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Question 5:

Which of the following number is the product of two integers whose sum is 11?

1. -28         2. -42          3. 12         4.26                 5. 32

Solution:

Let us assume the numbers are a and b.          a+b =11.

a and b can be positive/negative or 0.

Lets take option 1

Break 28 into prime factors….. -28 = (-1)*7*2*2

Combination of prime factors should give two numbers whose sum is 11

Possibilities are

(-14, 2) (-7,4), (14,-2)…………

None of the pairs give a difference of 11.

Lets take option 2

-42 = a*b

-42 =(-1)*2*7*3

Combination of prime factors should give two numbers whose sum is 11

(-14, 3) (-7,6), …………

-14 and 3 gives  11

Hence option 2 is the answer

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Question 6:

If the quotient a/b is positive, which of the following must be true?

1. a>0            2. b>0        3. ab>0            4. a-b>0         5. a+b>0

Solution ..refer to the solution for question 3

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Question 7:

If x and y are prime numbers. Which of the following cannot be the sum of x and y?

1.5     2.9      3.13     4.16       5.23

Solution ..refer to the solution for question 5

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Question 8:

If 0 <= x <=4 and y<12

Which of the following cannot be a value of  xy?

1.-2         2.0            3.6          4.24         5.48

Solution .refer the solution for question 3

Hope you understood both the methods. Even if you are out of touch with math ..

With logic you can ace the math section…

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