7TH GRADE MATH PROBLEMS AND SOLUTIONS

Question 1 :

.Find the value of 664.02 ÷ 9.3

(A) 71.4           (B) 19.2           (C) 6.11

Solution :

=  664.02 / 9.3  

In order to remove the decimal point, we have to multiply both numerator and denominator by 100.

  =  (664.02 / 9.3 ) ⋅ (100/100)

  =  66402/930

  =  714/10

=  71.4

Question 2 :

Let A be the sum of seven 7's. Let B be the sum of seven A's. Then find the value of B?

(A) 175         (B) 781        (C) 343

Solution :

A  =  Sum of seven 7's

A  =   (7 + 7 + 7 + 7 + 7 + 7 + 7)

A  =  49

B =  the sum of seven A's

B  =  7(49)

  =  343

So, the value of B is 343.

Question 3 :

In triangle ABC, BC = 4 and CA = 6. If the perimeter of the triangle is 4 times the length of side BC, what will be the length of AB?

(A) 8                     (B) 6                     (C) 7

Solution :

Perimeter of triangle  =  sum of length of all sides

4(BC)  =  AB + BC + CA

4(4)  =  AB + 4 + 6

16 = AB + 10

Subtract both sides by 10

16 - 10  =  AB

So, the value of AB is 6 cm

Question 4 :

David is 7 feet and 5 inches tall. His basketball hoop is 10 feet from ground. Given that there are 12 inches in a foot, how many inches should he jump to touch the hoop with his head?

(A) 35           (B) 31              (C) 38

Solution :

Height of David  =  7 feet 5 inches

David's height in inches  =  7(12) + 5

 =  89 inches

Height of basketball's hoop from ground  =  10 feet

=  10 (12)  =  120 inches

Number of inches which he has to cover  =  120 - 89

  =  31 inches

Question 5 :

Cameroon has 3 triangles and 6 pentagons. If R is being the total number of sides of the shapes and S be the number of shapes he has, then S + R is

(A) 48                       (B) 50                   (C) 52

Solution :

Number of triangles  =  3

Number of sides in 3 triangles  =  3(3)  =  9

Number of triangles  =  6

Number of sides in 6 pentagon  =  6(5)  =  30

R  =  9 + 30  =  39

S  =  39 + 9

  =  48

So, the answer is 48.

Question 6 :

Arnold would like to get at least 90% of math competition problems correct. However, he had only answered 20 out of 50 correctly so far this year. What is the least number of problems that he needs to solve in the future in order to reach 90%?

(A) 250         (B) 321                 (C) 310

Solution :

Percentage of marks he has scored so far 

=  (20/50) x 100

=  40%

40% of 50  =  20

90% of 50  = ?

  =  (90/100) x 50

  =  45 question

From this we come to know that he has to give correct answers for 45 question out of 50 questions.

Question 7 :

Find the ratio of 250g to 10kg

(A) 1:50               (B) 40:1                (C) 1:40

Solution :

In order to write the given numbers as ratio, we need to change the given quantities in the same kind.

10 kg  =  10 (1000)

 =  10000 grams

=  250 : 10000

=  250/10000

=  40 : 1

Question 8 :

In 2010, the population of a town is 1,50,000. If it is increased by 10% in the next year, find the population in 2011.

(A) 185000       (B) 165000          (C) 1,25,000

Solution :

Population in 2010 = 1,50,000

Increase in population = (10/100) ⋅ 150000

= 15000

Population in 2011 = 150000 + 15000

=  165000

Question 9 :

A company wishes to simultaneously promote two of its 6 department heads out of 6 to assistant managers. In how many ways these promotions can take place ?

Solution :

By using the concept combinations, we may solve this problem.

⁶ C ₂   =  (6 ⋅ 5) / 2

⁶ C ₂   =  30 / 2

  ⁶ C ₂   =  15

So, the answer is 15 ways.

Question 10 :

John got 66 marks out of 75 in Mathematics and 72 out of 80 in Science. In which subject did he score more ?

(A) 7                  (B) 3                     (C) 8

Solution :

Percentage of marks in the subject math

  =  (66/75) ⋅ 100

  =  88%

Percentage of marks in the subject science

  =  (72/80) ⋅ 100

  =  90%

He has scored more marks in science.

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