## GRADE 10 MATH PRACTICE QUESTIONS

Problem 1 :

AB and CD are two chords of a circle which intersect each other externally at P. If AB  =  4 cm, BP  =  5 cm, PD = 3 cm, then the length of CD is.

Solution : PA ⋅ PB  =  PC ⋅ PD

Let x be CD.

PA  =  9 cm, PB  =  5 cm, PC  =  3 + x and PD  =  3 cm

9 ⋅ 5  =  (3 + x) ⋅ 3

15  =  3 + x

x  =  15 - 3

x  =  12

So, CD is 12 cm.

Problem 2 :

Find a if the slope of the line joining (-6, 13) and (3, a) is -1/3

Solution :

Slope of the line  =  (y2 - y1)/(x2 - x1)

(a - 13) / (3 + 6)  =  -1/3

(a - 13) / 9  =  -1/3

a - 13  =  -3

a  =  -3 + 13

a  =  10

So, the value of a is 10.

Problem 3 :

Two dice are thrown together. What is the probability of getting a total 8 or a product 12.

Solution :

Sample space  =

{(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6) (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6) (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6) (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6) (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6) 6 (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6) }

n(s)  =  36

Let A and B are the events of the sum 8 or the product is 12.

A  =  {(2, 6) (3, 5) (4, 4) (5, 3) (6, 2)}

n(A)  =  5

p(A)  =  n(A)/n(S)

p(A)  =  5/36

B  =  { (2, 6) (3, 4) (4, 3)(6, 2) }

n(B)  =  4

p(B)  =  4/36

A n B  =   {(2, 6) (6, 2)}

n(A n B)  =  2

p(A n B)  =  n(A n B) / n(S)

p(A n B0  =  2/36

P(A u B)  =  p(A) + p(B) - p(A n B)

=  5/36 + 4/36 - 2/36

=  (9 - 2)/36

=  7/36

So, the probability of getting the sum 8 or the product as 12 is 7/36.

Problem 4 :

The line joining the points (-2, 7) and (3, 6) is parallel to the line joining (4, a) and (9, 1), find a.

Solution :

If two lines are parallel, they will have equal slopes.

Let the given points as A(-2, 7), B(3, 6), C(4, a) and D(9, 1)

Slope of line AB  =  Slope of line CD

Slope of the line joining two points  =  (y2 - y1)/(x2 - x1)

(6 - 7) / (3 + 2)  =  (1 - a)/(9 - 4)

-1/5  =  (1 - a)/5

-1  =  1 - a

a  =  1 + 1

a  =  2

So, value of a is 2.

Problem 5 :

Find the range of the heights of 12 girls in the class given in cm

120, 110, 150, 100, 130, 145, 150, 100, 140, 150, 135, 125

Solution :

Range  =  Large value - Small value

L  =  150 and S  =  100

Range  =  150 - 100

Range  =  50

Problem 6 :

The volume of a cylinder is 448π cm and its height is 7 cm.Find the radius of the cylinder.

Solution :

Volume of cylinder  = πr2 h

πr2 h  =  448 π

Height of the cylinder(h)  =  7 cm

r2 (7)  =  448

r2  =  64

r  =   8

So, the radius of the cylinder is 8 cm.

Problem 7 :

How many terms of a series

1 + 6 + 11 + ............

must be taken so that their sum is 970?

Solution :

Sn  =  970

(n/2) [2a + (n -1) d]  =  970

a  =  1, d  =  6 - 1  ==>  5

(n/2) [2(1) + (n -1) 5]  =  970

(n/2) [2+5n-5]  =  970

(n/2) (-3 + 5n)  =  970

-3n + 5n2  =  970(2)

5n- 3n  - 1940   =  0

((n - 20) (5n + 97)  =  0

n  =  20 and n  =  -97/5

So, 20 terms must be taken to have the sum 970.

Problem 8 :

Find the values of k for which the equation

12x²+4kx+3  =  0

has real and equal roots.

Solution :

Since the quadratic equation has real and equal roots

Discriminant  =  0

b2 - 4ac  =  0

a  =  12, b  =  4k and c  =  3

(4k)2 - 4(12)(3)  =  0

16k2 - 144  =  0

k2 - 9  =  0

k2 - 32  =  0

(k + 3) (k - 3)  =  0

k  =  -3 and k  =  3

So, the value of k is -3 or 3.

Problem 9 :

The sum of three numbers is 24.Among them one number is equal to half of the sum of the other two numbers but four times the difference of them. Find the numbers.

Solution :

Let x, y and z be the three numbers.

x+y+z  =  24  ---(1)

x  =  (y+z) / 2

2x-y-z  =  0 -----(2)

x  =  4(y-z)

x  =  4y-4z

x-4y+4z  =  0-----(3)

(1) + (2)

3x  =  24

x  =  8

By applying the value of x in (2), we get

16-y-z  =  0

y+z  =  16  ----(4)

By applying the value of x in (3), we get

8-4y+4z  =  0

y-z  =  2 ----(5)

(4) + (5)

2y  =  18

y  =  9

By applying the value of y in (5), we get

9-z  =  2

z  =  7

So, the required numbers are 8, 9 and 7.

Problem 10 :

A ladder 17 m long touches a window of a house 15 m above the ground. Determine the distance of the floor of the ladder from the house.

Solution : AC2  =  AB2 + BC2

172  =  152 + BC2

289 - 225  =  AC2

AC  =  8

So, the ladder is at the distance of 8 m. Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here.

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

WORD PROBLEMS

Word problems on simple equations

Word problems on linear equations

Algebra word problems

Word problems on trains

Area and perimeter word problems

Word problems on direct variation and inverse variation

Word problems on unit price

Word problems on unit rate

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

Double facts word problems

Trigonometry word problems

Percentage word problems

Profit and loss word problems

Markup and markdown word problems

Decimal word problems

Word problems on fractions

Word problems on mixed fractrions

One step equation word problems

Linear inequalities word problems

Ratio and proportion word problems

Time and work word problems

Word problems on sets and venn diagrams

Word problems on ages

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

Profit and loss shortcuts

Percentage shortcuts

Times table shortcuts

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

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

Sum of all three four digit numbers formed using 0, 1, 2, 3

Sum of all three four digit numbers formed using 1, 2, 5, 6

Featured Categories

Math Word Problems

SAT Math Worksheet

P-SAT Preparation

Math Calculators

Quantitative Aptitude

Transformations

Algebraic Identities

Trig. Identities

SOHCAHTOA

Multiplication Tricks

PEMDAS Rule

Types of Angles

Aptitude Test 