Grade 10 : CBSE : Mathematics Sample question paper (MCQ) with detailed answer - Term 1

 Class - X

Mathematics

Sample Question Paper - Term 1

1) The ratio of LCM and HCF of the least composite and the least prime numbers is _____

Solution

Least composite number = 4

Least prime number = 2

LCM of the least composite and the least prime numbers  

that is, LCM of 4 and 2 is 4

HCF of 4 and 2 is 2

ration of LCM and HCF = 4:2 = 2:1

Hence the answer is 2:1

2) The value of k for which the lines 5x+7y=3 and 15x+21y = k coincide is _______

Solution

consider the first equation 5x+7y=3

 observe the second equation, if we multiply the first equation by 3 then we will get the left-hand side of the second equation

that is, (5x+7y=3) * 3 

gives,  15x + 21y = 9

now let's compare the above equation with the second equation, 

we got k = 9

Hence the value of k is 9

Note: When two lines are coinciding with each other, then there could be no intercept difference between them. For example, 5x+7y=3 and 15x+21y = 9  are coinciding lines.

3) A girl walks 200m towards the East and then 150m towards the North. The distance of the girl from the starting point is _____

Solution 

Hence the distance of the girl from the starting point is 250m

4) The lengths of the diagonals of a rhombus are 24cm and 32cm, then the length of the altitude of the rhombus is _____

Solution

In Rhombus, all the sides are equal

Diagonals of the Rhombus are Perpendicular bisectors of each other.

AC is perpendicular to BD

so, OB = BD/2 = 32/2 = 16

      OA = AC/2 = 24/2 = 12

Now, in Right angle Triangle OAB , 

By Pythagorus theorem 

AB^2 = OA^2 + OB^2

          =  12^2 + 16^2

          = 144 + 256

AB^2 = 400

AB ( Base ) = 20

Area of Rhombus = 1/2 * diagonal 1 * diagonal 2

                              = 1/2 * 24 * 32  = 12 * 32

                              =  384 cm^2

Area using base and height :

 

Area of Rhombus = Base * Height

           384             =  20 * Height

height = 384/20 = 19.2Cm

Hence the altitude of the Rhombus = 19.2cm

5) Two fair coins are tossed. What is the probability of getting at the most one head?

Solution

Possible outcomes (Sample space S) = { HT , TT, HH , TH }

Number of possible outcomes [ n(S) = 4 ]

Favourable Outcome = Getting at the most one head = Outcomes with NO head + Outcome with 1 head

= { HT , TT , TH }

Number of favourable outcomes = 3

Probability = Number of favourable outcomes / Total number of outcomes

Probability of getting at the most one head = 3 / 4 or 3:4

Hence the answer is 3/4

6) ΔABC~ΔPQR. If AM and PN are altitudes of ΔABC and ΔPQR respectively and AB^2 : PQ^2 = 4 : 9, then AM:PN =

Solution

Given,  

               AB^2 : PQ^2 = 4 : 9

implies     AB^2 / PQ^2 = 4 / 9

               (AB/PQ)^2 = 2^2/3^2

                AB/PQ = 2/3

Also given, Triangle ABC and Triangle PQR are similar triangles then

Ratio of their altitudes = ratio of their sides

 AM : PN = AB : PQ

that is , AM : PN = 2 : 3

Hence the answer is 2 : 3