Here we bring you the CBSE Class 10 Mathematics Solved Practice Paper 2017-2018: Set-III. This practice paper is designed as per the latest examination format and consists of questions which are quite important from exam point of view.
Mathematics is one such subject in which cannot be learnt by just memorizing a set of formulae given in the book. Students are required to do a lot of practice to understand the concepts and be good at this subject. Therefore, solving more and more practice papers, guess papers and sample papers can be of huge benefit to make an effective preparation for the board exams.
CBSE Class 10 Mathematics Solved Practice Paper: Set-II, has been specially prepared by the subject experts after the brief analysis of previous year question papers and the latest examination format. Practicing this paper will definitely help to fine tune your preparations for the board exam 2018.
Structure of this paper is as below:
Practicing this paper will help the students in understanding the depth with which a topic should be studied in order to prepare in a more effective way to score a meritorious position.
Some sample questions and their solutions from the CBSE Class 10 Mathematics Solved Practice Paper are given below:
Q. If the HCF of 65 and 117 is expressible in the form of 65 m – 117, then find the value of m.
We have, 65 = 13 × 5
And 117 = 13 × 9
Hence, HCF = 13
According to question 65m – 117 = 13
⇒ 65m = 13 + 117 = 130
⇒ m = 130/65 = 2
Q. 144 cartons of Coke cans and 90 cartons of Pepsi cans are to be stacked in a canteen. If each stack is of the same height and is to contain cartons of the same drink, what would be the greatest number of cartons each stack would have ?
The greatest number of cartons in each stack is the HCF of 144 and 90 .
144 = 24 × 32
90 = 2 × 32 × 5
HCF = 2 × 32 = 18
∴ The greatest number of cartons = 18.
Q. Two tangents drawn at the end points of a chord XYof a circle of radius 10 cm, intersect at a point Z as shown in the figure given below. If the chord is 12 cm long then find the length of XZ and YZ.
Given, Radius of circle, OY = 10 cm
And, XY = 12 cm
XP = YP = 6 cm (OZ is a perpendicular bisector of XY)
In Δ OLB, using Pythagoras theorem we get:
OY2 = OP2 + PY2
⟹ 102 = OP2 + 62
⟹ OP2 = 100 ─ 36 = 64
⟹ OP = 8 cm
Let ZP = x and ZY = y
Then OZ = (x + 8) cm
And ZX = ZY = y (Tangents from an external point are equal in length)
Thus in Δ ZPY, we have from Pythagoras theorem,
ZY2 = ZP2 + PY2
⟹ y2 = x2 + 62
⟹ y2 = x2 + 36 ……(i)
Also in ΔOZY, we have from Pythagoras theorem,
OZ2 = ZY2 + OY2
⟹ (x + 8) 2 = y2 + 102
⟹ (x + 8) 2 = y2 + 100 …….(ii)
Putting value of y2 from (i) in equation (ii), we get:
(x + 8) 2 = x2 + 36 + 100
⟹ x2 + 64 + 16x = x2 + 36 + 100
⟹ 16x = 72
⟹ x = 9/2cm
y2 = x2 + 36
⟹ y2 = (9/2)2+ 36 = 225/4
⟹ y = 15/2cm
Therefore, ZX = ZY = y = 15/2 cm
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