Mensuration Model Questions Set 1 Section-Wise Topic Notes With Detailed Explanation And Example Questions
MOST IMPORTANT quantitative aptitude - 2 EXERCISES
The following question based on Mensuration topic of quantitative aptitude
(a) ${pr}/2$
(b) (p + n) r
(c) (2p – n) r
(d) None of the above
The correct answers to the above question in:
Answer: (a)
The n-sided polygon can be dinded into 'n' triangle with O, the Centre of the circle as one vertex for each triangle. The altitude of each triangle is r. Let the sides of the polygon be '$a_1 ', a_2 ----- a_n$.(Given $a_1 = a_2 = ------ a_n$)
∴ The area of polygon is ${nr}/2 = {pr}/2$
Area of polygon = ${a_1r}/2 + {a_2r}/2 + --- {a_nr}/2 = {pr}/2$
So, option (a) is correct.
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Read more model questions set 1 Based Quantitative Aptitude Questions and Answers
Question : 1
A boy has 3 library cards and 8 books of his interest in the library. Of these 8, he does not want to borrow chemistry part II unless Chemistry part I is also borrowed. In how many ways can he choose the three books to be borrowed ?
a) 26
b) 27
c) 41
d) 56
e) None of these
Answer »Answer: (c)
Two possibilities are there :
(i) Chemistry part I is available in 8 books with Chemistry part II.
or
(ii) Chemistry part II is not available in 8 books but Chemistry part I is available.
Total No. of ways = 1× 1 ×$^6C_1 + ^7C_3$ = 6 + ${7 × 6 × 5}/{3 × 2}$ = 6 + 35 = 41
Question : 2
In how many different ways can be letters of the word 'CYCLE' be arranged?
a) 30
b) 4240
c) 80
d) 120
e) None of these
Answer »Answer: (e)
CYCLE whereas C comes two times.
So, arrangements are = ${5!}/{2!}$ = ${5 × 4 × 3 × 2}/{2}$ = 60 ways
Question : 3
How many different ways can the letters in the word ATTEND be arranged?
a) 240
b) 120
c) 80
d) 60
e) None of these
Answer »Answer: (e)
There are 6 letter in the word 'ATTEND' whereas, T comes two times.
So, required number of ways = ${6!}/{2!}$ = ${720}/{2}$ = 360
Question : 4
Which one of the following is a Pythagorean triple in which one side differs from the hypotenuse by two units?
a) ($2n^2$ , 2n, 2n + 1)
b) (2n + 1, 4n, $2n^2$ + 2n)
c) (2n, 4n, $n^2$ + 1)
d) (2n, $n^2$ – 1, $n^2$ + 1) Where, n is a positive real number.
Answer »Answer: (d)
According to Pythagorean triplet.
The sum of square of base and perpendicular equal to square of hypotenuse.
By hit and trial method:—
$(2n)^2 + (n^2 – 1)^2 = (n^2 + 1)^2$
$4n^2 + n^4 + 1 – 2n^2 = n^4 + 2n^2$ + 1
$n^4 + 2n^2 + 1 = n^4 + 2n^2$ + 1
LHS = RHS
Question : 5
If area of a circle and a square are same, then what is the ratio of their perimeters ?
a) $√π/2$
b) 2 $√π$
c) $√π$
d) $√π/4$
Answer »Answer: (a)
Let radius of cirlce is ‘r’ and side of the square is a ∼ then, from question
$π r^2 = a^2 ⇒ r/a = 1/√π$
Now ∼ ${2πr}/{4a} = {2π}/{4√π} = √π/2$
Question : 6
If the length of the hypotenuse of a right angled triangle is 10 cm, then what is the maximum area of such a right angled triangle ?
a) 25 cm$^2$
b) 100 cm$^2$
c) 50 cm$^2$
d) 10 cm$^2$
Answer »Answer: (a)
Area will be maximum when P and B will be same
So $P^2 + P^2 = H^2 ⇒ P^2 = H^2/2$
⇒ P = $H/√2$
Area = $1/2 BP = 1/2 P^2 = 1/2. H^2/2 = H^2/4$
= ${100}/4 = 25 cm^2$
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