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Pradeep Sahajwani Classes (BASE-MAKER)
(NTSE, IMO, NSTSE, OLYMPIAD & Deep Foundation for Competitive Exams)
VIIIth – IXth – Xth CBSE ( Maths / Science )
S-20, Vivekanand Colony, Near Rajat Nagar, Bhopal, Mobile: 9479365082, 9893213700
VIIIth –DEMO PAPER(SA-I BASED)
Q1
Solve for x :
Q2
Solve :
Q3
Solve: 3𝑥 −
Q4
Find x,
Q5
Solve for a,
Q6
7𝑥−1
4
Q7
The denominator of a rational number is greater than its numerator by 8. If the numerator is
−
𝑥−1
2
6𝑥−2
4
1
3
−8𝑥 + 7 = 6𝑥 × 1
×6=1×6−
+
5
3
1
3
×6
=𝑥 −3
2𝑥 − 1 = 4𝑥
3𝑎−2
4
2𝑥 −
3𝑥−12
3
1−𝑥
2
−
2𝑎+3
3
=
2
=3−𝑎
10
3
3
2
increased by 17 and the denominator is decreased by 1, the new rational number is . Find
the original rational number.
Q8
Rinku has a total of Rs 590 as currency notes in the denominations of Rs 50, Rs 20 and Rs 10.
The ratio of the number of Rs 50 notes and Rs 20 notes is 3 : 5. If he has a total of 25 notes,
how many notes of each denomination he has ?
Q9
The perimeter of a rectangle is 52cm. How long is each side if the width is 2 cm more than
one-third of the length ?
Q10
The sum of three consecutive multiples of 5 is 210. Find these multiples .
Q11
If
Q12
The sum of two numbers is 45 & their difference is 17, Find the numbers ?
Q13
The difference of two numbers, one of which is one-fourth of the other, is 5 . Determine the
2 𝑟𝑑
3
of a number is 25 less than the original number, then find the number.
6
numbers
Q14
The sum of a two-digit number is 9. If the number is reduced by 63 the digits of
number gets interchanged. Find the number.
Q15
If ½ is subtracted from a number & the difference is multiplied by 4, the result is
10. Find the number.
PRADEEP SAHAJWANI CLASSES
Pradeep Sahajwani Classes (BASE-MAKER)
(NTSE, IMO, NSTSE, OLYMPIAD & Deep Foundation for Competitive Exams)
VIIIth – IXth – Xth CBSE ( Maths / Science )
S-20, Vivekanand Colony, Near Rajat Nagar, Bhopal, Mobile: 9479365082, 9893213700
ANSWERS
1
Q1. x= 2
11
Q2. x= 3
2
Q3. x=- 3
5
Q4. x=- 11
Q5. a=2
41
Q6. x= 11
𝑥
Q7. Let the required rational number be 𝑦
Conditon 1: y=x+8
Condition 2:
𝑥+17
𝑦−1
=
--------- (1)
3
2
(Put the value of y=x+8)
By solving the above equation; x=13
And from equation no. (1) y=21.
So the original rational no. =
13
21
(Ans).
Q8. Let the ratio be 3x:5x
So the total no. of notes= (3x)x50+(5x)x20+(25-8x)10=590
On solving the above equation we get x=2.
Hence the number of Rs50 notes=6 , Rs20 notes=10 and Rs10 notes=9 (Ans).
Q9. Perimeter of rectangle= 2(l+b) =52
𝑙
Here, b=3+2
𝑙
∴ 2[l+(3+2)] = 52
On solving the above equation we get, l=18 cm and b=8cm. (Ans).
Q10. Let the three consecutive multiples be x,x+5,x+10
∴x+x+5+x+10=210 solve the equation and get the valu of x i.e; x=65
So the three consecutive multiples are 65,70,75 (Ans).
PRADEEP SAHAJWANI CLASSES
Pradeep Sahajwani Classes (BASE-MAKER)
(NTSE, IMO, NSTSE, OLYMPIAD & Deep Foundation for Competitive Exams)
VIIIth – IXth – Xth CBSE ( Maths / Science )
S-20, Vivekanand Colony, Near Rajat Nagar, Bhopal, Mobile: 9479365082, 9893213700
Q11. Let the original no. be x
2
So, 3x=x-25
2
x-3x=25
x=75
Hence the required number is 75. (Ans).
Q12. Let the two numbers be x and 45-x.
∴x-(45-x) =17 solve the equation and get the value of x i.e; x=31
So the two numbers are 31 and (45-31)=14. (Ans).
𝑥
Q13. Let the two numbers be x and 4
𝑋
6
∴ x- 4 =5
8
Solve the equation and get the value of x i.e; x= 5
8
2
Hence the two numbers are 5 and 5 (Ans).
Q14. Let the unit place of the no. be x and tens place be (9-x).
10(9-x)+x = 90-10x+x = 90-9x
Now, the new number will be 10x+9-x = 9x+9
According to the given condition, (90-9x) – (9x+9) = 63
81-18x=63
X=1
Hence the original number is 90-(9x1) = 81 (Ans).
Q15. Let the number be x
1
So according to the given condition , (x-2 )4 = 10
On solving the above equation we get x=3
Hence the number= 3 (Ans).
PRADEEP SAHAJWANI CLASSES
Pradeep Sahajwani Classes (BASE-MAKER)
(NTSE, IMO, NSTSE, OLYMPIAD & Deep Foundation for Competitive Exams)
VIIIth – IXth – Xth CBSE ( Maths / Science )
S-20, Vivekanand Colony, Near Rajat Nagar, Bhopal, Mobile : 9479365082, 9893213700
CLASS IX DEMO PAPER (SA-I BASED)
Q1. (√7+√11) (√7-√11) is:(a)Irrational no. (b) whole no. (c) positive integer (d) negative integer
Q2. Remainder when x11 + 11 is divide by x+1 is :(a)10 (b) 11 (c) -10 (d) o
Q3. If the vertex of isosceles triangle is four times the sum of base angles then measure of vertex angle :(a)60 (b) 66 (c) 72 (d) 144
Q4. The coefficient o x3 in (3x3-1) (2x2-2) is:(a)3 (b) -3 (c) 6 (d) -6
1
Q5. If x= 1-√2 then the value of x4 + 𝑥 4 is :(a) 32 (b) 34 (c) -34 (d) 30
Q6. Simplification of (0.00032) −
2
5
gives :-
(a) 25 (b) 32 (c) -32 (d) 1
Q7. Ax = B , By=C and Cz=A then the value of xyz is :1
(a) 0 (b) 2
(c)1
(d) Can’t determined
Q8. In a given figure l ll m then value of p is:-
(a) 100 (b) 72 (c) 54 (d) 18
Q9. In a given figure AB ll EFllCD then the value of x and y:-
PRADEEP SAHAJWANI (BASE – MAKER) CLASSES VIIIth – IXth – Xth
Pradeep Sahajwani Classes (BASE-MAKER)
(NTSE, IMO, NSTSE, OLYMPIAD & Deep Foundation for Competitive Exams)
VIIIth – IXth – Xth CBSE ( Maths / Science )
S-20, Vivekanand Colony, Near Rajat Nagar, Bhopal, Mobile : 9479365082, 9893213700
(a) 32,22 (b) 22,35 (c) 42,35 (d) 50,22
Q10. In a given square ABCD AP=BQ=CR then find the value of RPQ:-
(a) 40 (b) 47 (c) 45 (d) 30
Q11. Check whether (2p+1) is a factor of q(p)=4p3+4p2-p-1 or not.
Q12. Prove that the angle between bisectors of two acute angle of right triangle is 135°.
Q13. The lengths of the sides of a triangle are in a ratio 3:4:5 and the perimeter is 144. Then find the area of
triangle.
Q14. Evaluate- (1/2)3 + (1/3)3+(-5/6)3, without solving directly.
Q15. If the bisector of exterior vertical angle of triangle is parallel to the base then show that the triangle is
isosceles.
1
2
Q16. If x=2+23 +23 then find the value of x3-6x2+6x.
Q17. Factorize (p+2q)2+101(p+2q)+100.
Q18. Prove that the triangle is equilateral if the medians are equal to each other.
PRADEEP SAHAJWANI (BASE – MAKER) CLASSES VIIIth – IXth – Xth
Pradeep Sahajwani Classes (BASE-MAKER)
(NTSE, IMO, NSTSE, OLYMPIAD & Deep Foundation for Competitive Exams)
VIIIth – IXth – Xth CBSE ( Maths / Science )
S-20, Vivekanand Colony, Near Rajat Nagar, Bhopal, Mobile : 9479365082, 9893213700
Q19. State RHS theorem and using it solve the problem “The image of object placed at point A before a plane
mirror l is seen at point B by an observer at D. Prove that the distance of image from the mirror is equal to the
distance of object from mirror”.
Q20. Draw the trapezium PQRS of which vertices P,Q,R and S are (4,6) (-2,3) (-2,5) and (4,-7) and find the area
also.
PRADEEP SAHAJWANI (BASE – MAKER) CLASSES VIIIth – IXth – Xth
Pradeep Sahajwani Classes (BASE-MAKER)
(NTSE, IMO, NSTSE, OLYMPIAD & Deep Foundation for Competitive Exams)
VIIIth – IXth – Xth CBSE ( Maths / Science )
S-20, Vivekanand Colony, Near Rajat Nagar, Bhopal, Mobile : 9479365082, 9893213700
ANSWERS
Q1 (d) –ve integer.
Q2 (a) 10.
Q3 (d) 44.
Q4 (c) 6.
Q5 (b) 34.
Q6 (a) 25.
Q7 (c) 1.
Q8 (c) 54.
Q9 (b) 22,35.
Q10
(c) 45.
1
Q11
Using factor theorem put p= - 2 in p(x)
1
1
1
1
P(- 2 ) = 4(- 2 )3 + 4(- 2 )2 - (-2) -1 =0
So (2p+1) is a factor.
Q12
Let the ∆ABC right angle at A and BO and CO are bisector. Now in
∆ABC,
A+B+C=180°
(Angle sum property)
∴2x+2y+90° = 180°
x+y = 45°
------------------ (1)
Now in ∆BOC
x+y+BOC =180°
45 + BOC =180°
∴ BOC = 135° (Ans).
Q13 Let 3x, 4x , 5x are sides of a triangle.
So, 3x+4x+5x=144
12x=144
=> x=12
Sides will be - 3x12=36 , 4x12=48 , 5x12=60.
PRADEEP SAHAJWANI (BASE – MAKER) CLASSES VIIIth – IXth – Xth
Pradeep Sahajwani Classes (BASE-MAKER)
(NTSE, IMO, NSTSE, OLYMPIAD & Deep Foundation for Competitive Exams)
VIIIth – IXth – Xth CBSE ( Maths / Science )
S-20, Vivekanand Colony, Near Rajat Nagar, Bhopal, Mobile : 9479365082, 9893213700
Now using heron’s formula.
Area of ∆= √𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐)
s=
36+48+60
=
2
72cm.
∴ √72(72 − 36)(72 − 48)(72 − 60)
864cm2
Alternate method : Check 36 , 48 , 60
Pythagoras triplet 602=362+482
So area =
1
2
1
x36x48
2
= 864cm2
base x height
Q14. Use identity a3+b3+c3-3abc.
If a+b+c = 0
1 1
2 3
5
6
Check + +(- )
1
1
5
So, (2)3 + (3)3 + (− 6)3
1
1
5
∴ 3x 2 x 3 x (− 6) = −
5
12
.
Q15.
Let ABC is a triangle , AE ll BC and AE is a bisector of DAC.
We need to prove AE ll BC
DAE = ABC
(Corresponding angle)
EAC = ACB
But DAE =EAC
(Alternate angle)
(Given)
ABC =ACB
∴ ABC is Isosceles triangle.
PRADEEP SAHAJWANI (BASE – MAKER) CLASSES VIIIth – IXth – Xth
Pradeep Sahajwani Classes (BASE-MAKER)
(NTSE, IMO, NSTSE, OLYMPIAD & Deep Foundation for Competitive Exams)
VIIIth – IXth – Xth CBSE ( Maths / Science )
S-20, Vivekanand Colony, Near Rajat Nagar, Bhopal, Mobile : 9479365082, 9893213700
1
2
Q16 x= 2+ 23 + 23
1
2
⇒ x-2 = 23 + 23
------------ (1)
cubing both sides,
1
2
⇒ (x-2)3 = (23 + 23 )3
1
2
1
2
1
2
⇒ x3 - 23 – 3x ×2(x-2) = (23 )3 + (23 )3 + 3x23 x23 (23 + 23 )
1
2
⇒ x3 – 8 – 6x(x-2) = 2+ 22 + 3x2(23 + 23 )
1
2
⇒ 2+4+6(23 + 23 )
1
2
⇒ 6+6(23 + 23 )
⇒ x3 – 6x2 + 12x – 8 = 6+6(x-2)
(From 1)
⇒ x3 – 6x2 + 12x – 8= 6+6x-12
⇒ x3 – 6x2 + 6x=14-12
⇒ x3 – 6x2 + 6x=2 (Ans).
Q17 Let (p+2q) =x
⇒ x2+101x +100
⇒ x2+100x+x+100
⇒ x(x+100) + 1(x+100)
⇒ (x+1) (x+100)
(p+2q+1) (p+2q+100). (Ans).
Q18
Let ABC is a ∆
PRADEEP SAHAJWANI (BASE – MAKER) CLASSES VIIIth – IXth – Xth
Pradeep Sahajwani Classes (BASE-MAKER)
(NTSE, IMO, NSTSE, OLYMPIAD & Deep Foundation for Competitive Exams)
VIIIth – IXth – Xth CBSE ( Maths / Science )
S-20, Vivekanand Colony, Near Rajat Nagar, Bhopal, Mobile : 9479365082, 9893213700
AD=BE=CF
(Given)
We need to prove ∆ABC is equilateral ∆
We know that point of intersection of medians G is centroid
2
3
1
3
BG= BE, GE = BE
2
1
CG= 3CF, FG= 3CF
But BE=CF (Given)
∴ BG=CG , GE=FG
BGF= CGE
∴∆BGF ≅ ∆CGE
(Vertically Opposite angles)
(SAS)
∴ BF = CE
(c.p.c.t.)
𝑜𝑟 2BF=2CE
(Median divides opposite sides in equal parts)
or 𝐴𝐵 = 𝐴𝐶
𝑠𝑖𝑚𝑖𝑙𝑎𝑟𝑙𝑦, 𝐴𝐷 = 𝐵𝐶 ;
Hence, ABC is equilateral triangle.
Q19
Statement of RHS theorem.
Now AQR= DQR
(Laws of reflection)
Since, AB ll QR
∴B=DQR
A=AQR
( Corresponding angles)
(Alternate angles)
∴A=B
⇒ QA = QB
𝑎𝑔𝑎𝑖𝑛, QPB = QPA
PQ=PQ
(Each 90 )
(Common)
PRADEEP SAHAJWANI (BASE – MAKER) CLASSES VIIIth – IXth – Xth
Pradeep Sahajwani Classes (BASE-MAKER)
(NTSE, IMO, NSTSE, OLYMPIAD & Deep Foundation for Competitive Exams)
VIIIth – IXth – Xth CBSE ( Maths / Science )
S-20, Vivekanand Colony, Near Rajat Nagar, Bhopal, Mobile : 9479365082, 9893213700
∴ ∆QPB ≅ ∆QAP (RHS)
∴ AP=BP
(c.p.c.t.)
𝐻𝑒𝑛𝑐𝑒 𝑝𝑟𝑜𝑣𝑒𝑑.
Q20
1
Area of trapezium = 2(b1+b2)xh
1
x(2+13)
2
x6
45 𝑠𝑞 𝑢𝑛𝑖𝑡𝑠. (𝐴𝑛𝑠).
PRADEEP SAHAJWANI (BASE – MAKER) CLASSES VIIIth – IXth – Xth
Pradeep Sahajwani Classes (BASE-MAKER)
(NTSE, IMO, NSTSE, OLYMPIAD & Deep Foundation for Competitive Exams)
VIIIth – IXth – Xth CBSE ( Maths / Science )
S-20, Vivekanand Colony, Near Rajat Nagar, Bhopal, Mobile : 9479365082, 9893213700
X-DEMO PAPER (SA-I BASED)
Q1. If sec4θ= cosec(θ-20°) then value of θ:(a) 69 (b)88 (c)66 (d) 63
Q2. Which one is not polynomial
(a) 6√𝑥 9 +2x+1
2
(b) 28x3-31x2+1 (c) 𝑥 2 +x-1 (d) a and c are not polynomial.
Q3. Sin2θ+cos2θ=1 is true only for
(a) 0<θ<90
(b) θ=0 or 90
(c) 0≤θ≤90
(d) 0≤θ≤60
Q4. The curves for more than and less than gives meet at (54,96). If mode is 60 then mean is
(a) 50 (b) 52.5 (c) 48 (d) 51
Q5. If sinθ1+sinθ2+sinθ3=3 then cos𝜃1+cos𝜃2+cos𝜃3=
(a)3 (b)1 (c)0 (d)2
Q6. If P and Q are acute angles of right angle triangle PQR at R then sin2P+sin2Q=
(a) 0 (b) -1 (c) cosR (d) sinR
Q7.
In a given figure ABC is a triangle PQRS i9s rectangle BC=12cm QR=6cm O is mid
point of BC. Altitude AO=24cm then area of rectangle PQRS
(a) 54cm2 (b) 56cm2 (c) 60cm2 (d) 72cm2
Q8.
In a figure A=90 and ADBC then which statement is true
(a) ∆BDC~∆BCA (b) ∆ADC~∆ADB (c) ∆CDA~∆CBA (d) ∆BDA~∆ADC
PRADEEP SAHAJWANI (BASE – MAKER) CLASSES VIIIth – IXth – Xth
Pradeep Sahajwani Classes (BASE-MAKER)
(NTSE, IMO, NSTSE, OLYMPIAD & Deep Foundation for Competitive Exams)
VIIIth – IXth – Xth CBSE ( Maths / Science )
S-20, Vivekanand Colony, Near Rajat Nagar, Bhopal, Mobile : 9479365082, 9893213700
Q9. Find the largest +ve integer that divide 398,436 and 542 leaving remainder 7,11 and 15 respectively.


Q10. If , are zeroes of polynomial x2+8x+6 then from quadratic polunomial whose zeroes are 1+ and 1+
Q11. If one zero is 7 times of ither find p . 3x2-8x+2p+1.
Q12. Prove that 2(sin4θ+cos6θ)-3(sin4θ+cos4θ)+1=0
Q13. Find the values of a and b if lines of the given equation coincide 2x+3y=9.with (a+b)x + (2a-b)y – 3(a+b+1)=0
Q14. In a field it was decided to plant 24 mango tree, 48 apple tree and 60 orange tree in distinct rows. Find the
minimum no. of trees in each rows and total no of rows
Q15. The sum of total ages of father and son is 70 years. If the father was to live till his son’s age equals to his
present age, the total of their ages would be 130 years. Find the present age of father and son.
Q16.
In a fig DEFG is a square. In a right angle triangle ABC at A=90.
Prove that DE2=BDxEC.
Q17.
In a given figure find ST.
PRADEEP SAHAJWANI (BASE – MAKER) CLASSES VIIIth – IXth – Xth
Pradeep Sahajwani Classes (BASE-MAKER)
(NTSE, IMO, NSTSE, OLYMPIAD & Deep Foundation for Competitive Exams)
VIIIth – IXth – Xth CBSE ( Maths / Science )
S-20, Vivekanand Colony, Near Rajat Nagar, Bhopal, Mobile : 9479365082, 9893213700
Q18. The sum of two digit no. and no. obtained by reversing the digit is 66. If the digit of no. differ by 2. Find the
no.
Q19.
In a given figure DE ll BC and AD:DB=5:4. Find the ratio of area of ∆DEF and ∆CFB.
Q20. The mean of 1,3,4,5,7 and 4 is m. the no.s 3,2,2,4,3,3 and p have mean m-1 and median q. then find the
value of p+q.
PRADEEP SAHAJWANI (BASE – MAKER) CLASSES VIIIth – IXth – Xth
Pradeep Sahajwani Classes (BASE-MAKER)
(NTSE, IMO, NSTSE, OLYMPIAD & Deep Foundation for Competitive Exams)
VIIIth – IXth – Xth CBSE ( Maths / Science )
S-20, Vivekanand Colony, Near Rajat Nagar, Bhopal, Mobile : 9479365082, 9893213700
ANSWERS
Q1 (c) 66
2
Q2 (c) 𝑥 2+x-1
Q3 (c) 0≤θ≤90°
Q4 (d) 51
Q5 (c) 0
Q6 (d) SinR
Q7 (a) 54cm2
Q8 (d) ∆BDA ~ ∆ADC
Q9 Required no. is HCF of (398-7 , 436-11 , 542-15)
(391,425,725)
17 (Ans).
Q10 x2- (sum)x + product




x2 – (1+ +1+ ) x + (1+)(1+ )
x2−(2+
2 +2
 
)x + (2+ + )

 
(𝛼+𝛽)2 −2𝛼𝛽
𝛼 2 +𝛽 2
)x+(2 +
)
𝛼𝛽
𝛼𝛽
x2 – 2+(
(−82 )−2x6
)x
6
x2 −2+(
x2 -
32
x
3
+
32
3
+
(−82 )−2×6
6
(Ans)
Q11 Let zeroes are ,7
Sum of zeroes =
−𝑏
𝑎
−8
+ 7 = - ( 3 )
8
8 = 3
PRADEEP SAHAJWANI (BASE – MAKER) CLASSES VIIIth – IXth – Xth
Pradeep Sahajwani Classes (BASE-MAKER)
(NTSE, IMO, NSTSE, OLYMPIAD & Deep Foundation for Competitive Exams)
VIIIth – IXth – Xth CBSE ( Maths / Science )
S-20, Vivekanand Colony, Near Rajat Nagar, Bhopal, Mobile : 9479365082, 9893213700
1
7
= 3 ∴= 3
𝑐
Now 𝛼𝛽=𝑎
1
3
7
×3=
2p+1
3
2
p=3 (Ans).
Q12 (Sin2θ+Cos2θ)2 = Sin4θ+Cos4θ+2Sin2θCos2θ
∴ Sin4θ+Cos4θ= (Sin2θ+Cos2θ)2 − 2Sin2θCos2θ
⇒Now , 2(Sin6θ+Cos6θ) – 3(Sin4θ+Cos4θ)+1
⇒ 2{(Sin2θ)3+(Cos2θ)3 }- 3{(Sin2θ+Cos2θ)2} – 2Sin2θCos2θ
Using a3+b3=(a+b)(a2-ab+b2)
⇒ 2[(Sin2θ+Cos2θ)(Sin4θ+Cos4θ-Sin2θCos2θ)-3(1-2Sin2θCos2θ)+1]
⇒ 2[1×{(Sin2θ+Cos2θ)2 −2Sin2θCos2θ − Sin2θCos2θ} −3+6Sin2θCos2θ+1
⇒ 2- 6Sin2θCos2θ-3+6Sin2θCos2θ+1
0 Hence proved.
𝑎
𝑏
𝑐
Q13 Using 𝑎1 = 𝑏1 = 𝑐1
2
2
2
⇒ Pair a1x+b1y+c1=0
⇒ a2x+b2y+cz=0
Now for given question
2
3
9
⇒ a+b = 2a−b = −3(a+b+1)
2
3
=
a + b 2a − b
3
9
=
2a − b −3(a + b + 1)
a=5b
1
1
=
2a − b −a − b − 1
3a= − 1
PRADEEP SAHAJWANI (BASE – MAKER) CLASSES VIIIth – IXth – Xth
Pradeep Sahajwani Classes (BASE-MAKER)
(NTSE, IMO, NSTSE, OLYMPIAD & Deep Foundation for Competitive Exams)
VIIIth – IXth – Xth CBSE ( Maths / Science )
S-20, Vivekanand Colony, Near Rajat Nagar, Bhopal, Mobile : 9479365082, 9893213700
a=
−1
3
−1
15
and b=
(Ans).
Q14. HCF of (24, 48, 60) = 12
⇒ Min. no. of tress = 12
⇒ Rows for mango trees =2
⇒ Rows for orange tree = 4
⇒ Rows for apple tree=5
Total 11 rows.
Q15 Let the age of father = x
and the age of son = y
⇒x+y =70
Now, x+ (x-y) + y + (x-y) =130
⇒ 3x-y=130
Solve for x and y
⇒ x=50 ; y=20
Hence the age of father is 50 and age of son in 20. (Ans).
Q16 In ∆ABC A=90°
⇒ B+C=90°
--------(1)
Also in ∆BGD
⇒ B+G =90°
∴ C=G
------(2)
(From 1 and 2)
⇒ BDG = CEF
(Each 90)
∴ ∆BGD~ ∆FCE
BD
⇒ FE =
BG
FC
=
GD
CE
⇒ BD x CE= FE x GD
PRADEEP SAHAJWANI (BASE – MAKER) CLASSES VIIIth – IXth – Xth
Pradeep Sahajwani Classes (BASE-MAKER)
(NTSE, IMO, NSTSE, OLYMPIAD & Deep Foundation for Competitive Exams)
VIIIth – IXth – Xth CBSE ( Maths / Science )
S-20, Vivekanand Colony, Near Rajat Nagar, Bhopal, Mobile : 9479365082, 9893213700
But FE=GD=DE
∴DE2=BD x CE
Hence proved.
Q17
In ∆PQR and ∆PTS
⇒ P=P
(Common)
⇒ Q=T (Given)
⇒ ∆PQR ~ ∆PTS
PT
(AA similarity criterion)
PQ
∴ ST = QR
⇒
10
ST
=
15
9
⇒ ST= 6cm (Ans).
Q18 Let the two digit no. 10x+y
According to the question
⇒ x-y=2
------(1)
⇒ (10x+y) + (10y+x) =66
⇒ x+y=8
-------- (2)
Solve (1) and (2) .
⇒ x=2 ; y=4
Two digit no. is 24 or 42 (Ans).
Q19
DE ll BC
⇒ ADE=ABC
⇒ A=A
(Common)
PRADEEP SAHAJWANI (BASE – MAKER) CLASSES VIIIth – IXth – Xth
Pradeep Sahajwani Classes (BASE-MAKER)
(NTSE, IMO, NSTSE, OLYMPIAD & Deep Foundation for Competitive Exams)
VIIIth – IXth – Xth CBSE ( Maths / Science )
S-20, Vivekanand Colony, Near Rajat Nagar, Bhopal, Mobile : 9479365082, 9893213700
∴ ∆ADE ~∆ABC
AD
DE
AD
5
AD
5
(AA similarity criterion)
⇒ AB = BC
---------(1)
⇒ DB = 4
(Given)
⇒ AB = 9
DE
5
Or, BC = 9
(From 1)
Now in ∆DEF and ∆CFB
⇒ 1=3
(Alternate angles)
⇒2=4
(Vertically opp. angles)
⇒ ∆DFE~∆CFB
∴
Ar. ∆DEF
Ar. ∆CFB
Q20 m=
=
(AA similarity criterion)
5 2
9
DE2
BC2
=( ) =
25
81
(Ans).
1+3+4+5+7+4
6
m= 4
again using formula of median,
3+2+2+4+3+3+p
=
7
17+p
7
m-1
=4−1
17+p = 21
p=4
Now,
2,2,3,3,3,p.
Median =
3rd +4th
2
3+3
2
N th
2
N
2
( ) +( +1)
th
2
=q
=q
PRADEEP SAHAJWANI (BASE – MAKER) CLASSES VIIIth – IXth – Xth
Pradeep Sahajwani Classes (BASE-MAKER)
(NTSE, IMO, NSTSE, OLYMPIAD & Deep Foundation for Competitive Exams)
VIIIth – IXth – Xth CBSE ( Maths / Science )
S-20, Vivekanand Colony, Near Rajat Nagar, Bhopal, Mobile : 9479365082, 9893213700
q=3
so required p+q = 3+4 = 7 (Ans).
PRADEEP SAHAJWANI (BASE – MAKER) CLASSES VIIIth – IXth – Xth