Advance Maths Questions for SSC CGL CAT

Advance Maths Questions for SSC CGL CAT

Advance Maths Questions for SSC CGL CAT



Q1. Consider an obtuse-angled triangles with sides 8 cm, 15 cm and x cm. If x is an integer then how many such triangles exist?

A. 5
B. 21
C. 10
D. 15

Q2. In the given figure , PQR is a triangle and quadrilateral ABCD is inscribed in it . QD = 2 cm , QC = 5 cm , CR = 3 cm , BR = 4 cm , PB = 6 cm , PA = 5 cm and AD = 3 cm . What is the area ( in cm² ) of the quadrilateral ABCD?

In the given figure , PQR is a triangle and quadrilateral ABCD is inscribed in it . QD = 2 cm , QC = 5 cm , CR = 3 cm , BR = 4 cm , PB = 6 cm , PA = 5 cm and AD = 3 cm . What is the area ( in cm² ) of the quadrilateral ABCD?

A. 17√21
B. 17/5
C. 17√21/5
D. √21/5

Q3. The quadratic equation x2 + mx + n = 0 has roots that are twice those of x2 + px + m = 0, and none of m, n and p is zero. What is the value of n/p?
(A) 2
(B) 16
(C) 4
(D) 8

Q4. Akash when going slower by 15 Km/hr, reaches late by 45 hours. If he goes faster by 10 Km/hr from his original speed, he reaches early by by 20 hours than the original time. Find the distance he covers.
आकाश जब 15 किमी/घंटा तक धीमी गति से चलता है, तो 45 घंटे देरी से पहुंचता है. यदि वह अपनी मूल गति से 10 किमी/घंटा तेजी से चलता है, तो वह मूल समय की तुलना में 20 घंटे जल्दी पहुंचता है. उसके द्वारा तय की जाने वाली दूरी ज्ञात कीजिये?
(A) 8750 Km/किमी
(B) 9750 Km/किमी
(C) 1000 Km/किमी
(D) 3750 Km/किमी

Q5. If x = √3/2 then find the value of √(1+x) - √(1-x)
(A) 1
(B) 0
(C) √3
(D) √2

Q6. A right angled triangle with sides AB = AC = 24 , is cut into three parts of equal area as shown in the figure given below . Find the sum of perimeters of all the triangles in the figure.
A right angled triangle with sides AB = AC = 24 , is cut into three parts of equal area as shown in the figure given below . Find the sum of perimeters of all the triangles in the figure.
(A) $144 + 40\sqrt{2} + 48\sqrt{5}$
(B) $144 + 80\sqrt{2} + 48\sqrt{5}$
(C) $72 + 40\sqrt{2} + 24\sqrt{5}$
(D) $72 + 40\sqrt{2} + 48\sqrt{5}$


Q7. PQRS is a square inside an isosceles right angled triangle ABC , such that points P, Q and S lie on the sides AB, BC and AC, respectively. If BQ = 8cm, PB = 4cm and PR not parallel to BC, what is the area of the triangle outside the square?
PQRS एक वर्ग है जो समदविबाह समकोण त्रिभुज के अंदर बनाया गया है । जहाँ P, Q और S क्रमशः भुजा AB, BC और AC पर है । यदि BQ = 8cm, PB = 4cm और PR, BC के समांतर नही है तब वर्ग के बाहर त्रिभुज के भाग का क्षेत्रफल ज्ञात करे।
PQRS is a square inside an isosceles right angled triangle ABC , such that points P, Q and S lie on the sides AB, BC and AC, respectively. If BQ = 8cm, PB = 4cm and PR not parallel to BC, what is the area of the triangle outside the square?  PQRS एक वर्ग है जो समदविबाह समकोण त्रिभुज के अंदर बनाया गया है । जहाँ P, Q और S क्रमशः भुजा AB, BC और AC पर है । यदि BQ = 8cm, PB = 4cm और PR, BC के समांतर नही है तब वर्ग के बाहर त्रिभुज के भाग का क्षेत्रफल ज्ञात करे। 1 ) 48cm b ) 96 cm c ) 144cm d ) 64cm

(A) 48 cm
(B) 96 cm
(C) 144 cm
(D) 64 cm

Q8. In the given fig . ABCD is a square, and BM = DE, calculate the value of x ?
दिए गए चित्र में , ABCD एक वर्ग है । और BM = DE तब x का मान ज्ञात करे।
In the given fig . ABCD is a square, and BM = DE, calculate the value of x ?   दिए गए चित्र में , ABCD एक वर्ग है । और BM = DE तब x का मान ज्ञात करे।   a ) 40 b ) 20 c ) 25 d ) 30

(A) 40
(B) 20
(C) 25
(D) 30


Solution

1. Solution

If c is the side opposite the obtuse angle & hence the longest, it should satisfy the following two conditions:

a2 + b2 < c2 and c < a + b

If 15 is the longest side, then:

82 + x2 < 152
64 + x2 < 225
x2 < 161
x < 12.69

So, keeping with our other range, x could be 8, 9, 10, 11 or 12.

If we make x our longest side, we have:

82 + 152 < x2
64 + 225 < x2
289 < x2
17 < x

x could be 18, 19, 20, 21 or 22.

So 10 possible values for x.

Ans. C


2. Solution

Area of the ΔPQR = √(14×6×4×4) = 8√21

By using sin formula for area = 1/2abSinθ

Ratio of area of ΔAPB and ΔPQR = 5×6/10×10 = 30/100 = 3/10 = 3×8/10×8 = 24/80

Ratio of area of ΔQDC and ΔPQR = 5×2/10×8 = 10/80

Ratio of area of ΔBRC and ΔPQR = 3×4/10×8 = 12/80

So the ratio of area of □ABCD and ΔPQR = (80 - 24 - 10 - 12)/80 = 34/80

Now the area of 80 unit ΔPQR = 8√21

So 1 unit = 8√21/80 = √21/10

So the area of 34 unit □ABCD = 34×√21/10

= 17√21/5 Ans.
In the given figure , PQR is a triangle and quadrilateral ABCD is inscribed in it . QD = 2 cm , QC = 5 cm , CR = 3 cm , BR = 4 cm , PB = 6 cm , PA = 5 cm and AD = 3 cm . What is the area ( in cm² ) of the quadrilateral ABCD?



3. Solution

If the roots of x2 + px + m = 0 are α and β
then roots of x2 + mx + n = 0 will be 2α and 2β
then sum of roots of first equation (α + β) = -p
and multiplication of roots of first equation (α×β) = m
again sum of roots of second equation 2(α + β) = -m
and multiplication of roots of second equation 4α×β = n
from above equations 2(-p) = -m
=> 2p = m
and 4m = n
=> 8p = n
=>n/p = 8

4. Solution

Let the actual speed is "v" distance "d"
According to the conditions:
d/(v - 15) = d/v + 45
=> 15d/v(v -15) = 45  .....(i)
d/(v + 10) = d/v - 20
=> 10d/v(v +10) = 20  .....(ii)
divide (i) by (ii)
15(v + 10)/10(v - 15) = 45/20
(v + 10)/(v - 15) = 3/2
v = 65
from equation (i)
15d/65×50 = 45
d = 9750
So the total distance will be 9750 km


5. Solution

Give that x = √3/2
If z = √(1+x) - √(1-x)
z2 = (1+x) + (1-x) - 2√(1-x2)
z2 = 2 - 2√(1-3/4)
z2 = 2 - 2×1/2
z = 1
Ans: 1


6. Solution


Given that area of all traingle is same so base of all traingle will be equal because hieght is same
∴ BD = DE = EC = BC/3 = $8\sqrt{2}$ (∵ ∠BAC = 90°)
Given that ∠BAC = 90° and AB = AC = 24
∴ ∠ABC = ∠ACB = 45°
Now according to cosine rule in ΔACE
${AE}^2 = {24}^2 + {(8\sqrt{2})}^2 - 2×24×8\sqrt{2}\cos45°$
${AE}^2 = 320$
$AE = 8\sqrt{5}$
Same as $AD = 8\sqrt{5}$
Now sum of perimeter of all triangle
ΔABD + ΔADE + ΔAEC + ΔABE + ΔACD + ΔABC
= $144 + 80\sqrt{2} + 48\sqrt{5}$
Ans: B


7. Solution

Given that ΔABC is isosceles right angled triangle and BQ = 8cm PB = 4cm
∴ ∠BAC = ∠ACB = 45° (∵ ΔABC is isosceles right angled triangle)
$PQ = 4\sqrt{5}$ (by Pythagoras theorem)
Now using sine rule in ΔAPS
$\frac{PS}{\sin45°} = \frac{AS}{\sin∠APS}$

$\frac{PS}{1/\sqrt{2}} = \frac{AS}{\cos∠BPQ}$ (∵ ∠QPS = 90°)

$\frac{PS}{1/\sqrt{2}} = \frac{AS}{PB/PQ}$

$AS = 4\sqrt{2}$ (∵ PS = PQ and PB = 4)

Now using cosine rule in ΔAPS
${PS}^2 = {AP}^2 + {AS}^2 - 2×AP×AS×\cos45°$
$80 = {AP}^2 + 32 - 8AP$
${AP}^2 - 8AP - 48 = 0$
AP = 12 or -4 (-4 is not valid)
∴ AB = 12 +4 = 16
Now area of ΔABC = $\frac{1}{2}×16×16 = 128$
and area of square PQRS = ${(4\sqrt{5})}^2 = 80$
∴ The area of the triangle outside the square = 128 - 80 = 48
Ans: (A)


8. Solution

Given that ABCD is a square and BM = DE.
दिए गए चित्र में , ABCD एक वर्ग है । और BM = DE तब x का मान ज्ञात करे।

ΔBAM  ΔDAM (by Side Angle Side postulate)
∴ BM = DM = DE (∵ BM = DE given)
Now in ΔDME
∠DME = ∠DEM = 40° (∵ DM = DE)
ΔMBC  ΔMDC (by Side Side Side postulate)
∴ ∠MBC = ∠MDC = x
Now in ΔDME
∠ODE = 180 - 40 - 40 - x
∠ODE = 100 - x
 in ΔBCO
∠BOC = (90 -x) (∵ ∠BCO = 90°)
∠BOC = ∠DOE (pairs of opposite angles)
Now in ΔDOE
(100 -x) + (90 -x) + 40 = 180
2x = 230 - 180
x = 25°
Ans: (C)










Advance Maths Questions for SSC CGL CAT Advance Maths Questions for SSC CGL CAT Reviewed by Super Pathshala on 13:18 Rating: 5

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